Quantum_Physics_For

Physics as a general discipline has no limits, from the very huge (galaxy-wide) to the very small (atoms and smaller). This book is about the very small side of things — that’s the specialty of quantum physics. When you quantize something, you can’t go any smaller; you’re dealing with discrete units.
About the Author: Steven Holzner is an award-winning author of technical and science books (like Physics For Dummies and Differential Equations For Dummies). He graduated from MIT and did his PhD in physics at Cornell University, where he was on the teaching faculty for 10 years. He’s also been on the faculty of MIT. Steve also teaches corporate groups around the country.
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wave function particle equate Energy operator Quantum Physics Angular Momentum State Schrdinger equate Harmonic oscillate
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wave function particle equate Energy operator Quantum Physics Angular Momentum State Schrdinger equate Harmonic oscillate
Quantum	Physics	For	Dummies,	Revised	Edition
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10	9	8	7	6	5	4	3	2	1
About	the	Author
Steven	Holzner	is	an	award-winning	author	of	technical	and	science	books	(like
Physics	For	Dummies	and	Differential	Equations	For	Dummies).	He	graduated
from	MIT	and	did	his	PhD	in	physics	at	Cornell	University,	where	he	was	on	the
teaching	faculty	for	10	years.	He's	also	been	on	the	faculty	of	MIT.	Steve	also
teaches	corporate	groups	around	the	country.
Author's	Acknowledgments
I'd	particularly	like	to	thank	the	people	at	Wiley:	Tracy	Boggier,	Tim	Gallan,
and	Danielle	Voirol.
Dedication
To	Nancy,	of	course!
Publisher's	Acknowledgments
Were	proud	of	this	book;	please	send	us	your	comments	through	our	Dummies
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Quantum	Physics	For	Dummies
Table	of	Contents
Introduction
About	This	Book
Conventions	Used	in	This	Book
Foolish	Assumptions
How	This	Book	Is	Organized
Part	I:	Small	World,	Huh?	Essential	Quantum	Physics
Part	II:	Bound	and	Undetermined:	Handling	Particles	in	Bound	States
Part	III:	Turning	to	Angular	Momentum	and	Spin
Part	IV:	Multiple	Dimensions:	Going	3D	with	Quantum	Physics
Part	V:	Group	Dynamics:	Introducing	Multiple	Particles
Part	VI:	The	Part	of	Tens
Icons	Used	in	This	Book
Where	to	Go	from	Here
Part	I:	Small	World,	Huh?	Essential	Quantum	Physics
Chapter	1:	Discoveries	and	Essential	Quantum	Physics
Being	Discrete:	The	Trouble	with	Black-Body	Radiation
First	attempt:	Wien's	Formula
Second	attempt:	Rayleigh-Jeans	Law
An	intuitive	(quantum)	leap:	Max	Planck's	spectrum
The	First	Pieces:	Seeing	Light	as	Particles
Solving	the	photoelectric	effect
Scattering	light	off	electrons:	The	Compton	effect
Proof	positron?	Dirac	and	pair	production
A	Dual	Identity:	Looking	at	Particles	as	Waves
You	Can't	Know	Everything	(But	You	Can	Figure	the	Odds)
The	Heisenberg	uncertainty	principle
Rolling	the	dice:	Quantum	physics	and	probability
Chapter	2:	Entering	the	Matrix:	Welcome	to	State	Vectors
Creating	Your	Own	Vectors	in	Hilbert	Space
Making	Life	Easier	with	Dirac	Notation
Abbreviating	state	vectors	as	kets
Writing	the	Hermitian	conjugate	as	a	bra
Multiplying	bras	and	kets:	A	probability	of	1
Covering	all	your	bases:	Bras	and	kets	as	basis-less	state	vectors
Understanding	some	relationships	using	kets
Grooving	with	Operators
Hello,	operator:	How	operators	work
I	expected	that:	Finding	expectation	values
Looking	at	linear	operators
Forward	and	Backward:	Finding	the	Commutator
Commuting
Finding	anti-Hermitian	operators
Starting	from	Scratch	and	Ending	Up	with	Heisenberg
Eigenvectors	and	Eigenvalues:	They're	Naturally	Eigentastic!
Understanding	how	they	work
Finding	eigenvectors	and	eigenvalues
Preparing	for	the	Inversion:	Simplifying	with	Unitary	Operators
Comparing	Matrix	and	Continuous	Representations
Going	continuous	with	calculus
Doing	the	wave
Part	II:	Bound	and	Undetermined:	Handling	Particles	in
Bound	States
Chapter	3:	Getting	Stuck	in	Energy	Wells
Looking	into	a	Square	Well
Trapping	Particles	in	Potential	Wells
Binding	particles	in	potential	wells
Escaping	from	potential	wells
Trapping	Particles	in	Infinite	Square	Potential	Wells
Finding	a	wave-function	equation
Determining	the	energy	levels
Normalizing	the	wave	function
Adding	time	dependence	to	wave	functions
Shifting	to	symmetric	square	well	potentials
Limited	Potential:	Taking	a	Look	at	Particles	and	Potential	Steps
Assuming	the	particle	has	plenty	of	energy
Assuming	the	particle	doesn't	have	enough	energy
Hitting	the	Wall:	Particles	and	Potential	Barriers
Getting	through	potential	barriers	when	E	>	V0
Getting	through	potential	barriers,	even	when	E	<	V0
Particles	Unbound:	Solving	the	Schrdinger	Equation	for	Free	Particles
Getting	a	physical	particle	with	a	wave	packet
Going	through	a	Gaussian	example
Chapter	4:	Back	and	Forth	with	Harmonic	Oscillators
Grappling	with	the	Harmonic	Oscillator	Hamiltonians
Going	classical	with	harmonic	oscillation
Understanding	total	energy	in	quantum	oscillation
Creation	and	Annihilation:	Introducing	the	Harmonic	Oscillator	Operators
Mind	your	p's	and	q's:	Getting	the	energy	state	equations
Finding	the	Eigenstates
Using	a	and	a	directly
Finding	the	harmonic	oscillator	energy	eigenstates
Putting	in	some	numbers
Looking	at	Harmonic	Oscillator	Operators	as	Matrices
A	Jolt	of	Java:	Using	Code	to	Solve	the	Schrdinger	Equation	Numerically
Making	your	approximations
Building	the	actual	code
Running	the	code
Part	III:	Turning	to	Angular	Momentum	and	Spin
Chapter	5:	Working	with	Angular	Momentum	on	the	Quantum	Level
Ringing	the	Operators:	Round	and	Round	with	Angular	Momentum
Finding	Commutators	of	Lx,	Ly,	and	Lz
Creating	the	Angular	Momentum	Eigenstates
Finding	the	Angular	Momentum	Eigenvalues
Deriving	eigenstate	equations	with	max	and	min
Getting	rotational	energy	of	a	diatomic	molecule
Finding	the	Eigenvalues	of	the	Raising	and	Lowering	Operators
Interpreting	Angular	Momentum	with	Matrices
Rounding	It	Out:	Switching	to	the	Spherical	Coordinate	System
The	eigenfunctions	of	Lz	in	spherical	coordinates
The	eigenfunctions	of	L2	in	spherical	coordinates
Chapter	6:	Getting	Dizzy	with	Spin
The	Stern-Gerlach	Experiment	and	the	Case	of	the	Missing	Spot
Getting	Down	and	Dirty	with	Spin	and	Eigenstates
Halves	and	Integers:	Saying	Hello	to	Fermions	and	Bosons
Spin	Operators:	Running	Around	with	Angular	Momentum
Working	with	Spin	1/2	and	Pauli	Matrices
Spin	1/2	matrices
Pauli	matrices
Part	IV:	Multiple	Dimensions:	Going	3D	with	Quantum
Physics
Chapter	7:	Rectangular	Coordinates:	Solving	Problems	in	Three
Dimensions
The	Schrdinger	Equation:	Now	in	3D!
Solving	Three-Dimensional	Free	Particle	Problems
The	x,	y,	and	z	equations
Finding	the	total	energy	equation
Adding	time	dependence	and	getting	a	physical	solution
Getting	Squared	Away	with	3D	Rectangular	Potentials
Determining	the	energy	levels
Normalizing	the	wave	function
Using	a	cubic	potential
Springing	into	3D	Harmonic	Oscillators
Chapter	8:	Solving	Problems	in	Three	Dimensions:	Spherical
Coordinates
A	New	Angle:	Choosing	Spherical	Coordinates	Instead	of	Rectangular
Taking	a	Good	Look	at	Central	Potentials	in	3D
Breaking	down	the	Schrdinger	equation
The	angular	part	of	(r,	,	)
The	radial	part	of	(r,	,	)
Handling	Free	Particles	in	3D	with	Spherical	Coordinates
The	spherical	Bessel	and	Neumann	functions
The	limits	for	small	and	large	
Handling	the	Spherical	Square	Well	Potential
Inside	the	square	well:	0	<	r	<	a
Outside	the	square	well:	r	>	a
Getting	the	Goods	on	Isotropic	Harmonic	Oscillators
Chapter	9:	Understanding	Hydrogen	Atoms
Coming	to	Terms:	The	Schrdinger	Equation	for	the	Hydrogen	Atom
Simplifying	and	Splitting	the	Schrdinger	Equation	for	Hydrogen
Solving	for	(R)
Solving	for	(r)
Solving	the	radial	Schrdinger	equation	for	small	r
Solving	the	radial	Schrdinger	equation	for	large	r
You	got	the	power:	Putting	together	the	solution	for	the	radial	equation
Fixing	f(r)	to	keep	it	finite
Finding	the	allowed	energies	of	the	hydrogen	atom
Getting	the	form	of	the	radial	solution	of	the	Schrdinger	equation
Some	hydrogen	wave	functions
Calculating	the	Energy	Degeneracy	of	the	Hydrogen	Atom
Quantum	states:	Adding	a	little	spin
On	the	lines:	Getting	the	orbitals
Hunting	the	Elusive	Electron
Chapter	10:	Handling	Many	Identical	Particles
Many-Particle	Systems,	Generally	Speaking
Considering	wave	functions	and	Hamiltonians
A	Nobel	opportunity:	Considering	multi-electron	atoms
A	Super-Powerful	Tool:	Interchange	Symmetry
Order	matters:	Swapping	particles	with	the	exchange	operator
Classifying	symmetric	and	antisymmetric	wave	functions
Floating	Cars:	Tackling	Systems	of	Many	Distinguishable	Particles
Juggling	Many	Identical	Particles
Losing	identity
Symmetry	and	antisymmetry
Exchange	degeneracy:	The	steady	Hamiltonian
Name	that	composite:	Grooving	with	the	symmetrization	postulate
Building	Symmetric	and	Antisymmetric	Wave	Functions
Working	with	Identical	Noninteracting	Particles
Wave	functions	of	two-particle	systems
Wave	functions	of	three-or-more-particle	systems
It's	Not	Come	One,	Come	All:	The	Pauli	Exclusion	Principle
Figuring	out	the	Periodic	Table
Part	V:	Group	Dynamics:	Introducing	Multiple	Particles
Chapter	11:	Giving	Systems	a	Push:	Perturbation	Theory
Introducing	Time-Independent	Perturbation	Theory
Working	with	Perturbations	to	Nondegenerate	Hamiltonians
A	little	expansion:	Perturbing	the	equations
Matching	the	coefficients	of		and	simplifying
Finding	the	first-order	corrections
Finding	the	second-order	corrections
Perturbation	Theory	to	the	Test:	Harmonic	Oscillators	in	Electric	Fields
Finding	exact	solutions
Applying	perturbation	theory
Working	with	Perturbations	to	Degenerate	Hamiltonians
Testing	Degenerate	Perturbation	Theory:	Hydrogen	in	Electric	Fields
Chapter	12:	Wham-Blam!	Scattering	Theory
Introducing	Particle	Scattering	and	Cross	Sections
Translating	between	the	Center-of-Mass	and	Lab	Frames
Framing	the	scattering	discussion
Relating	the	scattering	angles	between	frames
Translating	cross	sections	between	the	frames
Trying	a	lab-frame	example	with	particles	of	equal	mass
Tracking	the	Scattering	Amplitude	of	Spinless	Particles
The	incident	wave	function
The	scattered	wave	function
Relating	the	scattering	amplitude	and	differential	cross	section
Finding	the	scattering	amplitude
The	Born	Approximation:	Rescuing	the	Wave	Equation
Exploring	the	far	limits	of	the	wave	function
Using	the	first	Born	approximation
Putting	the	Born	approximation	to	work
Part	VI:	The	Part	of	Tens
Chapter	13:	Ten	Quantum	Physics	Tutorials
An	Introduction	to	Quantum	Mechanics
Quantum	Mechanics	Tutorial
Grains	of	Mystique:	Quantum	Physics	for	the	Layman
Quantum	Physics	Online	Version	2.0
Todd	K.	Timberlake's	Tutorial
Physics	24/7's	Tutorial
Stan	Zochowski's	PDF	Tutorials
Quantum	Atom	Tutorial
College	of	St.	Benedict's	Tutorial
A	Web-Based	Quantum	Mechanics	Course
Chapter	14:	Ten	Quantum	Physics	Triumphs
Wave-Particle	Duality
The	Photoelectric	Effect
Postulating	Spin
Differences	between	Newton's	Laws	and	Quantum	Physics
Heisenberg	Uncertainty	Principle
Quantum	Tunneling
Discrete	Spectra	of	Atoms
Harmonic	Oscillator
Square	Wells
Schrdinger's	Cat
Introduction
Physics	as	a	general	discipline	has	no	limits,	from	the	very	huge	(galaxy-wide)
to	the	very	small	(atoms	and	smaller).	This	book	is	about	the	very	small	side	of
things		that's	the	specialty	of	quantum	physics.	When	you	quantize
something,	you	can't	go	any	smaller;	you're	dealing	with	discrete	units.
Classical	physics	is	terrific	at	explaining	things	like	heating	cups	of	coffee	or
accelerating	down	ramps	or	cars	colliding,	as	well	as	a	million	other	things,	but
it	has	problems	when	things	get	very	small.	Quantum	physics	usually	deals	with
the	micro	world,	such	as	what	happens	when	you	look	at	individual	electrons
zipping	around.	For	example,	electrons	can	exhibit	both	particle	and	wave-like
properties,	much	to	the	consternation	of	experimenters		and	it	took	quantum
physics	to	figure	out	the	full	picture.
Quantum	physics	also	introduced	the	uncertainty	principle,	which	says	you
can't	know	a	particle's	exact	position	and	momentum	at	the	same	time.	And	the
field	explains	the	way	that	the	energy	levels	of	the	electrons	bound	in	atoms
work.	Figuring	out	those	ideas	all	took	quantum	physics,	as	physicists	probed
ever	deeper	for	a	way	to	model	reality.	Those	topics	are	all	coming	up	in	this
book.
About	This	Book
Because	uncertainty	and	probability	are	so	important	in	quantum	physics,	you
can't	fully	appreciate	the	subject	without	getting	into	calculus.	This	book
presents	the	need-to-know	concepts,	but	you	don't	see	much	in	the	way	of
thought	experiments	that	deal	with	cats	or	parallel	universes.	I	focus	on	the
math	and	how	it	describes	the	quantum	world.
I've	taught	physics	to	many	thousands	of	students	at	the	university	level,	and
from	that	experience,	I	know	most	of	them	share	one	common	trait:	Confusion
as	to	what	they	did	to	deserve	such	torture.
Quantum	Physics	For	Dummies,	Revised	Edition	largely	maps	to	a	college
course,	but	this	book	is	different	from	standard	texts.	Instead	of	writing	it	from
the	physicist's	or	professor's	point	of	view,	I've	tried	to	write	it	from	the
reader's	point	of	view.	In	other	words,	I've	designed	this	book	to	be	crammed
full	of	the	good	stuff		and	only	the	good	stuff.	Not	only	that,	but	you	can
discover	ways	of	looking	at	things	that	professors	and	teachers	use	to	make
figuring	out	problems	simple.
Although	I	encourage	you	to	read	this	book	from	start	to	finish,	you	can	also
leaf	through	this	book	as	you	like,	reading	the	topics	that	you	find	interesting.
Like	other	For	Dummies	books,	this	one	lets	you	skip	around	as	you	like	as
much	as	possible.	You	don't	have	to	read	the	chapters	in	order	if	you	don't	want
to.	This	is	your	book,	and	quantum	physics	is	your	oyster.
Conventions	Used	in	This	Book
Some	books	have	a	dozen	dizzying	conventions	that	you	need	to	know	before
you	can	even	start.	Not	this	one.	Here's	all	you	need	to	know:
	I	put	new	terms	in	italics,	like	this,	the	first	time	they're	discussed;	I	follow
them	with	a	definition.
	Vectors		those	items	that	have	both	a	magnitude	and	a	direction		are
given	in	bold,	like	this:	B.
	Web	addresses	appear	in	monofont.
Foolish	Assumptions
I	don't	assume	that	you	have	any	knowledge	of	quantum	physics	when	you	start
to	read	this	book.	However,	I	do	make	the	following	assumptions:
	You're	taking	a	college	course	in	quantum	physics,	or	you're	interested	in
how	math	describes	motion	and	energy	on	the	atomic	and	subatomic	scale.
	You	have	some	math	prowess.	In	particular,	you	know	some	calculus.	You
don't	need	to	be	a	math	pro,	but	you	should	know	how	to	perform	integration
and	deal	with	differential	equations.	Ideally,	you	also	have	some	experience
with	Hilbert	space.
	You	have	some	physics	background	as	well.	You've	had	a	year's	worth	of
college-level	physics	(or	understand	all	that's	in	Physics	For	Dummies)	before
you	tackle	this	one.
How	This	Book	Is	Organized
Quantum	physics		the	study	of	very	small	objects		is	actually	a	very	big
topic.	To	handle	it,	quantum	physicists	break	the	world	down	into	different
parts.	Here	are	the	various	parts	that	are	coming	up	in	this	book.
Part	I:	Small	World,	Huh?	Essential	Quantum
Physics
Part	I	is	where	you	start	your	quantum	physics	journey,	and	you	get	a	good
overview	of	the	topic	here.	I	survey	quantum	physics	and	tell	you	what	it's	good
for	and	what	kinds	of	problems	it	can	solve.	You	also	get	a	good	foundation	in
the	math	that	you	need	for	the	rest	of	the	book,	such	as	state	vectors	and
quantum	matrix	manipulations.	Knowing	this	stuff	prepares	you	to	handle	the
other	parts.
Part	II:	Bound	and	Undetermined:	Handling
Particles	in	Bound	States
Particles	can	be	trapped	inside	potentials;	for	instance,	electrons	can	be	bound
in	an	atom.	Quantum	physics	excels	at	predicting	the	energy	levels	of	particles
bound	in	various	potentials,	and	that's	what	Part	II	covers.	You	see	how	to
handle	particles	bound	in	square	wells	and	in	harmonic	oscillators.
Part	III:	Turning	to	Angular	Momentum	and	Spin
Quantum	physics	lets	you	work	with	the	micro	world	in	terms	of	the	angu-lar
momentum	of	particles,	as	well	as	the	spin	of	electrons.	Many	famous
experiments		such	as	the	Stern-Gerlach	experiment,	in	which	beams	of
particles	split	in	magnetic	fields		are	understandable	only	in	terms	of
quantum	physics,	and	you	get	all	the	details	here.
Part	IV:	Multiple	Dimensions:	Going	3D	with
Quantum	Physics
In	the	first	three	parts,	all	the	quantum	physics	problems	are	one-dimensional
to	make	life	a	little	easier	while	you're	understanding	how	to	solve	those
problems.	In	Part	IV,	you	branch	out	to	working	with	three-dimensional
problems	in	both	rectangular	and	spherical	coordinate	systems.	Taking	things
from	1D	to	3D	gives	you	a	better	picture	of	what	happens	in	the	real	world.
Part	V:	Group	Dynamics:	Introducing	Multiple
Particles
In	this	part,	you	work	with	multiple-particle	systems,	such	as	atoms	and	gases.
You	see	how	to	handle	many	electrons	in	atoms,	particles	interacting	with	other
particles,	and	particles	that	scatter	off	other	particles.
Dealing	with	multiple	particles	is	all	another	step	in	modeling	reality		after	all,
systems	with	only	a	single	particle	don't	take	you	very	far	in	the	real	world,
which	is	built	of	mega,	mega	systems	of	particles.	In	Part	V,	you	see	how
quantum	physics	can	handle	the	situation.
Part	VI:	The	Part	of	Tens
You	see	the	Part	of	the	Tens	in	all	For	Dummies	books.	This	part	is	made	up	of
fast-paced	lists	of	ten	items	each.	You	get	to	see	some	of	the	ten	best	online
tutorials	on	quantum	physics	and	a	discussion	of	quantum	physics'	ten	greatest
triumphs.
Icons	Used	in	This	Book
You	find	a	handful	of	icons	in	this	book,	and	here's	what	they	mean:
	This	icon	flags	particularly	good	advice,	especially	when	you're	solving
problems.
	This	icon	marks	something	to	remember,	such	as	a	law	of	physics	or	a
particularly	juicy	equation.
	This	icon	means	that	what	follows	is	technical,	insider	stuff.	You	don't
have	to	read	it	if	you	don't	want	to,	but	if	you	want	to	become	a	quantum
physics	pro	(and	who	doesn't?),	take	a	look.
	This	icon	helps	you	avoid	mathematical	or	conceptual	slip-ups.
Where	to	Go	from	Here
All	right,	you're	all	set	and	ready	to	go.	You	can	jump	in	anywhere	you	like.	For
instance,	if	you're	sure	electron	spin	is	going	to	be	a	big	topic	of	conversation	at
a	party	this	weekend,	check	out	Chapter	6.	And	if	your	upcoming	vacation	to
Geneva,	Switzerland,	includes	a	side	trip	to	your	new	favorite	particle
accelerator		the	Large	Hadron	Collider		you	can	flip	to	Chapter	12	and	read
up	on	scattering	theory.	But	if	you	want	to	get	the	full	story	from	the	beginning,
jump	into	Chapter	1	first		that's	where	the	action	starts.
Part	I
Small	World,	Huh?	Essential
Quantum	Physics
In	this	part	.	.	.
This	part	is	designed	to	give	you	an	introduction	to	the	ways	of	quantum
physics.	You	see	the	issues	that	gave	rise	to	quantum	physics	and	the	kinds	of
solutions	it	provides.	I	also	introduce	you	to	the	kind	of	math	that	quantum
physics	requires,	including	the	notion	of	state	vectors.
Chapter	1
Discoveries	and	Essential	Quantum
Physics
In	This	Chapter
	Putting	forth	theories	of	quantization	and	discrete	units
	Experimenting	with	waves	acting	as	particles
	Experimenting	with	particles	acting	as	waves
	Embracing	uncertainty	and	probability
According	to	classical	physics,	particles	are	particles	and	waves	are	waves,	and
never	the	twain	shall	mix.	That	is,	particles	have	an	energy	E	and	a	momentum
vector	p,	and	that's	the	end	of	it.	And	waves,	such	as	light	waves,	have	an
amplitude	A	and	a	wave	vector	k	(where	the	magnitude	of	k	=	
,	where		is
the	wavelength)	that	points	in	the	direction	the	wave	is	traveling.	And	that's	the
end	of	that,	too,	according	to	classical	physics.
But	the	reality	is	different		particles	turn	out	to	exhibit	wave-like	properties,
and	waves	exhibit	particle-like	properties	as	well.	The	idea	that	waves	(like
light)	can	act	as	particles	(like	electrons)	and	vice	versa	was	the	major
revelation	that	ushered	in	quantum	physics	as	such	an	important	part	of	the
world	of	physics.	This	chapter	takes	a	look	at	the	challenges	facing	classical
physics	around	the	turn	of	the	20th	century		and	how	quantum	physics
gradually	came	to	the	rescue.	Up	to	that	point,	the	classical	way	of	looking	at
physics	was	thought	to	explain	just	about	everything.	But	as	those	pesky
experimental	physicists	have	a	way	of	doing,	they	came	up	with	a	bunch	of
experiments	that	the	theoretical	physicists	couldn't	explain.
That	made	the	theoretical	physicists	mad,	and	they	got	on	the	job.	The	problem
here	was	the	microscopic	world		the	world	that's	too	tiny	to	see.	On	the	larger
scale,	classical	physics	could	still	explain	most	of	what	was	going	on		but	when
it	came	to	effects	that	depended	on	the	micro-world,	classical	physics	began	to
break	down.	Taking	a	look	at	how	classical	physics	collapsed	gives	you	an
introduction	to	quantum	physics	that	shows	why	people	needed	it.
Being	Discrete:	The	Trouble	with
Black-Body	Radiation
One	of	the	major	ideas	of	quantum	physics	is,	well,	quantization		measuring
quantities	in	discrete,	not	continuous,	units.	The	idea	of	quantized	energies
arose	with	one	of	the	earliest	challenges	to	classical	physics:	the	problem	of
black-body	radiation.
When	you	heat	an	object,	it	begins	to	glow.	Even	before	the	glow	is	visible,	it's
radiating	in	the	infrared	spectrum.	The	reason	it	glows	is	that	as	you	heat	it,
the	electrons	on	the	surface	of	the	material	are	agitated	thermally,	and
electrons	being	accelerated	and	decelerated	radiate	light.
Physics	in	the	late	19th	and	early	20th	centuries	was	concerned	with	the
spectrum	of	light	being	emitted	by	black	bodies.	A	black	body	is	a	piece	of
material	that	radiates	corresponding	to	its	temperature		but	it	also	absorbs
and	reflects	light	from	its	surroundings.	To	make	matters	easier,	physics
postulated	a	black	body	that	reflected	nothing	and	absorbed	all	the	light	falling
on	it	(hence	the	term	black	body,	because	the	object	would	appear	perfectly
black	as	it	absorbed	all	light	falling	on	it).	When	you	heat	a	black	body,	it	would
radiate,	emitting	light.
Well,	it	was	hard	to	come	up	with	a	physical	black	body		after	all,	what
material	absorbs	light	100	percent	and	doesn't	reflect	anything?	But	the
physicists	were	clever	about	this,	and	they	came	up	with	the	hollow	cavity	you
see	in	Figure	1-1,	with	a	hole	in	it.
When	you	shine	light	on	the	hole,	all	that	light	would	go	inside,	where	it	would
be	reflected	again	and	again		until	it	got	absorbed	(a	negligible	amount	of
light	would	escape	through	the	hole).	And	when	you	heated	the	hollow	cavity,
the	hole	would	begin	to	glow.	So	there	you	have	it		a	pretty	good
approximation	of	a	black	body.
Figure	1-1:	A	black	body.
You	can	see	the	spectrum	of	a	black	body	(and	attempts	to	model	that
spectrum)	in	Figure	1-2,	for	two	different	temperatures,	T1	and	T2.	The	problem
was	that	nobody	was	able	to	come	up	with	a	theoretical	explanation	for	the
spectrum	of	light	generated	by	the	black	body.	Everything	classical	physics
could	come	up	with	went	wrong.
Figure	1-2:	Black-body	radiation	spectrum.
First	attempt:	Wien's	Formula
The	first	one	to	try	to	explain	the	spectrum	of	a	black	body	was	Wilhelm	Wien,
in	1889.	Using	classical	thermodynamics,	he	came	up	with	this	formula:
where	u	(,	T)	is	the	intensity	distribution	of	the	light	spectrum	at	frequency	
of	a	black	body	at	the	temperature	T,	and	A	and		are	constants	which	can	be
measured	in	experiments.	(The	spectrum	is	given	by	u[,	T],	which	is	the
energy	density	of	the	emitted	light	as	a	function	of	frequency	and	temperature.)
This	equation,	Wien's	formula,	worked	fine	for	high	frequencies,	as	you	can	see
in	Figure	1-2;	however,	it	failed	for	low	frequencies.
Second	attempt:	Rayleigh-Jeans	Law
Next	up	in	the	attempt	to	explain	the	black-body	spectrum	was	the	Rayleigh-
Jeans	Law,	introduced	around	1900.	This	law	predicted	that	the	spectrum	of	a
black	body	was
where	k	is	Boltzmann's	constant	(approximately	1.3807		1023	JK1).
However,	the	Rayleigh-Jeans	Law	had	the	opposite	problem	of	Wien's	law:
Although	it	worked	well	at	low	frequencies	(see	Figure	1-2),	it	didn't	match	the
higher-frequency	data	at	all		in	fact,	it	diverged	at	higher	frequencies.	This
was	called	the	ultraviolet	catastrophe	because	the	best	predictions	available
diverged	at	high	frequencies	(corresponding	to	ultraviolet	light).	It	was	time	for
quantum	physics	to	take	over.
An	intuitive	(quantum)	leap:	Max	Planck's
spectrum
The	black-body	problem	was	a	tough	one	to	solve,	and	with	it	came	the	first
begin-nings	of	quantum	physics.	Max	Planck	came	up	with	a	radical	suggestion
	what	if	the	amount	of	energy	that	a	light	wave	can	exchange	with	matter
wasn't	continuous,	as	postulated	by	classical	physics,	but	discrete?	In	other
words,	Planck	postulated	that	the	energy	of	the	light	emitted	from	the	walls	of
the	black-body	cavity	came	only	in	integer	multiples	like	this,	where	h	is	a
universal	constant:
With	this	theory,	crazy	as	it	sounded	in	the	early	1900s,	Planck	converted	the
continuous	integrals	used	by	Rayleigh-Jeans	to	discrete	sums	over	an	infinite
number	of	terms.	Making	that	simple	change	gave	Planck	the	following
equation	for	the	spectrum	of	black-body	radiation:
This	equation	got	it	right		it	exactly	describes	the	black-body	spectrum,	both
at	low	and	high	(and	medium,	for	that	matter)	frequencies.
This	idea	was	quite	new.	What	Planck	was	saying	was	that	the	energy	of	the
radiating	oscillators	in	the	black	body	couldn't	take	on	just	any	level	of	energy,
as	classical	physics	allows;	it	could	take	on	only	specific,	quantized	energies.	In
fact,	Planck	hypothesized	that	that	was	true	for	any	oscillator		that	its	energy
was	an	integral	multiple	of	h.
	And	so	Planck's	equation	came	to	be	known	as	Planck's	quantization	rule,
and	h	became	Planck's	constant:	h	=	6.626		1034	Joule-seconds.	Saying
that	the	energy	of	all	oscillators	was	quantized	was	the	birth	of	quantum
physics.
One	has	to	wonder	how	Planck	came	up	with	his	theory,	because	it's	not	an
obvious	hypothesis.	Oscillators	can	oscillate	only	at	discrete	energies?	Where
did	that	come	from?	In	any	case,	the	revolution	was	on		and	there	was	no
stopping	it.
The	First	Pieces:	Seeing	Light	as
Particles
Light	as	particles?	Isn't	light	made	up	of	waves?	Light,	it	turns	out,	exhibits
properties	of	both	waves	and	particles.	This	section	shows	you	some	of	the
evidence.
Solving	the	photoelectric	effect
The	photoelectric	effect	was	one	of	many	experimental	results	that	made	up	a
crisis	for	classical	physics	around	the	turn	of	the	20th	century.	It	was	also	one
of	Einstein's	first	successes,	and	it	provides	proof	of	the	quantization	of	light.
Here's	what	happened.
When	you	shine	light	onto	metal,	as	Figure	1-3	shows,	you	get	emitted
electrons.	The	electrons	absorb	the	light	you	shine,	and	if	they	get	enough
energy,	they're	able	to	break	free	of	the	metal's	surface.	According	to	classical
physics,	light	is	just	a	wave,	and	it	can	exchange	any	amount	of	energy	with	the
metal.	When	you	beam	light	on	a	piece	of	metal,	the	electrons	in	the	metal
should	absorb	the	light	and	slowly	get	up	enough	energy	to	be	emitted	from	the
metal.	The	idea	was	that	if	you	were	to	shine	more	light	onto	the	metal,	the
electrons	should	be	emitted	with	a	higher	kinetic	energy.	And	very	weak	light
shouldn't	be	able	to	emit	electrons	at	all,	except	in	a	matter	of	hours.
But	that's	not	what	happened		electrons	were	emitted	as	soon	as	someone
shone	light	on	the	metal.	In	fact,	no	matter	how	weak	the	intensity	of	the
incident	light	(and	researchers	tried	experiments	with	such	weak	light	that	it
should	have	taken	hours	to	get	any	electrons	emitted),	electrons	were	emitted.
Immediately.
Figure	1-3:	The	photo-electric	effect.
Experiments	with	the	photoelectric	effect	showed	that	the	kinetic	energy,	K,	of
the	emitted	electrons	depended	only	on	the	frequency		not	the	intensity		of
the	incident	light,	as	you	can	see	in	Figure	1-4.
Figure	1-4:	Kinetic	energy	of	emitted	electrons	versus	frequency	of	the	incident	light.
In	Figure	1-4,	0	is	called	the	threshold	frequency,	and	if	you	shine	light	with	a
frequency	below	this	threshold	on	the	metal,	no	electrons	are	emitted.	The
emitted	electrons	come	from	the	pool	of	free	electrons	in	the	metal	(all	metals
have	a	pool	of	free	electrons),	and	you	need	to	supply	these	electrons	with	an
energy	equivalent	to	the	metal's	work	function,	W,	to	emit	the	electron	from	the
metal's	surface.
The	results	were	hard	to	explain	classically,	so	enter	Einstein.	This	was	the
beginning	of	his	heyday,	around	1905.	Encouraged	by	Planck's	success	(see	the
preceding	section),	Einstein	postulated	that	not	only	were	oscillators	quantized
but	so	was	light		into	discrete	units	called	photons.	Light,	he	suggested,	acted
like	particles	as	well	as	waves.
So	in	this	scheme,	when	light	hits	a	metal	surface,	photons	hit	the	free
electrons,	and	an	electron	completely	absorbs	each	photon.	When	the	energy,
h,	of	the	photon	is	greater	than	the	work	function	of	the	metal,	the	electron	is
emitted.	That	is,
h	=	W	+	K
where	W	is	the	metal's	work	function	and	K	is	the	kinetic	energy	of	the	emitted
electron.	Solving	for	K	gives	you	the	following:
K	=	h		W
You	can	also	write	this	in	terms	of	the	threshold	frequency	this	way:
K	=	h(		0)
So	apparently,	light	isn't	just	a	wave;	you	can	also	view	it	as	a	particle,	the
photon.	In	other	words,	light	is	quantized.
That	was	also	quite	an	unexpected	piece	of	work	by	Einstein,	although	it	was
based	on	the	earlier	work	of	Planck.	Light	quantized?	Light	coming	in	discrete
energy	packets?	What	next?
Scattering	light	off	electrons:	The	Compton	effect
To	a	world	that	still	had	trouble	comprehending	light	as	particles	(see	the
preceding	section),	Arthur	Compton	supplied	the	final	blow	with	the	Compton
effect.	His	experiment	involved	scattering	photons	off	electrons,	as	Figure	1-5
shows.
Figure	1-5:	Light	incident	on	an	electron	at	rest.
Incident	light	comes	in	with	a	wavelength	of		and	hits	the	electron	at	rest.
After	that	happens,	the	light	is	scattered,	as	you	see	in	Figure	1-6.
Figure	1-6:	Photon	scattering	off	an	electron.
Classically,	here's	what	should've	happened:	The	electron	should've	absorbed
the	incident	light,	oscillated,	and	emitted	it		with	the	same	wavelength	but
with	an	intensity	depending	on	the	intensity	of	the	incident	light.	But	that's	not
what	happened		in	fact,	the	wavelength	of	the	light	is	actually	changed	by	,
called	the	wavelength	shift.	The	scattered	light	has	a	wavelength	of		+			in
other	words,	its	wavelength	has	increased,	which	means	the	light	has	lost
energy.	And		depends	on	the	scattering	angle,	,	not	on	the	intensity	of	the
incident	light.
Arthur	Compton	could	explain	the	results	of	his	experiment	only	by	making	the
assumption	that	he	was	actually	dealing	with	two	particles		a	photon	and	an
electron.	That	is,	he	treated	light	as	a	discrete	particle,	not	a	wave.	And	he
made	the	assumption	that	the	photon	and	the	electron	collided	elastically	
that	is,	that	both	total	energy	and	momentum	were	conserved.
Making	the	assumption	that	both	the	light	and	the	electron	were	particles,
Compton	then	derived	this	formula	for	the	wavelength	shift	(it's	an	easy
calculation	if	you	assume	that	the	light	is	represented	by	a	photon	with	energy
E	=	h	and	that	its	momentum	is	p	=	E/c):
where	h	is	Planck's	constant,	me	is	the	mass	of	an	electron,	c	is	the	speed	of
light,	and		is	the	scattering	angle	of	the	light.
You	also	see	this	equation	in	the	equivalent	form:
where	c	is	the	Compton	wavelength	of	an	electron,	
.	And	experiment
confirms	this	relation		both	equations.
Note	that	to	derive	the	wavelength	shift,	Compton	had	to	make	the	assumption
that	here,	light	was	acting	as	a	particle,	not	as	a	wave.	That	is,	the	particle
nature	of	light	was	the	aspect	of	the	light	that	was	predominant.
Proof	positron?	Dirac	and	pair	production
In	1928,	the	physicist	Paul	Dirac	posited	the	existence	of	a	positively	charged
anti-electron,	the	positron.	He	did	this	by	taking	the	newly	evolving	field	of
quantum	physics	to	new	territory	by	combining	relativity	with	quantum
mechanics	to	create	relativistic	quantum	mechanics		and	that	was	the	theory
that	predicted,	through	a	plus/minus-sign	interchange		the	existence	of	the
positron.
It	was	a	bold	prediction		an	anti-particle	of	the	electron?	But	just	four	years
later,	physicists	actually	saw	the	positron.	Today's	high-powered	elementary
particle	physics	has	all	kinds	of	synchrotrons	and	other	particle	accelerators	to
create	all	the	elementary	particles	they	need,	but	in	the	early	20th	century,	this
wasn't	always	so.
In	those	days,	physicists	relied	on	cosmic	rays		those	particles	and	high-
powered	photons	(called	gamma	rays)	that	strike	the	Earth	from	outer	space	
as	their	source	of	particles.	They	used	cloud-chambers,	which	were	filled	with
vapor	from	dry	ice,	to	see	the	trails	such	particles	left.	They	put	their	chambers
into	magnetic	fields	to	be	able	to	measure	the	momentum	of	the	particles	as
they	curved	in	those	fields.
In	1932,	a	physicist	noticed	a	surprising	event.	A	pair	of	particles,	oppositely
charged	(which	could	be	determined	from	the	way	they	curved	in	the	magnetic
field)	appeared	from	apparently	nowhere.	No	particle	trail	led	to	the	origin	of
the	two	particles	that	appeared.	That	was	pair-production		the	conversion	of
a	high-powered	photon	into	an	electron	and	positron,	which	can	happen	when
the	photon	passes	near	a	heavy	atomic	nucleus.
So	experimentally,	physicists	had	now	seen	a	photon	turning	into	a	pair	of
particles.	Wow.	As	if	everyone	needed	more	evidence	of	the	particle	nature	of
light.	Later	on,	researchers	also	saw	pair	annihilation:	the	conversion	of	an
electron	and	positron	into	pure	light.
Pair	production	and	annihilation	turned	out	to	be	governed	by	Einstein's	newly
introduced	theory	of	relativity		in	particular,	his	most	famous	formula,	E	=
mc2,	which	gives	the	pure	energy	equivalent	of	mass.	At	this	point,	there	was	an
abundance	of	evidence	of	the	particle-like	aspects	of	light.
A	Dual	Identity:	Looking	at	Particles
as	Waves
In	1923,	the	physicist	Louis	de	Broglie	suggested	that	not	only	did	waves
exhibit	particle-like	aspects	but	the	reverse	was	also	true		all	material
particles	should	display	wave-like	properties.
How	does	this	work?	For	a	photon,	momentum	
,	where		is	the
photon's	frequency	and		is	its	wavelength.	And	the	wave	vector,	k,	is	equal	to
k	=	p/,	where		=	h/2.	De	Broglie	said	that	the	same	relation	should	hold	for
all	material	particles.	That	is,
De	Broglie	presented	these	apparently	surprising	suggestions	in	his	Ph.D.
thesis.	Researchers	put	these	suggestions	to	the	test	by	sending	a	beam
through	a	dual-slit	apparatus	to	see	whether	the	electron	beam	would	act	like	it
was	made	up	of	particles	or	waves.	In	Figure	1-7,	you	can	see	the	setup	and	the
results.
Figure	1-7:	An	electron	beam	going	through	two	slits.
In	Figure	1-7a,	you	can	see	a	beam	of	electrons	passing	through	a	single	slit
and	the	resulting	pattern	on	a	screen.	In	Figure	1-7b,	the	electrons	are	passing
through	a	second	slit.	Classically,	you'd	expect	the	intensities	of	Figure	1-7a
and	1-7b	simply	to	add	when	both	slits	are	open:
I	=	I1	+	I2
But	that's	not	what	happened.	What	actually	appeared	was	an	interference
pattern	when	both	slits	were	open	(Figure	1-7c),	not	just	a	sum	of	the	two	slits'
electron	intensities.
The	result	was	a	validation	of	de	Broglie's	invention	of	matter	waves.
Experiment	bore	out	the	relation	that	
,	and	de	Broglie	was	a	success.
	The	idea	of	matter	waves	is	a	big	part	of	what's	coming	up	in	the	rest	of
the	book.	In	particular,	the	existence	of	matter	waves	says	that	you	add	the
waves'	amplitude,	1(r,	t)	and	2(r,	t),	not	their	intensities,	to	sum	them:
(r,	t)	=	1(r,	t)	+	2(r,	t)
You	square	the	amplitude	to	get	the	intensity,	and	the	phase	difference
between	1(r,	t)	and	2(r,	t)	is	what	actually	creates	the	interference	pattern
that's	observed.
You	Can't	Know	Everything	(But	You
Can	Figure	the	Odds)
	So	particles	apparently	exhibit	wave-like	properties,	and	waves	exhibit
particle-like	properties.	But	if	you	have	an	electron,	which	is	it		a	wave	or
a	particle?	The	truth	is	that	physically,	an	electron	is	just	an	electron,	and
you	can't	actually	say	whether	it's	a	wave	or	a	particle.	The	act	of
measurement	is	what	brings	out	the	wave	or	particle	properties.	You	see
more	about	this	idea	throughout	the	book.
Quantum	mechanics	lives	with	an	uncertain	picture	quite	happily.	That	view
offended	many	eminent	physicists	of	the	time		notably	Albert	Einstein,	who
said,	famously,	"God	does	not	play	dice."	In	this	section,	I	discuss	the	idea	of
uncertainty	and	how	quantum	physicists	work	in	probabilities	instead.
The	Heisenberg	uncertainty	principle
The	fact	that	matter	exhibits	wave-like	properties	gives	rise	to	more	trouble	
waves	aren't	localized	in	space.	And	knowing	that	inspired	Werner	Heisenberg,
in	1927,	to	come	up	with	his	celebrated	uncertainty	principle.
You	can	completely	describe	objects	in	classical	physics	by	their	momentum
and	position,	both	of	which	you	can	measure	exactly.	In	other	words,	classical
physics	is	completely	deterministic.
On	the	atomic	level,	however,	quantum	physics	paints	a	different	picture.	Here,
the	Heisenberg	uncertainty	principle	says	that	there's	an	inherent	uncertainty
in	the	relation	between	position	and	momentum.	In	the	x	direction,	for	example,
that	looks	like	this:
where	x	is	the	measurement	uncertainty	in	the	particle's	x	position,	
	is	its
measurement	uncertainty	in	its	momentum	in	the	x	direction	and		=	h/2.
That	is	to	say,	the	more	accurately	you	know	the	position	of	a	particle,	the	less
accurately	you	know	the	momentum,	and	vice	versa.	This	relation	holds	for	all
three	dimensions:
And	the	Heisenberg	uncertainty	principle	is	a	direct	consequence	of	the	wave-
like	nature	of	matter,	because	you	can't	completely	pin	down	a	wave.
	Quantum	physics,	unlike	classical	physics,	is	completely	undeterministic.
You	can	never	know	the	precise	position	and	momentum	of	a	particle	at	any
one	time.	You	can	give	only	probabilities	for	these	linked	measurements.
Rolling	the	dice:	Quantum	physics	and	probability
In	quantum	physics,	the	state	of	a	particle	is	described	by	a	wave	function,	(r,
t).	The	wave	function	describes	the	de	Broglie	wave	of	a	particle,	giving	its
amplitude	as	a	function	of	position	and	time.	(See	the	earlier	section	"A	Dual
Identity:	Looking	at	Particles	as	Waves"	for	more	on	de	Broglie.)
	Note	that	the	wave	function	gives	a	particle's	amplitude,	not	intensity;	if
you	want	to	find	the	intensity	of	the	wave	function,	you	have	to	square	it:	|
(r,	t)|2.	The	intensity	of	a	wave	is	what's	equal	to	the	probability	that	the
particle	will	be	at	that	position	at	that	time.
That's	how	quantum	physics	converts	issues	of	momentum	and	position	into
probabilities:	by	using	a	wave	function,	whose	square	tells	you	the	probability
density	that	a	particle	will	occupy	a	particular	position	or	have	a	particular
momentum.	In	other	words,	|(r,	t)|2d3r	is	the	probability	that	the	particle	will
be	found	in	the	volume	element	d3r,	located	at	position	r	at	time	t.
Besides	the	position-space	wave	function	(r,	t),	there's	also	a	momentum-
space	version	of	the	wave	function:	(p,	t).
This	book	is	largely	a	study	of	the	wave	function		the	wave	functions	of	free
particles,	the	wave	functions	of	particles	trapped	inside	potentials,	of	identical
particles	hitting	each	other,	of	particles	in	harmonic	oscillation,	of	light
scattering	from	particles,	and	more.	Using	this	kind	of	physics,	you	can	predict
the	behavior	of	all	kinds	of	physical	systems.
Chapter	2
Entering	the	Matrix:	Welcome
to	State	Vectors
In	This	Chapter
	Creating	state	vectors
	Using	Dirac	notation	for	state	vectors
	Working	with	bras	and	kets
	Understanding	matrix	mechanics
	Getting	to	wave	mechanics
Quantum	physics	isn't	just	about	playing	around	with	your	particle	accelerator
while	trying	not	to	destroy	the	universe.	Sometimes,	you	get	to	do	things	that
are	a	little	more	mundane,	like	turn	lights	off	and	on,	perform	a	bit	of	calculus,
or	play	with	dice.
If	you're	actually	doing	physics	with	those	dice	(beyond	hurling	them	across	the
room),	the	lab	director	won't	even	get	mad	at	you.	In	quantum	physics,
absolute	measurements	are	replaced	by	probabilities,	so	you	may	use	dice	to
calculate	the	probabilities	that	various	numbers	will	come	up.	You	can	then
assemble	those	values	into	a	vector	(single-column	matrix)	in	Hilbert	space	(a
type	of	infinitely	dimensional	vector	space	with	some	properties	that	are
especially	valuable	in	quantum	physics).
This	chapter	introduces	how	you	deal	with	probabilities	in	quantum	physics,
starting	by	viewing	the	various	possible	states	a	particle	can	occupy	as	a	vector
	a	vector	of	probability	states.	From	there,	I	help	you	familiarize	yourself	with
some	mathematical	notations	common	in	quantum	physics,	including	bras,	kets,
matrices,	and	wave	functions.	Along	the	way,	you	also	get	to	work	with	some
important	operators.
Creating	Your	Own	Vectors	in
Hilbert	Space
In	quantum	physics,	probabilities	take	the	place	of	absolute	measurements.	Say
you've	been	experimenting	with	rolling	a	pair	of	dice	and	are	trying	to	figure
the	relative	probability	that	the	dice	will	show	various	values.	You	come	up	with
a	list	indicating	the	relative	probability	of	rolling	a	2,	3,	4,	and	so	on,	all	the	way
up	to	12:
Sum	of	the	Dice	Relative	Probability	
(Number	of	Ways	of	Rolling	a	Particular	Total)
2
1
3
2
4
3
5
6
6
5
7
6
8
5
9
4
10 3
11 2
12 1
In	other	words,	you're	twice	as	likely	to	roll	a	3	than	a	2,	you're	four	times	as
likely	to	roll	a	5	than	a	2,	and	so	on.	You	can	assemble	these	relative
probabilities	into	a	vector	(if	you're	thinking	of	a	"vector"	from	physics,	think	in
terms	of	a	column	of	the	vector's	components,	not	a	magnitude	and	direction)
to	keep	track	of	them	easily:
Okay,	now	you're	getting	closer	to	the	way	quantum	physics	works.	You	have	a
vector	of	the	probabilities	that	the	dice	will	occupy	various	states.	However,
quantum	physics	doesn't	deal	directly	with	probabilities	but	rather	with
probability	amplitudes,	which	are	the	square	roots	of	the	probabilities.	To	find
the	actual	probability	that	a	particle	will	be	in	a	certain	state,	you	add	wave
functions		which	are	going	to	be	represented	by	these	vectors		and	then
square	them	(see	Chapter	1	for	info	on	why).	So	take	the	square	root	of	all
these	entries	to	get	the	probability	amplitudes:
That's	better,	but	adding	the	squares	of	all	these	should	add	up	to	a	total
probability	of	1;	as	it	is	now,	the	sum	of	the	squares	of	these	numbers	is	36,	so
divide	each	entry	by	361/2,	or	6:
So	now	you	can	get	the	probability	amplitude	of	rolling	any	combination	from	2
to	12	by	reading	down	the	vector		the	probability	amplitude	of	rolling	a	2	is
1/6,	of	rolling	a	3	is	
	and	so	on.
Making	Life	Easier	with	Dirac
Notation
When	you	have	a	state	vector	that	gives	the	probability	amplitude	that	a	pair	of
dice	will	be	in	their	various	possible	states,	you	basically	have	a	vector	in	dice
space		all	the	possible	states	that	a	pair	of	dice	can	take,	which	is	an	11-
dimensional	space.	(See	the	preceding	section	for	more	on	state	vectors.)
But	in	most	quantum	physics	problems,	the	vectors	can	be	infinitely	large		for
example,	a	moving	particle	can	be	in	an	infinite	number	of	states.	Handling
large	arrays	of	states	isn't	easy	using	vector	notation,	so	instead	of	explicitly
writing	out	the	whole	vector	each	time,	quantum	physics	usually	uses	the
notation	developed	by	physicist	Paul	Dirac		the	Dirac	or	bra-ket	notation.
Abbreviating	state	vectors	as	kets
Dirac	notation	abbreviates	the	state	vector	as	a	ket,	like	this:	|	>.	So	in	the
dice	example,	you	can	write	the	state	vector	as	a	ket	this	way:
Here,	the	components	of	the	state	vector	are	represented	by	numbers	in	11-
dimensional	dice	space.	More	commonly,	however,	each	component	represents
a	function	of	position	and	time,	something	like	this:
	In	general,	a	set	of	vectors	N	in	Hilbert	space	is	linearly	independent	if
the	only	solution	to	the	following	equation	is	that	all	the	coefficients	an	=	0:
That	is,	as	long	as	you	can't	write	any	one	vector	as	a	linear	combination	of	the
others,	the	vectors	are	linearly	independent	and	so	form	a	valid	basis	in	Hilbert
space.
Writing	the	Hermitian	conjugate	as	a	bra
For	every	ket,	there's	a	corresponding	bra.	(The	terms	come	from	bra-ket,	or
bracket,	which	should	be	clearer	in	the	upcoming	section	titled	"Grooving	with
Operators.")	A	bra	is	the	Hermitian	conjugate	of	the	corresponding	ket.
Suppose	you	start	with	this	ket:
The	*	symbol	means	the	complex	conjugate.	(A	complex	conjugate	flips	the	sign
connecting	the	real	and	imaginary	parts	of	a	complex	number.)	So	the
corresponding	bra,	which	you	write	as	<|,	equals	|>T*.	The	bra	is	this	row
vector:
	Note	that	if	any	of	the	elements	of	the	ket	are	complex	numbers,	you
have	to	take	their	complex	conjugate	when	creating	the	associated	bra.	For
instance,	if	your	complex	number	in	the	ket	is	a	+	bi,	its	complex	conjugate
in	the	bra	is	a		bi.
Multiplying	bras	and	kets:	A	probability	of	1
You	can	take	the	product	of	your	ket	and	bra,	denoted	as	<|>,	like	this:
This	is	just	matrix	multiplication,	and	the	result	is	the	same	as	taking	the	sum	of
the	squares	of	the	elements:
	And	that's	the	way	it	should	be,	because	the	total	probability	should	add
up	to	1.	Therefore,	in	general,	the	product	of	the	bra	and	ket	equals	1:
If	this	relation	holds,	the	ket	|>	is	said	to	be	normalized.
Covering	all	your	bases:	Bras	and	kets	as	basis-
less	state	vectors
The	reason	ket	notation,	|>,	is	so	popular	in	quantum	physics	is	that	it	allows
you	to	work	with	state	vectors	in	a	basis-free	way.	In	other	words,	you're	not
stuck	in	the	position	basis,	the	momentum	basis,	or	the	energy	basis.	That's
helpful,	because	most	of	the	work	in	quantum	physics	takes	place	in	abstract
calculations,	and	you	don't	want	to	have	to	drag	all	the	components	of	state
vectors	through	those	calculations	(often	you	can't		there	may	be	infinitely
many	possible	states	in	the	problem	you're	dealing	with).
For	example,	say	that	you're	representing	your	states	using	position	vectors	in
a	three-dimensional	Hilbert	space		that	is,	you	have	x,	y,	and	z	axes,	forming	a
position	basis	for	your	space.	That's	fine,	but	not	all	your	calculations	have	to	be
done	using	that	position	basis.
You	may	want	to,	for	example,	represent	your	states	in	a	three-dimensional
momentum	space,	with	three	axes	in	Hilbert	space,	px,	py,	and	pz.	Now	you'd
have	to	change	all	your	position	vectors	to	momentum	vectors,	adjusting	each
component,	and	keep	track	of	what	happens	to	every	component	through	all
your	calculations.
So	Dirac's	bra-ket	notation	comes	to	the	rescue	here		you	use	it	to	perform	all
the	math	and	then	plug	in	the	various	components	of	your	state	vectors	as
needed	at	the	end.	That	is,	you	can	perform	your	calculations	in	purely	symbolic
terms,	without	being	tied	to	a	basis.
And	when	you	need	to	deal	with	the	components	of	a	ket,	such	as	when	you
want	to	get	physical	answers,	you	can	also	convert	kets	to	a	different	basis	by
taking	the	ket's	components	along	the	axes	of	that	basis.	Generally,	when	you
have	a	vector|>,	you	can	express	it	as	a	sum	over	N	basis	vectors,	|i,	like	so:
where	N	is	the	dimension	of	the	Hilbert	space,	and	i	is	an
integer	that	labels	the	basis	vectors.
Understanding	some	relationships	using	kets
Ket	notation	makes	the	math	easier	than	it	is	in	matrix	form	because	you	can
take	advantage	of	a	few	mathematical	relationships.	For	example,	here's	the	so-
called	Schwarz	inequality	for	state	vectors:
This	says	that	the	square	of	the	absolute	value	of	the	product	of	two	state
vectors,	|<|>|2,	is	less	than	or	equal	to	<|><|>.	This	turns	out	the	be
the	analog	of	the	vector	inequality:
So	why	is	the	Schwarz	inequality	so	useful?	It	turns	out	that	you	can	derive	the
Heisenberg	uncertainty	principle	from	it	(see	Chapter	1	for	more	on	this
principle).
Other	ket	relationships	can	also	simplify	your	calculations.	For	instance,	two
kets,	|>	and	|>,	are	said	to	be	orthogonal	if
And	two	kets	are	said	to	be	orthonormal	if	they	meet	the	following	conditions:
	
	
	
With	this	information	in	mind,	you're	now	ready	to	start	working	with
operators.
Grooving	with	Operators
What	about	all	the	calculations	that	you're	supposed	to	be	able	to	perform	with
kets?	Taking	the	product	of	a	bra	and	a	ket,	<|>,	is	fine	as	far	as	it	goes,	but
what	about	extracting	some	physical	quantities	you	can	measure?	That's	where
operators	come	in.
Hello,	operator:	How	operators	work
Here's	the	general	definition	of	an	operator	A	in	quantum	physics:	An	operator
is	a	mathematical	rule	that,	when	operating	on	a	ket,	|>,	transforms	that	ket
into	a	new	ket,	|'>	in	the	same	space	(which	could	just	be	the	old	ket
multiplied	by	a	scalar).	So	when	you	have	an	operator	A,	it	transforms	kets	like
this:
For	that	matter,	the	same	operator	can	also	transform	bras:
	Here	are	several	examples	of	the	kinds	of	operators	you'll	see:
	Hamiltonian	(H):	Applying	the	Hamiltonian	operator	(which	looks	different
for	every	different	physical	situation)	gives	you	E,	the	energy	of	the	particle
represented	by	the	ket	|>;	E	is	a	scalar	quantity:
	Unity	or	identity	(I):	The	unity	or	identity	operator	leaves	kets	unchanged:
	Gradient	():	The	gradient	operator	works	like	this:
	Linear	momentum	(P):	The	linear	momentum	operator	looks	like	this	in
quantum	mechanics:
	Laplacian	
:	You	use	the	Laplacian	operator,	which	is	much	like	a	second-
order	gradient,	to	create	the	energy-finding	Hamiltonian	operator:
	In	general,	multiplying	operators	together	is	not	the	same	independent
of	order,	so	for	the	operators	A	and	B,	AB		BA.
And	an	operator	A	is	said	to	be	linear	if
I	expected	that:	Finding	expectation	values
Given	that	everything	in	quantum	physics	is	done	in	terms	of	probabilities,
making	predictions	becomes	very	important.	And	the	biggest	such	prediction	is
the	expectation	value.	The	expectation	value	of	an	operator	is	the	average	value
that	you	would	measure	if	you	performed	the	measurement	many	times.	For
example,	the	expectation	value	of	the	Hamiltonian	operator	(see	the	preceding
section)	is	the	average	energy	of	the	system	you're	studying.
	The	expectation	value	is	a	weighted	average	of	the	probabilities	of	the
system's	being	in	its	various	possible	states.	Here's	how	you	find	the
expectation	value	of	an	operator	A:
Note	that	because	you	can	express	<|	as	a	row	operator	and	|>	as	a	column
vector,	you	can	express	the	operator	A	as	a	square	matrix.
For	example,	suppose	you're	working	with	a	pair	of	dice	and	the	probabilities	of
all	the	possible	sums	(see	the	earlier	section	"Creating	Your	Own	Vectors	in
Hilbert	Space").	In	this	dice	example,	the	expectation	value	is	a	sum	of	terms,
and	each	term	is	a	value	that	can	be	displayed	by	the	dice,	multiplied	by	the
probability	that	that	value	will	appear.
The	bra	and	ket	will	handle	the	probabilities,	so	it's	up	to	the	operator	that	you
create	for	this		call	it	the	Roll	operator,	R		to	store	the	dice	values	(2
through	12)	for	each	probability.	Therefore,	the	operator	R	looks	like	this:
So	to	find	the	expectation	value	of	R,	you	need	to	calculate	<|R|>.	Spelling
that	out	in	terms	of	components	gives	you	the	following:
Doing	the	math,	you	get
So	the	expectation	value	of	a	roll	of	the	dice	is	7.	Now	you	can	see	where	the
terms	bra	and	ket	come	from		they	"bracket"	an	operator	to	give	you
expectation	values.	In	fact,	the	expectation	value	is	such	a	common	thing	to	find
that	you'll	often	find	<|R|>	abbreviated	as	<R>,	so
Looking	at	linear	operators
An	operator	A	is	said	to	be	linear	if	it	meets	the	following	condition:
For	instance,	the	expression	|><|	is	actually	a	linear	operator.	In	order	for
us	to	see	this	we	shall	need	to	know	just	a	little	more	about	what	happens	when
we	take	the	products	of	bras	and	kets.	Firstly,	if	we	take	the	product	of	the	bra,
<|,	with	the	ket,	c|>,	where	c	is	a	complex	number,	then	we	get	the	answer,
Secondly,	if	we	take	the	product	of	the	bra,	<|,	with	the	sum	of	two	kets,	|1>
+|	2>,	then	we	get	the	answer,
Now	that	we	know	this	we	can	test	to	see	if	|><|	is	actually	a	linear	operator.
OK	then,	let's	apply	|><|	to	a	linear	combination	of	bras,	like	so,
where	c1	and	c2	are	complex	numbers.	Now	that	you	know	how	the	product	of	a
bra	with	a	sum	of	two	kets	goes,	you	can	say,
Then,	as	you	know,	<|c|>	=	c<|>,	you	can	finally	write	this	as,
This	is	exactly	what	a	linear	operator	should	do		if	you	replace	A	in	the	above
equation	defining	a	linear	operator,	with	|><|,	then	the	result	is	the	same	as
the	one	you	just	found.	So	|><|	is	indeed	a	linear	operator		though	I	would
agree,	it's	a	pretty	funny	looking	one!Going	Hermitian	with	Hermitian
Operators	and	Adjoints
The	Hermitian	adjoint		also	called	the	adjoint	or	Hermitian	conjugate		of	an
operator	A	is	denoted	A.	To	find	the	Hermitian	adjoint,	follow	these	steps:
1.	Replace	complex	constants	with	their	complex	conjugates.
The	Hermitian	adjoint	of	a	complex	number	is	the	complex	conjugate	of	that
number:
2.	Replace	kets	with	their	corresponding	bras,	and	replace	bras	with
their	corresponding	kets.
	You	have	to	exchange	the	bras	and	kets	when	finding	the	Hermitian
adjoint	of	an	operator,	so	finding	the	Hermitian	adjoint	of	an	operator	is	not
just	the	same	as	mathematically	finding	its	complex	conjugate.
3.	Replace	operators	with	their	Hermitian	adjoints.
In	quantum	mechanics,	operators	that	are	equal	to	their	Hermitian	adjoints
are	called	Hermitian	operators.	In	other	words,	an	operator	is	Hermitian	if
Hermitian	operators	appear	throughout	the	book,	and	they	have	special
properties.	For	instance,	the	matrix	that	represents	them	may	be
diagonalized		that	is,	written	so	that	the	only	nonzero	elements	appear
along	the	matrix's	diagonal.	Also,	the	expectation	value	of	a	Hermitian
operator	is	guaranteed	to	be	a	real	number,	not	complex	(see	the	earlier
section	"I	expected	that:	Finding	expectation	values").
4.	Write	your	final	equation.
	Here	are	some	relationships	concerning	Hermitian	adjoints:
	
	
	
	
	
Forward	and	Backward:	Finding	the
Commutator
	The	measure	of	how	different	it	is	to	apply	operator	A	and	then	B,	versus
B	and	then	A,	is	called	the	operators'	commutator.	Here's	how	you	define
the	commutator	of	operators	A	and	B:
[A,	B]	=	AB		BA
Commuting
Two	operators	commute	with	each	other	if	their	commutator	is	equal	to	zero.
That	is,	it	doesn't	make	any	difference	in	what	order	you	apply	them:
[A,	B]	=	0
Note	in	particular	that	any	operator	commutes	with	itself:
[A,	A]	=	0
And	it's	easy	to	show	that	the	commutator	of	A,	B	is	the	negative	of	the
commutator	of	B,	A:
[A,	B]	=	[B,	A]
It's	also	true	that	commutators	are	linear:
[A,	B	+	C	+	D	+	...]	=	[A,	B]	+	[A,	C]	+	[A,	D]	+	...
And	the	Hermitian	adjoint	of	a	commutator	works	this	way:
You	can	also	find	the	anticommutator,	{A,	B}:
{A,	B}	=	AB	+	BA
Finding	anti-Hermitian	operators
Here's	another	one:	What	can	you	say	about	the	Hermitian	adjoint	of	the
commutator	of	two	Hermitian	operators?	Here's	the	answer.	First,	write	the
adjoint:
The	definition	of	commutators	tells	you	the	following:
You	know	(AB)	=	BA	(see	the	earlier	section	"Going	Hermitian	with	Hermitian
Operators	and	Adjoints"	for	properties	of	adjoints).	Therefore,
But	for	Hermitian	operators,	A	=	A,	so	remove	the		symbols:
But	BA		AB	is	just	[A,	B],	so	you	have	the	following:
	A	and	B	here	are	Hermitian	operators.	When	you	take	the	Hermitian
adjoint	of	an	expression	and	get	the	same	thing	back	with	a	negative	sign	in
front	of	it,	the	expression	is	called	anti-Hermitian,	so	the	commutator	of	two
Hermitian	operators	is	anti-Hermitian.	(And	by	the	way,	the	expectation
value	of	an	anti-Hermitian	operator	is	guaranteed	to	be	completely
imaginary.)
Starting	from	Scratch	and	Ending
Up	with	Heisenberg
If	you've	read	through	the	last	few	sections,	you're	now	armed	with	all	this	new
technology:	Hermitian	operators	and	commutators.	How	can	you	put	it	to
work?	You	can	come	up	with	the	Heisenberg	uncertainty	relation	starting
virtually	from	scratch.
Here's	a	calculation	that	takes	you	from	a	few	basic	definitions	to	the
Heisenberg	uncertainty	relation.	This	kind	of	calculation	shows	how	much
easier	it	is	to	use	the	basis-less	bra	and	ket	notation	than	the	full	matrix	version
of	state	vectors.	This	isn't	the	kind	of	calculation	that	you'll	need	to	do	in	class,
but	follow	it	through		knowing	how	to	use	kets,	bras,	commutators,	and
Hermitian	operators	is	vital	in	the	coming	chapters.
The	uncertainty	in	a	measurement	of	the	Hermitian	operator	named	A	is
formally	given	by
That	is,	A	is	equal	to	the	square	root	of	the	expectation	value	of	A2	minus	the
squared	expectation	value	of	A.	If	you've	taken	any	math	classes	that	dealt	with
statistics,	this	formula	may	be	familiar	to	you.	Similarly,	the	uncertainty	in	a
measurement	using	Hermitian	operator	B	is
Now	consider	the	operators	A	and	B	(not	the	uncertainties	A	and	B
anymore),	and	assume	that	applying	A	and	B	as	operators	gives	you
measurement	values	like	this:
Like	any	operator,	using	A	and	B	can	result	in	new	kets:
Here's	the	key:	The	Schwarz	inequaility	(from	the	earlier	section
"Understanding	some	relationships	using	kets")	gives	you
So	you	can	see	that	the	inequality	sign,	,	which	plays	a	big	part	in	the
Heisenberg	uncertainty	relation,	has	already	crept	into	the	calculation.
Because	A	and	B	are	Hermitian,	<|>	is	equal	to	<|A2|>	and	<|>	is
equal	to	<|B2|>.	Because	A	=	A	(the	definition	of	a	Hermitian	operator),
you	can	see	that
This	means	that
That	is,	<|>	is	equal	to	<A2>	and	<|	>	is	equal	to	<B2>.	So	you	can
rewrite	the	Schwarz	inequality	like	this:
Okay,	where	has	this	gotten	you?	It's	time	to	be	clever.	Note	that	you	can	write
AB	as
Here,	{A,	B}	=	AB	+	BA	is	the	anticommutator	of	the	operators	A	and
B.	Because	[A,	B]	=	[A,	B]	(the	constants	<A>	and	<B>	subtract	out),	you
can	rewrite	this	equation:
Here's	where	the	math	gets	intense.	Take	a	look	at	what	you	know	so	far:
	The	commutator	of	two	Hermitian	operators,	[A,	B],	is	anti-Hermitian.
	The	expectation	value	of	an	anti-Hermitian	is	imaginary.
	{A,	B}	is	Hermitian.
	The	expectation	value	of	a	Hermitian	is	real.
All	this	means	that	you	can	view	the	expectation	value	of	the	equation	as	the
sum	of	real	({A,	B})	and	imaginary	([A,	B])	parts,	so
And	because	the	second	term	on	the	right	is	positive	or	zero,	you	can	say	that
the	following	is	true:
Whew!	But	now	compare	this	equation	to	the	relationship	from	the	earlier	use
of	the	Schwarz	inequality:
Combining	the	two	equations	gives	you	this:
This	has	the	look	of	the	Heisenberg	uncertainty	relation,	except	for	the	pesky
expectation	value	brackets,	<	>,	and	the	fact	that	A	and	B	appear	squared
here.	You	want	to	reproduce	the	Heisenberg	uncertainty	relation	here,	which
looks	like	this:
Okay,	so	how	do	you	get	the	left	side	of	the	equation	from	<A2><B2>	to
AB?	Because	an	earlier	equation	tells	you	that	A	=	A		<A>,	you	know	the
following:
Taking	the	expectation	value	of	the	last	term	in	this	equation,	you	get	this
result:
Square	the	earlier	equation	A	=	(<A2>		<A>2)1/2	to	get	the	following:
And	comparing	that	equation	to	the	before	it,	you	conclude	that
Cool.	That	result	means	that	
	becomes
This	inequality	at	last	means	that
Well,	well,	well.	So	the	product	of	two	uncertainties	is	greater	than	or	equal	to
1/2	the	absolute	value	of	the	commutator	of	their	respective	operators?	Wow.	Is
that	the	Heisenberg	uncertainty	relation?	Well,	take	a	look.	In	quantum
mechanics,	the	momentum	operator	looks	like	this:
And	the	operator	for	the	momentum	in	the	x	direction	is
So	what's	the	commutator	of	the	X	operator	(which	just	returns	the	x	posi-
tion	of	a	particle)	and	Px?	[X,	Px]	=	i,	so	from	
	you	get
this	next	equation	(remember,	x	and	px	here	are	the	uncertainties	in	x	and
px,	not	the	operators):
Hot	dog!	That	is	the	Heisenberg	uncertainty	relation.	(Notice	that	by	deriving	it
from	scratch,	however,	you	haven't	actually	constrained	the	physical	world
through	the	use	of	abstract	mathematics		you've	merely	proved,	using	a	few
basic	assumptions,	that	you	can't	measure	the	physical	world	with	perfect
accuracy.)
Eigenvectors	and	Eigenvalues:
They're	Naturally	Eigentastic!
As	you	know	if	you've	been	following	along	in	this	chapter,	applying	an	operator
to	a	ket	can	result	in	a	new	ket:
	To	make	things	easier,	you	can	work	with	eigenvectors	and	eigenvalues
(eigen	is	German	for	"innate"	or	"natural").	For	example,	|>	is	an
eigenvector	of	the	operator	A	if
	The	number	a	is	a	complex	constant
	
Note	what's	happening	here:	Applying	A	to	one	of	its	eigenvectors,	|>,	gives
you	|>	back,	multiplied	by	that	eigenvector's	eigenvalue,	a.
Although	a	can	be	a	complex	constant,	the	eigenvalues	of	Hermitian	operators
are	real	numbers,	and	their	eigenvectors	are	orthogonal	(that	is,	<|>	=	0).
Casting	a	problem	in	terms	of	eigenvectors	and	eigenvalues	can	make	life	a	lot
easier	because	applying	the	operator	to	its	eigenvectors	merely	gives	you	the
same	eigenvector	back	again,	multiplied	by	its	eigenvalue		there's	no	pesky
change	of	state,	so	you	don't	have	to	deal	with	a	different	state	vector.
Take	a	look	at	this	idea,	using	the	R	operator	from	rolling	the	dice,	which	is
expressed	this	way	in	matrix	form	(see	the	earlier	section	"I	expected	that:
Finding	expectation	values"	for	more	on	this	matrix):
The	R	operator	works	in	11-dimensional	space	and	is	Hermitian,	so	there'll	be
11	orthogonal	eigenvectors	and	11	corresponding	eigenvalues.
Because	R	is	a	diagonal	matrix,	finding	the	eigenvectors	is	easy.	You	can	take
unit	vectors	in	the	eleven	different	directions	as	the	eigenvectors.	Here's	what
the	first	eigenvector,	1,	would	look	like:
And	here's	what	the	second	eigenvector,	2,	would	look	like:
And	so	on,	up	to	11:
Note	that	all	the	eigenvectors	are	orthogonal.
And	the	eigenvalues?	They're	the	numbers	you	get	when	you	apply	the	R
operator	to	an	eigenvector.	Because	the	eigenvectors	are	just	unit	vectors	in	all
11	dimensions,	the	eigenvalues	are	the	numbers	on	the	diagonal	of	the	R
matrix:	2,	3,	4,	and	so	on,	up	to	12.
Understanding	how	they	work
	The	eigenvectors	of	a	Hermitian	operator	define	a	complete	set	of
orthonormal	vectors		that	is,	a	complete	basis	for	the	state	space.	When
viewed	in	this	"eigenbasis,"	which	is	built	of	the	eigenvectors,	the	operator
in	matrix	format	is	diagonal	and	the	elements	along	the	diagonal	of	the
matrix	are	the	eigenvalues.
This	arrangement	is	one	of	the	main	reasons	working	with	eigenvectors	is	so
useful;	your	original	operator	may	have	looked	something	like	this	(Note:	Bear
in	mind	that	the	elements	in	an	operator	can	also	be	functions,	not	just
numbers):
By	switching	to	the	basis	of	eigenvectors	for	the	operator,	you	diagonalize	the
matrix	into	something	more	like	what	you've	seen,	which	is	much	easier	to	work
with:
You	can	see	why	the	term	eigen	is	applied	to	eigenvectors		they	form	a
natural	basis	for	the	operator.
If	two	or	more	of	the	eigenvalues	are	the	same,	that	eigenvalue	is	said	to	be
degenerate.	So	for	example,	if	three	eigenvalues	are	equal	to	6,	then	the
eigenvalue	6	is	threefold	degenerate.
	Here's	another	cool	thing:	If	two	Hermitian	operators,	A	and	B,
commute,	and	if	A	doesn't	have	any	degenerate	eigenvalues,	then	each
eigenvector	of	A	is	also	an	eigenvector	of	B.	(See	the	earlier	section
"Forward	and	Backward:	Finding	the	Commutator"	for	more	on
commuting.)
Finding	eigenvectors	and	eigenvalues
So	given	an	operator	in	matrix	form,	how	do	you	find	its	eigenvectors	and
eigenvalues?	This	is	the	equation	you	want	to	solve:
And	you	can	rewrite	this	equation	as	the	following:
I	represents	the	identity	matrix,	with	1s	along	its	diagonal	and	0s	otherwise:
	The	solution	to	(A		aI)	|>	=	0	exists	only	if	the	determinant	of	the
matrix	A		aI	is	0:
det(A		aI)	=	0
Finding	eigenvalues
Any	values	of	a	that	satisfy	the	equation	det(A		aI)	=	0	are	eigenvalues	of	the
original	equation.	Try	to	find	the	eigenvalues	and	eigenvectors	of	the	following
matrix:
First,	convert	the	matrix	into	the	form	A		aI:
Next,	find	the	determinant:
det(A		aI)	=	(1		a)(4		a)	+	2
det(A		aI)	=	a2	+	5a	+	6
And	this	can	be	factored	as	follows:
det(A		aI)	=	a2	+	5a	+	6	=	(a	+	2)(a	+	3)
You	know	that	det(A		aI)	=	0,	so	the	eigenvalues	of	A	are	the	roots	of	this
equation;	namely,	a1	=	2	and	a2	=	3.
Finding	eigenvectors
How	about	finding	the	eigenvectors?	To	find	the	eigenvector	corresponding	to
a1	(see	the	preceding	section),	substitute	a1		the	first	eigenvalue,	2		into
the	matrix	in	the	form	A		aI:
So	you	have
Because	every	row	of	this	matrix	equation	must	be	true,	you	know	that	1	=	2.
And	that	means	that,	up	to	an	arbitrary	constant,	the	eigenvector
corresponding	to	a1	is	the	following:
Drop	the	arbitrary	constant,	and	just	write	this	as	a	matrix:
How	about	the	eigenvector	corresponding	to	a2?	Plugging	a2,	3,	into	the	matrix
in	A	aI	form,	you	get	the	following:
Then	you	have
So	21		2	=	0,	and	1	=	2		2.	And	that	means	that,	up	to	an	arbitrary
constant,	the	eigenvector	corresponding	to	a2	is
Drop	the	arbitrary	constant:
So	the	eigenvalues	of	this	next	matrix	operator
are	a1	=	2	and	a2	=	3.	And	the	eigenvector	corresponding	to	a1	is
The	eigenvector	corresponding	to	a2	is
Preparing	for	the	Inversion:
Simplifying	with	Unitary	Operators
Applying	the	inverse	of	an	operator	undoes	the	work	the	operator	did:
A1A	=	AA1	=	I
Sometimes,	finding	the	inverse	of	an	operator	is	helpful,	such	as	when	you	want
to	solve	equations	like	Ax	=	y.	Solving	for	x	is	easy	if	you	can	find	the	inverse	of
A:	x	=	A1y.
However,	finding	the	inverse	of	a	large	matrix	often	isn't	easy,	so	quantum
physics	calculations	are	sometimes	limited	to	working	with	unitary	operators,
U,	where	the	operator's	inverse	is	equal	to	its	adjoint,	U1	=	U.	(To	find	the
adjoint	of	an	operator,	A,	you	find	the	transpose	by	interchanging	the	rows	and
columns,	AT.	Then	take	the	complex	conjugate,	AT*	=	A.)	This	gives	you	the
following	equation:
The	product	of	two	unitary	operators,	U	and	V,	is	also	unitary	because
When	you	use	unitary	operators,	kets	and	bras	transform	this	way:
	
	
And	you	can	transform	other	operators	using	unitary	operators	like	this:
Note	that	the	preceding	equations	also	mean	the	following:
	
	
	
	Here	are	some	properties	of	unitary	transformations:
	If	an	operator	is	Hermitian,	then	its	unitary	transformed	version,	A'	=	UAU,
is	also	Hermitian.
	The	eigenvalues	of	A	and	its	unitary	transformed	version,	A'	=	UAU,	are	the
same.
	Commutators	that	are	equal	to	complex	numbers	are	unchanged	by	unitary
transformations:	[A',	B']	=	[A,	B].
Comparing	Matrix	and	Continuous
Representations
Werner	Heisenberg	developed	the	matrix-oriented	view	of	quantum	physics
that	you've	been	using	so	far	in	this	chapter.	It's	sometimes	called	matrix
mechanics.	The	matrix	representation	is	fine	for	many	problems,	but	sometimes
you	have	to	go	past	it,	as	you're	about	to	see.
One	of	the	central	problems	of	quantum	mechanics	is	to	calculate	the	energy
levels	of	a	system.	The	energy	operator	is	called	the	Hamilitonian,	H,	and
finding	the	energy	levels	of	a	system	breaks	down	to	finding	the	eigenvalues	of
the	problem:
Here,	E	is	an	eigenvalue	of	the	H	operator.
Here's	the	same	equation	in	matrix	terms:
The	allowable	energy	levels	of	the	physical	system	are	the	eigenvalues	E,	which
satisfy	this	equation.	These	can	be	found	by	solving	the	characteristic
polynomial,	which	derives	from	setting	the	determinant	of	the	above	matrix	to
zero,	like	so
That's	fine	if	you	have	a	discrete	basis	of	eigenvectors		if	the	number	of
energy	states	is	finite.	But	what	if	the	number	of	energy	states	is	infinite?	In
that	case,	you	can	no	longer	use	a	discrete	basis	for	your	operators	and	bras
and	kets		you	use	a	continuous	basis.
Going	continuous	with	calculus
Representing	quantum	mechanics	in	a	continuous	basis	is	an	invention	of	the
physicist	Erwin	Schrdinger.	In	the	continuous	basis,	summations	become
integrals.	For	example,	take	the	following	relation,	where	I	is	the	identity
matrix:
It	becomes	the	following:
And	every	ket	|>	can	be	expanded	in	a	basis	of	other	kets,	|n>,	like	this:
Doing	the	wave
Take	a	look	at	the	position	operator,	R,	in	a	continuous	basis.	Applying	this
operator	gives	you	r,	the	position	vector:
In	this	equation,	applying	the	position	operator	to	a	state	vector	returns	the
locations,	r,	that	a	particle	may	be	found	at.	You	can	expand	any	ket	in	the
position	basis	like	this:
And	this	becomes
	Here's	a	very	important	thing	to	understand:	(r)	=	<r|>	is	the	wave
function	for	the	state	vector	|>		it's	the	ket's	representation	in	the
position	basis.	Or	in	common	terms,	it's	just	a	function	where	the	quantity	|
(r)|2d3r	represents	the	probability	that	the	particle	will	be	found	in	the
region	d3r	at	r.
The	wave	function	is	the	foundation	of	what's	called	wave	mechanics,	as
opposed	to	matrix	mechanics.	What's	important	to	realize	is	that	when	you	talk
about	representing	physical	systems	in	wave	mechanics,	you	don't	use	the
basis-less	bras	and	kets	of	matrix	mechanics;	rather,	you	usually	use	the	wave
function		that	is,	bras	and	kets	in	the	position	basis.
Therefore,	you	go	from	talking	about	|>	to	<r|>,	which	equals	(r).	This
wave	function	appears	a	lot	in	the	coming	chapters,	and	it's	just	a	ket	in	the
position	basis.	So	in	wave	mechanics,	H|>	=	E|	>	becomes	the	following:
You	can	write	this	as	the	following:
But	what	is	<r|H|>?	It's	equal	to	H(r).	The	Hamiltonian	operator,	H,	is	the
total	energy	of	the	system,	kinetic	(p2/2m)	plus	potential	(V(r))	so	you	get	the
following	equation:
But	the	momentum	operator	is
Therefore,	substituting	the	momentum	operator	for	p	gives	you	this:
Using	the	Laplacian	operator,	you	get	this	equation:
You	can	rewrite	this	equation	as	the	following	(called	the	Schrdinger
equation):
So	in	the	wave	mechanics	view	of	quantum	physics,	you're	now	working	with	a
differential	equation	instead	of	multiple	matrices	of	elements.	This	all	came
from	working	in	the	position	basis,	(r)	=	<r|>	instead	of	just	|>.
The	quantum	physics	in	the	rest	of	the	book	is	largely	about	solving	this
differential	equation	for	a	variety	of	potentials,	V(r).	That	is,	your	focus	is	on
finding	the	wave	function	that	satisfies	the	Schrdinger	equation	for	various
physical	systems.	When	you	solve	the	Schrdinger	equation	for	(r),	you	can
find	the	allowed	energy	states	for	a	physical	system,	as	well	as	the	probability
that	the	system	will	be	in	a	certain	position	state.
Note	that,	besides	wave	functions	in	the	position	basis,	you	can	also	give	a	wave
function	in	the	momentum	basis,	(p),	or	in	any	number	of	other	bases.
	The	Heisenberg	technique	of	matrix	mechanics	is	one	way	of	working
with	quantum	physics,	and	it's	best	used	for	physical	systems	with	well-
defined	energy	states,	such	as	harmonic	oscillators.	The	Schrdinger	way	of
looking	at	things,	wave	mechanics,	uses	wave	functions,	mostly	in	the
position	basis,	to	reduce	questions	in	quantum	physics	to	a	differential
equation.
Part	II
Bound	and	Undetermined:	Handling
Particles	in	Bound	States
In	this	part	.	.	.
This	part	is	where	you	get	the	lowdown	on	one	of	quantum	physics'	favorite
topics:	solving	the	energy	levels	and	wave	functions	for	particles	trapped	in
various	bound	states.	For	example,	you	may	have	a	particle	trapped	in	a	square
well,	which	is	much	like	having	a	pea	in	a	box.	Or	you	may	have	a	particle	in
harmonic	oscillation.	Quantum	physics	is	expert	at	handling	those	kinds	of
situations.
Chapter	3
Getting	Stuck	in	Energy	Wells
In	This	Chapter
	Understanding	potential	wells
	Working	with	infinite	square	wells
	Determining	energy	levels
	Trapping	particles	with	potential	barriers
	Handling	free	particles
What's	that,	Lassie?	Stuck	in	an	energy	well?	Go	get	help!	In	this	chapter,	you
get	to	see	quantum	physics	at	work,	solving	problems	in	one	dimension.	You	see
particles	trapped	in	potential	wells	and	solve	for	the	allowable	energy	states
using	quantum	physics.	That	goes	against	the	grain	in	classical	physics,	which
doesn't	restrict	trapped	particles	to	any	particular	energy	spectrum.	But	as	you
know,	when	the	world	gets	microscopic,	quantum	physics	takes	over.
The	equation	of	the	moment	is	the	Schrdinger	equation	(derived	in	Chapter
2),	which	lets	you	solve	for	the	wave	function,	(x),	and	the	energy	levels,	E:
Looking	into	a	Square	Well
A	square	well	is	a	potential	(that	is,	a	potential	energy	well)	that	forms	a	square
shape,	as	you	can	see	in	Figure	3-1.
Figure	3-1:	A	square	well.
The	potential,	or	V(x),	goes	to	infinity	at	x	<	0	and	x	>	a	(where	x	is	distance),
like	this:
	V(x)	=	,	where	x	<	0
	V(x)	=	0,	where	0		x		a
	V(x)	=	,	where	x	>	a
Using	square	wells,	you	can	trap	particles.	If	you	put	a	particle	into	a	square
well	with	a	limited	amount	of	energy,	it'll	be	trapped	because	it	can't	overcome
the	infinite	potential	at	either	side	of	the	square	well.	Therefore,	the	particle
has	to	move	inside	the	square	well.
So	does	the	particle	just	sort	of	roll	around	on	the	bottom	of	the	square	well?
Not	exactly.	The	particle	is	in	a	bound	state,	and	its	wave	function	depends	on
its	energy.	The	wave	function	isn't	complicated:
So	you	have	the	allowed	wave	functions	for	the	states	n	=	1,	2,	3,	and	so	on.
The	energy	of	the	allowable	bound	states	are	given	by	the	following	equation:
The	rest	of	this	chapter	shows	you	how	to	solve	problems	like	this	one.
Trapping	Particles	in	Potential	Wells
Take	a	look	at	the	potential	in	Figure	3-2.	Notice	the	dip,	or	well,	in	the
potential,	which	means	that	particles	can	be	trapped	in	it	if	they	don't	have	too
much	energy.
The	particle's	kinetic	energy	summed	with	its	potential	energy	is	a	constant,
equal	to	its	total	energy:
If	its	total	energy	is	less	than	V1,	the	particle	will	be	trapped	in	the	potential
well,	you	see	in	Figure	3-2;	to	get	out	of	the	well,	the	particle's	kinetic	energy
would	have	to	become	negative	to	satisfy	the	equation,	which	is	impossible.
Figure	3-2:	A	potential	well.
In	this	section,	you	take	a	look	at	the	various	possible	states	that	a	particle	with
energy	E	can	take	in	the	potential	given	by	Figure	3-2.	Quantum-mechanically
speaking,	those	states	are	of	two	kinds		bound	and	unbound.	This	section
looks	at	them	in	overview.
Binding	particles	in	potential	wells
Bound	states	happen	when	the	particle	isn't	free	to	travel	to	infinity		it's	as
simple	as	that.	In	other	words,	the	particle	is	confined	to	the	potential	well.
A	particle	traveling	in	the	potential	well	you	see	in	Figure	3-2	is	bound	if	its
energy,	E,	is	less	than	both	V1	and	V2.	In	that	case,	the	particle	moves	(in	a
classical	approximation)	between	x1	and	x2.	A	particle	trapped	in	such	a	well	is
represented	by	a	wave	function,	and	you	can	solve	the	Schrdinger	equation
for	the	allowed	wave	functions	and	the	allowed	energy	states.	You	need	to	use
two	boundary	conditions	(the	Schrdinger	equation	is	a	second-order
differential	equation)	to	solve	the	problem	completely.
	Bound	states	are	discrete		that	is,	they	form	an	energy	spectrum	of
discrete	energy	levels.	The	Schrdinger	equation	gives	you	those	states.	In
addition,	in	one-dimensional	problems,	the	energy	levels	of	a	bound	state
are	not	degenerate		that	is,	no	two	energy	levels	are	the	same	in	the
entire	energy	spectrum.
Escaping	from	potential	wells
If	a	particle's	energy,	E,	is	greater	than	the	potential	V1	in	Figure	3-2,	the
particle	can	escape	from	the	potential	well.	There	are	two	possible	cases:	V1	<
E	<	V2	and	E	>	V2.	This	section	looks	at	them	separately.
Case	1:	Energy	between	the	two	potentials	(V1	<	E	<	V2)
If	V1	<	E	<	V2,	the	particle	in	the	potential	well	has	enough	energy	to	overcome
the	barrier	on	the	left	but	not	on	the	right.	The	particle	is	thus	free	to	move	to
negative	infinity,	so	its	allowed	x	region	is	between		and	x1.
Here,	the	allowed	energy	values	are	continuous,	not	discrete,	because	the
particle	isn't	completely	bound.	The	energy	eigenvalues	are	not	degenerate	
that	is,	no	two	energy	eigenvalues	are	the	same	(see	Chapter	2	for	more	on
eigenvalues).
The	Schrdinger	equation	is	a	second-order	differential	equation,	so	it	has	two
linearly	independent	solutions;	however,	in	this	case,	only	one	of	those	solutions
is	physical	and	doesn't	diverge.
The	wave	equation	in	this	case	turns	out	to	oscillate	for	x	<	x2	and	to	decay
rapidly	for	x	>	x2.
Case	2:	Energy	greater	than	the	higher	potential	(E	>	V2)
If	E	>	V2,	the	particle	isn't	bound	at	all	and	is	free	to	travel	from	negative
infinity	to	positive	infinity.
The	energy	spectrum	is	continuous	and	the	wave	function	turns	out	to	be	a	sum
of	a	function	moving	to	the	right	and	one	moving	to	the	left.	The	energy	levels
of	the	allowed	spectrum	are	therefore	doubly	degenerate.
That's	all	the	overview	you	need		time	to	start	solving	the	Schrdinger
equation	for	various	different	potentials,	starting	with	the	easiest	of	all:	infinite
square	wells.
Trapping	Particles	in	Infinite	Square
Potential	Wells
Infinite	square	wells,	in	which	the	walls	go	to	infinity,	are	a	favorite	in	physics
problems.	You	explore	the	quantum	physics	take	on	these	problems	in	this
section.
Finding	a	wave-function	equation
Take	a	look	at	the	infinite	square	well	that	appears	back	in	Figure	3-1.	Here's
what	that	square	well	looks	like:
	V(x)	=	,	where	x	<	0
	V(x)	=	0,	where	0		x		a
	V(x)	=	,	where	x	>	a
The	Schrdinger	equation	looks	like	this	in	three	dimensions:
Writing	out	the	Schrdinger	equation	gives	you	the	following:
You're	interested	in	only	one	dimension		x	(distance)		in	this	chapter,	so	the
Schrdinger	equation	looks	like
Because	V(x)	=	0	inside	the	well,	the	equation	becomes
And	in	problems	of	this	sort,	the	equation	is	usually	written	as
where	
	(k	is	the	wave	number).
So	now	you	have	a	second-order	differential	equation	to	solve	for	the	wave
function	of	a	particle	trapped	in	an	infinite	square	well.
You	get	two	independent	solutions	because	this	equation	is	a	second-order
differential	equation:
1(x)	=	A	sin(kx)
2(x)	=	B	cos(kx)
A	and	B	are	constants	that	are	yet	to	be	determined.
	The	general	solution	of	
	is	the	sum	of	
(x)	=	A	sin(kx)	+	B	cos(kx)
Determining	the	energy	levels
The	equation	(x)	=	A	sin(kx)	+	B	cos(kx)	tells	you	that	you	have	to	use	the
boundary	conditions	to	find	the	constants	A	and	B	(the	preceding	section
explains	how	to	derive	the	equation).	What	are	the	boundary	conditions?	The
wave	function	must	disappear	at	the	boundaries	of	an	infinite	square	well,	so
	(0)	=	0
	(a)	=	0
The	fact	that	(0)	=	0	tells	you	right	away	that	B	must	be	zero	because
cos(0)	=	1.	And	the	fact	that	(a)	=	0	tells	you	that	(a)	=	A	sin(ka)	=	0.
Because	sine	is	zero	when	its	argument	is	a	multiple	of	,	this	means	that
ka	=	n	n	=	1,	2,	3	...
Note	that	although	n	=	0	is	technically	a	solution,	it	yields	(x)	=	0	for	all	x,
which	is	not	normalizable,	so	it's	not	a	physical	solution		the	physical	solutions
begin	with	n	=	1.
This	equation	can	also	be	written	as
And	because	k2	=	2mE/2,	you	have	the	following	equation,	where	n	=	1,	2,	3,	...
	those	are	the	allowed	energy	states.	These	are	quantized	states,	
corresponding	to	the	quantum	numbers	1,	2,	3,	and	so	on:
Note	that	the	first	physical	state	corresponds	to	n	=	1,	which	gives	you	this	next
equation:
This	is	the	lowest	physical	state	that	the	particles	can	occupy.	Just	for	kicks,	put
some	numbers	into	this,	assuming	that	you	have	an	electron,	mass	9.11		1031
kilograms,	confined	to	an	infinite	square	well	of	width	of	the	order	of	the	Bohr
radius	(the	average	radius	of	an	electron's	orbit	in	a	hydrogen	atom);	let's	say	a
=	3.00	x	1010	meters.
	gives	you	this	energy	for	the	ground	state:
That's	a	very	small	amount,	about	4.2	electron	volts	(eV		the	amount	of
energy	one	electron	gains	falling	through	1	volt).	Even	so,	it's	already	on	the
order	of	the	energy	of	the	ground	state	of	an	electron	in	the	ground	state	of	a
hydrogen	atom	(13.6	eV),	so	you	can	say	you're	certainly	in	the	right	quantum
physics	ballpark	now.
Normalizing	the	wave	function
Okay,	you	have	this	for	the	wave	equation	for	a	particle	in	an	infinite	square
well:
The	wave	function	is	a	sine	wave,	going	to	zero	at	x	=	0	and	x	=	a.	You	can	see
the	first	two	wave	functions	plotted	in	Figure	3-3.
Figure	3-3:	Wave	functions	in	a	square	well.
Normalizing	the	wave	function	lets	you	solve	for	the	unknown	constant	A.	In	a
normalized	function,	the	probability	of	finding	the	particle	between	x	and	dx,	|
(x)|2dx,	adds	up	to	1	when	you	integrate	over	the	whole	square	well,	x	=	0	to	x
=	a:
Substituting	for	(x)	gives	you	the	following:
Here's	what	the	integral	in	this	equation	equals:
So	from	the	previous	equation,	
.	Solve	for	A:
Therefore,	here's	the	normalized	wave	equation	with	the	value	of	A	plugged	in:
And	that's	the	normalized	wave	function	for	a	particle	in	an	infinite	square	well.
Adding	time	dependence	to	wave	functions
Now	how	about	seeing	how	the	wave	function	for	a	particle	in	an	infinite	square
well	evolves	with	time?	The	Schrdinger	equation	looks	like	this:
You	can	also	write	the	Schrdinger	equation	this	way,	where	H	is	the	Hermitian
Hamiltonian	operator:
H(r)	=	E(r)
That's	actually	the	time-independent	Schrdinger	equation.	The	time-
dependent	Schrdinger	equation	looks	like	this:
Combining	the	preceding	three	equations	gives	you	the	following,	which	is
another	form	of	the	time-dependent	Schrdinger	equation:
And	because	you're	dealing	with	only	one	dimension,	x,	this	equation	becomes
This	is	simpler	than	it	looks,	however,	because	the	potential	doesn't	change
with	time.	In	fact,	because	E	is	constant,	you	can	rewrite	the	equation	as
That	equation	makes	life	a	lot	simpler		it's	easy	to	solve	the	time-dependent
Schrdinger	equation	if	you're	dealing	with	a	constant	potential.	In	this	case,
the	solution	is
	Neat.	When	the	potential	doesn't	vary	with	time,	the	solution	to	the	time-
dependent	Schrdinger	equation	simply	becomes	(x),	the	spatial	part,
multiplied	by	eiEt/,	the	time-dependent	part.
So	when	you	add	in	the	time-dependent	part	to	the	time-independent	wave
function,	you	get	the	time-dependent	wave	function,	which	looks	like	this:
The	energy	of	the	nth	quantum	state	is
Therefore,	the	result	is
Shifting	to	symmetric	square	well	potentials
The	standard	infinite	square	well	looks	like	this:
	V(x)	=	,	where	x	<	0
	V(x)	=	0,	where	0		x		a
	V(x)	=	,	where	x	>	a
But	what	if	you	want	to	shift	things	so	that	the	square	well	is	symmetric	around
the	origin	instead?	That	is,	you	move	the	square	well	so	that	it	extends	from
a/2	to	a/2?	Here's	what	the	new	infinite	square	well	looks	like	in	this	case:
	V(x)	=	,	where	x	<	a/2
	V(x)	=	0,	where	a/2		x		a/2
	V(x)	=	,	where	x	>	a/2
You	can	translate	from	this	new	square	well	to	the	old	one	by	adding	a/2	to	x,
which	means	that	you	can	write	the	wave	function	for	the	new	square	well	in
this	equation	like	the	following:
Doing	a	little	trig	gives	you	the	following	equations:
So	as	you	can	see,	the	result	is	a	mix	of	sines	and	cosines.	The	bound	states	are
these,	in	increasing	quantum	order:
	
	
	
	
And	so	on.
Note	that	the	cosines	are	symmetric	around	the	origin:	(x)	=	(x).	The	sines
are	anti-symmetric:	(x)	=	(x).
Limited	Potential:	Taking	a	Look	at
Particles	and	Potential	Steps
Truly	infinite	potentials	(which	I	discuss	in	the	previous	sections)	are	hard	to
come	by.	In	this	section,	you	look	at	some	real-world	examples,	where	the
potential	is	set	to	some	finite	V0,	not	infinity.	For	example,	take	a	look	at	the
situation	in	Figure	3-4.	There,	a	particle	is	traveling	toward	a	potential	step.
Currently,	the	particle	is	in	a	region	where	V	=	0,	but	it'll	soon	be	in	the	region
V	=	V0.
Figure	3-4:	A	potential	step,	E	>	V0.
There	are	two	cases	to	look	at	here	in	terms	of	E,	the	energy	of	the	particle:
	E	>	V0:	Classically,	when	E	>	V0,	you	expect	the	particle	to	be	able	to
continue	on	to	the	region	x	>	0.
	E	<	V0:	When	E	<	V0,	you'd	expect	the	particle	to	bounce	back	and	not	be
able	to	get	to	the	region	x	>	0	at	all.
In	this	section,	you	start	by	taking	a	look	at	the	case	where	the	particle's
energy,	E,	is	greater	than	the	potential	V0,	as	shown	in	Figure	3-4;	then	you
take	a	look	at	the	case	where	E	<	V0.
Assuming	the	particle	has	plenty	of	energy
Start	with	the	case	where	the	particle's	energy,	E,	is	greater	than	the	potential
V0.	From	a	quantum	physics	point	of	view,	here's	what	the	Schrdinger
equation	would	look	like:
	For	the	region	x	<	0:	
Here,	
.
	For	the	region	x	>	0:	
In	this	equation,	
.
In	other	words,	k	is	going	to	vary	by	region,	as	you	see	in	Figure	3-5.
Figure	3-5:	The	value	of	k	by	region,	where	E	>	V0.
Treating	the	first	equation	as	a	second-order	differential	equation,	you	can	see
that	the	most	general	solution	is	the	following:
1(x)	=	Aeik1x	+	Beik1x,	where	x	<	0
And	for	the	region	x	>	0,	solving	the	second	equation	gives	you	this:
2(x)	=	Ceik2x	+	Deik2x,	where	x	>	0
Note	that	eikx	represents	plane	waves	traveling	in	the	+x	direction,	and	eikx
represents	plane	waves	traveling	in	the	x	direction.
What	this	solution	means	is	that	waves	can	hit	the	potential	step	from	the	left
and	be	either	transmitted	or	reflected.	Given	that	way	of	looking	at	the
problem,	you	may	note	that	the	wave	can	be	reflected	only	going	to	the	right,
not	to	the	left,	so	D	must	equal	zero.	That	makes	the	wave	equation	become	the
following:
	Where	x	<	0:	1(x)	=	Aeik1x	+	Beik1x
	Where	x	>	0:	2(x)	=	Ceik2x
The	term	Aeik1x	represents	the	incident	wave,	Beik1x	is	the	reflected	wave,	and
Ceik2x	is	the	transmitted	wave.
Calculating	the	probability	of	reflection	or	transmission
You	can	calculate	the	probability	that	the	particle	will	be	reflected	or
transmitted	through	the	potential	step	by	calculating	the	reflection	and
transmission	coefficients.	These	are	defined	in	terms	of	something	called	the
current	density	J(x);	this	is	given	in	terms	of	the	wave	function	by
If	Jr	is	the	reflected	current	density,	and	Ji,	is	the	incident	current	density,	then
R,	the	reflection	coefficient	is
T,	the	transmission	coefficient,	is
You	now	have	to	calculate	Jr,	Ji,	and	Jt.	Actually,	that's	not	so	hard		start	with
Ji.	Because	the	incident	part	of	the	wave	is	i(x)	=	Aeik1x,	the	incident	current
density	is
And	this	just	equals	
.	Jr	and	Jt	work	in	the	same	way:
So	you	have	this	for	the	reflection	coefficient:
T,	the	transmission	coefficient,	is
Finding	A,	B,	and	C
So	how	do	you	figure	out	the	constants	A,	B,	and	C?	You	do	that	as	you	figure
out	the	coefficients	with	the	infinite	square	well	potential		with	boundary
conditions	(see	the	earlier	section	"Trapping	Particles	in	Infinite	Square	Well
Potentials").	However,	here,	you	can't	necessarily	say	that	(x)	goes	to	zero,
because	the	potential	is	no	longer	infinite.	Instead,	the	boundary	conditions	are
that	(x)	and	d(x)/dx	are	continuous	across	the	potential	step's	boundary.	In
other	words,
	1(0)	=	2(0)
	
You	know	the	following:
	Where	x	<	0:	1(x)	=	Aeik1x	+	Beik1x
	Where	x	>	0:	2(x)	=	Ceik2x
Therefore,	plugging	these	two	equations	into	1(0)	=	2(0)	gives	you	A	+	B	=
C.
And	plugging	them	into	
	gives	you
k1A		k1B	=	k2C
Solving	for	B	in	terms	of	A	gives	you	this	result:
Solving	for	C	in	terms	of	A	gives	you
You	can	then	calculate	A	from	the	normalization	condition	of	the	wave	function:
But	you	don't	actually	need	A,	because	it	drops	out	of	the	ratios	for	the
reflection	and	transmission	coefficients,	R	and	T.	In	particular,
Therefore,
That's	an	interesting	result,	and	it	disagrees	with	classical	physics,	which	says
that	there	should	be	no	particle	reflection	at	all.	As	you	can	see,	if	k1		k2,	then
there	will	indeed	be	particle	reflection.
Note	that	as	k1	goes	to	k2,	R	goes	to	0	and	T	goes	to	1,	which	is	what	you'd
expect.
So	already	you	have	a	result	that	differs	from	the	classical		the	particle	can	be
reflected	at	the	potential	step.	That's	the	wave-like	behavior	of	the	particle
coming	into	play	again.
Assuming	the	particle	doesn't	have	enough	energy
Okay,	now	try	the	case	where	E	<	V0	when	there's	a	potential	step,	as	shown	in
Figure	3-6.	In	this	case,	the	particle	doesn't	have	enough	energy	to	make	it	into
the	region	x	>	0,	according	to	classical	physics.	See	what	quantum	physics	has
to	say	about	it.
Figure	3-6:	A	potential	step,	E	<	V0.
Tackle	the	region	x	<	0	first.	There,	the	Schrdinger	equation	would	look
like	this:
where	
.
You	know	the	solution	to	this	from	the	previous	discussion	on	potential	steps
(see	"Limited	Potential:	Taking	a	Look	at	Particles	and	Potential	Steps"):
Okay,	but	what	about	the	region	x	>	0?	That's	a	different	story.	Here's	the
Schrdinger	equation:
where	
.
But	hang	on;	E		V0	is	less	than	zero,	which	would	make	k	imaginary,	which	is
impossible	physically.	So	change	the	sign	in	the	Schrdinger	equation	from	plus
to	minus:
And	use	the	following	for	k2	(note	that	this	is	positive	if	E	<	V0):
Okay,	so	now	you	have	to	solve	the	differential
.	There	are	two	linearly	independent
solutions:
	(x)	=	Cek2x
	(x)	=	Dek2x
And	the	general	solution	to	
	is
However,	wave	functions	must	be	finite	everywhere,	and	the	second	term	is
clearly	not	finite	as	x	goes	to	infinity,	so	D	must	equal	zero	(note	that	if	x	goes	to
negative	infinity,	the	first	term	also	diverges,	but	because	the	potential	step	is
limited	to	x	>	0,	that	isn't	a	problem).	Therefore,	here's	the	solution	for	x	>	0:
So	your	wave	functions	for	the	two	regions	are
1(x)	=	Aeik1x	+	Beik1x	x	<	0
2(x)	=	Cek2x	x	>	0
Putting	this	in	terms	of	the	incident,	reflected,	and	transmitted	wave	functions,
i(x),	r(x),	and	t(x),	you	have	the	following:
	i(x)	=	Aeik1x
	r(x)	=	Beik1x
	t(x)	=	Cek2x
Finding	transmission	and	reflection	coefficients
Now	you	can	figure	out	the	reflection	and	transmission	coefficients,	R	and	T	(as
you	do	for	the	case	E	>	V0	in	the	earlier	section	"Assuming	the	particle	has
plenty	of	energy"):
Actually,	this	is	very	easy	in	this	case;	take	a	look	at	Jt:
But	because	t(x)	=	Cek2x,	t(x)	is	completely	real,	which	means	that	in	this
case,	the	following	is	true:
And	this	equation,	of	course,	is	equal	to	zero.
So	Jt	=	0;	therefore,	T	=	0.	If	T	=	0,	then	R	must	equal	1.	That	means	that	you
have	a	complete	reflection,	just	as	in	the	classical	solution.
The	nonzero	solution:	Finding	a	particle	in	x	>	0
Despite	the	complete	reflection,	there's	a	difference	between	the	mathematical
and	classical	solution:	There	actually	is	a	nonzero	chance	of	finding	the	particle
in	the	region	x	>	0.	To	see	that,	take	a	look	at	the	probability	density	for	x	>	0,
which	is
P(x)	=	|t(x)|2
Plugging	in	for	the	wave	function	t(x)	gives	you
P(x)	=	|t(x)|2	=	|C|2e2k2x
You	can	use	the	continuity	conditions	to	solve	for	C	in	terms	of	A:
	1(0)	=	2(0)
	
Using	the	continuity	conditions	gives	you	the	following:
This	does	fall	quickly	to	zero	as	x	gets	large,	but	near	x	=	0,	it	has	a	nonzero
value.
You	can	see	what	the	probability	density	looks	like	for	the	E	<	V0	case	of	a
potential	step	in	Figure	3-7.
Figure	3-7:	The	value	of	k	by	region,	E	<	V0.
Okay,	you've	taken	care	of	infinite	square	wells	and	potential	steps.	Now	what
about	the	case	where	the	potential	step	doesn't	extend	out	to	infinity	but	is
itself	bounded?	That	brings	you	to	potential	barriers,	which	I	discuss	in	the	next
section.
Hitting	the	Wall:	Particles	and
Potential	Barriers
What	if	the	particle	could	work	its	way	through	a	potential	step		that	is,	the
step	was	of	limited	extent?	Then	you'd	have	a	potential	barrier,	which	is	set	up
something	like	this:
	V(x)	=	0,	where	x	<	0
	V(x)	=	V0,	where	0		x		a
	V(x)	=	0,	where	x	>	a
You	can	see	what	this	potential	looks	like	in	Figure	3-8.
Figure	3-8:	A	potential	barrier	E	>	V0.
In	solving	the	Schrdinger	equation	for	a	potential	barrier,	you	have	to
consider	two	cases,	corresponding	to	whether	the	particle	has	more	or	less
energy	than	the	potential	barrier.	In	other	words,	if	E	is	the	energy	of	the
incident	particle,	the	two	cases	to	consider	are	E	>	V0	and	E	<	V0.	This	section
starts	with	E	>	V0.
Getting	through	potential	barriers	when	E	>	V0
In	the	case	where	E	>	V0,	the	particle	has	enough	energy	to	pass	through	the
potential	barrier	and	end	up	in	the	x	>	a	region.	This	is	what	the	Schrdinger
equation	looks	like:
	For	the	region	x	<	0:	
where	
	For	the	region	0		x		a:	
where	
	For	the	region	x	>	a:	
where	
The	solutions	for	1(x),	2(x),	and	3(x)	are	the	following:
	Where	x	<	0:	1(x)	=	Aeik1x	+	Beik1x
	Where	0		x		a:	2(x)	=	Ceik2x	+	Deik2x
	Where	x	>	a:	3(x)	=	Eeik1x	+	Feik1x
In	fact,	because	there's	no	leftward	traveling	wave	in	the	x	>	a	region,	F	=	0,
so	3(x)	=	Eeik1x.
So	how	do	you	determine	A,	B,	C,	D,	and	E?	You	use	the	continuity	conditions,
which	work	out	here	to	be	the	following:
Okay,	from	these	equations,	you	get	the	following:
	A	+	B	=	C	+	D
	ik1(A		B)	=	ik2(C		D)
	Ceik2a	+	Deik2a	=	Eeik1a
	ik2Ceik2a		ik2Deik2a	=	ik1Eeik1a
So	putting	all	of	these	equations	together,	you	get	this	for	the	coefficient	E	in
terms	of	A:
Wow.	So	what's	the	transmission	coefficient,	T?	Well,	T	is
And	this	works	out	to	be
Whew!	Note	that	as	k1	goes	to	k2,	T	goes	to	1,	which	is	what	you'd	expect.
So	how	about	R,	the	reflection	coefficient?	I'll	spare	you	the	algebra;	here's
what	R	equals:
You	can	see	what	the	E	>	V0	probability	density,	|(x)|2,	looks	like	for	the
potential	barrier	in	Figure	3-9.
Figure	3-9:	|(x)|2	for	a	potential	barrier	E	>	V0.
That	completes	the	potential	barrier	when	E	>	V0.
Getting	through	potential	barriers,	even	when	E	<
V0
What	happens	if	the	particle	doesn't	have	as	much	energy	as	the	potential	of
the	barrier?	In	other	words,	you're	now	facing	the	situation	you	see	in	Figure	3-
10.
Figure	3-10:	A	potential	barrier	E	<	V0.
Now	the	Schrdinger	equation	looks	like	this:
	For	the	region	x	<	0:	1(x)	=	Aeik1x	+	Beik1x
	For	the	region	0		x		a:	
where	
.
But	now	E		V0	is	less	than	0,	which	would	make	k	imaginary.	And	that's
impossible	physically.	So	change	the	sign	in	the	Schrdinger	equation	from
plus	to	minus:
And	use	this	for	k2:	
.
	For	the	region	x	>	a:	
where	
.
All	this	means	that	the	solutions	for	1(x),	2(x),	and	3(x)	are	the	following:
	Where	x	<	0:	1(x)	=	Aeik1x	+	Beik1x
	Where	0		x		a:	2(x)	=	Cek2x	+	Dek2x
	Where	x	>	a:	3(x)	=	Eeik1x	+	Feik1x
In	fact,	there's	no	leftward	traveling	wave	in	the	region	x	>	a;	F	=	0,	so	3(x)
is	3(x)	=	Eeik1x.
This	situation	is	similar	to	the	case	where	E	>	V0,	except	for	the	region	0		x	
a.	The	wave	function	oscillates	in	the	regions	where	it	has	positive	energy,	x	<	0
and	x	>	a,	but	is	a	decaying	exponential	in	the	region	0		x		a.
You	can	see	what	the	probability	density,	|(x)|2,	looks	like	in	Figure	3-11.
Figure	3-11:	|(x)|2	for	a	potential	barrier	E	<	V0.
Finding	the	reflection	and	transmission	coefficients
How	about	the	reflection	and	transmission	coefficients,	R	and	T?	Here's	what
they	equal:
As	you	may	expect,	you	use	the	continuity	conditions	to	determine	A,	B,	and	E:
	1(0)	=	2(0)
	
	2(a)	=	3(a)
	
A	fair	bit	of	algebra	and	trig	is	involved	in	solving	for	R	and	T;	here's	what	R
and	T	turn	out	to	be:
Despite	the	equation's	complexity,	it's	amazing	that	the	expression	for	T	can	be
nonzero.	Classically,	particles	can't	enter	the	forbidden	zone	0		x		a	because
E	<	V0,	where	V0	is	the	potential	in	that	region;	they	just	don't	have	enough
energy	to	make	it	into	that	area.
Tunneling	through
Quantum	mechanically,	the	phenomenon	where	particles	can	get	through
regions	that	they're	classically	forbidden	to	enter	is	called	tunneling.	Tunneling
is	possible	because	in	quantum	mechanics,	particles	show	wave	properties.
Tunneling	is	one	of	the	most	exciting	results	of	quantum	physics		it	means
that	particles	can	actually	get	through	classically	forbidden	regions	because	of
the	spread	in	their	wave	functions.	This	is,	of	course,	a	microscopic	effect	
don't	try	to	walk	through	any	closed	doors		but	it's	a	significant	one.	Among
other	effects,	tunneling	makes	transistors	and	integrated	circuits	possible.
You	can	calculate	the	transmission	coefficient,	which	tells	you	the	probability
that	a	particle	gets	through,	given	a	certain	incident	intensity,	when	tunneling
is	involved.	Doing	so	is	relatively	easy	in	the	preceding	section	because	the
barrier	that	the	particle	has	to	get	through	is	a	square	barrier.	But	in	general,
calculating	the	transmission	coefficient	isn't	so	easy.	Read	on.
Getting	the	transmission	with	the	WKB	approximation
The	way	you	generally	calculate	the	transmission	coefficient	is	to	break	up	the
potential	you're	working	with	into	a	succession	of	square	barriers	and	to	sum
them.	That's	called	the	Wentzel-Kramers-Brillouin	(WKB)	approximation	
treating	a	general	potential,	V(x),	as	a	sum	of	square	potential	barriers.
The	result	of	the	WKB	approximation	is	that	the	transmission	coefficient	for	an
arbitrary	potential,	V(x),	for	a	particle	of	mass	m	and	energy	E	is	given	by	this
expression	(that	is,	as	long	as	V(x)	is	a	smooth,	slowly	varying	function):
So	now	you	can	amaze	your	friends	by	calculating	the	probability	that	a	particle
will	tunnel	through	an	arbitrary	potential.	It's	the	stuff	science	fiction	is	made
of		well,	on	the	microscopic	scale,	anyway.
Particles	Unbound:	Solving	the
Schrdinger	Equation	for	Free
Particles
What	about	particles	outside	any	square	well		that	is,	free	particles?	There
are	plenty	of	particles	that	act	freely	in	the	universe,	and	quantum	physics	has
something	to	say	about	them.
Here's	the	Schrdinger	equation:
What	if	the	particle	were	a	free	particle,	with	V(x)	=	0?	In	that	case,	you'd	have
the	following	equation:
And	you	can	rewrite	this	as
where	the	wave	number,	k,	is	
.
You	can	write	the	general	solution	to	this	Schrdinger	equation	as
(x)	=	Aeikx	+	Beikx
If	you	add	time-dependence	to	the	equation,	you	get	this	time-dependent	wave
function:
That's	a	solution	to	the	Schrdinger	equation,	but	it	turns	out	to	be	unphysical.
To	see	this,	note	that	for	either	term	in	the	equation,	you	can't	normalize	the
probability	density,	|(x)|2	(see	the	earlier	section	titled	"Normalizing	the	wave
function"	for	more	on	normalizing):
|(x)|2	=	|A|2	or	|B|2
What's	going	on	here?	The	probability	density	for	the	position	of	the	particle	is
uniform	throughout	all	x!	In	other	words,	you	can't	pin	down	the	particle	at	all.
This	is	a	result	of	the	form	of	the	time-dependent	wave	function,	which	uses	an
exact	value	for	the	wave	number,	k		and	p	=	k	and	E	=	k2/2m.	So	what	that
equation	says	is	that	you	know	E	and	p	exactly.	And	if	you	know	p	and	E	exactly,
that	causes	a	large	uncertainty	in	x	and	t		in	fact,	x	and	t	are	completely
uncertain.	That	doesn't	correspond	to	physical	reality.
For	that	matter,	the	wave	function	(x),	as	it	stands,	isn't	something	you	can
normalize.	Trying	to	normalize	the	first	term,	for	example,	gives	you	this
integral:
And	for	the	first	term	of	(x,	t),	this	is
And	the	same	is	true	of	the	second	term	in	(x,	t).
So	what	do	you	do	here	to	get	a	physical	particle?	The	next	section	explains.
Getting	a	physical	particle	with	a	wave	packet
	If	you	have	a	number	of	solutions	to	the	Schrdinger	equation,	any	linear
combination	of	those	solutions	is	also	a	solution.	So	that's	the	key	to	getting
a	physical	particle:	You	add	various	wave	functions	together	so	that	you	get
a	wave	packet,	which	is	a	collection	of	wave	functions	of	the	form	ei(kx		Et/
)	such	that	the	wave	functions	interfere	constructively	at	one	location	and
interfere	destructively	(go	to	zero)	at	all	other	locations:
This	is	usually	written	as	a	continuous	integral:
What	is	(k,	t)?	It's	the	amplitude	of	each	component	wave	function,	and	you
can	find	(k,	t)	from	the	Fourier	transform	of	the	equation:
Because	k	=	p/,	you	can	also	write	the	wave	packet	equations	like	this,	in
terms	of	p,	not	k:
Well,	you	may	be	asking	yourself	just	what's	going	on	here.	It	looks	like	(x,	t)
is	defined	in	terms	of	(p,	t),	but	(p,	t)	is	defined	in	terms	of	(x,	t).	That	looks
pretty	circular.
The	answer	is	that	the	two	previous	equations	aren't	definitions	of	(x,	t)	or
(p,	t);	they're	just	equations	relating	the	two.	You're	free	to	choose	your	own
wave	packet	shape	yourself		for	example,	you	may	specify	the	shape
of	(p,	t),	and	
	would	let	you	find	(x,	t).
Going	through	a	Gaussian	example
Here's	an	example	in	which	you	get	concrete,	selecting	an	actual	wave	packet
shape.	Choose	a	so-called	Gaussian	wave	packet,	which	you	can	see	in	Figure	3-
12		localized	in	one	place,	zero	in	the	others.
Figure	3-12:	A	Gaussian	wave	packet.
The	amplitude	(k)	you	may	choose	for	this	wave	packet	is
You	start	by	normalizing	(k)	to	determine	what	A	is.	Here's	how	that	works:
Substituting	in	(k)	gives	you	this	equation:
Doing	the	integral	(that	means	looking	it	up	in	math	tables)	gives	you	the
following:
Therefore,	
.
So	here's	your	wave	function:
This	little	gem	of	an	integral	can	be	evaluated	to	give	you	the	following:
So	that's	the	wave	function	for	this	Gaussian	wave	packet	(Note:	The
exp[x2/a2]	is	the	Gaussian	part	that	gives	the	wave	packet	the	distinctive	shape
that	you	see	in	Figure	3-12)		and	it's	already	normalized.
Now	you	can	use	this	wave	packet	function	to	determine	the	probability	that
the	particle	will	be	in,	say,	the	region	0		x		a/2.	The	probability	is
In	this	case,	the	integral	is
And	this	works	out	to	be
So	the	probability	that	the	particle	will	be	in	the	region	0		x		a/2	is	1/3.	Cool!
Chapter	4
Back	and	Forth	with	Harmonic
Oscillators
In	This	Chapter
	Hamiltonians:	Looking	at	total	energy
	Solving	for	energy	states	with	creation	and	annihilation	operators
	Understanding	the	matrix	version	of	harmonic	oscillator	operators
	Writing	computer	code	to	solve	the	Schrdinger	equation
Harmonic	oscillators	are	physics	setups	with	periodic	motion,	such	as	things
bouncing	on	springs	or	tick-tocking	on	pendulums.	You're	probably	already
familiar	with	harmonic	oscillator	problems	in	the	macroscopic	arena,	but	now
you're	going	microscopic.	There	are	many,	many	physical	cases	that	can	be
approximated	by	harmonic	oscillators,	such	as	atoms	in	a	crystal	structure.
In	this	chapter,	you	see	both	exact	solutions	to	harmonic	oscillator	problems	as
well	as	computational	methods	for	solving	them.	Knowing	how	to	solve	the
Schrdinger	equation	using	computers	is	a	useful	skill	for	any	quantum	physics
expert.
Grappling	with	the	Harmonic
Oscillator	Hamiltonians
Okay,	time	to	start	talking	Hamiltonians	(and	I'm	not	referring	to	fans	of	the
U.S.	Founding	Father	Alexander	Hamilton).	The	Hamiltonian	will	let	you	find
the	energy	levels	of	a	system.
Going	classical	with	harmonic	oscillation
In	classical	terms,	the	force	on	an	object	in	harmonic	oscillation	is	the	following
(this	is	Hooke's	law):
F	=	kx
In	this	equation,	k	is	the	spring	constant,	measured	in	Newtons/meter,	and	x	is
displacement.	The	key	point	here	is	that	the	restoring	force	on	whatever	is	in
harmonic	motion	is	proportional	to	its	displacement.	In	other	words,	the	farther
you	stretch	a	spring,	the	harder	it'll	pull	back.
Because	F	=	ma,	where	m	is	the	mass	of	the	particle	in	harmonic	motion	and	a
is	its	instantaneous	acceleration,	you	can	substitute	for	F	and	write	this
equation	as
ma	+	kx	=	0
Here's	the	equation	for	instantaneous	acceleration,	where	x	is	displacement
and	t	is	time:
So	substituting	for	a,	you	can	rewrite	the	force	equation	as
Dividing	by	the	mass	of	the	particle	gives	you	the	following:
If	you	take	k/m	=	2	(where		is	the	angular	frequency),	this	becomes
You	can	solve	this	equation	for	x,	where	A	and	B	are	constants:
x	=	A	sint	+	B	cost
Therefore,	the	solution	is	an	oscillating	one	because	it	involves	sines	and
cosines,	which	represent	periodic	waveforms.
Understanding	total	energy	in	quantum	oscillation
Now	look	at	harmonic	oscillators	in	quantum	physics	terms.	The	Hamiltonian
(H)	is	the	sum	of	kinetic	and	potential	energies		the	total	energy	of	the
system:
H	=	KE	+	PE
For	a	harmonic	oscillator,	here's	what	these	energies	are	equal	to:
	The	kinetic	energy	at	any	one	moment	is	the	following,	where	p	is	the
particle's	momentum	and	m	is	its	mass:
	The	particle's	potential	energy	is	equal	to	the	following,	where	k	is	the
spring	constant	and	x	is	displacement:
(Note:	The	k	is	replaced	because	2	=	k/m.)
Therefore,	in	quantum	physics	terms,	you	can	write	the	Hamiltonian	as	H	=	KE
+	PE,	or
where	P	and	X	are	the	momentum	and	position	operators.
You	can	apply	the	Hamiltonian	operator	to	various	eigenstates	(see	Chapter	2
for	more	on	eigenstates),	|>,	of	the	harmonic	oscillator	to	get	the	total	energy,
E,	of	those	eigenstates:
The	problem	now	becomes	one	of	finding	the	eigenstates	and	eigenvalues.
However,	this	doesn't	turn	out	to	be	an	easy	task.	Unlike	the	potentials	V(x)
covered	in	Chapter	3,	V(x)	for	a	harmonic	oscillator	is	more	complex,
depending	as	it	does	on	x2.
So	you	have	to	be	clever.	The	way	you	solve	harmonic	oscillator	problems	in
quantum	physics	is	with	operator	algebra		that	is,	you	introduce	a	new	set	of
operators.	And	they're	coming	up	now.
Creation	and	Annihilation:
Introducing	the	Harmonic	Oscillator
Operators
	Creation	and	annihilation	may	sound	like	big	make-or-break	the	universe
kinds	of	ideas,	but	they	play	a	starring	role	in	the	quantum	world	when
you're	working	with	harmonic	oscillators.	You	use	the	creation	and
annihilation	operators	to	solve	harmonic	oscillator	problems	because	doing
so	is	a	clever	way	of	handling	the	tougher	Hamiltonian	equation	(see	the
preceding	section).	Here's	what	these	two	new	operators	do:
	Creation	operator:	The	creation	operator	raises	the	energy	level	of	an
eigenstate	by	one	level,	so	if	the	harmonic	oscillator	is	in	the	fourth	energy
level,	the	creation	operator	raises	it	to	the	fifth	level.
	Annihilation	operator:	The	annihilation	operator	does	the	reverse,
lowering	eigenstates	one	level.
These	operators	make	it	easier	to	solve	for	the	energy	spectrum	without	a	lot	of
work	solving	for	the	actual	eigenstates.	In	other	words,	you	can	understand	the
whole	energy	spectrum	by	looking	at	the	energy	difference	between
eigenstates.
Mind	your	p's	and	q's:	Getting	the	energy	state
equations
Here's	how	people	usually	solve	for	the	energy	spectrum.	First,	you	introduce
two	new	operators,	p	and	q,	which	are	dimensionless;	they	relate	to	the	P
(momentum)	and	X	(position)	operators	this	way:
	
	
You	use	these	two	new	operators,	p	and	q,	as	the	basis	of	the	annihilation
operator,	a,	and	the	creation	operator,	a:
	
	
Now	you	can	write	the	harmonic	oscillator	Hamiltonian	like	this,	in	terms	of	a
and	a:
As	for	creating	new	operators	here,	the	quantum	physicists	went	crazy,	even
giving	a	name	to	aa:	the	N	or	number	operator.	So	here's	how	you	can	write
the	Hamiltonian:
The	N	operator	returns	the	number	of	the	energy	level	of	the	harmonic
oscillator.	If	you	denote	the	eigenstates	of	N	as	|n>,	you	get	this,	where	n	is	the
number	of	the	nth	state:
N|n>	=	n|n>
Because	H	=	(N	+	1/2),	and	because	H|n>	=	En	|n>,	then	by	comparing	the
previous	two	equations,	you	have
Amazingly,	that	gives	you	the	energy	eigenvalues	of	the	nth	state	of	a	quantum
mechanical	harmonic	oscillator.	So	here	are	the	energy	states:
	The	ground	state	energy	corresponds	to	n	=	0:
	The	first	excited	state	is
	The	second	excited	state	has	an	energy	of
And	so	on.	That	is,	the	energy	levels	are	discrete	and	nondegenerate	(not
shared	by	any	two	states).	Thus,	the	energy	spectrum	is	made	up	of	equidistant
bands.
Finding	the	Eigenstates
When	you	have	the	eigenstates	(see	Chapter	2	to	find	out	all	about
eigenstates),	you	can	determine	the	allowable	states	of	a	system	and	the
relative	probability	that	the	system	will	be	in	any	of	those	states.
The	commutator	of	operators	A,	B	is	[A,	B]	=	AB		BA,	so	note	that	the
commutator	of	a	and	a	is	the	following:
This	is	equal	to	the	following:
This	equation	breaks	down	to	[a,	a]	=	1.	And	putting	together	this	equation
with	
Finding	the	energy	of	a|n>
Okay,	with	the	commutator	relations,	you're	ready	to	go.	The	first	question	is	if
the	energy	of	state	|n>	is	En,	what	is	the	energy	of	the	state	a|n>?	Well,	to	find
this	rearrange	the	commutator	[a,	H]	=	a	to	get	Ha	=	aH		a.	Then	use
this	to	write	the	action	of		on	a|n>	like	this:
H(a|n>)
=	(aH		a)|n>
=	(En		)(a|n>)
So	a|n>	is	also	an	eigenstate	of	the	harmonic	oscillator,	with	energy	En		,
not	En.	That's	why	a	is	called	the	annihilation	or	lowering	operator:	It	lowers
the	energy	level	of	a	harmonic	oscillator	eigenstate	by	one	level.
Finding	the	energy	of	a|n>
So	what's	the	energy	level	of	a|n>?	You	can	write	that	can	like	this:
All	this	means	that	a|n>	is	an	eigenstate	of	the	harmonic	oscillator,	with
energy	En	+	,	not	just	En		that	is,	the	a	operator	raises	the	energy	level	of
an	eigenstate	of	the	harmonic	oscillator	by	one	level.
Using	a	and	a	directly
If	you've	been	following	along	from	the	preceding	section,	you	know	that
H(a|n>)	=	(En		)(a|n>)	and	H(a|n>)	=	(En	+	)(a|n>).	You	can	derive
the	following	from	the	these	equations:
	a|n>	=	C|n		1>
	a|n>	=	D|n	+	1>
C	and	D	are	positive	constants,	but	what	do	they	equal?	The	states	|n		1>	and
|n	+	1>	have	to	be	normalized,	which	means	that	<n		1|n		1>	=	<n	+	1|n	+
1>	=	1.	So	take	a	look	at	the	quantity	using	the	C	operator:
(<n|a)(a|n>)	=	C2<n		1|n		1>
And	because	|n		1>	is	normalized,	<n		1|n		1>	=	1:
(<n|a)(a|n>)	=	C2
<n|aa|n>	=	C2
But	you	also	know	that	aa	=	N,	the	energy	level	operator,	so	you	get	the
following	equation:
<n|N|n>	=	C2
N|n>	=	n|n>,	where	n	is	the	energy	level,	so
n<n|n>	=	C2
However,	<n|n>	=	1,	so
n	=	C2
n	=	C
This	finally	tells	you,	from	a|n>	=	C|n		1>,	that
a|n>	=	n	|n		1>
That's	cool		now	you	know	how	to	use	the	lowering	operator,	a,	on
eigenstates	of	the	harmonic	oscillator.
What	about	the	raising	operator,	a?	First	we	rearrange	the	commutator	[a,	H]
=		a,	to	get
Ha	=	aH	+	a
Then	you	follow	the	same	course	of	reasoning	you	take	with	the	a	operator	to
show	the	following:
a|n>	=	(n	+	1)1/2	|n	+	1>
So	at	this	point,	you	know	what	the	energy	eigenvalues	are	and	how	the	raising
and	lowering	operators	affect	the	harmonic	oscillator	eigenstates.	You've	made
quite	a	lot	of	progress,	using	the	a	and	a	operators	instead	of	trying	to	solve
the	Schrdinger	equation.
Finding	the	harmonic	oscillator	energy	eigenstates
The	charm	of	using	the	operators	a	and	a	is	that	given	the	ground	state,	|0>,
those	operators	let	you	find	all	successive	energy	states.	If	you	want	to	find	an
excited	state	of	a	harmonic	oscillator,	you	can	start	with	the	ground	state,	|0>,
and	apply	the	raising	operator,	a.	For	example,	you	can	do	this:
	
	
	
	
And	so	on.	In	general,	you	have	this	relation:
Working	in	position	space
Okay,	
	is	fine	as	far	as	it	goes		but	just	what	is	|0>?	Can't
you	get	a	spatial	eigenstate	of	this	eigenvector?	Something	like	0(x),	not
just|0>?	Yes,	you	can.	In	other	words,	you	want	to	find	<x|0>	=	0(x).	So	you
need	the	representations	of	a	and	a	in	position	space.
The	p	operator	is	defined	as
Because	
,	you	can	write
And	writing	x0	=	
,	this	becomes
Okay,	what	about	the	a	operator?	You	know	that
And	that
Therefore,
You	can	also	write	this	equation	as
Okay,	so	that's	a	in	the	position	representation.	What's	a?	That	turns	out	to	be
this:
Now's	the	time	to	be	clever.	You	want	to	solve	for	|0>	in	the	position	space,	or
<x|0>.	Here's	the	clever	part		when	you	use	the	lowering	operator,	a,	on|0>,
you	have	to	get	0	because	there's	no	lower	state	than	the	ground	state,	so	a|0>
=	0.	And	applying	the	<x|	bra	gives	you	<x|a|0>	=	0.
That's	clever	because	it's	going	to	give	you	a	homogeneous	differential
equation	(that	is,	one	that	equals	zero).	First,	you	substitute	for	a:
Multiplying	both	sides	by	
	gives	you	the	following
The	solution	to	this	compact	differential	equation	is
That's	a	gaussian	function,	so	the	ground	state	of	a	quantum	mechanical
harmonic	oscillator	is	a	gaussian	curve,	as	you	see	in	Figure	4-1.
Figure	4-1:	The	ground	state	of	a	quantum	mechanical	harmonic	oscillator.
Finding	the	wave	function	of	the	ground	state
As	a	gaussian	curve,	the	ground	state	of	a	quantum	oscillator	is	0(x)	=	A
exp(x2/2x02).	How	can	you	figure	out	A?	Wave	functions	must	be	normalized,	so
the	following	has	to	be	true:
Substituting	for	0(x)	gives	you	this	next	equation:
You	can	evaluate	this	integral	to	be
Therefore,
This	means	that	the	wave	function	for	the	ground	state	of	a	quantum
mechanical	harmonic	oscillator	is
Cool.	Now	you've	got	an	exact	wave	function.
A	little	excitement:	Finding	the	first	excited	state
Okay,	the	preceding	section	shows	you	what	0(x)	looks	like.	What	about	the
first	excited	state,	1(x)?	Well,	as	you	know,	1(x)	=	<x|1>	and	|1>	=	a|0>,	so
1(x)	=	<x|a|0>
And	you	know	that	a	is	the	following:
Therefore,	1(x)	=	<x|a|0>	becomes
And	because	0(x)	=	<x|0>,	you	get	the	following	equation:
You	also	know	the	following:
Therefore,	
	becomes
What's	1(x)	look	like?	You	can	see	a	graph	of	1(x)	in	Figure	4-2,	where	it	has
one	node	(transition	through	the	x	axis).
Figure	4-2:	The	first	excited	state	of	a	quantum	mechanical	harmonic	oscillator.
Finding	the	second	excited	state
All	right,	how	about	finding	2(x)	and	so	on?	You	can	find	2(x)	from	this
equation:
Substituting	for	a,	the	equation	becomes
Using	hermite	polynomials	to	find	any	excited	state
You	can	generalize	the	differential	equation	for	n(x)	like	this:
To	solve	this	general	differential	equation,	you	make	use	of	the	fact	that
Hn(x)	is	the	nth	hermite	polynomial,	which	is	defined	this	way:
Holy	mackerel!	What	do	the	hermite	polynomials	look	like?	Here's	H0(x),	H1(x),
and	so	on:
	H0(x)	=	1
	H1(x)	=	2x
	H2(x)	=	4x2		2
	H3(x)	=	8x3		12x
	H4(x)	=	16x4		48x2	+	12
	H5(x)	=	32x5		160x3	+	120x
What	does	this	buy	you?	You	can	express	the	wave	functions	for	quantum
mechanical	harmonic	oscillators	like	this,	using	the	hermite	polynomials	Hn(x):
And	that's	what	the	wave	function	looks	like	for	a	quantum	mechanical
harmonic	oscillator.	Who	knew	it	would've	involved	hermite	polynomials?
You	can	see	what	2(x)	looks	like	in	Figure	4-3;	note	that	there	are	two	nodes
here		in	general,	n(x)	for	the	harmonic	oscillator	will	have	n	nodes.
Figure	4-3:	The	second	excited	state	of	a	quantum	mechanical	harmonic	oscillator.
Putting	in	some	numbers
The	preceding	section	gives	you	n(x),	and	you've	already	solved	for	En,	so
you're	on	top	of	harmonic	oscillators.	Take	a	look	at	an	example.
Say	that	you	have	a	proton	undergoing	harmonic	oscillation	with		=	4.58	
1021	sec1,	as	shown	in	Figure	4-4.
Figure	4-4:	A	proton	undergoing	harmonic	oscillation.
What	are	the	energies	of	the	various	energy	levels	of	the	proton?	You	know	that
in	general,
So	here	are	the	energies	of	the	proton,	in	megaelectron	volts	(MeV):
	
	
	
	
And	so	on.
Now	what	about	the	wave	functions?	The	general	form	of	n(x)	is
where	
Convert	all	length	measurements	into	femtometers	(1	fm	=	1		1015	m),	giving
you	x0	=	3.71	fm.	Here's	0(x),	where	x	is	measured	in	femtometers:
Here	are	a	couple	more	wave	functions:
	
	
Looking	at	Harmonic	Oscillator
Operators	as	Matrices
Because	the	harmonic	oscillator	has	regularly	spaced	energy	levels,	people
often	view	it	in	terms	of	matrices,	which	can	make	things	simpler.	For	example,
the	following	may	be	the	ground	state	eigenvector	(note	that	it's	an	infinite
vector):
And	this	may	be	the	first	excited	state:
And	so	on.	The	N	operator,	which	just	returns	the	energy	level,	would	then	look
like	this:
So	N|2>	gives	you
This	is	equal	to
In	other	words,	N|2>	=	2|2>.
How	about	the	a	(lowering)	operator?	That	looks	like	this:
In	this	representation,	what	is	a|1>?	In	general,	a|n>	=	n1/2|n		1>,	so	a|1>
should	equal|0>.	Take	a	look:
This	matrix	multiplication	equals	the	following:
In	other	words,	a|1>	=|0>,	just	as	expected.
So	how	about	the	a	(raising)	operator?	Here's	how	it	works	in	general:	a|n>	=
(n	+	1)1/2|n	+	1>.	In	matrix	terms,	a	looks	like	this:
For	example,	you	expect	that	a|1>	=	 |2>.	Does	it?	The	matrix	multiplication
is
This	equals	the	following:
So	a|1>	=	 |2>,	as	it	should.
How	about	taking	a	look	at	the	Hamiltonian,	which	returns	the	energy	of	an
eigenstate,	H|n>	=	En|n>?	In	matrix	form,	the	Hamiltonian	looks	like	this:
So	if	you	prefer	the	matrix	way	of	looking	at	things,	that's	how	it	works	for	the
harmonic	oscillator.
A	Jolt	of	Java:	Using	Code	to	Solve
the	Schrdinger	Equation
Numerically
Here's	the	one-dimensional	Schrdinger	equation:
And	for	harmonic	oscillators,	you	can	write	the	equation	like	this,	where	
:
In	general,	as	the	potential	V(x)	gets	more	and	more	complex,	using	a	computer
to	solve	the	Schrdinger	equation	begins	to	look	more	and	more	attractive.	In
this	section,	I	show	you	how	to	do	just	that	for	the	harmonic	oscillator
Schrdinger	equation.
Making	your	approximations
In	computer	terms,	you	can	approximate	(x)	as	a	collection	of	points,	1,	2,
3,	4,	5,	and	so	on,	as	you	see	in	Figure	4-5.
Figure	4-5:	Dividing	(x)	along	the	x	axis.
Each	point	along	(x)		1,	2,	3,	4,	5,	and	so	on		is	separated	from	its
neighbor	by	a	distance,	h0,	along	the	x	axis.	And	because	d/dx	is	the	slope	of
(x),	you	can	make	the	approximation	that
In	other	words,	the	slope,	d/dx,	is	approximately	equal	to	y/x,	which	is
equal	to	n	+	1		n	(=	y)	divided	by	h0	(=	x).
You	can	rearrange	the	equation	to	this:
That's	a	crude	approximation	for	n	+	1,	given	n.	So,	for	example,	if	you	know
4,	you	can	find	the	approximate	value	of	5,	if	you	know	d/dx	in	the	region	of
4.
You	can,	of	course,	find	better	approximations	for	n	+	1.	In	particular,
physicists	often	use	the	Numerov	algorithm	when	solving	the	Schrdinger
equation,	and	that	algorithm	gives	you	n	+	1	in	terms	of	n	and	n		1.	Here's
what	the	Numerov	algorithm	says:
In	this	equation,	for	the	harmonic	oscillator,	
and	the	boundary	conditions	are	()	=	()	=	0.	Wow.	Imagine	having	to
calculate	this	by	hand.	Why	not	leave	it	up	to	the	computer?
For	a	proton	undergoing	harmonic	oscillation	with		=	4.58		1021	sec1,	the
exact	ground	state	energy	is
You	solve	this	problem	exactly	earlier	in	this	chapter.	The	following	sections
have	you	try	to	get	this	same	result	using	the	Numerov	algorithm	and	a
computer.
Building	the	actual	code
To	calculate	the	ground	state	energy	of	the	harmonic	oscillator	using	the
Numerov	algorithm,	this	section	uses	the	Java	programming	language,	which
you	can	get	for	free	from	java.sun.com.
Here's	how	you	use	the	program:	You	choose	a	trial	value	of	the	energy	for	the
ground	state,	E0,	and	then	calculate	(x)	at	,	which	should	be	zero		and	if
it's	not,	you	can	adjust	your	guess	for	E0	and	try	again.	You	keep	going	until
()	=	0	(or	if	not	actually	0,	a	very	small	number	in	computer	terms)		and
when	it	does,	you	know	you've	guessed	the	correct	energy.
Approximating	()
How	do	you	calculate	()?	After	all,	infinity	is	a	pretty	big	number,	and	the
computer	is	going	to	have	trouble	with	that.	In	practical	terms,	you	have	to	use
a	number	that	approximates	infinity.	In	this	case,	you	can	use	the	classical
turning	points	of	the	proton		the	points	where	all	the	proton's	energy	is
potential	energy	and	it	has	stopped	moving	in	preparation	for	reversing	its
direction.
At	the	turning	points,	xt,	
	(that	is,	all	the	energy	is	in	potential	energy),
so
And	this	is	on	the	order	of	5	femtometers	(fm),	so	you	assume	that	(x)	should
surely	be	zero	at,	say,	15	fm.	Here's	the	interval	over	which	you	calculate
(x):
	xmin	=	15	fm
	xmax	=	15	fm
Divide	this	30	fm	interval	into	200	segments,	making	the	width	of	each
segment,	h0,	equal	to	(xmax		xmin)/200	=	h0	=	0.15	fm.
Okay,	you're	making	progress.	You'll	start	by	assuming	that	(xmin)	=	0,	guess	a
value	of	E0,	and	then	calculate	(xmax)	=	200	(because	there	are	200	segments,
at	x	=	xmax,	n	=	200),	which	should	equal	zero	when	you	get	E0.
Here's	what	the	results	tell	you:
	Correct:	If	abs(200)	is	zero,	or	in	practical	terms,	less	than,	say,	your
maximum	allowed	value	of	max	=	1		108,	then	you're	done		the	E0	you
guessed	was	correct.
	Too	high:	If	abs(200)	is	larger	than	your	maximum	allowed	,	max	(=	1	
108),	and	200	is	positive,	the	energy	you	chose	for	E0	was	too	high.	You	have
to	subtract	a	small	amount	of	energy,	E		say	1		107	MeV		from	your
guess	for	the	energy;	then	calculate	abs(200)	again	and	see	whether	it's	still
higher	than	your	maximum	allowed	,	max.	If	so,	you	have	to	repeat	the
process	again.
	Too	low:	If	abs(200)	is	larger	than	your	maximum	allowed	,	max	(=	1		10
8),	and	200	is	negative,	the	energy	you	chose	for	E0	was	too	low.	You	have	to
add	a	small	amount	of	energy,	E,	to	your	guess	for	the	energy;	then
calculate	abs(200)	again	and	see	whether	it's	still	higher	than	your	maximum
allowed	,	max.	If	so,	you	have	to	repeat	the	process.
So	how	do	you	calculate	200?	Given	two	starting	values,	0	and	1,	use	the
Numerov	algorithm:
Keep	calculating	successive	points	along	(x):	2,	3,	4,	and	so	on.	The	last
point	is	200.
Okay,	you're	on	our	way.	You're	going	to	start	the	code	with	the	assumption
that	0	=	0	and	1	is	a	very	small	number	(you	can	choose	any	small	number
you	like).	Because	you	know	that	the	exact	ground	level	energy	is	actually	1.50
MeV,	start	the	code	with	the	guess	that	E0	=	1.4900000	MeV	and	let	the
computer	calculate	the	actual	value	using	increments	of	E	=	1		107	MeV.
Note	also	this	equation	depends	on	kn(x)2,	kn		1(x)2,	and	kn	+	1(x)2.	Here's	how
you	can	find	these	values,	where	Ecurrent	is	the	current	guess	for	the	ground
state	energy	(substitute	n,	n		1,	and	n	+	1	for	j):
And	you	know	that		=	4.58		1021	sec1,	so
	
	
Therefore,	kj2(xj)	=	0.05Ecurrent		5.63		103xj2,	where	xj	for	a	particular
segment	j	is	xj	=	jh0	+	xmin.
Writing	the	code
Okay,	now	I'm	going	to	put	together	all	the	info	from	the	preceding	section	into
some	Java	code.	You	start	with	a	Java	class,	se	(for	Schrdinger	Equation),	in	a
file	you	name	se.java:
	
public	class	se
.
.
.
}
Then	you	set	up	the	variables	and	constants	you'll	need,	including	an	array	for
the	values	you	calculate	for		(because	to	find	n	+	1,	you'll	have	had	to	store
the	already-calculated	values	of	n	and	n		1):
	
public	class	se
{
	
double	psi[];
double	ECurrent;
double	Emin	=	1.490;
double	xMin	=	-15.;
double	xMax	=	15.;
double	hZero;
double	EDelta	=	0.0000001;
double	maxPsi	=	0.00000001;
int	numberDivisions	=	200;
.
.
.
	
}
The	se	class's	constructor	gets	run	first,	so	you	initialize	values	there,	including
0	(=	(xmin)	=	0)	and	1	(any	small	number	you	want)	to	get	the	calculation
going.	In	the	main	method,	called	after	the	constructor,	you	create	an	object	of
the	se	class	and	call	it	calculate	method	to	get	things	started:
	
public	class	se
{
	
double	psi[];
double	ECurrent;
double	Emin	=	1.490;
double	xMin	=	-15.;
double	xMax	=	15.;
double	hZero;
double	EDelta	=	0.0000001;
double	maxPsi	=	0.00000001;
int	numberDivisions	=	200;
	
public	se()
{
ECurrent	=	Emin;
psi	=	new	double[numberDivisions	+	1];
psi[0]	=	0;
psi[1]	=	-0.000000001;
psi[numberDivisions]	=	1.0;
hZero	=	(xMax	-	xMin)	/	numberDivisions;
}
	
public	static	void	main(String	[]	argv)
	
{
se	de	=	new	se();
de.calculate();
}
.
.
.
}
The	real	work	takes	place	in	the	calculate	method,	where	you	use	the	current
guess	for	the	energy	and	calculate	200:
	If	abs(200)	is	less	than	your	maximum	allowed	value	of	,	max,	you've	found
the	answer		your	current	guess	for	the	energy	is	right	on.
	If	abs(200)	is	greater	than	max	and	200	is	positive,	you	have	to	subtract	E
from	your	current	guess	for	the	energy	and	try	again.
	If	abs(200)	is	greater	than	max	and	200	is	negative,	you	have	to	add	E	to
your	current	guess	for	the	energy	and	then	try	again.
Here's	what	all	this	looks	like	in	code:
	
public	void	calculate()
{
while(Math.abs(psi[numberDivisions])>	maxPsi){
for	(int	i	=	1;	i	<numberDivisions;	i++){
psi[i	+	1]	=	calculateNextPsi(i);
}
if	(psi[numberDivisions]>	0.0)	{
ECurrent	=	ECurrent		EDelta;
}
else	{
ECurrent	=	ECurrent	+	EDelta;
}
System.out.println(Psi200:		+	psi[numberDivisions]	+		E:		+	round(ECurrent));
}
System.out.println(\nThe	ground	state	energy	is		+	round(ECurrent)	+		MeV.);
}
Note	that	the	next	value	of		(that	is,	n	+	1)	is	calculated	with	a	method	named
calculateNextPsi.	Here's	where	you	use	the	Numerov	algorithm		given	n,	n
	1,	you	can	calculate	n	+	1:
	
public	double	calculateNextPsi(int	n)
{
double	KSqNMinusOne	=	calculateKSquared(n	-	1);
double	KSqN	=	calculateKSquared(n);
double	KSqNPlusOne	=	calculateKSquared(n	+	1);
double	nextPsi	=	2.0	*(1.0	-	(5.0	*	hZero	*	hZero	*	KSqN	/	12.0))	*	psi[n];
nextPsi	=	nextPsi	-	(1.0	+	(hZero	*	hZero	*	KSqNMinusOne	/	12.0))	*	psi[n	-	1];
nextPsi	=	nextPsi	/(1.0	+	(hZero	*	hZero	*	KSqNPlusOne	/	12.0));
return	nextPsi;
}
Finally,	note	that	to	calculate	n	+	1,	you	need	kn,	kn		1,	and	kn	+	1,	which	you
find	with	a	method	named	calculateKSquared,	which	uses	the	numeric	values
you've	already	figured	out	for	this	problem:
	
public	double	calculateKSquared(int	n)
{
double	x	=	(hZero	*	n)	+	xMin;
return	(((0.05)	*	ECurrent)	-	((x	*	x)	*	5.63e-3));
}
Whew.	Here's	the	whole	program,	se.java:
	
public	class	se
{
	
double	psi[];
double	ECurrent;
double	Emin	=	1.490;
double	xMin	=	-15.;
double	xMax	=	15.;
double	hZero;
double	EDelta	=	0.0000001;
double	maxPsi	=	0.00000001;
int	numberDivisions	=	200;
	
public	se()
{
ECurrent	=	Emin;
psi	=	new	double[numberDivisions	+	1];
psi[0]	=	0;
psi[1]	=	-0.000000001;
psi[numberDivisions]	=	1.0;
hZero	=	(xMax	-	xMin)	/	numberDivisions;
}
	
public	static	void	main(String	[]	argv)
{
se	de	=	new	se();
de.calculate();
}
	
public	void	calculate()
{
while(Math.abs(psi[numberDivisions])>	maxPsi){
for	(int	i	=	1;	i	<numberDivisions;	i++){
psi[i	+	1]	=	calculateNextPsi(i);
}
if	(psi[numberDivisions]>	0.0)	{
ECurrent	=	ECurrent	-	EDelta;
}
else	{
ECurrent	=	ECurrent	+	EDelta;
}
System.out.println(Psi200:		+	psi[numberDivisions]	+		E:		+	round(ECurrent));
}
System.out.println(\nThe	ground	state	energy	is		+	round(ECurrent)	+		MeV.);
	
}
	
public	double	calculateKSquared(int	n)
{
double	x	=	(hZero	*	n)	+	xMin;
return	(((0.05)	*	ECurrent)	-	((x	*	x)	*	5.63e-3));
}
	
public	double	calculateNextPsi(int	n)
{
double	KSqNMinusOne	=	calculateKSquared(n	-	1);
double	KSqN	=	calculateKSquared(n);
double	KSqNPlusOne	=	calculateKSquared(n	+	1);
double	nextPsi	=	2.0	*(1.0	-	(5.0	*	hZero	*	hZero	*	KSqN	/	12.0))	*	psi[n];
nextPsi	=	nextPsi	-	(1.0	+	(hZero	*	hZero	*	KSqNMinusOne	/	12.0))	*	psi[n	-	1];
nextPsi	=	nextPsi	/(1.0	+	(hZero	*	hZero	*	KSqNPlusOne	/	12.0));
return	nextPsi;
}
	
public	double	round(double	val)
{
double	divider	=	100000;
val	=	val	*	divider;
double	temp	=	Math.round(val);
return	(double)temp	/	divider;
}
}
Okay,	now	you	can	compile	the	code	with	javac,	the	Java	compiler	(if	javac	isn't
in	your	computer's	path,	be	sure	to	add	the	correct	path	to	your	command-line
command,	such	as	C:>C:\java\bin\javac	se.java).
	
C:>javac	se.java
This	creates	se.class	from	se.java,	and	you	can	run	se.class	with	Java	itself
(adding	the	correct	path	if	needed):
	
C:>java	se
Running	the	code
When	you	run	the	java	code	for	the	harmonic	oscillator	Schrdinger	equation,
it	displays	the	successive	values	of	200	as	it	adjusts	the	current	guess	for	the
energy	as	it	narrows	in	on	the	right	answer		which	it	displays	at	the	end	of
the	run.	Here's	what	you	see:
	
C:>java	se
PSI200:	-1.0503644097337778E-4	E:	1.49
PSI200:	-1.050354423295303E-4	E:	1.49
PSI200:	-1.0503444368533108E-4	E:	1.49
PSI200:	-1.0503344504260495E-4	E:	1.49
.
.
.
PSI200:	-6.12820872814324E-8	E:	1.50066
PSI200:	-6.031127521356655E-8	E:	1.50066
PSI200:	-5.934046348307554E-8	E:	1.50066
PSI200:	-5.836965180600015E-8	E:	1.50066
PSI200:	-5.739883979461778E-8	E:	1.50066
PSI200:	-5.6428029151212084E-8	E:	1.50066
PSI200:	-5.5457218252899224E-8	E:	1.50066
PSI200:	-5.4486408066519986E-8	E:	1.50066
PSI200:	-5.351559702201636E-8	E:	1.50066
PSI200:	-5.254478723976338E-8	E:	1.50066
PSI200:	-5.157397714326237E-8	E:	1.50066
PSI200:	-5.060316801012202E-8	E:	1.50066
PSI200:	-4.963235841725704E-8	E:	1.50066
PSI200:	-4.866154915227413E-8	E:	1.50066
PSI200:	-4.7690740419271214E-8	E:	1.50066
PSI200:	-4.6719932089691944E-8	E:	1.50066
PSI200:	-4.574912368974434E-8	E:	1.50066
PSI200:	-4.4778315322587505E-8	E:	1.50066
PSI200:	-4.380750790476514E-8	E:	1.50066
PSI200:	-4.28367005783992E-8	E:	1.50066
PSI200:	-4.186589345217578E-8	E:	1.50066
PSI200:	-4.0895085873184064E-8	E:	1.50066
PSI200:	-3.992427935226201E-8	E:	1.50066
PSI200:	-3.8953472673066213E-8	E:	1.50066
PSI200:	-3.79826665057731E-8	E:	1.50066
PSI200:	-3.701186038502826E-8	E:	1.50066
PSI200:	-3.604105453620266E-8	E:	1.50066
PSI200:	-3.507024949509914E-8	E:	1.50066
PSI200:	-3.4099444217875174E-8	E:	1.50066
PSI200:	-3.312863911389194E-8	E:	1.50066
PSI200:	-3.2157834719961815E-8	E:	1.50066
PSI200:	-3.1187030089902856E-8	E:	1.50066
PSI200:	-3.021622619594536E-8	E:	1.50066
	
	
PSI200:	-2.9245421985136167E-8	E:	1.50066
PSI200:	-2.8274618172375295E-8	E:	1.50066
PSI200:	-2.7303815344369703E-8	E:	1.50066
PSI200:	-2.633301196069577E-8	E:	1.50066
PSI200:	-2.5362208888510866E-8	E:	1.50066
PSI200:	-2.439140632085814E-8	E:	1.50066
PSI200:	-2.342060424823075E-8	E:	1.50066
PSI200:	-2.244980221960756E-8	E:	1.50066
PSI200:	-2.147900005347249E-8	E:	1.50067
PSI200:	-2.0508198285622532E-8	E:	1.50067
PSI200:	-1.9537397616823192E-8	E:	1.50067
PSI200:	-1.8566596602866105E-8	E:	1.50067
PSI200:	-1.7595795286272332E-8	E:	1.50067
PSI200:	-1.6624994703779555E-8	E:	1.50067
PSI200:	-1.565419461892862E-8	E:	1.50067
PSI200:	-1.4683394780836424E-8	E:	1.50067
PSI200:	-1.3712594592034165E-8	E:	1.50067
PSI200:	-1.2741795159638587E-8	E:	1.50067
PSI200:	-1.177099622966848E-8	E:	1.50067
PSI200:	-1.0800197142733883E-8	E:	1.50067
PSI200:	-9.82939798529632E-9	E:	1.50067
	
The	ground	state	energy	is	1.50067	MeV.
And	there	you	have	it		the	program	approximates	the	ground	state	energy	as
1.50067	MeV,	pretty	darn	close	to	the	value	you	calculated	theoretically	in	the
earlier	section	"Making	your	approximations":	1.50	MeV.
Very	cool.
Part	III
Turning	to	Angular	Momentum	and
Spin
In	this	part	.	.	.
Things	that	spin	and	rotate		that's	the	topic	of	this	part.	Quantum	physics	has
all	kinds	of	things	to	say	about	how	angular	momentum	and	spin	are	quantized,
and	you	see	it	all	in	this	part.
Chapter	5
Working	with	Angular	Momentum	on
the	Quantum	Level
In	This	Chapter
	Angular	momentum
	Angular	momentum	and	the	Hamiltonian
	Matrix	representation	of	angular	momentum
	Eigenfunctions	of	angular	momentum
In	classical	mechanics,	you	may	measure	angular	momentum	by	attaching	a
golf	ball	to	a	string	and	whirling	it	over	your	head.	In	quantum	mechanics,	think
in	terms	of	a	single	molecule	made	up	of	two	bound	atoms	rotating	around	each
other.	That's	the	level	at	which	quantum	mechanical	effects	become	noticeable.
And	at	that	level,	it	turns	out	that	angular	momentum	is	quantized.	And	since
that	has	tangible	results	in	many	cases,	such	as	the	spectrum	of	excited	atoms,
it's	an	important	topic.
Besides	having	kinetic	and	potential	energy,	particles	can	also	have	rotational
energy.	Here's	what	the	Hamiltonian	(total	energy		see	Chapter	4)	looks	like:
Here,	L	is	the	angular	momentum	operator	and	I	is	the	rotation	moment	of
inertia.	What	are	the	eigenstates	of	angular	momentum?	If	L	is	the	angular
momentum	operator,	and	l	is	an	eigenvalue	of	L,	then	you	could	write	the	
following:
But	that	turns	out	to	be	incomplete	because	angular	momentum	is	a	vector	in
three-dimensional	space		and	it	can	be	pointing	any	direction.	Angular
momentum	is	typically	given	by	a	magnitude	and	a	component	in	one	direction,
which	is	usually	the	Z	direction.	So	in	addition	to	the	magnitude	l,	you	also
specify	the	component	of	L	in	the	Z	direction,	Lz	(the	choice	of	Z	is	arbitrary	
you	can	just	as	easily	use	the	X	or	Y	direction).
If	the	quantum	number	of	the	Z	component	of	the	angular	momentum	is
designated	by	m,	then	the	complete	eigenstate	is	given	by	|l,	m>,	so	the
equation	becomes	the	following:
That's	the	kind	of	discussion	about	eigenstates	that	I	cover	in	this	chapter,	and
I	begin	with	a	discussion	of	angular	momentum.
Ringing	the	Operators:	Round	and
Round	with	Angular	Momentum
Take	a	look	at	Figure	5-1,	which	depicts	a	disk	rotating	in	3D	space.	Because
you're	working	in	3D,	you	have	to	go	with	vectors	to	represent	both	magnitude
and	direction.
Figure	5-1:	A	rotating	disk	with	angular	momentum	vector	L.
As	you	can	see,	the	disk's	angular	momentum	vector,	L,	points	perpendicular	to
the	plane	of	rotation.	Here,	you	can	apply	the	right-hand	rule:	If	you	wrap	your
right	hand	in	the	direction	something	is	rotating,	your	thumb	points	in	the
direction	of	the	L	vector.
Having	the	L	vector	point	out	of	the	plane	of	rotation	has	some	advantages.	For
example,	if	something	is	rotating	at	a	constant	angular	speed,	the	L	vector	will
be	constant	in	magnitude	and	direction		which	makes	more	sense	than	having
the	L	vector	rotating	in	the	plane	of	the	disk's	rotation	and	constantly	changing
direction.
Because	L	is	a	3D	vector,	it	can	point	in	any	direction,	which	means	that	it	has
x,	y,	and	z	components,	Lx,	Ly,	and	Lz	(which	aren't	vectors,	just	magnitudes).
You	can	see	Lz	in	Figure	5-1.
L	is	the	vector	product	of	position	R	and	linear	momentum	P,	so	(L	=	R		P).
You	can	also	write	Lx,	Ly,	and	Lz	at	any	given	moment	in	terms	of	operators	like
this,	where	Px,	Py,	and	Pz	are	the	momentum	operators	(which	return	the
momentum	in	the	x,	y,	and	z	directions)	and	X,	Y,	and	Z	are	the	position
operators	(which	return	the	position	in	the	x,	y,	and	z	directions):
	Lx	=	YPz		ZPy
	Ly	=	ZPx		XPz
	Lz	=	XPy		YPx
You	can	write	the	momentum	operators	Px,	Py,	and	Pz	as
In	the	same	way	you	can	represent	the	position	operators	by	their	equivalent
coordinates,	i.e.
	
	
	
Then	if	we	substitute	these	operator	representations	into	the	equations	for	Lx,
Ly,	and	Lz,	you	get,
Finding	Commutators	of	L
x
,	L
y
,	and	L
z
First	examine	Lx,	Ly,	and	Lz	by	taking	a	look	at	how	they	commute;	if	they
commute	(for	example,	if	[Lx,	Ly]	=	0),	then	you	can	measure	any	two	of	them
(Lx	and	Ly,	for	example)	exactly.	If	not,	then	they're	subject	to	the	uncertainty
relation,	and	you	can't	measure	them	simultaneously	exactly.
Okay,	so	what's	the	commutator	of	Lx	and	Ly?	Using	Lx	=	YPz		ZPy	and	Ly	=
ZPx		XPz,	you	can	write	the	following:
[Lx,	Ly]	=	[YPz		ZPy,	ZPx		XPz]
You	can	write	this	equation	as
[Lx,	Ly]	=	[YPz,	ZPx]		[YPz,	XPz]		[ZPy,	ZPx]	+	[ZPy,	XPz]
=	
=	i(XPy		YPx)
But	XPy		YPx	=	Lz,	so	[Lx,	Ly]	=	iLz.	So	Lx	and	Ly	don't	commute,	which
means	that	you	can't	measure	them	both	simultaneously	with	complete
precision.	You	can	also	show	that	[Ly,	Lz]	=	iLx	and	[Lz,	Lx]	=	iLy.
	Because	none	of	the	components	of	angular	momentum	commute	with
each	other,	you	can't	measure	any	two	simultaneously	with	complete
precision.	Rats.
That	also	means	that	the	Lx,	Ly,	and	Lz	operators	can't	share	the	same
eigenstates.	So	what	can	you	do?	How	can	you	find	an	operator	that	shares
eigenstates	with	the	various	components	of	L	so	that	you	can	write	the
eigenstates	as	|l,	m>?
The	usual	trick	here	is	that	the	square	of	the	angular	momentum,	L2,	is	a	scalar,
not	a	vector,	so	it'll	commute	with	the	Lx,	Ly,	and	Lz	operators,	no	problem:
	[L2,	Lx]	=	0
	[L2,	Ly]	=	0
	[L2,	Lz]	=	0
Okay,	cool,	you're	making	progress.	Because	Lx,	Ly,	and	Lz	don't	commute,	you
can't	create	an	eigenstate	that	lists	quantum	numbers	for	any	two	of	them.	But
because	L2	commutes	with	them,	you	can	construct	eigenstates	that	have
eigenvalues	for	L2	and	any	one	of	Lx,	Ly,	and	Lz.	By	convention,	the	direction
that's	usually	chosen	is	Lz.
Creating	the	Angular	Momentum
Eigenstates
Now's	the	time	to	create	the	actual	eigenstates,	|l,	m>,	of	angular	momentum
states	in	quantum	mechanics.	When	you	have	the	eigenstates,	you'll	also	have
the	eigenvalues,	and	when	you	have	the	eigenvalues,	you	can	solve	the
Hamiltonian	and	get	the	allowed	energy	levels	of	an	object	with	angular
momentum.
	Don't	make	the	assumption	that	the	eigenstates	are	|l,	m>;	rather,	say
they're	|,	>,	where	the	eigenvalue	of	L2	is	L2|,	>	=	2|,	>.	So	the
eigenvalue	of	L2	is	2,	where	you	have	yet	to	solve	for	.	Similarly,	the
eigenvalue	of	Lz	is	Lz|,	>	=	|,	>.
To	proceed	further,	you	have	to	introduce	raising	and	lowering	operators	(as
you	do	with	the	harmonic	oscillator	in	Chapter	4).	That	way,	you	can	solve	for
the	ground	state	by,	for	example,	applying	the	lowering	operator	to	the	ground
state	and	setting	the	result	equal	to	zero		and	then	solving	for	the	ground
state	itself.
In	this	case,	the	raising	operator	is	L+	and	the	lowering	operator	is	L.	These
operators	raise	and	lower	the	Lz	quantum	number.	In	a	way	analogous	to	the
raising	and	lowering	operators	in	Chapter	4,	you	can	define	the	raising	and
lowering	operators	this	way:
	Raising:	L+	=	Lx	+	iLy
	Lowering:	L	=	Lx		iLy
These	two	equations	mean	that
You	can	also	see	that
That	means	the	following	are	all	equal	to	L2:
	L2	=	L+L	+	Lz2		L
	L2	=	LL+	+	Lz2	+	Lz
	L2	=	_1	(L+L	+	LL+)	+	Lz2	2
You	can	also	see	that	these	equations	are	true:
	[L2,	L]	=	0
	[L+,	L]	=	2Lz
	[Lz,	L]	=	L
Okay,	now	you	can	put	all	this	to	work.	You're	getting	to	the	good	stuff.
Take	a	look	at	the	operation	of	L+	on	|,	>:
L+|,	>	=	?
To	see	what	L+|,	>	is,	start	by	applying	the	Lz	operator	on	it	like	this:
Lz	L+|,	>	=	?
From	[Lz,	L]	=	L,	you	can	see	that	Lz	L+		L+	Lz	=	L+,	so
Lz	L+	|,	>	=	L+	Lz|,	>	+	L+	|,	>
And	because	Lz|,	>	=	|,	>,	you	have	the	following:
Lz	L+|,	>	=	(	+	1)L+|,	>
This	equation	means	that	the	eigenstate	L+|,	>	is	also	an	eigenstate	of	the	Lz
operator,	with	an	eigenvalue	of	(	+	1).	Or	in	a	more	comprehensible	way:
L+|,	>	=	c|,		+	1>
where	c	is	a	constant	you	find	later	in	"Finding	the	Eigenvalues	of	the	Raising
and	Lowering	Operators."
So	the	L+	operator	has	the	effect	of	rasing	the		quantum	number	by	1.
Similarly,	the	lowering	operator	does	this:
L|,	>	=	d|,			1>
Now	take	a	look	at	what	L2L+|,	>	equals:
L2	L+	|,	>	=	?
Because	L2	is	a	scalar,	it	commutes	with	everything.	L2	L+		L+	L2	=	0,	so	this	is
true:
L2	L+|,	>	=	L+	L2|,	>
And	because	L2|,	>	=	2|,	>,	you	have	the	following	equation:
L2	L+|,	>	=	2	L+|,	>
Similarly,	the	lowering	operator,	L,	gives	you	this:
L2	L|,	>	=	2	L|,	>
So	the	results	of	these	equations	mean	that	the	L	operators	don't	change	the	
eigenvalue	of	|,	>	at	all.
Okay,	so	just	what	are		and	?	Read	on.
Finding	the	Angular	Momentum
Eigenvalues
The	eigenvalues	of	the	angular	momentum	are	the	possible	values	the	angular
momentum	can	take,	so	they're	worth	finding.	Let's	take	a	look	at	how	to	do
just	that.
Deriving	eigenstate	equations	with	max	and	min
Note	that	L2		Lz2	=	Lx2	+	Ly2,	which	is	a	positive	number,	so	L2		Lz2		0.	That
means	that
And	substituting	in	L2|,	>	=	2|,	>	and	Lz2|,	>	=	|,	>,	and	using
the	fact	that	the	eigenstates	are	normalized,	gives	you	this:
Therefore,			2.	So	there's	a	maximum	possible	value	of	,	which	you	can	call
max.
You	can	be	clever	now,	because	there	has	to	be	a	state	|,	max>	such	that	you
can't	raise		any	more.	Thus,	if	you	apply	the	raising	operator,	you	get	zero:
L+|,	max>	=	0
Applying	the	lowering	operator	to	this	also	gives	you	zero:
LL+|,	max>	=	0
And	because	LL+	=	L2		Lz2		Lz,	that	means	the	following	is	true:
(L2		Lz2		Lz)|	,	max>	=	0
Putting	in	L2|,	max>	=	2	and	
	gives	you	this:
(		max2		max)2	=	0
	=	max(max	+	1)	=	0
Cool,	now	you	know	what		is.	At	this	point,	it's	usual	to	rename	max	as	l	and	
as	m,	so	|,	>	becomes	|l,	m>	and
	L2|l,	m>	=	l(l	+	1)	2|l,	m>
	Lz|l,	m>	=	m|l,	m>
You	can	say	even	more.	In	addition	to	a	max,	there	must	also	be	a	min	such	that
when	you	apply	the	lowering	operator,	L,	you	get	zero,	because	you	can't	go
any	lower	than	min:
L|l,	min>	=	0
And	you	can	apply	L+	on	this	as	well:
L+L|l,	min>	=	0
From	LL+	=	L2		Lz2	+	Lz,	you	know	that
(L2		Lz2	+	Lz)|,	min>	=	0
which	gives	you	the	following:
(		min2	+	min)2	=	0
		min2	+	min	=	0
	=	min2		min
	=	min(min		1)
And	comparing	this	equation	to		=	max(max	+	1)	=	0	gives	you
max	=	min
Note	that	because	you	reach	|,	min>	by	n	successive	applications	of	L	on	|,
max>,	you	get	the	following:
max	=	min	+	n
Coupling	these	two	equations	gives	you
max	=	n/2
Therefore,	max	can	be	either	an	integer	or	half	an	integer	(depending	on
whether	n	is	even	or	odd).
	Because	l	=	max,	m	=	,	and	n	is	a	positive	number,	you	can	find	that	l	
m		l.	So	now	you	have	it:
	The	eigenstates	are	|l,	m>.
	The	quantum	number	of	the	total	angular	momentum	is	l.
	The	quantum	number	of	the	angular	momentum	along	the	z	axis	is	m.
	L2|l,	m>	=	2l(l	+	1)|l,	m>,	where	l	=	0,	1/2,	1,	3/2,	...
	Lz|l,	m>	=	m|l,	m>,	where	m	=	l,	(l		1),	...,	l		1,	l.
	l		m		l.
For	each	l,	there	are	2l	+	1	values	of	m.	For	example,	if	l	=	2,	then	m	can	equal
2,	1,	0,	1,	or	2.	If	l	=	5/2,	then	m	can	equal	5/2,	3/2,	1/2,	1/2,	3/2,	and	5/2.
You	can	see	a	representative	L	and	Lz	in	Figure	5-2.	L	is	the	total	angular
momentum	and	Lz	is	the	projection	of	that	total	angular	momentum	on	the
z	axis.
Figure	5-2:	L	and	Lz.
Getting	rotational	energy	of	a	diatomic	molecule
Here's	an	example	that	involves	finding	the	rotational	energy	spectrum	of	a
diatomic	molecule.	Figure	5-3	shows	the	setup:	A	rotating	diatomic	molecule	is
composed	of	two	atoms	with	masses	m1	and	m2.	The	first	atom	rotates	at	r	=	r1,
and	the	second	atom	rotates	at	r	=	r2.	What's	the	molecule's	rotational	energy?
Figure	5-3:	A	rotating	diatomic	molecule.
The	Hamiltonian	(as	you	can	see	at	the	chapter	intro)	is
I	is	the	rotational	moment	of	inertia,	which	is
I	=	m1r12	+	m2r22	=	r2
where	r	=	|r1		r2|	and	
Because	L	=	I,	L	=	r2.	Therefore,	the	Hamiltonian	becomes
So	applying	the	Hamiltonian	to	the	eigenstates,	|l,	m>,	gives	you	the	following:
And	as	you	know,	L2|l,	m>	=	l(l	+	1)2|l,	m>,	so	this	equation	becomes
And	because	H|l,	m>	=	E|l,	m>,	you	can	see	that
And	that's	the	energy	as	a	function	of	l,	the	angular	momentum	quantum
number.
Finding	the	Eigenvalues	of	the
Raising	and	Lowering	Operators
This	section	looks	at	finding	the	eigenvalues	of	the	raising	and	lowering	angular
momentum	operators,	which	raise	and	lower	a	state's	z	component	of	angular
momentum.
Start	by	taking	a	look	at	L+,	and	plan	to	solve	for	c:
L+|l,	m>	=	c|l,	m	+	1>
So	L+|l,	m>	gives	you	a	new	state,	and	multiplying	that	new	state	by	its
transpose	should	give	you	c2:
(L+|l,	m>)L+|l,	m>	=	c2
To	see	this	equation,	note	that	(L+|l,	m>)L+|l,	m>	=	c2<l,	m	+	1|l,	m	+	1>	=	c2.
On	the	other	hand,	also	note	that(L+|l,	m>)L+|l,	m>	=	<	l,	m|L+	L|l,	m>,	so	you
have
<l,	m|L+	L|l,	m>	=	c2
What	do	you	do	about	L+	L?	Well,	you	see	earlier	in	the	chapter,	in	"Creating
the	Angular	Momentum	Eigenstates,"	that	this	is	true:	L+L	=	L2		Lz2	+	Lz.	So
your	equation	becomes	the	following:
Great!	That	means	that	c	is	equal	to
So	what	is
?	Applying	the	L2	and	Lz	operators	gives	you	this
value	for	c:
c	=	[l(l	+	1)		m(m	+	1)]1/2
And	that's	the	eigenvalue	of	L+,	which	means	you	have	this	relation:
L+|l,	m>	=	[l(l	+	1)		m(m	+	1)]1/2|l,	m	+	1>
Similarly,	you	can	show	that	L	gives	you	the	following:
L|l,	m>	=	[l(l	+	1)		m(m		1)]1/2|l,	m		1>
Interpreting	Angular	Momentum
with	Matrices
Chapter	4	covers	a	matrix	interpretation	of	harmonic	oscillator	states	and
operators,	and	you	can	handle	angular	momentum	the	same	way	(which	often
makes	understanding	what's	going	on	with	angular	momentum	easier).	You	get
to	take	a	look	at	the	matrix	representation	of	angular	momentum	on	a	quantum
level	now.
Consider	a	system	with	angular	momentum,	with	the	total	angular	momentum
quantum	number	l	=	1.	That	means	that	m	can	take	the	values	1,	0,	and	1.	So
you	can	represent	the	three	possible	angular	momentum	states	like	this:
Okay,	so	what	are	the	operators	you've	seen	in	this	chapter	in	matrix
representation?	For	example,	what	is	L2?	You	can	write	L2	this	way	in	matrix
form:
Okay,	<1,	1|L2|1,	1>	=	l(l	+	1)2	=	22;	<1,	1|L2|1,	0>	=	0;	<1,	0|L2|1,	0>	=
22;	and	so	on;	Therefore,	the	preceding	matrix	becomes	the	following:
And	you	can	also	write	this	as
So	in	matrix	form,	the	equation	L2|1,	1>	=	22|1,	1>	becomes
How	about	the	L+	operator?	As	you	probably	know	(from	the	preceding
section),	L+|l,	m>	=	[l(l	+	1)		m(m	+	1)]1/2|l,	m	+	1>.	In	this	example,	l	=	1
and	m	=	1,	0,	and	1.	So	you	have	the	following:
	L+|1,	1>	=	0
	
	
So	the	L+	operator	looks	like	this	in	matrix	form:
Therefore,	L+|1,	0>	would	be
And	this	equals
In	other	words,	
.
Okay,	what	about	L?	You	know	that
L|l,	m>	=	[l(l	+	1)		m(m		1)]1/2|l,	m		1>.
In	this	example,	l	=	1	and	m	=	1,	0,	and	1.
So	that	means	the	following:
	
	
	L|1,	1>	=	0
So	the	L	operator	looks	like	this	in	matrix	form:
That	means	that	L	|1,	1>	would	be
This	equals
Which	tells	you	that
Just	as	you'd	expect.
Okay,	you've	found	L2,	L+,	and	L.	Finding	the	matrix	representation	of	Lz	is
simple	because
	|1,	1>	=	Lz|1,	1>
	0	=	Lz|1,	0>
	|1,	1>	=	Lz|1,	1>
So	you	have	that
Thus	Lz	|1,	1>	equals
And	this	equals
So	Lz|1,	1>	=	|1,	1>.
Now	what	about	finding	the	Lx	and	Ly	operators?	That's	not	as	hard	as	you	may
think,	because
and
Take	a	look	at	Lx	first.	L+	equals
And	L	equals
So	Lx	equals:
Okay,	now	what	about	Ly?	
,	so:
Cool.	This	is	going	pretty	well		how	about	calculating	[Lx,	Ly]?	To	do	that,	you
need	to	calculate	[Lx,	Ly]	=	LxLy		LyLx.	First	find	LxLy:
This	equals
And	similarly,	LyLx	equals
And	this	equals
So
And	this	equals
But	because
You	can	write	the	commutator,	[Lx,	Ly]	like	this:
This	is	just	the	old	result	that	we	know	and	love,	so	it	all	checks	out!
Rounding	It	Out:	Switching	to	the
Spherical	Coordinate	System
So	far,	this	chapter	has	been	dealing	with	angular	momentum	in	terms	of	bras
and	kets,	such	as:
The	charm	of	bras	and	kets	is	that	they	don't	limit	you	to	any	specific	system	of
representation	(see	Chapter	2).	So	you	have	the	general	eigenstates,	but	what
are	the	actual	eigenfunctions	of	Lz	and	L2?	That	is,	you're	going	to	try	to	find
the	actual	functions	that	you	can	use	with	the	angular	momentum	operators
like	L2	and	Lz.
To	find	the	actual	eigenfunctions	(not	just	the	eigenstates),	you	turn	from
rectangular	coordinates,	x,	y,	and	z,	to	spherical	coordinates	because	it'll	make
the	math	much	simpler	(after	all,	angular	momentum	is	about	things	going
around	in	circles).	Figure	5-4	shows	the	spherical	coordinate	system.
Figure	5-4:	The	spherical	coordinate	system.
In	the	rectangular	(Cartesian)	coordinate	system,	you	use	x,	y,	and	z	to	orient
yourself.	In	the	spherical	coordinate	system,	you	also	use	three	quantities:	r,	,
and	,	as	Figure	5-4	shows.	You	can	translate	between	the	spherical	coordinate
system	and	the	rectangular	one	this	way:	The	r	vector	is	the	vector	to	the
particle	that	has	angular	momentum,		is	the	angle	of	r	from	the	z	axis,	and		is
the	angle	of	r	from	the	x	axis.
	x	=	r	sin	cos
	y	=	r	sin	sin
	z	=	r	cos
Consider	the	equations	for	angular	momentum:
When	you	take	the	angular	momentum	equations	with	the	spherical-
coordinate-system	conversion	equations,	you	can	derive	the	following:
	
	
	
	
Okay,	these	equations	look	pretty	involved.	But	there's	one	thing	to	notice:
They	depend	only	on		and	,	which	means	their	eigenstates	depend	only	on	
and	,	not	on	r.	So	the	eigenfunctions	of	the	operators	in	the	preceding	list	can
be	denoted	like	this:
<,	|l,	m>
Traditionally,	you	give	the	name	Ylm(,	)	to	the	eigenfunctions	of	angular
momentum	in	spherical	coordinates,	so	you	have	the	following:
Ylm(,	)	=	<,	|l,	m>
All	right,	time	to	work	on	finding	the	actual	form	of	Ylm(,	).	You	know	that
when	you	use	the	L2	and	Lz	operators	on	angular	momentum	eigenstates,	you
get	this:
L2|l,	m>	=	l(l	+	1)	2|l,	m>
Lz|l,	m>	=	m|l,	m>
So	the	following	must	be	true:
	L2Ylm(,	)	=	l(l	+	1)2Ylm(,	)
	LzYlm(,	)	=	mYlm(,)
In	fact,	you	can	go	further.	Note	that	Lz	depends	only	on	,	which	suggests	that
you	can	split	Ylm(,)	up	into	a	part	that	depends	on		and	a	part	that	depends
on	.	Splitting	Ylm(,)	up	into	parts	looks	like	this:
Ylm(,)	=	lm()m()
That's	what	makes	working	with	spherical	coordinates	so	helpful		you	can
split	the	eigenfunctions	up	into	two	parts,	one	that	depends	only	on		and	one
part	that	depends	only	on	.
The	eigenfunctions	of	Lz	in	spherical	coordinates
Start	by	finding	the	eigenfunctions	of	Lz	in	spherical	coordinates.	In	spherical
coordinates,	the	Lz	operator	looks	like	this:
So	LzYlm(,	)	=	Lzlm()m()	is
which	is	the	following:
And	because	LzYlm(,)	=	mYlm(,),	this	equation	can	be	written	in	this
version:
Cancelling	out	terms	from	the	two	sides	of	this	equation	gives	you	this
differential	equation:
This	looks	easy	to	solve,	and	the	solution	is	just
m()	=	Ceim
where	C	is	a	constant	of	integration.
You	can	determine	C	by	insisting	that	m()	be	normalized		that	is,	that	the
following	hold	true:
which	gives	you
So	m()	is	equal	to	this:
You're	making	progress		you've	been	able	to	determine	the	form	of	m(),	so
Ylm(,)	=	lm()	m(),	which	equals
That's	great		you're	halfway	there,	but	you	still	have	to	determine	the	form	of
	lm(),	the	eigenfunction	of	L2.	That's	coming	up	next.
The	eigenfunctions	of	L2	in	spherical	coordinates
Now	you're	going	to	tackle	the	eigenfunction	of	L2,	lm().	You	already	know
that	in	spherical	coordinates,	the	L2	operator	looks	like	this:
That's	quite	an	operator.	And	you	know	that
So	applying	the	L2	operator	to	Ylm(,)	gives	you	the	following:
And	because	L2Ylm(,	)	=	l(l	+	1)2Ylm(,	)	=	l(l	+	1)2lm()m(),	this
equation	becomes
Wow,	what	have	you	gotten	in	to?	Cancelling	terms	and	subtracting	the	right-
hand	side	from	the	left	finally	gives	you	this	differential	equation:
Combining	terms	and	dividing	by	eim	gives	you	the	following:
Holy	cow!	Isn't	there	someone	who's	tried	to	solve	this	kind	of	differential
equation	before?	Yes,	there	is.	This	equation	is	a	Legendre	differential
equation,	and	the	solutions	are	well-known.	(Whew!)	In	general,	the	solutions
take	this	form:
lm()	=	ClmPlm(cos)
where	Plm(cos)	is	the	Legendre	function.
	So	what	are	the	Legendre	functions?	You	can	start	by	separating	out	the
m	dependence,	which	works	this	way	with	the	Legendre	functions:
where	Pl(x)	is	called	a	Legendre	polynomial	and	is	given	by	the	Rodrigues	
formula:
You	can	use	this	equation	to	derive	the	first	few	Legendre	polynomials	like	this:
	P0(x)	=	1
	P1(x)	=	x
	P2(x)	=	1/2	(3x2		1)
	P3(x)	=	1/2	(5x3		3x)
	P4(x)	=	1/8	(35x4		30x2	+	3)
	P5(x)	=	1/8	(63x5		70x3	+	15x)
and	so	on.	That's	what	the	first	few	Pl(x)	polynomials	look	like.	So	what	do	the
associated	Legendre	functions,	Plm(x)	look	like?	You	can	also	calculate	them.
You	can	start	off	with	Pl0(x),	where	m	=	0.	Those	are	easy	because	Pl0(x)	=	Pl(x),
so
	P10(x)	=	x
	P20(x)	=	1/2	(3x2		1)
	P30(x)	=	1/2	(5x3		3x)
Also,	you	can	find	that
	P11(x)	=	(1		x2)1/2
	P21(x)	=	3x(1		x2)1/2
	P22(x)	=	3(1		x2)
	
	P32(x)	=	15x(1		x2)
	P33(x)	=	15x(1		x2)3/2
These	equations	give	you	an	overview	of	what	the	Plm	functions	look	like,	which
means	you're	almost	done.	As	you	may	recall,	lm(),	the		part	of	Ylm(,	),	is
related	to	the	Plm	functions	like	this:
lm()	=	ClmPlm(cos)
And	now	you	know	what	the	Plm	functions	look	like,	but	what	do	Clm,	the
constants,	look	like?	As	soon	as	you	have	those,	you'll	have	the	complete
angular	momentum	eigenfunctions,	Ylm(,	),	because	Ylm(,	)	=
lm()m().
You	can	go	about	calculating	the	constants	Clm	the	way	you	always	calculate
such	constants	of	integration	in	quantum	physics		you	normalize	the
eigenfunctions	to	1.	For	Ylm(,)	=	lm()m(),	that	looks	like	this:
Substitute	the	following	three	quantities	in	this	equation:
	Ylm(,	)	=	lm()	m()
	
	lm()	=	ClmPlm(cos)
You	get	the	following:
The	integral	over		gives	2,	so	this	becomes
You	can	evaluate	the	integral	to	this:
So	in	other	words:
Which	means	that
So	Ylm(,	)	=	lm()m(),	which	is	the	angular	momentum	eigenfunction	in
spherical	coordinates,	is
The	functions	given	by	this	equation	are	called	the	normalized	spherical
harmonics.	Here	are	what	the	first	few	normalized	spherical	harmonics	look
like:
	
	
	
In	fact,	you	can	use	these	relations	to	convert	the	spherical	harmonics	to
rectangular	coordinates:
	
	
	
Substituting	these	equations	into
	gives	you	the
spherical	harmonics	in	rectangular	coordinates:
	
	
	
	
	
	
Chapter	6
Getting	Dizzy	with	Spin
In	This	Chapter
	Discovering	spin	with	the	Stern-Gerlach	experiment
	Looking	at	eigenstates	and	spin	notation
	Understanding	fermions	and	bosons
	Comparing	the	spin	operators	with	angular	momentum	operators
	Working	with	spin	1/2	and	Pauli	matrices
Physicists	have	suggested	that	orbital	angular	momentum	is	not	the	only	kind	of
angular	momentum	present	in	an	atom		electrons	could	also	have	intrinsic
built-in	angular	momentum.	This	kind	of	built-in	angular	momentum	is	called
spin.	Whether	or	not	electrons	actually	spin	will	never	be	known		they're	as
close	to	point-like	particles	as	you	can	come,	without	any	apparent	internal
structure.	Yet	the	fact	remains	that	they	have	intrinsic	angular	momentum.	And
that's	what	this	chapter	is	about		the	intrinsic,	built-in	quantum	mechanical
spin	of	subatomic	particles.
The	Stern-Gerlach	Experiment	and
the	Case	of	the	Missing	Spot
The	Stern-Gerlach	experiment	unexpectedly	revealed	the	existence	of	spin
back	in	1922.	Physicists	Otto	Stern	and	Walther	Gerlach	sent	a	beam	of	silver
atoms	through	the	poles	of	a	magnet		whose	magnetic	field	was	in	the	z
direction		as	you	can	see	in	Figure	6-1.
Figure	6-1:	The	Stern-Gerlach	experiment.
Because	46	of	silver's	47	electrons	are	arranged	in	a	symmetrical	cloud,	they
contribute	nothing	to	the	orbital	angular	momentum	of	the	atom.	The	47th
electron	can	be	in
	The	5s	state,	in	which	case	its	angular	momentum	is	l	=	0	and	the	z
component	of	that	angular	momentum	is	0
	The	5p	state,	in	which	case	its	angular	momentum	is	l	=	1,	which	means	that
the	z	component	of	its	angular	momentum	can	be	1,	0,	or	1
That	means	that	Stern	and	Gerlach	expected	to	see	one	or	three	spots	on	the
screen	you	see	at	right	in	Figure	6-1,	corresponding	to	the	different	states	of
the	z	component	of	angular	momentum.
But	famously,	they	saw	only	two	spots.	This	puzzled	the	physics	community	for
about	three	years.	Then,	in	1925,	physicists	Samuel	A.	Goudsmit	and	George	E.
Uhlenbeck	suggested	that	electrons	contained	intrinsic	angular	momentum	
and	that	intrinsic	angular	momentum	is	what	gave	them	a	magnetic	moment
that	interacted	with	the	magnetic	field.	After	all,	it	was	apparent	that	some
angular	momentum	other	than	orbital	angular	momentum	was	at	work	here.
And	that	built-in	angular	momentum	came	to	be	called	spin.
The	beam	of	silver	atoms	divides	in	two,	depending	on	the	spin	of	the	47th
electron	in	the	atom,	so	there	are	two	possible	states	of	spin,	which	came	to	be
known	as	up	and	down.
Spin	is	a	purely	quantum	mechanical	effect,	and	there's	no	real	classical
analog.	The	closest	you	can	come	is	to	liken	spin	to	the	spin	of	the	Earth	as	it
goes	around	the	sun		that	is,	the	Earth	has	both	spin	(because	it's	rotating	on
its	axis)	and	orbital	angular	momentum	(because	it's	revolving	around	the	sun).
But	even	this	picture	doesn't	wholly	explain	spin	in	classical	terms,	because	it's
conceivable	that	you	could	stop	the	Earth	from	spinning.	But	you	can't	stop
electrons	from	possessing	spin,	and	that	also	goes	for	other	subatomic	particles
that	possess	spin,	such	as	protons.
	Spin	doesn't	depend	on	spatial	degrees	of	freedom;	even	if	you	were	to
have	an	electron	at	rest	(which	violates	the	uncertainty	principle),	it	would
still	possess	spin.
Getting	Down	and	Dirty	with	Spin
and	Eigenstates
Spin	throws	a	bit	of	a	curve	at	you.	When	dealing	with	orbital	angular
momentum	(see	Chapter	5),	you	can	build	angular	momentum	operators
because	orbital	angular	momentum	is	the	product	of	momentum	and	radius.
But	spin	is	built	in;	there's	no	momentum	operator	involved.	So	here's	the	crux:
You	cannot	describe	spin	with	a	differential	operator,	as	you	can	for	orbital
angular	momentum.
In	Chapter	5,	I	show	how	orbital	angular	momentum	can	be	reduced	to	these
differential	operators:
	
	
	
And	you	can	find	eigenfunctions	for	angular	momentum,	such	as	Y20:
But	because	you	can't	express	spin	using	differential	operators,	you	can't	find
eigenfunctions	for	spin	as	you	do	for	angular	momentum.	So	that	means	that
you're	left	with	the	bra	and	ket	way	of	looking	at	things	(bras	and	kets	aren't
tied	to	any	specific	representation	in	spatial	terms).
In	Chapter	5,	you	also	take	a	look	at	things	in	angular	momentum	terms,
introducing	the	eigenstates	of	orbital	angular	momentum	like	this:	|l	,	m>
(where	l	is	the	angular	momentum	quantum	number	and	m	is	the	quantum
number	of	the	z	component	of	angular	momentum).
You	can	use	the	same	notation	for	spin	eigenstates.	As	with	orbital	angular
momentum,	you	can	use	a	total	spin	quantum	number	and	a	quantum	number
that	indicates	the	spin	along	the	z	axis	(Note:	There's	no	true	z	axis	built	in
when	it	comes	to	spin		you	introduce	a	z	axis	when	you	apply	a	magnetic	field;
by	convention,	the	z	axis	is	taken	to	be	in	the	direction	of	the	applied	magnetic
field).
	The	letters	given	to	the	total	spin	quantum	number	and	the	z-axis
component	of	the	spin	are	s	and	m	(you	sometimes	see	them	written	as	s
and	ms).	In	other	words,	the	eigenstates	of	spin	are	written	as	|s,	m>.
So	what	possible	values	can	s	and	m	take?	That's	coming	up	next.
Halves	and	Integers:	Saying	Hello	to
Fermions	and	Bosons
In	analogy	with	orbital	angular	momentum,	you	can	assume	that	m	(the	z-axis
component	of	the	spin)	can	take	the	values	s,	s	+	1,	...,	s		1,	and	s,	where	s	is
the	total	spin	quantum	number.	For	electrons,	Stern	and	Gerlach	observed	two
spots,	so	you	have	2s	+	1	=	2,	which	means	that	s	=	1/2.	And	therefore,	m	can
be	+1/2	or	1/2.	So	here	are	the	possible	eigenstates	for	electrons	in	terms	of
spin:
|1/2,	1/2>
|1/2,	1/2>
So	do	all	subatomic	particles	have	s	=	1/2?	Nope.	Here	are	their	options:
	Fermions:	In	physics,	particles	with	half-integer	spin	are	called	fermions.
They	include	electrons,	protons,	neutrons,	and	so	on,	even	quarks.	For
example,	electrons,	protons,	and	neutrons	have	spin	s	=	1/2,	and	delta
particles	have	s	=	3/2.
	Bosons:	Particles	with	integer	spin	are	called	bosons.	They	include	photons,
pi	mesons,	and	so	on;	even	the	postulated	particles	involved	with	the	force	of
gravity,	gravitons,	are	supposed	to	have	integer	spin.	For	example,	pi	mesons
have	spin	s	=	0,	photons	have	s	=	1,	and	so	forth.
So	for	electrons,	the	spin	eigenstates	are	|1/2,	1/2>	and	|1/2,	1/2>.	For
photons,	the	eigenstates	are	|1,	1>,	|1,	0>,	and	|1,	1>.	Therefore,	the	possible
eigenstates	depend	on	the	particle	you're	working	with.
Spin	Operators:	Running	Around
with	Angular	Momentum
Because	spin	is	a	type	of	built-in	angular	momentum,	the	spin	operators	have	a
lot	in	common	with	the	orbital	angular	momentum	operators.	In	Chapter	5,	I
discuss	the	orbital	angular	momentum	operators	L2	and	Lz,	and	as	you	may
expect,	there	are	analogous	spin	operators,	S2	and	Sz.	However,	these
operators	are	just	operators;	they	don't	have	a	differential	form	like	the	orbital
angular	momentum	operators	do.
In	fact,	all	the	orbital	angular	momentum	operators,	such	as	Lx,	Ly,	and	Lz,
have	analogs	here:	Sx,	Sy,	and	Sz.	The	commutation	relations	among	Lx,	Ly,
and	Lz	are	the	following:
	[Lx,	Ly]	=	iLz
	[Ly,	Lz]	=	iLx
	[Lz,	Lx]	=	iLy
And	they	work	the	same	way	for	spin:
	[Sx,	Sy]	=	iSz
	[Sy,	Sz]	=	iSx
	[Sz,	Sx]	=	iSy
The	L2	operator	gives	you	the	following	result	when	you	apply	it	to	an	orbital
angular	momentum	eigenstate:
L2|l,	m>	=	l(l	+	1)2|l,	m>
And	just	as	you'd	expect,	the	S2	operator	works	in	an	analogous	fashion:
S2|s,	m>	=	s(s	+	1)2|s,	m>
The	Lz	operator	gives	you	this	result	when	you	apply	it	to	an	orbital	angular
momentum	eigenstate	(see	Chapter	5):
Lz|l,	m>	=	m|l,	m>
And	by	analogy,	the	Sz	operator	works	this	way:
Sz|s,	m>	=	m|s,	m>
What	about	the	raising	and	lowering	operators,	L+	and	L?	Are	there	analogs
for	spin?	In	angular	momentum	terms,	L+	and	L	work	like	this:
	
	
There	are	spin	raising	and	lowering	operators	as	well,	S+	and	S,	and	they	work
like	this:
	
	
In	the	next	section,	I	take	a	special	look	at	particles	with	spin	1/2.
Working	with	Spin	1/2	and	Pauli
Matrices
Spin	1/2	particles	(fermions)	need	a	little	extra	attention.	The	eigenvalues	of
the	S2	operator	here	are
And	the	eigenvalues	of	the	Sz	operator	are
You	can	represent	these	two	equations	graphically	as	shown	in	Figure	6-2,
where	the	two	spin	states	have	different	projections	along	the	z	axis.
Figure	6-2:	Spin	magnitude	and	z	projection.
Spin	1/2	matrices
Time	to	take	a	look	at	the	spin	eigenstates	and	operators	for	particles	of	spin
1/2	in	terms	of	matrices.	There	are	only	two	possible	states,	spin	up	and	spin
down,	so	this	is	easy.	First,	you	can	represent	the	eigenstate	|1/2,	1/2>	like
this:
And	the	eigenstate	|1/2,	1/2>	looks	like	this:
Now	what	about	spin	operators	like	S2?	The	S2	operator	looks	like	this	in	matrix
terms:
And	this	works	out	to	be	the	following:
Similarly,	you	can	represent	the	Sz	operator	this	way:
This	works	out	to
Using	the	matrix	version	of	Sz,	for	example,	you	can	find	the	z	component	of
the	spin	of,	say,	the	eigenstate	|1/2,	1/2>.	Finding	the	z	component	looks	like
this:
Sz	|1/2,	1/2>
Putting	this	in	matrix	terms	gives	you	this	matrix	product:
Here's	what	you	get	by	performing	the	matrix	multiplication:
And	putting	this	back	into	ket	notation,	you	get	the	following:
How	about	the	raising	and	lowering	operators	S+	and	S?	The	S+	operator	looks
like	this:
And	the	lowering	operator	looks	like	this:
So,	for	example,	you	can	figure	out	what	S+|1/2,	1/2>	is.	Here	it	is	in	matrix
terms:
Performing	the	multiplication	gives	you	this:
Or	in	ket	form,	it's	S+|1/2,	1/2>	=	|1/2,	1/2>.	Cool.
Pauli	matrices
Sometimes,	you	see	the	operators	Sx,	Sy,	and	Sz	written	in	terms	of	Pauli	
matrices,	x,	y,	and	z.	Here's	what	the	Pauli	matrices	look	like:
Now	you	can	write	Sx,	Sy,	and	Sz	in	terms	of	the	Pauli	matrices	like	this:
Whoo!	And	that	concludes	your	look	at	spin.
Part	IV
Multiple	Dimensions:	Going	3D	with
Quantum	Physics
In	this	part	.	.	.
The	previous	parts	deal	mostly	with	particles	in	one-dimensional	systems.	This
part	expands	that	coverage	to	three	dimensions,	like	in	the	real	world.	You	see
how	to	handle	quantum	physics	in	three-dimensional	coordinates		whether
rectangular	or	spherical		which	lays	the	groundwork	for	working	with
electrons	in	atoms.
Chapter	7
Rectangular	Coordinates:	Solving
Problems	in	Three	Dimensions
In	This	Chapter
	Exploring	the	Schrdinger	equation	in	the	x,	y,	and	z	dimensions
	Working	with	free	particles	in	3D
	Getting	into	rectangular	potentials
	Seeing	harmonic	oscillators	in	3D	space
One-dimensional	problems	are	all	very	well	and	good,	but	the	real	world	has
three	dimensions.	This	chapter	is	all	about	leaving	one-dimensional	potentials
behind	and	starting	to	take	a	look	at	spinless	quantum	mechanical	particles	in
three	dimensions.
Here,	you	work	with	three	dimensions	in	rectangular	coordinates,	starting	with
a	look	at	the	Schrdinger	equation	in	glorious,	real-life	3D.	You	then	delve	into
free	particles,	box	potentials,	and	harmonic	oscillators.	Note:	By	the	way,	the
next	chapter	uses	spherical	coordinates	because	some	problems	are	better	in
one	system	than	the	other.	Problems	with	spherical	symmetry	are	best	handled
in	spherical	coordinates,	for	example.
The	Schrdinger	Equation:	Now	in
3D!
In	one	dimension,	the	time-dependent	Schrdinger	equation	(of	the	type	in
Chapters	3	and	4	that	let	you	find	the	wave	function)	looks	like	this:
And	you	can	generalize	that	into	three	dimensions	like	this:
Using	the	Laplacian	operator,	you	can	recast	this	into	a	more	compact	form.
Here's	what	the	Laplacian	looks	like:
And	here's	the	3D	Schrdinger	equation	using	the	Laplacian:
To	solve	this	equation,	when	the	potential	doesn't	vary	with	time,	break	out	the
time-dependent	part	of	the	wave	function:
Here,	(x,	y,	z)	is	the	solution	of	the	time-independent	Schrdinger	equation,
and	E	is	the	energy:
So	far,	so	good.	But	now	you've	run	into	a	wall		the	expression	
is	in	general	very	hard	to	deal	with,	so	the	current	equation	is	in	general	very
hard	to	solve.
So	what	should	you	do?	Well,	you	can	focus	on	the	case	in	which	the	equation	is
separable		that	is,	where	you	can	separate	out	the	x,	y,	and	z	dependence	and
find	the	solution	in	each	dimension	separately.	In	other	words,	in	separable
cases,	the	potential,	V(x,	y,	z),	is	actually	the	sum	of	the	x,	y,	and	z	potentials:
V(x,	y,	z)	=	Vx(x)	+	Vy(y)	+	Vz(z)
Now	you	can	break	the	Hamiltonian	in	
	into	three	Hamilitonians,	Hx,	Hy,	and	Hz:
(Hx	+	Hy	+	Hz)(x,	y,	z)	=	E(x,	y,	z)
where
	
	
	
When	you	divide	up	the	Hamiltonian	as	in	(Hx	+	Hy	+	Hz)(x,	y,	z)	=	E(x,	y,
z),	you	can	also	divide	up	the	wave	function	that	solves	that	equation.	In
particular,	you	can	break	the	wave	function	into	three	parts,	one	for	x,	y,	and	z:
(x,	y,	z)	=	X(x)Y(y)Z(z)
Where	X(x),	Y(y),	and	Z(z)	are	functions	of	the	coordinates	x,	y,	and	z	and	are
not	to	be	confused	with	the	position	operators.	This	separation	of	the	wave
function	into	three	parts	is	going	to	make	life	considerably	easier,	because	now
you	can	break	the	Hamiltonian	up	into	three	separate	operators	added
together:
E	=	Ex	+	Ey	+	Ez
So	you	now	have	three	independent	Schrdinger	equations	for	the	three
dimensions:
	
	
	
This	system	of	independent	differential	equations	looks	a	lot	easier	to	solve	than
(Hx	+	Hy	+	Hz)(x,	y,	z)	=	E(x,	y,	z).	In	essence,	you've	broken	the	three-
dimensional	Schrdinger	equation	into	three	one-dimensional	Schrdinger
equations.	That	makes	solving	3D	problems	tractable.
Solving	Three-Dimensional	Free
Particle	Problems
Consider	the	free	particle	you	see	in	three	dimensions	in	Figure	7-1.
Figure	7-1:	A	free	particle	in	3D.
Because	the	particle	is	traveling	freely,	V(x)	=	V(y)	=	V(z)	=	0.	So	the	three
independent	Schrdinger	equations	for	the	three	dimensions	covered	in	the
preceding	section	become	the	following:
	
	
	
If	you	rewrite	these	equations	in	terms	of	the	wave	number,	k,	where
,	then	these	equations	become	the	following:
	
	
	
In	this	section,	you	take	a	look	at	the	solutions	to	these	equations,	find	the	total
energy,	and	add	time	dependence.
The	x,	y,	and	z	equations
Take	a	look	at	the	x	equation	for	the	free	particle,	
.	You	can	write
its	general	solution	as
X(x)	=	Axeikx	x
Y(y)	=	Ayeiky	y
Z(z)	=	Azeikz	z
where	Ax,	Ay,	and	Az	are	constants.
Because	(x,	y,	z)	=	X(x)Y(y)Z(z),	you	get	this	for	(x,	y,	z):
where	A=	Ax	Ay	Az.	The	part	in	the	parentheses	in	the	exponent	is	the	dot
product	of	the	vectors	k	and	r,	k		r.	That	is,	if	the	vector	a	=	(ax,	ay,	az)	in
terms	of	components	and	the	vector	b	=	(bx,	by,	bz),	then	the	dot	product	of	a
and	b	is	a		b	=	(axbx,	ayby,	azbz).	So	here's	how	you	can	rewrite	the	(x,	y,	z)
equation:
Finding	the	total	energy	equation
The	total	energy	of	the	free	particle	is	the	sum	of	the	energy	in	three
dimensions:
E	=	Ex	+	Ey	+	Ez
With	a	free	particle,	the	energy	of	the	x	component	of	the	wave	function	is
.	And	this	equation	works	the	same	way	for	the	y	and	z	components,	so
here's	the	total	energy	of	the	particle:
Note	that	kx2	+	ky2	+	kz2	is	the	square	of	the	magnitude	of	k		that	is,	k	.	k=	k2
Therefore,	you	can	write	the	equation	for	the	total	energy	as
		Note	that	because	E	is	a	constant,	no	matter	where	the	particle	is
pointed,	all	the	eigenfunctions	of	
,	
,	and
	are	infinitely	degenerate	as	you	vary	kx,	ky,	and	kz.
Adding	time	dependence	and	getting	a	physical
solution
You	can	add	time	dependence	to	the	solution	for	(x,	y,	z),	giving	you	(x,	y,
z,	t),	if	you	remember	that,	for	a	free	particle,	
.	That	equation
gives
you	this	form	for	(x,	y,	z,	t):
Because	
,	the	equation	turns	into
In	fact,	now	that	the	right	side	of	the	equation	is	in	terms	of	the	radius	vector	r,
you	can	make	the	left	side	match:
That's	the	solution	to	the	Schrdinger	equation,	but	it's	unphysical	(as	I	discuss
for	the	one-dimensional	Schrdinger	equation	for	a	free	particle	in	Chapter	3).
Why?	Trying	to	normalize	this	equation	in	three	dimensions,	for	example,	gives
you	the	following,	where	A	is	a	constant:
Thus,	the	integral	diverges	and	you	can't	normalize	(r,	t)	as	I've	written	it.	So
what	do	you	do	here	to	get	a	physical	particle?
	The	key	to	solving	this	problem	is	realizing	that	if	you	have	a	number	of
solutions	to	the	Schrdinger	equation,	then	any	linear	combination	of	those
solutions	is	also	a	solution.	In	other	words,	you	add	various	wave	functions
together	so	that	you	get	a	wave	packet,	which	is	a	collection	of	wave
functions	of	the	form	eik		r	such	that
	The	wave	functions	interfere	constructively	at	one	location.
	They	interfere	destructively	(go	to	zero)	at	all	other	locations.
Look	at	the	time-independent	version:
However,	for	a	free	particle,	the	energy	states	are	not	separated	into	distinct
bands;	the	possible	energies	are	continuous,	so	people	write	this	summation	as
an	integral:
So	what	is	(k)?	It's	the	three-dimensional	analog	of	(k)	that	you	find	in
Chapter	3;	that	is,	it's	the	amplitude	of	each	component	wave	function.	You	can
find	(k)	from	the	Fourier	transform	of	1(x)	=	Aeik1x	+	Beik1x	(where	x	<	0)
like	this:
In	practice,	you	choose	(k)	yourself.	Look	at	an	example,	using	the	following
form	for	(k),	which	is	for	a	Gaussian	wave	packet	(Note:	The	exponential	part
is	what	makes	this	a	Gaussian	wave	form):
where	a	and	A	are	constants.	You	can	begin	by	normalizing	(k)	to	determine
what	A	is.	Here's	how	that	works:
Okay.	Performing	the	integral	gives	you
which	means	that	the	wave	function	is
You	can	evaluate	this	equation	to	give	you	the	following,	which	is	what	the	time-
independent	wave	function	for	a	Gaussian	wave	packet	looks	like	in	3D:
Okay,	that's	how	things	look	when	V(r)	=	0.	But	can't	you	solve	some	problems
when	V(r)	is	not	equal	to	zero?	Yep,	you	sure	can.	Check	out	the	next	section.
Getting	Squared	Away	with	3D
Rectangular	Potentials
This	section	takes	a	look	at	a	3D	potential	that	forms	a	box,	as	you	see	in	Figure
7-2.	You	want	to	get	the	wave	functions	and	the	energy	levels	here.
Figure	7-2:	A	box	potential	in	3D.
Inside	the	box,	say	that	V(x,	y,	z)	=	0,	and	outside	the	box,	say	that	V(x,	y,	z)	=
.	So	you	have	the	following:
Dividing	V(x,	y,	z)	into	Vx(x),	Vy(y),	and	Vz(z)	gives	you
	
	
	
Okay,	because	the	potential	goes	to	infinity	at	the	walls	of	the	box,	the	wave
function,	(x,	y,	z),	must	go	to	zero	at	the	walls,	so	that's	your	constraint.	In	3D,
the	Schrdinger	equation	looks	like	this	in	three	dimensions:
Writing	this	out	gives	you	the	following:
Take	this	dimension	by	dimension.	Because	the	potential	is	separable,	you	can
write	(x,	y,	z)	as	(x,	y,	z)	=	X(x)Y(y)Z(z).	Inside	the	box,	the	potential	equals
zero,	so	the	Schrdinger	equation	looks	like	this	for	x,	y,	and	z:
	
	
	
The	next	step	is	to	rewrite	these	equations	in	terms	of	the	wave	number,	k.
Because	
,	you	can	write	the	Schrdinger	equations	for	x,	y,	and	z	as
the	following	equations:
	
	
	
Start	by	taking	a	look	at	the	equation	for	x.	Now	you	have	something	to	work
with		a	second	order	differential	equation,	
.	Here	are	the	two
independent	solutions	to	this	equation,	where	A	and	B	are	yet	to	be
determined:
	X1(x)	=	A	sin(kx)
	X2(x)	=	B	cos(kx)
So	the	general	solution	of	
	is	the	sum	of	the	last	two	equations:
X(x)	=	A	sin(kx)	+	B	cos(kx)
Great.	Now	take	a	look	at	determining	the	energy	levels.
Determining	the	energy	levels
To	be	able	to	determine	the	energy	levels	of	a	particle	in	a	box	potential,	you
need	an	exact	value	for	X(x)		not	just	one	of	the	terms	of	the	constants	A	and
B.	You	have	to	use	the	boundary	conditions	to	find	A	and	B.	What	are	the
boundary	conditions?	The	wave	function	must	disappear	at	the	boundaries	of
the	box,	so
	X(0)	=	0
	X(Lx)	=	0
So	the	fact	that	(0)	=	0	tells	you	right	away	that	B	must	be	0,	because	cos(0)
=	1.	And	the	fact	that	X(Lx)	=	0	tells	you	that	X(Lx)	=	A	sin(kxLx)	=	0.	Because
the	sine	is	0	when	its	argument	is	a	multiple	of	,	this	means	that
And	because	
,	it	means	that
That's	the	energy	in	the	x	component	of	the	wave	function,	corresponding	to
the	quantum	numbers	1,	2,	3,	and	so	on.	The	total	energy	of	a	particle	of	mass
m	inside	the	box	potential	is	E	=	Ex	+	Ey	+	Ez.	Following	
,	you	have	this
for	Ey	and	Ez:
So	the	total	energy	of	the	particle	is	E	=	Ex	+	Ey	+	Ez,	which	equals	this:
And	there	you	have	the	total	energy	of	a	particle	in	the	box	potential.
Normalizing	the	wave	function
Now	how	about	normalizing	the	wave	function	(x,	y,	z)?	In	the	x	dimension,
you	have	this	for	the	wave	equation:
So	the	wave	function	is	a	sine	wave,	going	to	zero	at	x	=	0	and	x	=	Lz.	You	can
also	insist	that	the	wave	function	be	normalized,	like	this:
By	normalizing	the	wave	function,	you	can	solve	for	the	unknown	constant	A.
Substituting	for	X(x)	in	the	equation	gives	you	the	following:
Therefore,	
	becomes	
,	which	means	you
can	solve	for	A:
Great,	now	you	have	the	constant	A,	so	you	can	get	X(x):
Now	get	(x,	y,	z).	You	can	divide	the	wave	function	into	three	parts:
(x,	y,	z)	=	X(x)Y(y)Z(z)
By	analogy	with	X(x),	you	can	find	Y(y)	and	Z(z):
So	(x,	y,	z)	equals	the	following:
That's	a	pretty	long	wave	function.	In	fact,	when	you're	dealing	with	a	box
potential,	the	energy	looks	like	this:
Using	a	cubic	potential
When	working	with	a	box	potential,	you	can	make	things	simpler	by	assuming
that	the	box	is	actually	a	cube.	In	other	words,	L	=	Lx	=	Ly	=	Lz.	When	the	box
is	a	cube,	the	equation	for	the	energy	becomes
So,	for	example,	the	energy	of	the	ground	state,	where	nx	=	ny	=	nz	=	1,	is
given	by	the	following,	where	E111	is	the	ground	state:
Note	that	there's	some	degeneracy	in	the	energies;	for	example,	note	that
	E211	(nx	=	2,	ny	=	1,	nz	=	1)	is	
	E121	(nx	=	1,	ny	=	2,	nz	=	1)	is	
	E112	(nx	=	1,	ny	=	1,	nz	=	2)	is	
So	E211	=	E121	=	E112,	which	means	that	the	first	excited	state	is	threefold
degenerate,	matching	the	threefold	equivalence	in	dimensions.
In	general,	when	you	have	symmetry	built	into	the	physical	layout	(as	you	do
when	L	=	Lx	=	Ly	=	Lz),	you	have	degeneracy.
The	wave	function	for	a	cubic	potential	is	also	easier	to	manage	than	the	wave
function	for	a	general	box	potential	(where	the	sides	aren't	of	the	same	length).
Here's	the	wave	function	for	a	cubic	potential:
So,	for	example,	here's	the	wave	function	for	the	ground	state	(nx	=	1,	ny	=	1,
nz	=	1),	111(x,	y,	z):
And	here's	211(x,	y,	z):
And	121(x,	y,	z):
Springing	into	3D	Harmonic
Oscillators
In	one	dimension,	the	general	particle	harmonic	oscillator	(which	I	first
describe	in	Chapter	4)	looks	like	Figure	7-3,	where	the	particle	is	under	the
influence	of	a	restoring	force		here	illustrated	as	a	spring.
Figure	7-3:	A	harmonic	oscillator.
The	restoring	force	has	the	form	Fx	=	kxx	in	one	dimension,	where	kx	is	the
constant	of	proportionality	between	the	force	on	the	particle	and	the	location	of
the	particle.	The	potential	energy	of	the	particle	as	a	function
of	location	x	is	
.	This	is	also	sometimes	written	as
where	
.
In	this	section,	you	take	a	look	at	the	harmonic	oscillator	in	three	dimensions.	In
three	dimensions,	the	potential	looks	like	this:
Now	that	you	have	a	form	for	the	potential,	you	can	start	talking	in	terms	of
Schrdinger's	equation:
Substituting	in	for	the	three-dimension	potential,	V(x,	y,	z),	gives	you	this
equation:
Take	this	dimension	by	dimension.	Because	you	can	separate	the	potential	into
three	dimensions,	you	can	write	(x,	y,	z)	as	(x,	y,	z)	=	X(x)Y(y)Z(z).
Therefore,	the	Schrdinger	equation	looks	like	this	for	x:
You	solve	that	equation	in	Chapter	4,	where	you	get	this	next	solution:
where	
	and	nx	=	0,	1,	2,	and	so	on.	The	Hnx	term	indicates	a	hermite
polynomial,	which	looks	like	this:
	H0(x)	=	1
	H1(x)	=	2x
	H2(x)	=	4x2		2
	H3(x)	=	8x3		12x
	H4(x)	=	16x4		48x2	+	12
	H5(x)	=	32x5		160x3	+	120x
Therefore,	you	can	write	the	wave	function	like	this:
That's	a	relatively	easy	form	for	a	wave	function,	and	it's	all	made	possible	by
the	fact	that	you	can	separate	the	potential	into	three	dimensions.
What	about	the	energy	of	the	harmonic	oscillator?	The	energy	of	a	one-
dimensional	harmonic	oscillator	is	
.	And	by	analogy,	the	energy	of	a
three-dimensional	harmonic	oscillator	is	given	by
Note	that	if	you	have	an	isotropic	harmonic	oscillator,	where	x	=	y	=	z	=	,
the	energy	looks	like	this:
As	for	the	cubic	potential,	the	energy	of	a	3D	isotropic	harmonic	oscillator	is
degenerate.	For	example,	E112	=	E121	=	E211.	In	fact,	it's	possible	to	have	more
than	threefold	degeneracy	for	a	3D	isotropic	harmonic	oscillator		for	example,
E200	=	E020	=	E002	=	E110	=	E101	=	E011.
In	general,	the	degeneracy	of	a	3D	isotropic	harmonic	oscillator	is
where	n	=	nx	+	ny	+	nz.
Chapter	8
Solving	Problems	in	Three
Dimensions:	Spherical	Coordinates
In	This	Chapter
	Problems	in	spherical	coordinates
	Free	particles	in	spherical	coordinates
	Square	well	potentials
	Isotropic	harmonic	oscillators
In	your	other	life	as	a	sea	captain-slash-pilot,	you're	probably	pretty	familiar
with	latitude	and	longitude		coordinates	that	basically	name	a	couple	of
angles	as	measured	from	the	center	of	the	Earth.	Put	together	the	angle	east	or
west,	the	angle	north	or	south,	and	the	all-important	distance	from	the	center
of	the	Earth,	and	you	have	a	vector	that	gives	a	good	description	of	location	in
three	dimensions.	That	vector	is	part	of	a	spherical	coordinate	system.
Navigators	talk	more	about	the	pair	of	angles	than	the	distance	("Earth's
surface"	is	generally	specific	enough	for	them),	but	quantum	physicists	find
both	angles	and	radius	length	important.	Some	3D	quantum	physics	problems
even	allow	you	to	break	down	a	wave	function	into	two	parts:	an	angular	part
and	a	radial	part.
In	this	chapter,	I	discuss	three-dimensional	problems	that	are	best	handled
using	spherical	coordinates.	(For	3D	problems	that	work	better	in	rectangular
coordinate	systems,	see	Chapter	7.)
A	New	Angle:	Choosing	Spherical
Coordinates	Instead	of	Rectangular
Say	you	have	a	3D	box	potential,	and	suppose	that	the	potential	well	that	the
particle	is	trapped	in	looks	like	this,	which	is	suited	to	working	with	rectangular
coordinates:
Because	you	can	easily	break	this	potential	down	in	the	x,	y,	and	z	directions,
you	can	break	the	wave	function	down	that	way,	too,	as	you	see	here:
(x,	y,	z)	=	X(x)Y(y)Z(z)
Solving	for	the	wave	function	gives	you	the	following	normalized	result	in
rectangular	coordinates:
The	energy	levels	also	break	down	into	separate	contributions	from	all	three
rectangular	axes:
E	=	Ex	+	Ey	+	Ez
And	solving	for	E	gives	you	this	equation	(from	Chapter	7):
But	what	if	the	potential	well	a	particle	is	trapped	in	has	spherical	symmetry,
not	rectangular?	For	example,	what	if	the	potential	well	were	to	look	like	this,
where	r	is	the	radius	of	the	particle's	location	with	respect	to	the	origin	and
where	a	is	a	constant?
Clearly,	trying	to	stuff	this	kind	of	problem	into	a	rectangular-coordinates	kind
of	solution	is	only	asking	for	trouble,	because	although	you	can	do	it,	it	involves
lots	of	sines	and	cosines	and	results	in	a	pretty	complex	solution.	A	much	better
tactic	is	to	solve	this	kind	of	a	problem	in	the	natural	coordinate	system	in
which	the	potential	is	expressed:	spherical	coordinates.
	Figure	8-1	shows	the	spherical	coordinate	system	along	with	the
corresponding	rectangular	coordinates,	x,	y,	and	z.	In	the	spherical
coordinate	system,	you	locate	points	with	a	radius	vector	named	r,	which
has	three	components:
	An	r	component	(the	length	of	the	radius	vector)
		(the	angle	from	z	axis	to	the	the	r	vector)
		(the	angle	from	the	x	axis	to	the	the	r	vector)
Figure	8-1:	The	spherical	coordinate	system.
Taking	a	Good	Look	at	Central
Potentials	in	3D
This	chapter	focuses	on	problems	that	involve	central	potentials		that	is,
spherically	symmetrical	potentials,	of	the	kind	where	V(r)	=	V(r).	In	other
words,	the	potential	is	independent	of	the	vector	nature	of	the	radius	vector;
the	potential	depends	on	only	the	magnitude	of	vector	r	(which	is	r),	not	on	the
angle	of	r.
When	you	work	on	problems	that	have	a	central	potential,	you're	able	to
separate	the	wave	function	into	a	radial	part	(which	depends	on	the	form	of	the
potential)	and	an	angular	part,	which	is	a	spherical	harmonic.	Read	on.
Breaking	down	the	Schrdinger	equation
The	Schrdinger	equation	looks	like	this	in	three	dimensions,	where		is	the
Laplacian	operator	(see	Chapter	2	for	more	on	operators):
And	the	Laplacian	operator	looks	like	this	in	rectangular	coordinates:
In	spherical	coordinates,	it's	a	little	messy,	but	you	can	simplify	later.	Check	out
the	spherical	Laplacian	operator:
Here,	L2	is	the	square	of	the	orbital	angular	momentum:
So	in	spherical	coordinates,	the	Schrdinger	equation	for	a	central	potential
looks	like	this	when	you	substitute	in	the	terms:
	Take	a	look	at	the	preceding	equation.	The	first	term	actually
corresponds	to	the	radial	kinetic	energy		that	is,	the	kinetic	energy	of	the
particle	moving	in	the	radial	direction.	The	second	term	corresponds	to	the
rotational	kinetic	energy.	And	the	third	term	corresponds	to	the	potential
energy.
So	what	can	you	say	about	the	solutions	to	this	version	of	the	Schrdinger
equation?	You	can	note	that	the	first	term	depends	only	on	r,	as	does	the	third,
and	that	the	second	term	depends	only	on	angles.	So	you	can	break	the	wave
function,	(r)	=	(r,	,	),	into	two	parts:
	A	radial	part
	A	part	that	depends	on	the	angles
This	is	a	special	property	of	problems	with	central	potentials.
The	angular	part	of	(r,	,	)
When	you	have	a	central	potential,	what	can	you	say	about	the	angular	part	of
(r,	,	)?	The	angular	part	must	be	an	eigenfunction	of	L2,	and	as	I	show	in
Chapter	5,	the	eigenfunctions	of	L2	are	the	spherical	harmonics,	Ylm(,	)
(where	l	is	the	total	angular	momentum	quantum	number	and	m	is	the	z
component	of	the	angular	momentum's	quantum	number).	The	spherical
harmonics	equal
Here	are	the	first	several	normalized	spherical	harmonics:
	
	
	
	
	
	
That's	what	the	angular	part	of	the	wave	function	is	going	to	be:	a	spherical
harmonic.
The	radial	part	of	(r,	,	)
You	can	give	the	radial	part	of	the	wave	function	the	name	Rnl(r),	where	n	is	a
quantum	number	corresponding	to	the	quantum	state	of	the	radial	part	of	the
wave	function	and	l	is	the	total	angular	momentum	quantum	number.	The
radial	part	is	symmetric	with	respect	to	angles,	so	it	can't	depend	on	m,	the
quantum	number	of	the	z	component	of	the	angular	momentum.	In	other
words,	the	wave	function	for	particles	in	central	potentials	looks	like	the
following	equation	in	spherical	coordinates:
(r,	,	)	=	Rnl(r)Ylm(,	)
The	next	step	is	to	solve	for	Rnl(r)	in	general.	Substituting	(r,	,	)	from	the
preceding	equation	into	the	Schrdinger	equation,
,	gives	you
Okay,	what	can	you	make	of	this?	First,	note	(from	Chapter	5)	that	the	spherical
harmonics	are	eigenfunctions	of	L2	(that's	the	whole	reason	for	using	them),
with	eigenvalue	l(l	+	1)2:
So	the	last	term	in	this	equation	is	simply	l(l	+	1)2.	That	means	that
	takes	the	form
,	which	equals
	The	preceding	equation	is	the	one	you	use	to	determine	the	radial	part	of
the	wave	function,	Rnl(r).	It's	called	the	radial	equation	for	a	central
potential.
When	you	solve	the	radial	equation	for	Rnl(r),	you	can	then	find	(r,	,	)
because	you	already	know	Ylm(,):
(r,	,	)	=	Rnl(r)Ylm(,)
Thus,	this	chapter	simply	breaks	down	to	finding	the	solution	to	the	radial
equation.
Note:	Incidentally,	the	radial	equation	is	really	a	differential	equation	in	one
dimension:	the	r	dimension.	By	selecting	only	problems	that	contain	central
potentials,	you	reduce	the	general	problem	of	finding	the	wave	function
of	particles	trapped	in	a	three-dimensional	spherical	potential	to	a	one-
dimensional	differential	equation.
Handling	Free	Particles	in	3D	with
Spherical	Coordinates
In	this	section	and	the	next,	you	take	a	look	at	some	example	central	potentials
to	see	how	to	solve	the	radial	equation	(see	the	preceding	section	for	more	on
the	radial	part).	Here,	you	work	with	a	free	particle,	in	which	no	potential	at	all
constrains	the	particle.
The	wave	function	in	spherical	coordinates	takes	this	form:
And	you	know	all	about	Ylm(,	),	because	it	gives	you	the	spherical	harmonics.
The	problem	is	now	to	solve	for	the	radial	part,	Rnl(r).	Here's	the	radial
equation:
For	a	free	particle,	V(r)	=	0,	so	the	radial	equation	becomes
The	way	you	usually	handle	this	equation	is	to	substitute		for	kr,	where	k	=
(2mE)/	,	and	because	we	have	a	version	of	the	same	equation	for	each	n
index	it	is	convenient	to	simply	remove	it,	so	that	Rnl	(r)	becomes	Rl	(kr)	=	Rl
().	This	substitution	means	that
	becomes	the	following:
In	this	section,	you	see	how	the	spherical	Bessel	and	Neumann	functions	come
to	the	rescue	when	you're	dealing	with	free	particles.
The	spherical	Bessel	and	Neumann	functions
The	radial	part	of	the	equation,	
,
looks	tough,	but	the	solutions	turn	out	to	be	well-known		this	equation	is
called	the	spherical	Bessel	equation,	and	the	solution	is	a	combination	of	the
spherical	Bessel	functions	[jl()]	and	the	spherical	Neumann	functions	[nl()]:
Rl()	=	Aljl()	+	Blnl()
	where	Al	and	Bl	are	constants.	So	what	are	the	spherical	Bessel
functions	and	the	spherical	Neumann	functions?	The	spherical	Bessel
functions	are	given	by
Here's	what	the	first	few	iterations	of	jl()	look	like:
	
	
	
How	about	the	spherical	Neumann	functions?	The	spherical	Neumann
functions	are	given	by
Here	are	the	first	few	iterations	of	nl():
	
	
	
The	limits	for	small	and	large	
According	to	the	spherical	Bessel	equation,	the	radial	part	of	the	wave	function
for	a	free	particle	looks	like	this:
Rl()	=	Aljl()	+	Blnl()
Take	a	look	at	the	spherical	Bessel	functions	and	Neumann	functions	for	small
and	large	:
	Small	:	The	Bessel	functions	reduce	to	
.
The	Neumann	functions	reduce	to	
.
	Large	:	The	Bessel	functions	reduce	to	
.
The	Neumann	functions	reduce	to	
.
	Note	that	the	Neumann	functions	diverge	for	small	.	Therefore,	any
wave	function	that	includes	the	Neumann	functions	also	diverges,	which	is
unphysical.	So	the	Neumann	functions	aren't	acceptable	functions	in	the
wave	function.
That	means	the	wave	function	(r,	,	),	which	equals	Rnl(r)	Ylm(,	),	equals
the	following:
(r,	,	)	=	Al	jl(kr)	Ylm(,	)
where	k	=	(2mEn)1/2/.	Note	that	because	k	can	take	any	value,	the	energy
levels	are	continuous.
Handling	the	Spherical	Square	Well
Potential
Take	a	look	at	a	spherical	square	well	potential	of	the	kind	you	can	see	in
Figure	8-2	(I	introduce	square	wells	in	Chapter	3).	This	potential	traps	particles
inside	it	when	E	<	0	and	scatters	particles	when	E	>	0.	Mathematically,	you	can
express	the	square	well	potential	like	this:
Figure	8-2:	The	spherical	square	well	potential.
Note	that	this	potential	is	spherically	symmetric	and	varies	only	in	r,	not	in		or
.	You're	dealing	with	a	central	potential,	so	you	can	break	the	wave	function
into	an	angular	part	and	a	radial	part	(see	the	earlier	section	"Taking	a	Good
Look	at	Central	Potentials	in	3D").
This	section	has	you	take	a	look	at	the	radial	equation,	handling	the	two	cases
of	0	<	r	<	a	and	r	>	a	separately.
Inside	the	square	well:	0	<	r	<	a
For	a	spherical	square	well	potential,	here's	what	the	radial	equation	looks	like
for	the	region	0	<	r	<	a:
In	this	region,	V(r)	=	V0,	so	you	have
Taking	the	V0	term	over	to	the	right	gives	you	the	following:
And	here's	what	dividing	by	r	gives	you:
Then,	multiplying	by	2m/2,	you	get
Now	make	the	change	of	variable		=	kr,	where	k	=	(2m(E	+V0))1/2/,	so	that
Rnl(r)	becomes	Rl(kr)	=	Rl().	Using	this	substitution	means	that
	takes	the	following	form:
This	is	the	spherical	Bessel	equation	(just	as	you	see	for	the	free	particle	in
"Handling	Free	Particles	in	3D	with	Spherical	Coordinates").	This	time,	k	=
[2m(E	+V0)]1/2/,	not	(2mE)1/2/.	That	makes	sense,	because	now	the	particle	is
trapped	in	the	square	well,	so	its	total	energy	is	E	+	V0,	not	just	E.
The	solution	to	the	preceding	equation	is	a	combination	of	the	spherical	Bessel
functions	[jl()]	and	the	spherical	Neumann	functions	[nl()]:
Rl()	=	Aljl()	+	Blnl()
You	can	apply	the	same	constraint	here	that	you	apply	for	a	free	particle:	The
wave	function	must	be	finite	everywhere.	For	small	,	the	Bessel	functions	look
like	this:
And	for	small	,	the	Neumann	functions	reduce	to
So	the	Neumann	functions	diverge	for	small	,	which	makes	them	unacceptable
for	wave	functions	here.	That	means	that	the	radial	part	of	the	wave	function	is
just	made	up	of	spherical	Bessel	functions,	where	Al	is	a	constant:
Rl()	=	Aljl()
The	whole	wave	function	inside	the	square	well,	inside(r,	,	),	is	a	product	of
radial	and	angular	parts,	and	it	looks	like	this:
where	inside	=	r(2m(E	+V0))1/2/	and	Ylm(,	)	are	the	spherical	harmonics.
Outside	the	square	well:	r	>	a
Outside	the	square	well,	in	the	region	r	>	a,	the	particle	is	just	like	a	free
particle,	so	here's	what	the	radial	equation	looks	like:
You	solve	this	equation	earlier	in	"Handling	Free	Particles	in	3D	with	Spherical
Coordinates":	Because		=	kr,	where	k	=	(2mE)1/2/,	you	substitute		for	kr	so
that	Rnl(r)	becomes	Rl(kr)	=	Rl().	Using	this	substitution	means	that	the	radial
equation	takes	the	following	form:
The	solution	is	a	combination	of	spherical	Bessel	functions	and	spherical
Neumann	functions,	where	Bl	and	Cl	are	constants:
If	the	energy	E	<	0,	we	must	have	Cl	=	i	Bl"	so	that	the	wave	function	decays
exponentially	at	large	distances	r.	So	the	radial	solution	outside	the	square	well
looks	like	this,	where	outside	=	r(2mE)1/2/:
From	the	preceding	section,	you	know	that	the	wave	function	inside	the	square
well	is
So	how	do	you	find	the	constants	Al,	Bl,	and	Cl?	You	find	those	constants
through	continuity	constraints:	At	the	inside/outside	boundary,	where	r	=	a,	the
wave	function	and	its	first	derivative	must	be	continuous.	So	to	determine	the
constants	you	have	to	solve	these	two	equations:
	
	
Getting	the	Goods	on	Isotropic
Harmonic	Oscillators
This	section	takes	a	look	at	spherically	symmetric	harmonic	oscillators	in	three
dimensions.	In	one	dimension,	you	write	the	harmonic	oscillator	potential	like
this:
where	
	(here,	k	is	the	spring	constant;	that	is,	the	restoring	force	of	the
harmonic	oscillator	is	F	=	kx).	You	can	turn	these	two	equations	into	three-
dimensional	versions	of	the	harmonic	potential	by	replacing	x	with	r:
where	
.	Because	this	potential	is	spherically	symmetric,	the	wave
function	is	going	to	be	of	the	following	form:
where	you	have	yet	to	solve	for	the	radial	function	Rnl(r)	and	where	Ylm(,	)
describes	the	spherical	harmonics.
As	you	know,	the	radial	Schrdinger	equation	looks	like	this:
Substituting	for	V(r)	from	
	gives	you	the	following:
Well,	the	solution	to	this	equation	is	pretty	difficult	to	obtain,	and	you're	not
going	to	gain	anything	by	going	through	the	math	(pages	and	pages	of	it),	so
here's	the	solution:
where	exp(x)	=	ex	and
And	the	Lab(r)	functions	are	the	generalized	Laguerre	polynomials:
Wow.	Aren't	you	glad	you	didn't	slog	through	the	math?	Here	are	the	first	few
generalized	Laguerre	polynomials:
	
	
	
	
All	right,	you	have	the	form	for	Rnl(r).	To	find	the	complete	wave	function,
nlm(r,	,	),	you	multiply	by	the	spherical	harmonics,	Ylm(,	):
Now	take	a	look	at	the	first	few	wave	functions	for	the	isotropic	harmonic
oscillator	in	spherical	coordinates:
	
	
	
	
As	you	can	see,	when	you	have	a	potential	that	depends	on	r2,	as	with	harmonic
oscillators,	the	wave	function	gets	pretty	complex	pretty	fast.
The	energy	of	an	isotropic	3D	harmonic	oscillator	is	quantized,	and	you	can
derive	the	following	relation	for	the	energy	levels:
So	the	energy	levels	start	at	3/2	and	then	go	to	5/2,	7/2,	and	so	on.
Chapter	9
Understanding	Hydrogen	Atoms
In	This	Chapter
	The	Schrdinger	equation	for	hydrogen
	The	radial	wave	functions
	Energy	degeneracy
	Location	of	the	electron
Not	only	is	hydrogen	the	most	common	element	in	the	universe,	but	it's	also	the
simplest.	And	one	thing	quantum	physics	is	good	at	is	predicting	everything
about	simple	atoms.	This	chapter	is	all	about	the	hydrogen	atom	and	solving	the
Schrdinger	equation	to	find	the	energy	levels	of	the	hydrogen	atom.	For	such
a	small	little	guy,	the	hydrogen	atom	can	whip	up	a	lot	of	math		and	I	solve
that	math	in	this	chapter.
Using	the	Schrdinger	equation	tells	you	just	about	all	you	need	to	know	about
the	hydrogen	atom,	and	it's	all	based	on	a	single	assumption:	that	the	wave
function	must	go	to	zero	as	r	goes	to	infinity,	which	is	what	makes	solving	the
Schrdinger	equation	possible.	I	start	by	introducing	the	Schrdinger	equation
for	the	hydrogen	atom	and	take	you	through	calculating	energy	degeneracy
and	figuring	out	how	far	the	electron	is	from	the	proton.
Coming	to	Terms:	The	Schrdinger
Equation	for	the	Hydrogen	Atom
Hydrogen	atoms	are	composed	of	a	single	proton,	around	which	rotates	a
single	electron.	You	can	see	how	that	looks	in	Figure	9-1.
	Note	that	the	proton	isn't	at	the	exact	center	of	the	atom		the	center	of
mass	is	at	the	exact	center.	In	fact,	the	proton	is	at	a	radius	of	rp	from	the
exact	center,	and	the	electron	is	at	a	radius	of	re.
Figure	9-1:	The	hydrogen	atom.
So	what	does	the	Schrdinger	equation,	which	will	give	you	the	wave	equations
you	need,	look	like?	Well,	it	includes	terms	for	the	kinetic	and	potential	energy
of	the	proton	and	the	electron.	Here's	the	term	for	the	proton's	kinetic	energy:	
where	
.	Here,	xp	is	the	proton's	x	position,	yp	is	the	proton's
y	position,	and	zp	is	its	z	position.
The	Schrdinger	equation	also	includes	a	term	for	the	electron's	kinetic
energy:	
	where	
.	Here,	xe	is	the	electron's	x	position,	ye	is
the	electron's	y	position,	and	ze	is	its	z	position.
Besides	the	kinetic	energy,	you	have	to	include	the	potential	energy,	V(r),	in	the
Schrdinger	equation,	which	makes	the	time-independent	Schrdinger
equation	look	like	this:
where	(re,	rp)	is	the	electron	and	proton's	wave	function.
The	electrostatic	potential	energy,	V(r),	for	a	central	potential	is	given	by	the
following	formula,	where	r	is	the	radius	vector	separating	the	two	charges:
	As	is	common	in	quantum	mechanics,	you	use	CGS	(centimeter-gram-
second)
system	of	units,	where	
.
So	the	potential	due	to	the	electron	and	proton	charges	in	the	hydrogen	atom	is
Note	that	r	=	re		rp,	so	the	preceding	equation	becomes
which	gives	you	this	Schrdinger	equation:
Okay,	so	how	do	you	handle	this	equation?	Find	out	in	the	next	section.
Simplifying	and	Splitting	the
Schrdinger	Equation	for	Hydrogen
Here's	the	usual	quantum	mechanical	Schrdinger	equation	for	the	hydrogen
atom:
The	problem	is	that	you're	taking	into	account	the	distance	the	proton	is	from
the	center	of	mass	of	the	atom,	so	the	math	is	messy.	If	you	were	to	assume	that
the	proton	is	stationary	and	that	rp	=	0,	this	equation	would	break	down	to	the
following,	which	is	much	easier	to	solve:
Unfortunately,	that	equation	isn't	exact	because	it	ignores	the	movement	of	the
proton,	so	you	see	the	more-complete	version	of	the	equation	in	quantum
mechanics	texts.
To	simplify	the	usual	Schrdinger	equation,	you	switch	to	center-of-mass
coordinates.	The	center	of	mass	of	the	proton/electron	system	is	at	this
location:
And	the	vector	between	the	electron	and	proton	is
r	=	re		rp
Using	vectors	R	and	r	instead	of	re	and	rp	makes	the	Schrdinger	equation
easier	to	solve.	The	Laplacian	for	R	is	
.	And	the	Laplacian
for	r	is	
.
How	can	you	relate	
	and	
	to	the	usual	equation's	
	and	 ?	After	the
algebra	settles,	you	get
where	M	=	me	+	mp	is	the	total	mass	and	
	is	called	the	reduced	mass.
When	you	put	together	the	equations	for	the	center	of	mass,	the	vector
between	the	proton	and	the	electron,	the	total	mass,	and	m,	then	the	time-
independent	Schrdinger	equation	becomes	the	following:
Then,	given	the	vectors,	R	and	r	,	the	potential	is	given	by,
The	Schrdinger	equation	then	becomes
This	looks	easier		the	main	improvement	being	that	you	now	have	|r|	in	the
denominator	of	the	potential	energy	term	rather	than	|re		rp|.
Because	the	equation	contains	terms	involving	either	R	or	r	but	not	both,	the
form	of	this	equation	indicates	that	it's	a	separable	differential	equation.	And
that	means	you	can	look	for	a	solution	of	the	following	form:
(R,	r)	=	(R)(r)
Substituting	the	preceding	equation	into	the	one	before	it	gives	you	the
following:
And	dividing	this	equation	by	(R)(r)	gives	you
Well,	well,	well.	This	equation	has	terms	that	depend	on	either	(R)	or	(r)	but
not	both.	That	means	you	can	separate	this	equation	into	two	equations,	like
this	(where	the	total	energy,	E,	equals	ER	+	Er):
	
	
Multiplying	
	by	(R)	gives	you
And	multiplying	
	by	(r)	gives	you
Now	you	have	two	Schrdinger	equations.	The	next	two	sections	show	you	how
to	solve	them	independently.
Solving	for	(R)
In	
,	how	do	you	solve	for	(R),	which	is	the	wave
function	of	the	center	of	mass	of	the	electron/proton	system?	This	is	a
straightforward	differential	equation,	and	the	solution	is
(R)	=	Ceik		r
Here,	C	is	a	constant	and	k	is	the	wave	vector,	where	
.
	In	practice,	ER	is	so	small	that	people	almost	always	just	ignore	(R)	
that	is,	they	assume	it	to	be	1.	In	other	words,	the	real	action	is	in	(r),	not
in	(R);	(R)	is	the	wave	function	for	the	center	of	mass	of	the	hydrogen
atom,	and	(r)	is	the	wave	function	for	a	(fictitious)	particle	of	mass	m.
Solving	for	(r)
The	Schrdinger	equation	for	(r)	is	the	wave	function	for	a	made-up	particle
of	mass	m	(in	practice,	m		me	and	(r)	is	pretty	close	to	(re),	so	the	energy,
Er,	is	pretty	close	to	the	electron's	energy).	Here's	the	Schrdinger	equation	for
(r):
You	can	break	the	solution,	(r),	into	a	radial	part	and	an	angular	part	(see
Chapter	8):
(r)	=	Rnl(r)Ylm(,	)
The	angular	part	of	(r)	is	made	up	of	spherical	harmonics,	Ylm(,	),	so	that
part's	okay.	Now	you	have	to	solve	for	the	radial	part,	Rnl(r).	Here's	what	the
Schrdinger	equation	becomes	for	the	radial	part:
where	r	=	|r|.	To	solve	this	equation,	you	take	a	look	at	two	cases		where	r	is
very	small	and	where	r	is	very	large.	Putting	them	together	gives	you	the	rough
form	of	the	solution.
Solving	the	radial	Schrdinger	equation	for	small
r
For	small	r	,	the	terms	and	,	in	the	previous	equation,	become	much	smaller
than	the	rest,	so	we	neglect	them	and	write	the	radial	Schrdinger	as,
And	multiplying	by	2m/2,	you	get
The	solution	to	this	equation	is	proportional	to
Rnl(r)	~	Arl	+	Brl		1
Note,	however,	that	Rnl(r)	must	vanish	as	r	goes	to	zero		but	the	rl		1	term
goes	to	infinity.	And	that	means	that	B	must	be	zero,	so	you	have	this	solution
for	small	r:
Rnl(r)	~	rl
That	takes	care	of	small	r.	The	next	section	takes	a	look	at	very	large	r.
Solving	the	radial	Schrdinger	equation	for	large	r
For	very	large	r,	
	becomes
Because	the	electron	is	in	a	bound	state	in	the	hydrogen	atom,	E	<	0;	thus,	the
solution	to	the	preceding	equation	is	proportional	to
Rnl(r)	~	Aer	+	Ber
where	
.
Note	that	Rnl(r)	~	Aer	+	Ber	diverges	as	r	goes	to	infinity	because	of	the
Ber	term,	so	B	must	be	equal	to	zero.	That	means	that	Rnl(r)	~	er.	In	the	next
section,	you	put	the	solutions	for	small	r	and	large	r	together.
You	got	the	power:	Putting	together	the	solution
for	the	radial	equation
Putting	together	the	solutions	for	small	r	and	large	r	(see	the	preceding
sections),	the	Schrdinger	equation	gives	you	a	solution	to	the	radial
Schrdinger
equation	of	Rnl(r)	=	rlf(r)er,	where	f(r)	is	some	as-yet-undetermined	function
of	r.	Your	next	task	is	to	determine	f(r),	which	you	do	by	substituting	this
equation	into	the	radial	Schrdinger	equation,	giving	you	the	following:
Performing	the	substitution	gives	you	the	following	differential	equation:
Quite	a	differential	equation,	eh?	But	just	sit	back	and	relax		you	solve	it	with
a	power	series,	which	is	a	common	way	of	solving	differential	equations.	Here's
the	power-series	form	of	f(r)	to	use:
Substituting	the	preceding	equation	into	the	one	before	it	gives	you
Changing	the	index	of	the	second	term	from	k	to	k		1	gives	you
Because	each	term	in	this	series	has	to	be	zero,	you	have
Dividing	by	rk2	gives	you
This	equation	gives	the	recurrence	relation	of	the	infinite	series,
.	That	is,	if	you	have
one	coefficient,	you	can	get	the	next	one	using	this	equation.	What	does	that
buy	you?	Well,	take	a	look	at	the	ratio	of	ak/ak1:
Here's	what	this	ratio	approaches	as	k	goes	to	:
This	resembles	the	expansion	for	ex,	which	is
As	for	e2x,	the	ratio	of	successive	terms	is
And	in	the	limit	k		,	the	ratio	of	successive	expansion	coefficients	of	e2x
approaches	
:
That's	the	case	for	e2x.	For	f(r),	you	have
Comparing	these	two	equations,	it's	apparent	that
The	radial	wave	function,	Rnl(r),	looks	like	this:
where	
.
Plugging	the	form	you	have	for	f(r),	
,	into
	gives	you	the	following:
Rnl(r)	=	rle2rer
=	rler
Okay,	should	you	be	overjoyed?	Well,	no.	Here's	what	the	wave	function	(r)
looks	like:	(r)	=	Rnl(r)	Ylm(,	).	And	substituting	in	your	form	of	Rnl(r)	from
this	equation	gives	you
(r)	=	rler	Ylm(,	)
That	looks	fine		except	that	it	goes	to	infinity	as	r	goes	to	infinity.	You	expect
(r)	to	go	to	zero	as	r	goes	to	infinity,	so	this	version	of	Rnl(r)	=	rler	is	clearly
unphysical.	In	other	words,	something	went	wrong	somewhere.	How	can	you	fix
this	version	of	f(r)?
Fixing	f(r)	to	keep	it	finite
You	need	the	solution	for	the	radial	equation	to	go	to	zero	as	r	goes	to	infinity.
The	problem	of	having	(r)	go	to	infinity	as	r	goes	to	infinity	lies	in	the	form	you
assume	for	f(r)	in	the	preceding	section,	which	is
The	solution	is	to	say	that	this	series	must	terminate	at	a	certain	index,	which
you	call	N.	N	is	called	the	radial	quantum	number.	So	this	equation	becomes
the	following	(note	that	the	summation	is	now	to	N,	not	infinity):
For	this	series	to	terminate,	aN+1,	aN+2,	aN+3,	and	so	on	must	all	be	zero.	The
recurrence	relation	for	the	coefficients	ak	is
For	aN+1	to	be	zero,	the	factor	multiplying	ak1	must	be	zero	for	k	=	N	+	1,
which	means	that
Substituting	in	k	=	N	+	1	gives	you	
.	And	dividing	by	2	gives	you
.	Making	the	substitution	N	+	l	+	1		n,	where	n	is	called	the
principal	quantum	number,	gives	you
This	is	the	quantization	condition	that	must	be	met	if	the	series	for	f(r)	is	to	be
finite,	which	it	must	be,	physically:
Because	
,	the	equation	
	puts	constraints
on	the	allowable	values	of	the	energy.
Finding	the	allowed	energies	of	the	hydrogen	atom
The	quantization	condition	for	(r)	to	remain	finite	as	r	goes	to	infinity	is
where	
.	Substituting		into	the	quantization-condition	equation	gives
you	the	following:
Now	solve	for	the	energy,	E.	Squaring	both	sides	of	the	preceding	equation
gives	you
So	here's	the	energy,	E	(Note:	Because	E	depends	on	the	principal	quantum
number,	I've	renamed	it	En):
	Physicists	often	write	this	result	in	terms	of	the	Bohr	radius		the	orbital
radius	that	Niels	Bohr	calculated	for	the	electron	in	a	hydrogen	atom,	r0.
The
Bohr	radius	is	
.
And	in	terms	of	r0,	here's	what	En	equals:
The	ground	state,	where	n	=	1,	works	out	to	be	about	E	=	13.6	eV.
Notice	that	this	energy	is	negative	because	the	electron	is	in	a	bound	state	
you'd	have	to	add	energy	to	the	electron	to	free	it	from	the	hydrogen	atom.
Here	are	the	first	and	second	excited	states:
	First	excited	state,	n	=	2:	E	=	3.4	eV
	Second	excited	state,	n	=	3:	E	=	1.5	eV
Okay,	now	you've	used	the	quantization	condition,	which	is
to	determine	the	energy	levels	of	the	hydrogen	atom.
Getting	the	form	of	the	radial	solution	of	the
Schrdinger	equation
In	this	section,	you	complete	the	calculation	of	the	wave	functions.	Go	to	the
calculation	of	Rnl(r)	(see	the	earlier	section	titled	"You	got	the	power:	Putting
together	the	solution	for	the	radial	equation").	So	far,	you	know	that
,	where	
.	Therefore,
In	fact,	this	isn't	quite	enough;	the	preceding	equation	comes	from	solving	the
radial	Schrdinger	equation:
The	solution	is	only	good	to	a	multiplicative	constant,	so	you	add	such	a
constant,	Anl	(which	turns	out	to	depend	on	the	principal	quantum	number	n
and	the	angular	momentum	quantum	number	l),	like	this:
You	find	Anl	by	normalizing	Rnl(r).
Now	try	to	solve	for	Rnl(r)	by	just	flat-out	doing	the	math.	For	example,	try	to
find	R10(r).	In	this	case,	n	=	1	and	l	=	0.	Then,	because	N	+	l	+	1	=	n,	you	have
N	=	n		l		1.	So	N	=	0	here.	That	makes	Rnl(r)	look	like	this:
And	the	summation	in	this	equation	is	equal	to	
,	so
And	because	l	=	0,	rl	=	1,	so	R10(r)	=	A10er	a0,	where	
.	Therefore,	you
can	also	write	R10(r)	=	A10era0	as
where	r0	is	the	Bohr	radius.	To	find	A10	and	a0,	you	normalize	100(r,	,	)	to	1,
which	means	integrating	|100(r,	,	)|2d3r	over	all	space	and	setting	the	result
to	1.
Now	d3r	=	r2	sin	dr	d	d,	and	integrating	the	spherical	harmonics,	such	as
Y00,	over	a	complete	sphere,	
,	gives	you	1.	Therefore,	you're
left	with	the	radial	part	to	normalize:
Plugging	
	into	
	gives	you
You	can	solve	this	kind	of	integral	with	the	following	relation:
With	this	relation,	the	equation	
	becomes
Therefore,
This	is	a	fairly	simple	result.	Because	A10	is	just	there	to	normalize	the	result,
you	can	set	A10	to	1	(this	wouldn't	be	the	case	if	
	involved
multiple	terms).	Therefore,	
.	That's	fine,	and	it	makes	R10(r),	which	is
You	know	that	nlm(r,	,	)	=	Rnl(r)	Ylm(,	).
And	so	100(r,	,	)	becomes
Whew.	In	general,	here's	what	the	wave	function	nlm(r,	,	)	looks	like	for
hydrogen:
where	Lnl12l+1(2r/nr0)	is	a	generalized	Laguerre	polynomial.	Here	are	the	first
few	generalized	Laguerre	polynomials:
	L0b(r)	=	1
	L1b(r)	=	r	+	b	+	1
	
	
Some	hydrogen	wave	functions
So	what	do	the	hydrogen	wave	functions	look	like?	In	the	preceding	section,
you	find	that	100(r,	,	)	looks	like	this:
Here	are	some	other	hydrogen	wave	functions:
	
	
	
	
	
Note	that	nlm(r,	,	)	behaves	like	rl	for	small	r	and	therefore	goes	to	zero.
And	for	large	r,	nlm(r,	,	)	decays	exponentially	to	zero.	So	you've	solved	the
problem	you	had	earlier	of	the	wave	function	diverging	as	r	becomes	large	
and	all	because	of	the	quantization	condition,	which	cut	the	expression	for	f(r)
from	an	exponent	to	a	polynomial	of	limited	order.	Not	bad.
You	can	see	the	radial	wave	function	R10(r)	in	Figure	9-2.	R20(r)	appears	in
Figure	9-3.	And	you	can	see	R21(r)	in	Figure	9-4.
Figure	9-2:	The	radial	wave	function	R10(r).
Figure	9-3:	R20(r).
Figure	9-4:	R21(r).
Calculating	the	Energy	Degeneracy
of	the	Hydrogen	Atom
Each	quantum	state	of	the	hydrogen	atom	is	specified	with	three	quantum
numbers:	n	(the	principal	quantum	number),	l	(the	angular	momentum
quantum	number	of	the	electron),	and	m	(the	z	component	of	the	electron's
angular	momentum,	nlm[r,	,	]).	How	many	of	these	states	have	the	same
energy?	In	other	words,	what's	the	energy	degeneracy	of	the	hydrogen	atom	in
terms	of	the	quantum	numbers	n,	l,	and	m?
Well,	the	actual	energy	is	just	dependent	on	n,	as	you	see	earlier	in	the	section
titled	"Finding	the	allowed	energies	of	the	hydrogen	atom":
where	m	is	the	mass,	not	the	quantum	number.	That	means	the	E	is
independent	of	l	and	m.	So	how	many	states,	|n,	l,	m>,	have	the	same	energy
for	a	particular	value	of	n?	Well,	for	a	particular	value	of	n,	l	can	range	from
zero	to	n		1.	And	each	l	can	have	different	values	of	m,	so	the	total	degeneracy
is
The	degeneracy	in	m	is	the	number	of	states	with	different	values	of	m	that
have	the	same	value	of	l.	For	any	particular	value	of	l,	you	can	have	m	values	of
l,	l	+	1,	...,	0,	...,	l		1,	l.	And	that's	(2l	+	1)	possible	m	states	for	a	particular
value	of	l.	So	you	can	plug	in	(2l	+	1)	for	the	degeneracy	in	m:
And	this	series	works	out	to	be	just	n2.
So	the	degeneracy	of	the	energy	levels	of	the	hydrogen	atom	is	n2.	For
example,	the	ground	state,	n	=	1,	has	degeneracy	=	n2	=	1	(which	makes	sense
because	l,	and	therefore	m,	can	only	equal	zero	for	this	state).
For	n	=	2,	you	have	a	degeneracy	of	4:
	200(r,	,	)
	211(r,	,	)
	210(r,	,	)
	211(r,	,	)
Cool.
Quantum	states:	Adding	a	little	spin
You	may	be	asking	yourself		what	about	the	spin	of	the	electron?	Right	you
are!	The	spin	of	the	electron	does	provide	additional	quantum	states.	Up	to	now
in	this	section,	you've	been	treating	the	wave	function	of	the	hydrogen	atom	as
a	product	of	radial	and	angular	parts:
nlm(r,	,	)	=	Rnl(r)Ylm(,	)
Now	you	can	add	a	spin	part,	corresponding	to	the	spin	of	the	electron,	where	s
is	the	spin	of	the	electron	and	ms	is	the	z	component	of	the	spin:
The	spin	part	of	the	equation	can	take	the	following	values:
	|1/2,	1/2>
	|1/2,	1/2>
Hence,	nlm(r,	,	)	now	becomes	nlmms(r,	,	):
And	this	wave	function	can	take	two	different	forms,	depending	on	ms,	like	this:
	
	
In	fact,	you	can	use	the	spin	notation	(which	you	use	in	Chapter	6),	where
For	example,	for	|1/2,	1/2>,	you	can	write	the	wave	function	as
And	for	|1/2,	1/2>,	you	can	write	the	wave	function	as
What	does	this	do	to	the	energy	degeneracy?	If	you	include	the	spin	of	the
electron,	there	are	two	spin	states	for	every	state	|n,	l,	m>,	so	the	degeneracy
becomes
So	if	you	include	the	electron's	spin,	the	energy	degeneracy	of	the	hydrogen
atom	is	2n2.
In	fact,	you	can	even	add	the	spin	of	the	proton	to	the	wave	function	(although
people	don't	usually	do	that,	because	the	proton's	spin	interacts	only	weakly
with	magnetic	fields	applied	to	the	hydrogen	atom).	In	that	case,	you	have	a
wave	function	that	looks	like	the	following:
where	se	is	the	spin	of	the	electron,	mse	is	the	z	component	of	the	electron's
spin,	sp	is	the	spin	of	the	proton,	and	msp	is	the	z	component	of	the	proton's
spin.
If	you	include	the	proton's	spin,	the	wave	function	can	now	take	four	different
forms,	depending	on	ms,	like	this:
	
	
	
	
The	degeneracy	must	now	include	the	proton's	spin,	so	that's	a	factor	of	four
for	each	|n,	l,	m>:
On	the	lines:	Getting	the	orbitals
When	you	study	heated	hydrogen	in	spectroscopy,	you	get	a	spectrum
consisting	of	various	lines,	named	the	s	(for	sharp),	p	(for	principal),	d	(for
diffuse),	and	f	(for	fundamental)	lines.	And	other,	unnamed	lines	are	present	as
well		the	g,	h,	and	so	on.
The	s,	p,	d,	f,	and	the	rest	of	the	lines	turn	out	to	correspond	to	different
angular	momentum	states	of	the	electron,	called	orbitals.	The	s	state
corresponds	to	l	=	0;	the	p	state,	to	l	=	1;	the	d	state,	to	l	=	2;	the	f	state,	to	l	=
3;	and	so	on.	Each	of	these	angular	momentum	states	has	a	differently	shaped
electron	cloud	around	the	proton		that	is,	a	different	orbital.
Three	quantum	numbers		n,	l,	and	m		determine	orbitals.	For	example,	the
electron	cloud	for	the	|1,	0,	0>	state	(1s,	with	m	=	0)	appears	in	Figure	9-5.
Figure	9-5:	The	|1,	0,	0>	state.
The	|3,	2,	1>	state	(3d,	with	m	=	2)	appears	in	Figure	9-6.
Figure	9-6:	The	|3,	2,	1>	state.
The	|2,	1,	1>	state	(2p,	with	m	=	1)	appears	in	Figure	9-7.
Figure	9-7:	The	|2,	1,	1>	state.
Hunting	the	Elusive	Electron
Just	where	is	the	electron	at	any	one	time?	In	other	words,	how	far	is	the
electron	from	the	proton?	You	can	find	the	expectation	value	of	r,	that	is,	<r>,
to	tell	you.	If	the	wave	function	is	nlm(r,	,	),	then	the	following	expression
represents	the	probability	that	the	electron	will	be	found	in	the	spatial	element
d3r:
|nlm(r,	,	)|2d3r
In	spherical	coordinates,	d3r	=	r2	sin	dr	d	d.	So	you	can	write	|nlm(r,	,
)|2d3r	as
|nlm(r,	,	)|2r2	sin	dr	d	d
The	probability	that	the	electron	is	in	a	spherical	shell	of	radius	r	to	r	+	dr	is
therefore
And	because	nlm(r,	,	)	=	Rnl(r)Ylm(,	),	this	equation	becomes	the
following:
The	preceding	equation	is	equal	to
or	
Spherical	harmonics	are	normalized,	so	this	just	becomes
|Rnl(r)|2r2	dr
Okay,	that's	the	probability	that	the	electron	is	inside	the	spherical	shell	from	r
to	r	+	dr.	So	the	expectation	value	of	r,	which	is	<r>,	is
which	is
This	is	where	things	get	more	complex,	because	Rnl(r)	involves	the	Laguerre
polynomials.	But	after	a	lot	of	math,	here's	what	you	get:
where	r0	is	the	Bohr	radius:	
.	The	Bohr	radius	is	about	5.29		1011
meters,	so	the	expectation	value	of	the	electron's	distance	from	the	proton	is
<r>	=	[3n2		l(l	+	1)](2.65		1011)	meters
So,	for	example,	in	the	1s	state	(|1,	0,	0>),	the	expectation	value	of	r	is	equal	to
<r>1s	=	3(2.65		1011)	=	7.95		1011	meters
And	in	the	4p	state	(|	4,	1,	m>),
<r>4p	=	46(2.65		1011)	=	1.22		109	meters
And	that	concludes	this	chapter,	which	has	been	a	triumph	for	the	Schrdinger
equation.
Chapter	10
Handling	Many	Identical	Particles
In	This	Chapter
	Looking	at	wave	functions	and	Hamiltonians	in	many-particle	systems
	Working	with	identical	and	distinguishable	particles
	Identifying	and	creating	symmetric	and	antisymmetric	wave	functions
	Explaining	electron	shells	and	the	periodic	table
Hydrogen	atoms	(see	Chapter	9)	involve	only	a	proton	and	an	electron,	but	all
other	atoms	involve	more	electrons	than	that.	So	how	do	you	deal	with	multiple-
electron	atoms?	For	that	matter,	how	do	you	deal	with	multi-particle	systems,
such	as	even	a	simple	gas?
In	general,	you	can't	deal	with	problems	like	this		exactly,	anyway.	Imagine
the	complexity	of	just	two	electrons	moving	in	a	helium	atom		you'd	have	to
take	into	account	the	interaction	of	the	electrons	not	only	with	the	nucleus	of
the	atom	but	also	with	each	other		and	that	depends	on	their	relative
positions.	So	not	only	does	the	Hamiltonian	have	a	term	in	1/r1	for	the	potential
energy	of	the	first	electron	and	1/r2	for	the	second	electron,	but	it	also	has	a
term	in	
	for	the	potential	energy	that	comes	from	the	interaction	of	the	two
electrons.	And	that	makes	an	exact	wave	function	just	about	impossible	to	find.
However,	even	without	finding	exact	wave	functions,	you	can	still	do	a
surprising	amount	with	multi-particle	systems,	such	as	deriving	the	Pauli
exclusion	principle		which	says,	among	other	things,	that	no	two	electrons	can
be	in	the	exact	same	quantum	state.	In	fact,	you'll	probably	be	surprised	at	how
much	you	can	actually	say	about	multi-particle	systems	using	quantum
mechanics.	This	chapter	starts	with	an	introduction	to	many-particle	systems
and	goes	on	to	discuss	identical	particles,	symmetry	(and	anti-symmetry),	and
electron	shells.
Many-Particle	Systems,	Generally
Speaking
You	can	see	a	multi-particle	system	in	Figure	10-1,	where	a	number	of	particles
are	identified	by	their	position	(ignore	spin	for	the	moment).	This	section
explains	how	to	describe	that	system	in	quantum	physics	terms.
Figure	10-1:	A	multi-particle	system.
Considering	wave	functions	and	Hamiltonians
Begin	by	working	with	the	wave	function.	The	state	of	a	system	with	many
particles,	as	shown	in	Figure	10-1,	is	given	by	(r1,	r2,	r3,	...).	And	here's	the
probability	that	particle	1	is	in	d3r1,	particle	2	is	in	d3r2,	particle	3	is	in	d3r3,	and
so	on:
The	normalization	of	(r1,	r2,	r3,	...)	demands	that
Okay,	so	what	about	the	Hamiltonian,	which	gives	you	the	energy	states?	That
is,	what	is	H,	where	H(r1,	r2,	r3,	...)	=	E(r1,	r2,	r3,	...)?	When	you're	dealing
with	a	single	particle,	you	can	write	this	as
But	in	a	many	particle	system	the	Hamiltonian	must	represent	the	total	energy
of	all	particles,	not	just	one.The	total	energy	of	the	system	is	the	sum	of	the
energy	of	all	the	particles	(omitting	spin	for	the	moment),	so	here's	how	you
can	generalize	the	Hamiltonian	for	multi-particle	systems:
This,	in	turn,	equals	the	following:
Here,	mi	is	the	mass	of	the	ith	particle	and	V	is	the	multi-particle	potential.
A	Nobel	opportunity:	Considering	multi-electron
atoms
This	section	takes	a	look	at	how	the	Hamiltonian	wave	function	(see	the
preceding	section)	would	work	for	a	neutral,	multi-electron	atom.	A	multi-
electron	atom,	which	you	see	in	Figure	10-2,	is	the	most	common	multi-particle
system	that	quantum	physics	considers.	Here,	R	is	the	coordinate	of	the	nucleus
(relative	to	the	center	of	mass),	r1	is	the	coordinate	of	the	first	electron	(relative
to	the	center	of	mass),	r2	the	coordinate	of	the	second	electron,	and	so	on.
Figure	10-2:	A	multi-electron	atom.
If	you	have	Z	electrons,	the	wave	function	looks	like	(r1,	r2,	...,	rZ,	R).	And	the
kinetic	energy	of	the	electrons	and	the	nucleus	looks	like	this:
And	the	potential	energy	of	the	system	looks	like	this:
So	adding	the	two	preceding	equations,	here's	what	you	get	for	the	total
energy	(E	=	KE	+	PE)	of	a	multi-particle	atom:
Okay,	now	that	looks	like	a	proper	mess.	Want	to	win	the	Nobel	prize	in
physics?	Just	come	up	with	the	general	solution	to	the	preceding	equation.	As	is
always	the	case	when	you	have	a	multi-particle	system	in	which	the	particles
interact	with	each	other,	you	can't	split	this	equation	into	a	system	of	N
independent	equations.
	In	cases	where	the	N	particles	of	a	multi-particle	system	don't	interact
with	each	other,	where	you	can	disconnect	the	Schrdinger	equation	into	a
set	of	N	independent	equations,	solutions	may	be	possible.	But	when	the
particles	interact	and	the	Schrdinger	equation	depends	on	those
interactions,	you	can't	solve	that	equation	for	any	significant	number	of
particles.
However,	that	doesn't	mean	all	is	lost	by	any	means.	You	can	still	say	plenty
about	equations	like	this	one	if	you're	clever		and	it	all	starts	with	an
examination	of	the	symmetry	of	the	situation,	which	I	discuss	next.
A	Super-Powerful	Tool:	Interchange
Symmetry
Even	though	finding	general	solutions	for	equations	like	the	one	for	the	total
energy	of	a	multi-particle	atom	(in	the	preceding	section)	is	impossible,	you	can
still	see	what	happens	when	you	exchange	particles	with	each	other		and	the
results	are	very	revealing.	This	section	covers	the	idea	of	interchange
symmetry.
Order	matters:	Swapping	particles	with	the
exchange	operator
You	can	determine	what	happens	to	the	wave	function	when	you	swap	two
particles.	Whether	the	wave	function	is	symmetric	under	such	operations	gives
you	insight	into	whether	two	particles	can	occupy	the	same	quantum	state.	This
section	discusses	swapping	particles	and	looking	at	symmetric	and
antisymmetric	functions.
Take	a	look	at	the	general	wave	function	for	N	particles:
(r1,	r2,	...,	ri,	...,	rj,	...,	rN)
Note:	In	this	chapter,	I	talk	about	symmetry	in	terms	of	the	location	coordinate,
r,	to	keep	things	simple,	but	you	can	also	consider	other	quantities,	such	as
spin,	velocity,	and	so	on.	That	wouldn't	make	this	discussion	any	different,
because	you	can	wrap	all	of	a	particle's	quantum	measurements		location,
velocity,	speed,	and	so	on		into	a	single	quantum	state,	which	you	can	call	.
Doing	so	would	make	the	general	wave	function	for	N	particles	into	this:	(1,
2,	...,	i,	...,	j,	...,	N).	But	as	I	said,	this	section	just	considers	the	wave	function
(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	to	keep	things	simple.
Now	imagine	that	you	have	an	exchange	operator,	Pij,	that	exchanges	particles
i	and	j.	In	other	words,
Pij(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	(r1,	r2,	...,	rj,	...,	ri,	...,	rN)
And	Pij	=	Pji,	so
Pij(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	(r1,	r2,	...,	rj,	...,	ri,	...,	rN)
=	Pji(r1,	r2,	...,	ri,	...,	rj,	...,	rN)
Also,	note	that	applying	the	exchange	operator	twice	just	puts	the	two
exchanged	particles	back	where	they	were	originally,	so	Pij2	=	1.	Here's	what
that	looks	like:
Pij	Pij	(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	Pij(r1,	r2,	...,	rj,	...,	ri,	...,	rN)
=	(r1,	r2,	...,	ri,	...,	rj,	...,	rN)
However,	in	general,	Pij	and	Plm	(where	ij		lm)	do	not	commute.	That	is,	Pij
Plm		Plm	Pij	(ij		lm).	Therefore,	[Pij,	Plm]		0	(ij		lm).	For	example,	say	you
have	four	particles	whose	wave	function	is
Apply	the	exchange	operators	P12	and	P14	to	see	whether	P12	P14	equals	P14	P12.
Here's	P14	(r1,	r2,	r3,	r4):
And	here's	what	P12P14	(r1,	r2,	r3,	r4)	looks	like:
Okay.	Now	take	a	look	at	P14	P12	(r1,	r2,	r3,	r4).	Here's	P12	(r1,	r2,	r3,	r4):
And	here's	what	P14	P12	(r1,	r2,	r3,	r4)	looks	like:
As	you	can	see	by	comparing	
	and	this	last	equation,
P12	P14	(r1,	r2,	r3,	r4)		P14	P12	(r1,	r2,	r3,	r4).	In	other	words,	the	order	in	which
you	apply	exchange	operators	matters.
Classifying	symmetric	and	antisymmetric	wave
functions
Pij2	=	1	(see	the	preceding	section),	so	note	that	if	a	wave	function	is	an	
eigenfunction	of	Pij,	then	the	possible	eigenvalues	are	1	and	1.	That	is,
for	(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	an	eigenfunction	of	Pij	looks	like
Pij(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	(r1,	r2,	...ri,	...,	rj,	...,	rN)	or	(r1,	r2,	...,	ri,	...,
rj,	...,	rN)
That	means	there	are	two	kinds	of	eigenfunctions	of	the	exchange	operator:
	Symmetric	eigenfunctions:	Pijs(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	s(r1,	r2,	...,	ri,
...,	rj,	...,	rN)
	Antisymmetric	eigenfunctions:	Pija(r1,	r2,	...,	ri,	...,	rj,	...,	rN)	=	a(r1,	r2,
...,	ri,	...,	rj,	...,	rN)
Now	take	a	look	at	some	symmetric	and	some	antisymmetric	eigenfunctions.
How	about	this	one		is	it	symmetric	or	antisymmetric?
1(r1,	r2)	=	(r1		r2)2
You	can	apply	the	exchange	operator	P12:
P12	1(r1,	r2)	=	(r2		r1)2
Note	that	because	(r1		r2)2	=	(r2		r1)2,	1(r1,	r2)	is	a	symmetric	wave	function;
that's	because	P12	1(r1,	r2)	=	1(r1,	r2).
How	about	this	wave	function?
Again,	apply	the	exchange	operator,	P12:
Okay,	but	because	
,	you	know	that	P12	2(r1,	r2)	=	2(r1,	r2),
so	2(r1,	r2)	is	symmetric.
Here's	another	one:
Now	apply	P12:
How	does	that	equation	compare	to	the	original	one?	Well,
,	so	P12	3(r1,	r2)	=	3(r1,	r2).	Therefore,	3(r1,	r2)	is
antisymmetric.
What	about	this	one?
To	find	out,	apply	P12:
All	right		how's	this	compare	with	the	original	equation?
Okay		4(r1,	r2)	is	symmetric.
You	may	think	you	have	this	process	down	pretty	well,	but	what	about	this	next
wave	function?
Start	by	applying	P12:
So	how	do	these	two	equations	compare?
That	is,	5(r1,	r2)	is	neither	symmetric	nor	antisymmetric.	In	other	words,	5(r1,
r2)	is	not	an	eigenfunction	of	the	P12	exchange	operator.
Floating	Cars:	Tackling	Systems	of
Many	Distinguishable	Particles
All	right,	if	you've	been	reading	this	chapter	from	the	start,	you	pretty	much
have	the	idea	of	swapping	particles	down.	Now	you	look	at	systems	of	particles
that	you	can	distinguish		that	is,	systems	of	identifiably	different	particles.	As
you	see	in	this	section,	you	can	decouple	such	systems	into	linearly	independent
equations.
Suppose	you	have	a	system	of	many	different	types	of	cars	floating	around	in
space.	You	can	distinguish	all	those	cars	because	they're	all	different		they
have	different	masses,	for	one	thing.
Now	say	that	each	car	interacts	with	its	own	potential		that	is,	the	potential
that	any	one	car	sees	doesn't	depend	on	any	other	car.	That	means	that	the
potential	for	all	cars	is	just	the	sum	of	the	individual	potentials	each	car	sees,
which	looks	like	this,	assuming	you	have	N	cars:
Being	able	to	cut	the	potential	energy	up	into	a	sum	of	independent	terms	like
this	makes	life	a	lot	easier.	Here's	what	the	Hamiltonian	looks	like:
Notice	how	much	simpler	this	equation	is	than	the	Hamiltonian	for	the
hydrogen	atom	which	I	give	you	here:
Note	that	you	can	separate	the	previous	equation	for	the	potential	of	all	cars
into	N	different	equations:
And	the	total	energy	is	just	the	sum	of	the	energies	of	the	individual	cars:
And	the	wave	function	is	just	the	product	of	the	individual	wave	functions:
where	the		symbol	is	just	like	,	except	it	stands	for	a	product	of	terms,	not	a
sum,	and	ni	refers	to	all	the	quantum	numbers	of	the	ith	particle.
	As	you	can	see,	when	the	particles	you're	working	with	are
distinguishable	and	subject	to	independent	potentials,	the	problem	of
handling	many	of	them	becomes	simpler.	You	can	break	the	system	up	into
N	independent	one-particle	systems.	The	total	energy	is	just	the	sum	of	the
individual	energies	of	each	particle.	The	Schrdinger	equation	breaks	down
into	N	different	equations.	And	the	wave	function	ends	up	just	being	the
product	of	the	wave	functions	of	the	N	different	particles.
Take	a	look	at	an	example.	Say	you	have	four	particles,	each	with	a	different
mass,	in	a	square	well.	You	want	to	find	the	energy	and	the	wave	function	of
this	system.	Here's	what	the	potential	of	the	square	well	looks	like	this	for	each
of	the	four	noninteracting	particles:
Here's	what	the	Schrdinger	equation	looks	like:
You	can	separate	the	preceding	equation	into	four	one-particle	equations:
	
	
	
	
I've	already	solved	such	one-dimensional	problems	in	Chapter	3.	The	energy
levels	are
And	because	the	total	energy	is	the	sum	of	the	individual	energies	is	
,	the
energy	in	general	is
So	here's	the	energy	of	the	ground	state		where	all	particles	are	in	their
ground	states,	n1	=	n2	=	n3	=	n4	=	1:
For	a	one-dimensional	system	with	a	particle	in	a	square	well,	the	wave
function	is
The	wave	function	for	the	four-particle	system	is	just	the	product	of	the
individual	wave	functions,	so	it	looks	like	this:
For	example,	for	the	ground	state,	n1	=	n2	=	n3	=	n4	=	1,	you	have
So	as	you	can	see,	systems	of	N	independent,	distinguishable	particles	are	often
susceptible	to	solution		all	you	have	to	do	is	to	break	them	up	into	N
independent	equations.
Juggling	Many	Identical	Particles
When	the	particles	in	a	multi-particle	system	are	all	indistinguishable,	that's
when	the	real	adventure	begins.	When	you	can't	tell	the	particles	apart,	how
can	you	tell	which	one's	where?	This	section	explains	what	happens.
Losing	identity
Say	you	have	a	bunch	of	pool	balls	and	you	want	to	look	at	them	classically.	You
can	paint	each	pool	ball	differently,	and	then,	even	as	they	hurtle	around	the
pool	table,	you're	able	to	distinguish	them		seven	ball	in	the	corner	pocket,
and	that	sort	of	thing.	Classically,	identical	particles	retain	their	individuality.
You	can	still	tell	them	apart.
The	same	isn't	true	quantum	mechanically,	because	identical	quantum	particles
really	are	identical		you	can't	paint	them,	as	you	can	pool	balls.
For	example,	look	at	the	scenario	in	Figure	10-3.	There,	two	electrons	are
colliding	and	bouncing	apart.	Seems	like	keeping	track	of	the	two	electrons
would	be	easy.
Figure	10-3:	An	electron	colliding	with	another	electron.
But	now	look	at	the	scenario	in	Figure	10-4		the	electrons	could've	bounced
like	that,	not	like	the	bounce	shown	in	Figure	10-3.	And	you'd	never	know	it.
Figure	10-4:	An	electron	colliding	with	another	electron.
So	which	electron	is	which?	From	the	experimenter's	point	of	view,	you	can't
tell.	You	can	place	detectors	to	catch	the	electrons,	but	you	can't	determine
which	of	the	incoming	electrons	ended	up	in	which	detector,	because	of	the	two
possible	scenarios	in	Figures	10-3	and	10-4.
	Quantum	mechanically,	identical	particles	don't	retain	their	individuality
in	terms	of	any	measurable,	observable	quantity.	You	lose	the	individuality
of	identical	particles	as	soon	as	you	mix	them	with	similar	particles.	This
idea	holds	true	for	any	N-particle	system.	As	soon	as	you	let	N	identical
particles	interact,	you	can't	say	which	exact	one	is	at	r1	or	r2	or	r3	or	r4	and
so	on.
Symmetry	and	antisymmetry
In	practical	terms,	the	loss	of	individuality	among	identical	particles	means	that
the	probability	density	remains	unchanged	when	you	exchange	particles.	For
example,	if	you	were	to	exchange	electron	10,281	with	electron	59,830,	you'd
still	have	the	same	probability	that	an	electron	would	occupy	d3r10,281	and
d3r59,830.
Here's	what	this	idea	looks	like	mathematically	(r	and	s	are	the	location	and
spins	of	the	particles):
|(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)|2	=	|(r1s1,	r2s2,	...,	rjsj,	...,	risi,	...,	rNsN)|2
The	preceding	equation	means	that
(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	(r1s1,	r2s2,	...,	rjsj,	...,	risi,	...,	rNsN)
	So	the	wave	function	of	a	system	of	N	identical	particles	must	be	either
symmetric	or	antisymmetric	when	you	exchange	two	particles.	Spin	turns
out	to	be	the	deciding	factor:
	Antisymmetric	wave	function:	If	the	particles	have	half-odd-integral	spin
(1/2,	3/2,	and	so	on),	then	this	is	how	the	wave	function	looks	under
exchange	of	particles:
(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	(r1s1,	r2s2,	...,	rjsj,	...,	risi,	...,	rNsN)
	Symmetric	wave	function:	If	the	particles	have	integral	spin	(0,	1,	and	so
on),	this	is	how	the	wave	function	looks	under	exchange	of	particles:
(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	(r1s1,	r2s2,	...,	rjsj,	...,	risi,	...,	rNsN)
Having	symmetric	or	antisymmetric	wave	functions	leads	to	some	different
physical	behavior,	depending	on	whether	the	wave	function	is	symmetric	or
antisymmetric.
	In	particular,	particles	with	integral	spin,	such	as	photons	or	pi	mesons,
are	called	bosons.	And	particles	with	half-odd-integral	spin,	such	as
electrons,	protons,	and	neutrons,	are	called	fermions.	The	behavior	of
systems	of	fermions	is	very	different	from	the	behavior	of	systems	of
bosons.
Exchange	degeneracy:	The	steady	Hamiltonian
The	Hamiltonian,	which	you	can	represent	like	this
H(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)
doesn't	vary	under	exchange	of	two	identical	particles.	In	other	words,	the
Hamiltonian	is	invariant	here,	no	matter	how	many	identical	particles	you
exchange.	That's	called	exchange	degeneracy,	and	mathematically,	it	looks	like
this:
H(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	H(r1s1,	r2s2,	...,	rjsj,	...,	risi,	...,	rNsN)
That	means,	incidentally,	that	the	exchange	operator,	Pij,	is	an	invariant	of	the
motion	because	it	commutes	with	the	Hamiltonian:
[H,	Pij]	=	0
Name	that	composite:	Grooving	with	the
symmetrization	postulate
In	the	earlier	section	titled	"Symmetry	and	antisymmetry,"	I	show	that	the
wave	function	of	a	system	of	N	particles	is	either	symmetric	or	antisymmetric
under	the	exchange	of	two	particles:
	Symmetric:	(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	(r1s1,	r2s2,	...,	rjsj,	...,	risi,
...,	rNsN)
	Antisymmetric:	(r1s1,	r2s2,	...,	risi,	...,	rjsj,	...,	rNsN)	=	(r1s1,	r2s2,	...,	rjsj,
...,	risi,	...,	rNsN)
This	turns	out	to	be	the	basis	of	the	symmetrization	postulate,	which	says	that
in	systems	of	N	identical	particles,	only	states	that	are	symmetric	or
antisymmetric	exist		and	it	says	that	states	of	mixed	symmetry	don't	exist.
	The	symmetrization	postulate	also	says,	as	observed	from	nature,	that
	Particles	with	half-odd-integral	states	(1/2,	3/2,	5/2,	...)	are	fermions,	and
they	have	antisymmetric	states	under	the	interchange	of	two	particles.
	Particles	with	integral	spin	(0,	1,	2,	...)	are	bosons,	and	they	have	symmetric
states	under	the	interchange	of	two	particles.
So	the	wave	function	of	N	fermions	is	completely	antisymmetric,	and	the	wave
function	of	N	bosons	is	completely	symmetric.
Determining	whether	a	particle	is	a	fermion	or	a	boson	may	seem	like	an	easy
task		just	look	it	up.	Electrons,	protons,	and	neutrons	are	fermions,	for
example,	with	half-odd-integral	spin.	And	photons,	pi	mesons,	and	so	on	are
bosons,	with	integral	spins.
But	what	if	the	particle	you're	studying	is	a	composite	particle?	What	if,	for
example,	you	have	an	alpha	particle,	which	is	made	up	of	two	protons	and	two
neutrons?	Is	that	a	fermion	or	a	boson?
In	fact,	protons	and	neutrons	themselves	are	made	up	of	three	quarks,	and	pi
mesons	are	made	up	of	two		and	quarks	have	spin	1/2.
Composites	can	be	either	fermions	or	bosons		it	all	depends	on	whether	the
spin	of	the	composite	particle	ends	up	being	half-odd-integral	or	integral.	If	the
composite	particle's	spin	is	1/2,	3/2,	5/2,	and	so	on,	then	the	composite	particle
is	a	fermion.	If	the	composite	particle's	spin	is	0,	1,	2,	and	so	on,	then	the
composite	particle	is	a	boson.
	In	general,	if	the	composite	particle	is	made	up	of	an	odd	number	of
fermions,	then	it's	a	fermion.	Otherwise,	it's	a	boson.	So	for	example,
because	quarks	are	fermions	and	because	nucleons	such	as	protons	and
neutrons	are	made	up	of	three	quarks,	those	nucleons	end	up	being
fermions.	But	because	pi	mesons	are	made	up	of	two	quarks,	they	end	up
being	bosons.	The	alpha	particle,	which	consists	of	two	protons	and	two
neutrons,	is	a	boson.	You	can	even	consider	whole	atoms	to	be	composite
particles.	For	example,	consider	the	hydrogen	atom:	That	atom	is	made	up
of	a	proton	(a	fermion)	and	an	electron	(another	fermion),	so	that's	two
fermions.	And	that	makes	the	hydrogen	atom	a	boson.
Building	Symmetric	and
Antisymmetric	Wave	Functions
Many	of	the	wave	functions	that	are	solutions	to	physical	setups	like	the	square
well	aren't	inherently	symmetric	or	antisymmetric;	they're	simply	asymmetric.
In	other	words,	they	have	no	definite	symmetry.	So	how	do	you	end	up	with
symmetric	or	antisymmetric	wave	functions?
The	answer	is	that	you	have	to	create	them	yourself,	and	you	do	that	by	adding
together	asymmetric	wave	functions.	For	example,	say	that	you	have	an
asymmetric	wave	function	of	two	particles,	(r1s1,	r2s2).
	To	create	a	symmetric	wave	function,	add	together	(r1s1,	r2s2)	and	the
version	where	the	two	particles	are	swapped,	(r2s2,	r1s1).	Assuming	that
(r1s1,	r2s2)	and	(r2s2,	r1s1)	are	normalized,	you	can	create	a	symmetric
wave	function	using	these	two	wave	functions	this	way		just	by	adding	the
wave	functions:
	You	can	make	an	antisymmetric	wave	function	by	subtracting	the	two
wave	functions:
This	process	gets	rapidly	more	complex	the	more	particles	you	add,	however,
because	you	have	to	interchange	all	the	particles.	For	example,	what	would	a
symmetric	wave	function	based	on	the	asymmetric	three-particle	wave	function
(r1s1,	r2s2,	r3s3)	look	like?	Why,	it'd	look	like	this:
And	how	about	the	antisymmetric	wave	function?	That	looks	like	this:
And	in	this	way,	at	least	theoretically,	you	can	create	symmetric	and
antisymmetric	wave	functions	for	any	system	of	N	particles.
Working	with	Identical
Noninteracting	Particles
Working	with	identical	noninteracting	particles	makes	life	easier	because	you
can	treat	the	equations	individually	instead	of	combining	them	into	one	big
mess.	Say	you	have	a	system	of	N	identical	particles,	each	of	which	experiences
the	same	potential.	You	can	separate	the	Schrdinger	equation	into	N	identical
single-particle	equations:
And	the	total	energy	is	just	the	sum	of	the	energies	of	the	individual	particles:
But	now	look	at	the	wave	function	for	the	system.	Earlier	in	the	chapter	(see
"Floating	Cars:	Tackling	Systems	of	Many	Distinguishable	Particles"),	you
consider	the	wave	function	of	a	system	of	N	distinguishable	particles	and	come
up	with	the	product	of	all	the	individual	wave	functions:
However,	that	equation	doesn't	work	with	identical	particles	because	you	can't
say	that	particle	1	is	in	state	1(r1),	particle	2	is	in	state	2(r2),	and	so	on	
they're	identical	particles	here,	not	distinguishable	particles	as	before.
The	other	reason	this	equation	doesn't	work	here	is	that	it	has	no	inherent
symmetry		and	systems	of	N	identical	particles	must	have	a	definite
symmetry.	So	instead	of	simply	multiplying	the	wave	functions,	you	have	to	be	a
little	more	careful.
Wave	functions	of	two-particle	systems
How	do	you	create	symmetric	and	antisymmetric	wave	functions	for	a	two-
particle	system?	Start	with	the	single-particle	wave	functions	(see	the	earlier
section	"Building	Symmetric	and	Antisymmetric	Wave	Functions"):
	
	
By	analogy,	here's	the	symmetric	wave	function,	this	time	made	up	of	two
single-particle	wave	functions:
And	here's	the	antisymmetric	wave	function,	made	up	of	the	two	single-particle
wave	functions:
where	ni	stands	for	all	the	quantum	numbers	of	the	ith	particle.
Note	in	particular	that	a(r1s1,	r2s2)	=	0	when	n1	=	n2;	in	other	words,	the
antisymmetric	wave	function	vanishes	when	the	two	particles	have	the	same	set
of	quantum	numbers		that	is,	when	they're	in	the	same	quantum	state.	That
idea	has	important	physical	ramifications.
You	can	also	write	s(r1s1,	r2s2)	like	this,	where	P	is	the	permutation	operator,
which	takes	the	permutation	of	its	argument:
And	also	note	that	you	can	write	a(r1s1,	r2s2)	like	this:
where	the	term	(1)P	is	1	for	even	permutations	(where	you	exchange	both	r1s1
and	r2s2	and	also	n1	and	n2)	and	1	for	odd	permutations	(where	you	exchange
r1s1	and	r2s2	but	not	n1	and	n2;	or	you	exchange	n1	and	n2	but	not	r1s1	and	r2s2).
In	fact,	people	sometimes	write	a(r1s1,	r2s2)	in	determinant	form	like	this:
Note	that	this	determinant	is	zero	if	n1	=	n2.
Wave	functions	of	three-or-more-particle	systems
Now	you	get	to	put	together	the	wave	function	of	a	system	of	three	particles
from	single-particle	wave	functions.
The	symmetric	wave	function	looks	like	this:
And	the	antisymmetric	wave	function	looks	like	this:
This	asymmetric	wave	function	goes	to	zero	if	any	two	single	particles	have	the
same	set	of	quantum	numbers	(ni	=	nj,	i		j).
How	about	generalizing	this	to	systems	of	N	particles?	If	you	have	a	system	of
N	particles,	the	symmetric	wave	function	looks	like	this:
And	the	antisymmetric	wave	function	looks	like	this:
The	big	news	is	that	the	antisymmetric	wave	function	for	N	particles	goes	to
zero	if	any	two	particles	have	the	same	quantum	numbers	(ni	=	nj,	i		j).	And
that	has	a	big	effect	in	physics,	as	you	see	next.
It's	Not	Come	One,	Come	All:	The
Pauli	Exclusion	Principle
The	antisymmetric	wave	function	vanishes	if	any	two	particles	in	an	N-particle
system	have	the	same	quantum	numbers.	Because	fermions	are	the	type	of
particles	that	have	antisymmetric	wave	functions,	that's	the	equivalent	of
saying	that	in	a	system	of	N	particles,	no	two	fermions	can	have	the	same
quantum	numbers		that	is,	occupy	the	same	state.
That	idea,	which	Austrian	physicist	Wolfgang	Pauli	first	formulated	in	1925,	is
called	the	Pauli	exclusion	principle.	The	topic	of	discussion	at	that	time	was	the
atom,	and	the	Pauli	exclusion	principle	applied	to	the	electrons	(a	type	of
fermion),	which	are	present	in	all	atoms.
	The	Pauli	exclusion	principle	states	that	no	two	electrons	can	occupy	the
same	quantum	state	inside	a	single	atom.	And	that	result	is	important	for
the	structure	of	atoms.	Instead	of	just	piling	on	willy-nilly,	electrons	have	to
fill	quantum	states	that	aren't	already	taken.	The	same	isn't	true	for	bosons
	for	example,	if	you	have	a	heap	of	alpha	particles	(bosons),	they	can	all
be	in	the	same	quantum	state.	Not	so	for	fermions.
There	are	various	quantum	numbers	that	electrons	can	take	in	an	atom		n
(the	energy),	l	(the	angular	momentum),	m	(the	z	component	of	the	angular
momentum),	and	ms	(the	z	component	of	spin).	And	using	that	information,	you
can	construct	the	electron	structure	of	atoms.
Figuring	out	the	Periodic	Table
One	of	the	biggest	successes	of	the	Schrdinger	equation,	together	with	the
Pauli	exclusion	principle	(see	the	preceding	section),	is	explaining	the	electron
structure	of	atoms.
	The	electrons	in	an	atom	have	a	shell	structure,	and	they	fill	that
structure	based	on	the	Pauli	exclusion	principle,	which	maintains	that	no
two	electrons	can	have	the	same	state:
	The	major	shells	are	specified	by	the	principal	quantum	number,	n,
corresponding	to	the	distance	of	the	electron	from	the	nucleus.
	Shells,	in	turn,	have	subshells	based	on	the	orbital	angular	momentum
quantum	number,	l.
	In	turn,	each	subshell	has	subshells		called	orbitals		which	are	based	on
the	z	component	of	the	angular	momentum,	m.
So	each	shell	n	has	n		1	subshells,	corresponding	to	l	=	0,	1,	2,	...,	n		1.	And	in
turn,	each	subshell	has	2l	+	1	orbitals,	corresponding	to	m	=	1,	l	+	1,	...,	l		1,
l.
Much	as	with	the	hydrogen	atom,	the	various	subshells	(l	=	0,	1,	2,	3,	4,	and	so
on)	are	called	the	s,	p,	d,	f,	g,	h,	and	so	on	states.	So,	for	example,	for	a	given	n,
an	s	state	has	one	orbital	(m	=	0),	a	p	state	has	three	orbitals	(m	=	1,	0,	and
1),	a	d	state	has	five	orbitals	(m	=	2,	1,	0,	1,	and	2),	and	so	on.
In	addition,	due	to	the	z	component	of	the	spin,	ms,	each	orbital	can	contain
two	electrons		one	with	spin	up,	and	one	with	spin	down.
So	how	do	the	electrons,	as	fermions,	fill	the	structure	of	an	atom?	Electrons
can't	fill	a	quantum	state	that's	already	been	taken.	For	atoms	in	the	ground
state,	electrons	fill	the	orbitals	in	order	of	increasing	energy.	As	soon	as	all	of	a
subshell's	orbitals	are	filled,	the	next	electron	goes	on	to	the	next	subshell;	and
when	the	subshell	is	filled,	the	next	electron	goes	on	to	the	next	shell,	and	so
on.
Of	course,	as	you	fill	the	different	electron	shells,	subshells,	and	orbitals,	you
end	up	with	a	different	electron	structure.	And	because	interactions	between
electrons	form	the	basis	of	chemistry,	as	electrons	fill	the	successive	quantum
levels	in	various	atoms,	you	end	up	with	different	chemical	properties	for	those
atoms		which	set	up	the	period	(row)	and	group	(column)	organization	of	the
periodic	table.
Part	V
Group	Dynamics:	Introducing
Multiple	Particles
In	this	part	.	.	.
This	part	introduces	you	to	working	with	multiple	particles	at	the	same	time.
Now,	all	the	particles	in	the	system	can	interact	not	only	with	an	overall
potential	but	also	with	each	other.	You	see	how	to	deal	with	atoms	(electron
and	nucleus	systems)	here,	as	well	as	systems	of	many	atoms.	After	all,	the
whole	world	is	made	up	of	many-particle	systems.	Good	thing	quantum	physics
is	up	to	the	task.
Chapter	11
Giving	Systems	a	Push:	Perturbation
Theory
In	This	Chapter
	Nondegenerate	and	degenerate	perturbation	theory
	Perturbing	harmonic	oscillators
	The	Stark	effect	and	perturbing	hydrogen	atoms
Problems	in	quantum	physics	can	become	pretty	tough	pretty	fast		another
way	of	saying	that,	unfortunately,	you	just	can't	find	exact	solutions	to	many
quantum	physics	problems.	This	is	particularly	the	case	when	you	merge	two
kinds	of	systems.	For	example,	you	may	know	all	about	how	square	wells	work
and	all	about	how	electrons	in	magnetic	fields	work,	but	what	if	you	combine
the	two?	The	wave	functions	of	each	system,	which	you	know	exactly,	are	no
longer	applicable		you	need	some	sort	of	mix	instead.
Perturbation	theory	to	the	rescue!	This	theory	lets	you	handle	mixes	of
situations,	as	long	as	the	interference	isn't	too	strong.	In	this	chapter,	you
explore	time-independent	perturbation	theory	and	degenerate	and
nondegenerate	Hamiltonians.	You	also	look	at	some	examples	that	place
harmonic	oscillators	and	hydrogen	atoms	in	electric	fields.
Introducing	Time-Independent
Perturbation	Theory
	The	idea	behind	time-independent	perturbation	theory	is	that	you	start
with	a	known	system		one	whose	wave	functions	you	know	and	whose
energy	levels	you	know.	Everything	is	all	set	up	to	this	point.	Then	some
new	stimulus		a	perturbation		comes	along,	disturbing	the	status	quo.
For	example,	you	may	apply	an	electrostatic	or	magnetic	field	to	your
known	system,	which	changes	that	system	somewhat.
Perturbation	theory	lets	you	handle	situations	like	this		as	long	as	the
perturbation	isn't	too	strong.	In	other	words,	if	you	apply	a	weak	magnetic	field
to	your	known	system,	the	energy	levels	will	be	mostly	unchanged	but	with	a
correction.	(Note:	That's	why	it's	called	perturbation	theory	and	not	drastic-
interference	theory.)	The	change	you	make	to	the	setup	is	slight	enough	so	that
you	can	calculate	the	resulting	energy	levels	and	wave	functions	as	corrections
to	the	fundamental	energy	levels	and	wave	functions	of	the	unperturbed
system.
So	what	does	it	mean	to	talk	of	perturbations	in	physics	terms?	Say	that	you
have	this	Hamiltonian:
Here,	H0	is	a	known	Hamiltonian,	with	known	eigenfunctions	and	eigenvalues,
and	W	is	the	so-called	perturbation	Hamiltonian,	where	<<1	indicates	that
the	perturbation	Hamiltonian	is	small.
Finding	the	eigenstates	of	the	Hamiltonian	in	this	equation	is	what	solving
problems	like	this	is	all	about		in	other	words,	here's	the	problem	you	want	to
solve:
The	way	you	solve	this	equation	depends	on	whether	the	exact,	known	solutions
of	H0	are	degenerate	(that	is,	several	states	have	the	same	energy)	or
nondegenerate.	The	next	section	solves	the	nondegenerate	case.
Working	with	Perturbations	to
Nondegenerate	Hamiltonians
Start	with	the	case	in	which	the	unperturbed	Hamiltonian,	H0,	has
nondegenerate	solutions.	That	is,	for	every	state	|n>,	there's	exactly	one
energy,	En,
that	isn't	the	same	as	the	energy	for	any	other	state:	
	(just	as	a
one-to-one	function	has	only	one	x	value	for	any	y).	You	refer	to	these
nondegenerate	energy	levels	of	the	unperturbed	Hamiltonian	as	E(0)n	to
distinguish	them	from	the	corrections	that	the	perturbation	introduces,	so	the
equation	becomes
From	here	on,	I	refer	to	the	energy	levels	of	the	perturbed	system	as	En.
The	idea	behind	perturbation	theory	is	that	you	can	perform	expansions	based
on	the	parameter		(which	is	much,	much	less	than	1)	to	find	the	wave
functions	and	energy	levels	of	the	perturbed	system.	In	this	section,	you	go	up
to	terms	in	2	in	the	expansions.
A	little	expansion:	Perturbing	the	equations
To	find	the	energy	of	the	perturbed	system,	En,	start	with	the	energy	of	the
unperturbed	system:
En	=	E(0)n	+	...
Add	the	first-order	correction	to	the	energy,	E(1)n:
And	add	the	second-order	correction	to	the	energy,	2E(2)n,	as	well:
Now	what	about	the	wave	function	of	the	perturbed	system,	|n>?	Start	with
the	wave	function	of	the	unperturbed	system,	|n>:
Add	to	it	the	first-order	correction,	|(1)n>:
And	then	add	to	that	the	second-order	correction	to	the	wave	function,	2|
(2)n>:
Note	that	when	0,	
	becomes	the	unperturbed
energy:
En	=	E(0)n
And	
	becomes	the	unperturbed
wave	function:
So	your	task	is	to	calculate	E(1)n	and	E(2)n,	as	well	as	(1)n	and	(2)n.	So	how	do
you	do	that	in	general?	Time	to	start	slinging	some	math.	You	start	with	three
perturbed	equations:
	Hamiltonian:	
	Energy	levels:	
	Wave	functions:	
Combine	these	three	equations	to	get	this	jumbo	equation:
Matching	the	coefficients	of		and	simplifying
You	can	handle	the	jumbo	equation	in	the	preceding	section	by	setting	the
coefficients	of		on	either	side	of	the	equal	sign	equal	to	each	other.
Equating	the	zeroth	order	terms	in		on	either	side	of	this	equation,	here's
what	you	get:
Now	for	the	first-order	terms	in	;	equating	them	on	either	side	of	the	jumbo
equation	gives	you
Now	equate	the	coefficients	of	2	in	the	jumbo	equation,	giving	you
Okay,	that's	the	equation	you	derive	from	the	second	order	in	.	Now	you	have
to	solve	for	E(1)n,	E(2)n,	and	so	on	using	the	zeroth-order,	first-order,	and
second-order	equations.
Start	by	noting	that	the	unperturbed	wave	function,	|n>	isn't	going	to	be	very
different	from	the	perturbed	wave	function,	|n>,	because	the	perturbation	is
small.	That	means	that	
.	In	fact,	you	can	normalize	|n>	so	that	<n|
n>	is	exactly	equal	to	1:
Given	that	
,	the	equation	becomes
And	because	the	coefficients	of		must	both	vanish,	you	get	the	following:
This	equation	is	useful	for	simplifying	the	math.
Finding	the	first-order	corrections
After	matching	the	coefficients	of		and	simplifying	(see	the	preceding	section),
you	want	to	find	the	first-order	corrections	to	the	energy	levels	and	the	wave
functions.	Find	the	first-order	correction	to	the	energy,	E(1)n,	by
multiplying	
	by	<n|:
Then	the	first	term	can	be	neglected	and	we	can	use	our	simplification	above	to
write	the	first	order	energy	perturbation	as:
Swell,	that's	the	expression	you	use	for	the	first-order	correction,	E(1)n.
Now	look	into	finding	the	first-order	correction	to	the	wave	function,	|(1)n>.
You	can	multiply	the	wave-function	equation	by	this	next	expression,	which	is
equal	to	1:
So	you	have
Note	that	the	m	=	n	term	is	zero	because	<n|(1)n>	=	0.
So	what	is	<m|(1)n>?	You	can	find	out	by	multiplying	the	first-order
correction,	
,	by	<m|	to	give	you
And	substituting	that	into	
	gives	you
Okay,	that's	your	term	for	the	first-order	correction	to	the	wave	function,
|(1)n>.	From	
,	the	wave	function
looks	like	this,	made	up	of	of	zeroth-,	first-,	and	second-order	corrections:
Ignoring	the	second-order	correction	for	the	moment	and	substituting
	in	for	the	first-order	correction	gives	you	this	for
the	wave	function	of	the	perturbed	system,	to	the	first	order:
That's	the	wave	function	of	the	perturbed	system	in	terms	of	the	perturbation.
But	that's	still	only	the	first-order	correction.	How	about	the	second?	Read	on.
Finding	the	second-order	corrections
Now	find	the	second-order	corrections	to	the	energy	levels	and	the	wave
functions	(the	preceding	section	covers	first-order	corrections).	To	find	E(2)n,
multiply	both	sides	of	
	by	<n|:
This	looks	like	a	tough	equation	until	you	realize	that	<n|(1)n>	is	equal	to
zero,	so	you	get
Because	<n|(2)n>	is	also	equal	to	zero,	and	again	neglecting	the	first	term,
you	get
E(2)n	is	just	a	number,	so	you	have
And	of	course,	because	<n|n>	=	1,	you	have
Note	that	if	|(1)n>	is	an	eigenstate	of	W,	the	second-order	correction	equals
zero.
Okay,	so	E(2)n	=	<n|W|(1)n>.	How	can	you	make	that	simpler?	Well,
from	using	
.	Substituting	that	equation	into
	gives	you
Now	you	have	
	and	
.	Here's
the	total	energy	with	the	first-	and	second-order	corrections:
So	from	this	equation,	you	can	say
That	gives	you	the	first-	and	second-order	corrections	to	the	energy,	according
to	perturbation	theory.
Note	that	for	this	equation	to	converge,	the	term	in	the	summation	must	be
small.	And	note	in	particular	what	happens	to	the	expansion	term	if	the	energy
levels	are	degenerate:
In	that	case,	you're	going	to	end	up	with	an	E(0)n	that	equals	an	E(0)m,	which
means	that	the	energy-corrections	equation	blows	up,	and	this	approach	to
perturbation	theory	is	no	good		which	is	to	say	that	you	need	a	different
approach	to	perturbation	theory	(coming	up	later	in	"Working	with
Pertubations	to	Degenerate	Hamiltonians")	to	handle	systems	with	degenerate
energy	states.
In	the	next	section,	I	show	you	an	example	to	make	the	idea	of	perturbing
nondegenerate	Hamiltonians	more	real.
Perturbation	Theory	to	the	Test:
Harmonic	Oscillators	in	Electric
Fields
Consider	the	case	in	which	you	have	a	small	particle	oscillating	in	a	harmonic
potential,	back	and	forth,	as	Figure	11-1	shows.
Figure	11-1:	A	harmonic	oscillator.
Here's	the	Hamiltonian	for	that	particle,	where	the	particle's	mass	is	m,	its
location	is	x,	and	the	angular	frequency	of	the	motion	is	:
Now	assume	that	the	particle	is	charged,	with	charge	q,	and	that	you	apply	a
weak	electric	field,	,	as	Figure	11-2	shows.
Figure	11-2:	Applying	an	electric	field	to	a	harmonic	oscillator.
The	force	due	to	the	electric	field	in	this	case	is	the	perturbation,	and	the
Hamiltonian	becomes
In	this	section,	you	find	the	energy	and	wave	functions	of	the	perturbed	system
and	compare	them	to	the	exact	solutions.
Finding	exact	solutions
So	what	are	the	energy	eigenvalues	of	the	preceding	Hamiltonian	for	the
harmonic	oscillator	in	an	electric	field?	First	solve	for	the	eigenvalues	exactly;
then	use	perturbation	theory.	You	can	solve	for	the	exact	energy	eigenvalues	by
making	one	of	the	following	substitutions:
	
	
Substituting	the	equation	solved	for	x	into	
	gives
you
The	last	term	is	a	constant,	so	the	equation	is	of	the	form
where	
.	
	is	just	the	Hamiltonian	of	a	harmonic	oscillator
with	an	added	constant,	which	means	that	the	energy	levels	are	simply
Substituting	in	for	C	gives	you	the	exact	energy	levels:
Great		that's	the	exact	solution.
Applying	perturbation	theory
As	soon	as	you	have	the	exact	eigenvalues	for	your	charged	oscillator	(see	the
preceding	section),	you	have	something	to	compare	the	solution	from
perturbation	theory	to.	Now	you	can	find	the	energy	and	wave	functions	of	the
perturbed	system.
Energy	of	the	charged	oscillator
So	what	is	the	energy	of	the	charged	oscillator,	as	given	by	perturbation
theory?	You	know	that	the	corrected	energy	is	given	by
where	W	is	the	perturbation	term	in	the	Hamiltonian.	That	is,	here,	W	=	qx.
Now	take	a	look	at	the	corrected	energy	equation	using	qx	for	W.	The	first-
order	correction	is	
,	which,	using	W	=	qx,	becomes
<n|qx|n>	or	q<n|x|n>
But	<n|x|n>	=	0,	because	that's	the	expectation	value	of	x,	and	harmonic
oscillators	spend	as	much	time	in	negative	x	territory	as	in	positive	x	territory
	that	is,	the	average	value	of	x	is	zero.	So	the	first-order	correction	to	the
energy,	as	given	by	perturbation	theory,	is	zero.
Okay,	what's	the	second-order	correction	to	the	energy,	as	given	by
perturbation	theory?	Here	it	is:
And	because	W	=	qx,	you	have
Cast	this	in	terms	of	bras	and	kets	(see	Chapter	4),	changing	<m|	to	<m|	and	|
n>	to	|n>,	making	the	second-order	energy	correction	into	this	expression:
You	can	decipher	this	step	by	step.	First,	the	energy	is
That	makes	figuring	out	the	second-order	energy	a	little	easier.
Also,	the	following	expressions	turn	out	to	hold	for	a	harmonic	oscillator:
	
	
	
	
With	these	four	equations,	you're	ready	to	tackle	
,	the	second-
order	correction	to	the	energy.	Omitting	higher-power	terms,	the	summation	in
this	equation	becomes
And	substituting	in	the	for	E(0)n		E(0)n+1	and	E(0)n		E(0)n1	gives	you
Now,	substituting	in	for	<n	+	1|x|n>	and	<n		1|x|n>	gives	you
or
So	the	second-order	correction	is
Therefore,	according	to	perturbation	theory,	the	energy	of	the	harmonic
oscillator	in	the	electric	field	should	be
Compare	this	result	to	the	earlier	equation	for	the	exact	energy	levels,
		they're	the	same!	In	other	words,	perturbation	theory	has
given	you	the	same	result	as	the	exact	answer.	How's	that	for	agreement?
Of	course,	you	can't	expect	to	hit	the	same	answer	every	time	using
perturbation	theory,	but	this	result	is	impressive!
Wave	functions	of	the	charged	oscillator
Now	figure	out	what	the	charged	oscillator's	wave	function	looks	like	in	the
presence	of	the	electric	field.	Here's	the	wave	function	of	the	perturbed	system,
to	the	first	order:
Using	the	<n|	and	|n>	bras	and	kets	you're	used	to	for	harmonic	oscillators,
this	becomes
Because	W	=	qx,	this	becomes
Evidently,	as	with	the	energy,	only	two	terms	contribute,	because	<n|x|n>	=	0.
In	particular,	the	two	terms	that	contribute	are
	
	
Note	also	that	
	and	
.
These	four	equations	mean	that
Note	what	this	equation	means:	Adding	an	electric	field	to	a	quantum	harmonic
oscillator	spreads	the	wave	function	of	the	harmonic	oscillator.
Originally,	the	harmonic	oscillator's	wave	function	is	just	the	standard	harmonic
oscillator	wave	function,	|n>	=	|n>.	Applying	an	electric	field	spreads	the
wave	function,	adding	a	component	of	|n		1>,	which	is	proportional	to	the
electric	field,	,	and	the	charge	of	the	oscillator,	q,	like	this:
And	the	wave	function	also	spreads	to	the	other	adjacent	state,	|n	+	1>,	like
this:
	You	end	up	mixing	states.	That	blending	between	states	means	that	the
perturbation	you	apply	must	be	small	with	respect	to	the	separation
between	unperturbed	energy	states,	or	you	risk	blurring	the	whole	system
to	the	point	that	you	can't	make	any	predictions	about	what's	going	to
happen.
In	any	case,	that's	a	nice	result		blending	the	states	in	proportion	to	the
strength	of	the	electric	field	you	apply		and	it's	typical	of	the	result	you	get
with	perturbation	theory.
Okay,	that's	how	nondegenerate	perturbation	theory	works.	As	you	can	see,	it's
strongly	dependent	on	having	the	energy	states	separate	so	that	your	solution
can	blend	them.	But	what	happens	when	you	have	a	system	where	the	energies
are	degenerate?	You	take	a	look	at	that	in	the	next	section.
Working	with	Perturbations	to
Degenerate	Hamiltonians
This	section	tackles	systems	in	which	the	energies	are	degenerate.	Take	a	look
at	this	unperturbed	Hamiltonian:
In	other	words,	several	states	have	the	same	energy.	Say	the	energy	states	are
f-fold	degenerate,	like	this:
How	does	this	affect	the	perturbation	picture?	The	complete	Hamiltonian,	H,	is
made	up	of	the	original,	unperturbed	Hamiltonian,	H0,	and	the	perturbation
Hamiltonian,	H:
In	zeroth-order	approximation,	you	can	write	the	eigenfunction	|n>	as	a
combination	of	the	degenerate	states	|n>:
Note	that	in	what	follows,	you	assume	that	<n|n>	=	1	and	<m|n>	=	0	if	m
is	not	equal	to	n.	Also,	you	assume	that	the	|n>	are	normalized		that	is,	<n|
n>	=	1.
Plugging	this	zeroth-order	equation	into	the	complete	Hamiltonian	equation,
you	get
Now	multiplying	that	equation	by	<n|	gives	you
Using	the	fact	that	<n|n>	=	1	and	<m|n>	=	0	if	m	is	not	equal	to	n	gives
you
Physicists	often	write	that	equation	as
where	
.	And	people	also	write	that	equation	as
where	E(1)n	=	En		E(0)n.	That's	a	system	of	linear	equations,	and	the	solution
exists	only	when	the	determinant	to	this	array	is	nonvanishing:
The	determinant	of	this	array	is	an	fth	degree	equation	in	E(1)n,	and	it	has	f
different	roots,	E(1)n.	Those	f	different	roots	are	the	first-order	corrections
to	the	Hamiltonian.	Usually,	those	roots	are	different	because	of	the	applied
perturbation.	In	other	words,	the	perturbation	typically	gets	rid	of	the
degeneracy.
So	here's	the	way	you	find	the	eigenvalues	to	the	first	order		you	set	up	an	f-
by-f	matrix	of	the	perturbation	Hamiltonian,	H,	where	H	=	<n|H|n>:
Then	diagonalize	this	matrix	and	determine	the	f	eigenvalues	E(1)n	and	the
matching	eigenvectors:
Then	you	get	the	energy	eigenvalues	to	first	order	this	way:
And	the	eigenvectors	are
In	the	next	section,	you	look	at	an	example	to	clarify	this	idea.
Testing	Degenerate	Perturbation
Theory:	Hydrogen	in	Electric	Fields
In	this	section,	you	see	whether	degenerate	perturbation	theory	can	handle	the
hydrogen	atom,	which	has	energy	states	degenerate	in	different	angular
momentum	quantum	numbers,	when	you	remove	that	degeneracy	by	applying
an	electric	field.	This	setup	is	called	the	Stark	effect.
Specifically,	suppose	you	apply	an	electric	field,	,	to	a	hydrogen	atom	in	the	n
=	2	excited	state.	That	state	has	four	eigenfunctions	that	have	the	same	energy,
where	the	quantum	numbers	are	|nlm>	(note	that	you're	renaming	these
eigenfunctions	|1>,	|2>,	and	so	on	to	make	the	calculation	easier):
	|1>	=|200>
	|2>	=|211>
	|3>	=|210>
	|4>	=|21		1>
All	these	unperturbed	states	have	the	same	energy,	E	=	R/4,	where	R	is	the
Rydberg	constant,	13.6	eV.	But	at	least	some	of	these	states	will	have	their
energies	changed	when	you	apply	the	electric	field.
What	does	the	electric	field,	,	cause	the	perturbation	Hamiltonian,	Hp,	to
become?	Here's	the	perturbation	Hamiltonian:
Hp	=	ez
So	you	have	to	evaluate	this	equation	for	the	various	states.	For	example,	what
is	the	following	expression	equal	to,	where	<1|	=	<200|	and	|3>	=	|210>?
<1|Hp|3>
You	solve	for	the	unperturbed	hydrogen	wave	functions	in	Chapter	9.	In
general,	here's	what	the	wave	function	nlm(r,	,	)	looks	like	for	hydrogen:
where	Lnl12l+1(2r/nr0)	is	a	generalized	Laguerre	polynomial.	Doing	all	the
math	gives	you	the	following	result,	where	a0	is	the	Bohr	radius	of	the	atom:
The	<1|Hp|3>	is	just	one	term	you	have	to	compute,	of	course.	Here's	the	full
matrix	for	the	perturbation	Hamiltonian	connecting	all	states,	where	Hp	=
<|Hp|>:
Doing	the	math	gives	you	this	remarkably	simple	result:
Diagonalizing	this	matrix	gives	you	these	eigenvalues		the	first-order
corrections	to	the	unperturbed	energies:
	E(1)1	=	3ea0
	E(1)2	=	0
	E(1)3	=	3ea0
	E(1)4	=	0
where	E(1)1	is	the	first-order	correction	to	the	energy	of	the	|1>	eigenfunction,
E(1)2	is	the	first-order	correction	to	the	energy	of	the	|2>	eigenfunction,	and	so
on.	Adding	these	corrections	to	the	unperturbed	energy	for	the	n	=	2	state
gives	you	the	final	energy	levels:
	
	
	
	
where	R	is	the	Rydberg	constant.	Note	this	result:	The	Stark	effect	removes	the
energy	degeneracy	in	|200>	and	|210>	(the	|1>	and	|3>	eigenfunctions),	but
the	degeneracy	in	|211>	and	|21		1>	(the	|2>	and	|4>	eigenfunctions)
remains.
Chapter	12
Wham-Blam!	Scattering	Theory
In	This	Chapter
	Switching	between	lab	and	center-of-mass	frames
	Solving	the	Schrdinger	equation
	Finding	the	wave	function
	Putting	the	Born	approximation	to	work
Your	National	Science	Foundation	grant	finally	came	through,	and	you	built
your	new	synchrotron		a	particle	accelerator.	Electrons	and	anti-electrons
accelerate	at	near	the	speed	of	light	along	a	giant	circular	track	enclosed	in	a
vacuum	chamber	and	collide,	letting	you	probe	the	structure	of	the	high-energy
particles	you	create.	You're	sitting	at	the	console	of	your	giant	new	experiment,
watching	the	lights	flashing	and	the	signals	on	the	screens	approvingly.	Millions
of	watts	of	power	course	through	the	thick	cables,	and	the	radiation	monitors
are	beeping,	indicating	that	things	are	working.	Cool.
You're	accelerating	particles	and	smashing	them	against	each	other	to	observe
how	they	scatter.	But	this	is	slightly	more	complex	than	observing	how	pool
balls	collide.	Classically,	you	can	predict	the	exact	angle	at	which	colliding
objects	will	bounce	off	each	other	if	the	collision	is	elastic	(that	is,	momentum
and	kinetic	energy	are	both	conserved).	Quantum	mechanically,	however,	you
can	only	assign	probabilities	to	the	angles	at	which	things	scatter.
Physicists	use	large	particle	accelerators	to	discover	more	about	the	structure
of	matter,	and	that	study	is	central	to	modern	physics.	This	chapter	serves	as
an	introduction	to	that	field	of	study.	You	get	to	take	a	look	at	particle
scattering	on	the	subatomic	level.
Introducing	Particle	Scattering	and
Cross	Sections
Think	of	a	scattering	experiment	in	terms	of	particles	in	and	particles	out.	Look
at	Figure	12-1,	for	example.	In	the	figure,	particles	are	being	sent	in	a	stream
from	the	left	and	interacting	with	a	target;	most	of	them	continue	on
unscattered,	but	some	particles	interact	with	the	target	and	scatter.
Figure	12-1:	Scattering	from	a	target.
Those	particles	that	do	scatter	do	so	at	a	particular	angle	in	three	dimensions
	that	is,	you	give	the	scattering	angle	as	a	solid	angle,	d,	which	equals	sin
d	d,	where		and		are	the	spherical	angles	I	introduce	in	Chapter	8.
The	number	of	particles	scattered	into	a	specific	d	per	unit	time	is
proportional	to	a	very	important	quantity	in	scattering	theory:	the	differential
cross	section.
	The	differential	cross	section	is	given	by	
,	and	it's	a	measure	of	the
number	of	particles	per	second	scattered	into	d	per	incoming	flux.	The
incident	flux,	J	(also	called	the	current	density),	is	the	number	of	incident
particles
per	unit	area	per	unit	time.	So	
	is
where	N(,	)	is	the	number	of	particles	at	angles		and	.
The	differential	cross	section	
	has	the	dimensions	of	area,	so	calling
it	a	cross	section	is	appropriate.	The	cross	section	is	sort	of	like	the	size	of	the
bull's	eye	when	you're	aiming	to	scatter	incident	particles	through	a	specific
solid	angle.
	The	differential	cross	section	is	the	cross	section	for	scattering	to	a
specific	solid	angle.	The	total	cross	section,	,	is	the	cross	section	for
scattering	of	any	kind,	through	any	angle.	So	if	the	differential	cross	section
for	scattering	to	a	particular	solid	angle	is	like	the	bull's	eye,	the	total	cross
section	corresponds	to	the	whole	target.
You	can	relate	the	total	cross	section	to	the	differential	cross	section	by
integrating	the	following:
Translating	between	the	Center-of-
Mass	and	Lab	Frames
Now	you	can	start	getting	into	the	details	of	scattering,	beginning	with	a
discussion	of	the	center-of-mass	frame	versus	the	lab	frame.	Experiments	take
place	in	the	lab	frame,	but	you	do	scattering	calculations	in	the	center-of-mass
frame,	so	you	have	to	know	how	to	translate	between	the	two	frames.	This
section	explains	how	the	frames	differ	and	shows	you	how	to	relate	the
scattering	angles	and	cross	sections	when	you	change	frames.
Framing	the	scattering	discussion
Look	at	Figure	12-2		that's	scattering	in	the	lab	frame.	One	particle,	traveling
at	v1lab,	is	incident	on	another	particle	that's	at	rest	(v2lab	=	0)	and	hits	it.
After	the	collision,	the	first	particle	is	scattered	at	angle	1,	traveling	at	v'1lab,
and	the	other	particle	is	scattered	at	angle	2	and	velocity	v'2lab.
Figure	12-2:	Scattering	in	the	lab	frame.
Now	in	the	center-of-mass	frame,	the	center	of	mass	is	stationary	and	the
particles	head	toward	each	other	with	velocities	v1c	and	v2c,	respectively.	After
they	collide,	they	head	away	from	each	other	with	velocities	v'1c	and	v'2c,	at
angles		and			.
You	have	to	move	back	and	forth	between	these	two	frames		the	lab	frame
and	the	center-of-mass	frame		so	you	need	to	relate	the	velocities	and	angles
(in	a	nonrelativistic	way).
Relating	the	scattering	angles	between	frames
To	relate	the	angles	1	and	,	you	start	by	noting	that	you	can	connect	v1lab	and
v1c	using	the	velocity	of	the	center	of	mass,	vcm,	this	way:
v1lab	=	v1c	+	vcm
In	addition,	here's	what	can	say	about	the	velocity	of	particle	1	after	it	collides
with	particle	2:
v'1lab	=	v'1c	+	vcm
Now	you	can	find	the	components	of	these	velocities:
	v'1lab	cos1	=	v'1c	cos	+	vcm
	v'1lab	sin1	=	v'1c	sin
Dividing	the	equation	in	the	second	bullet	by	the	one	in	the	first	gives	you
But	wouldn't	it	be	easier	if	you	could	relate	1	and		by	something	that	didn't
involve	the	velocities,	only	the	masses,	such	as	the	following?
Well,	you	can.	To	see	that,	start	with
And	you	can	show	that
You	can	also	use	the	conservation	of	momentum	to	say	what	happens	after	the
collision.	In	fact,	because	the	center	of	mass	is	stationary	in	the	center-of-mass
frame,	the	total	momentum	before	and	after	the	collision	is	zero	in	that	frame,
like	this:
m1v1c	+	m2v2c	=	0
Therefore
And	after	the	collision,	m1v'1c	+	m2v'2c	=	0,	which	means	that
Also,	if	the	collision	is	elastic	(and	you	assume	all	collisions	are	elastic	in	this
chapter),	kinetic	energy	is	conserved	in	addition	to	momentum,	so	that	means
the	following	is	true:
Substituting	
	and	
	into	this	equation	gives	you
v'1c	=	v1c	and	v'2c	=	v2c
Given	these	two	equations,	you	can	redo	
	as
Dividing	the	magnitude	of	each	side	of	
	by	the	magnitude	of	the
above	equation	gives	you
And	because	you	saw	earlier	that	
,	substituting
	into	this	equation	gives	you	at	last
Okay,	that	relates	1	and	,	which	is	what	you	were	trying	to	do.	Using	the
relation	
,	you	can	rewrite	
	as	the
following:
You	can	also	relate	2	and	.	You	can	show	that	
,	which,	using	a	little
trig,	means	that
Okay,	now	you've	related	the	angles	between	the	lab	and	center-of-mass
frames.	How	about	relating	the	cross	sections	in	the	two	frames?	That's	in	the
next	section.
Translating	cross	sections	between	the	frames
The	preceding	section	relates	1	and		and	2		the	angles	of	the	scattered
particles	in	the	lab	frame	and	the	center-of-mass	frame.	Now	how	about
relating	the	differential	cross	section		the	bull's	eye	when	you're	aiming	to
scatter	the	particles	at	a	particular	angle		between	the	lab	and	center-of-
mass	frames?
The	differential	d	(total	cross	section)	is	infinitesimal	in	size,	and	it	stays	the
same	between	the	two	frames.	But	the	angles	that	make	up	d,	the	scattering
angle,	vary	when	you	translate	between	frames.	You	get	to	take	a	look	at	how
that	works	now,	relating	the	lab	differential	cross	section:
to	the	center-of-mass	differential	cross	section:
In	the	lab	frame,	d1	=	sin1	d1	d1.	And	in	the	center-of-mass	frame,	d	=
sin	d	d.	Because	dlab	=	dcm,	the	following	equation	is	true:
Putting	that	equation	with	the	equations	for	the	lab	frame	and	the	center-of-
mass	frame,	you	have
Because	you	have	cylindrical	symmetry	here,		=	1,	so
You've	already	seen	that	
,	so
.	Therefore
You	can	also	show	that
Trying	a	lab-frame	example	with	particles	of	equal
mass
Say	you	have	two	particles	of	equal	mass	colliding	in	the	lab	frame	(where	one
particle	starts	at	rest).	You	want	to	show	that	the	two	particles	end	up	traveling
at	right	angles	with	respect	to	each	other	in	the	lab	frame.
Note	that	if	m1	=	m2,	then	
	gives	tan(1)	=
tan(/2),	so	1	=	/2.	And	
becomes
Note	also	that	tan(2)	=	cot(/2),	or	tan(2)	=	tan(/2		/2).
You	know	that	1	=	/2,	and	tan(2)	=	tan(/2		/2)	tells	you	that	the	following
is	true:
2	=	/2		/2
So	substituting	1	=	/2	into	the	preceding	equation	gives	you
2	=	/2		1
2	+	1	=	/2
Therefore,	2	and	1,	the	angles	of	the	particles	in	the	lab	frame	after	the
collision,	add	up	to	/2		which	means	2	and	1	are	at	right	angles	with
respect	to	each	other.	Cool.
In	this	case,	you	can	use	the	relations	you've	already	derived	to	get	these
relations	in	the	special	case	where	m1	=	m2:
	
	
	
	
Tracking	the	Scattering	Amplitude
of	Spinless	Particles
In	the	earlier	section	"Translating	between	the	Center-of-Mass	and	Lab
Frames,"	you	see	how	to	translate	from	the	lab	frame	to	the	center-of-mass
frame	and	back	again,	and	those	translations	work	classically	as	well	as	in
quantum	physics	(as	long	as	the	speeds	involved	are	nonrelativistic).	Now	you
look	at	the	elastic	scattering	of	two	spinless	nonrelativistic	particles	from	the
time-independent	quantum	physics	point	of	view.
Assume	that	the	interaction	between	the	particles	depends	only	on	their
relative	distance,	|r1		r2|.	You	can	reduce	problems	of	this	kind	to	two
decoupled	problems	(see	Chapter	9	for	details).	The	first	decoupled	equation
treats	the	center	of	mass	of	the	two	particles	as	a	free	particle,	and	the	second
equation	is	for	an	effective	particle	of	mass	
.
The	first	decoupled	equation,	the	free-particle	equation	of	the	center	of	mass,	is
of	no	interest	to	you	in	scattering	discussions.	The	second	equation	is	the	one	to
concentrate	on,	where		=
:
You	can	use	the	preceding	equation	to	solve	for	the	probability	that	a	particle	is
scattered	into	a	solid	angle	d		and	you	give	this	probability	by	the
differential	cross	section,	
.
	In	quantum	physics,	wave	packets	represent	particles.	In	terms	of
scattering,	these	wave	packets	must	be	wide	enough	so	that	the	spreading
that	occurs	during	the	scattering	process	is	negligible	(however,	the	wave
packet	can't	be	so	spread	that	it	encompasses	the	whole	lab,	including	the
particle	detectors).	Here's	the	crux:	After	the	scattering,	the	wave	function
breaks	up	into	two	parts		an	unscattered	part	and	a	scattered	part.	That's
how	scattering	works	in	the	quantum	physics	world.
The	incident	wave	function
Assume	that	the	scattering	potential	V(r)	has	a	very	finite	range,	a.	Outside	that
range,	the	wave	functions	involved	act	like	free	particles.	So	the	incident
particle's	wave	function,	outside	the	limit	of	V(r)		that	is,	outside	the	range	a
from	the	other	particle		is	given	by	this	equation,	because	V(r)	is	zero:
where	
.
The	form	
	is	the	equation	for	a	plane	wave,	so	inc(r)	is	inc(r)	=
Aeik0r,	where	A	is	a	constant	and	k0		r	is	the	dot	product	between	the	incident
wave's	wave	vector	and	r.	In	other	words,	you're	treating	the	incident	particle
as	a	particle	of	momentum	
.
The	scattered	wave	function
After	the	scattering	of	the	spinless	particles,	the	nonscattered	wave	function
isn't	of	much	interest	to	you,	but	the	scattered	wave	function	is.	Although	the
incident	wave	function	has	the	form	inc(r)	=	Aeik0r,	the	scattered	wave
function	has	a	slightly	different	form:
The	f(,	)	part	is	called	the	scattering	amplitude,	and	your	job	is	to	find	it.
Here,	A	is	a	normalization	factor	and
where	E	is	the	energy	of	the	scattered	particle.
Relating	the	scattering	amplitude	and	differential
cross	section
The	scattering	amplitude	of	spinless	particles	turns	out	to	be	crucial	to
understanding	scattering	from	the	quantum	physics	point	of	view.	To	see	that,
take	a	look	at	the	current	densities,	Jinc	(the	flux	density	of	the	incident
particle)	and	Jsc	(the	current	density	for	the	scattered	particle):
	
	
Inserting	your	expressions	for	inc	and	sc	into	these	equations	gives	you	the
following,	where	f(,	)	is	the	scattering	amplitude:
	
	
Now	in	terms	of	the	current	density,	the	number	of	particles	dN(,	)	scattered
into	d	and	passing	through	an	area	dA	=	r2d	is
dN(,	)	=	Jscr2d
Plugging	in	
	into	the	preceding	equation	gives	you
Also,	recall	from	the	beginning	of	the	chapter	that	
.	You	get
And	here's	the	trick		for	elastic	scattering,	k	=	k0,	which	means	that	this	is
your	final	result:
	The	problem	of	determining	the	differential	cross	section	breaks	down	to
determining	the	scattering	amplitude.
Finding	the	scattering	amplitude
To	find	the	scattering	amplitude		and	therefore	the	differential	cross	section
	of	spinless	particles,	you	work	on	solving	the	Schrdinger	equation:
.	You	can	also	write	this	as
You	can	express	the	solution	to	that	differential	equation	as	the	sum	of	a
homogeneous	solution	and	a	particular	solution:
(r)	=	h(r)	+	p(r)
The	homogeneous	solution	satisfies	this	equation:
And	the	homogeneous	solution	is	a	plane	wave		that	is,	it	corresponds	to	the
incident	plane	wave:
To	take	a	look	at	the	scattering	that	happens,	you	have	to	find	the	particular
solution.	You	can	do	that	in	terms	of	Green's	functions,	so	the	solution	to
	is
where	
.
This	integral	breaks	down	to
You	can	solve	the	preceding	equation	in	terms	of	incoming	and/or	outgoing
waves.	Because	the	scattered	particle	is	an	outgoing	wave,	the	Green's	function
takes	this	form:
You	already	know	that
So	substituting	
	into	the	preceding	equation	gives	you
Wow,	that's	an	integral	equation	for	(r)	,	the	wave	equation		how	do	you	go
about	solving	this	whopper?	Why,	you	use	the	Born	approximation,	of	course.
The	Born	Approximation:	Rescuing
the	Wave	Equation
Okay,	your	dilemma	is	to	solve	the	following	equation	for	(r),	where	inc	=
Aeik0r:
You	can	do	that	with	a	series	of	successive	approximations,	called	the	Born
approximation	(this	is	a	famous	result).	To	start,	the	zeroth	order	Born
approximation	is	just	0(r)	=	inc(r).	And	substituting	this	zeroth-order	term,
0(r),	into	the	first	equation	in	this	section	gives	you	the	first-order	term:
which,	using	0	(r)	=	inc	(r)	gives	you
You	get	the	second-order	term	by	substituting	this	equation	into
:
And	substituting	
	into	the	preceding	equation	gives
you
The	pattern	continues	for	the	higher	terms,	which	you	can	find	by	plugging
lower-order	terms	into	higher	ones.
Exploring	the	far	limits	of	the	wave	function
Now	that	you've	used	the	Born	approximation	(see	the	preceding	section),	take
a	look	at	the	case	where	r	is	large		in	scattering	experiments,	r	>>	r',	where	r
is	the	distance	from	the	target	to	the	detector	and	r'	is	the	size	of
the	detector.	What	happens	to	
,	the
exact	integral	equation	for	the	wave	function,	when	r	>>	r'?	Here's	the	answer:
Because	r	>>	r',	you	can	say	that	k|r		r'|		kr		k		r',	where	k		r'	is	the	dot
product	of	k	and	r'	(k	is	the	wave	vector	of	the	scattered	particle).	And
Using	the	last	two	equations	in	
	gives	you
And	here
The	differential	cross	section	is	given	by	
,	which	in	this	case
becomes
Using	the	first	Born	approximation
If	the	potential	is	weak,	the	incident	plane	wave	is	only	a	little	distorted	and	the
scattered	wave	is	also	a	plane	wave.	That's	the	assumption	behind	the	first
Born	approximation,	which	you	take	a	look	at	here.	So	if	you	make	the
assumption	that	the	potential	is	weak,	you	can	determine	from	the	equation
	that
Okay,	so	what	is	f(,	)?	Well
And	this	equals	the	following,	where	q	=	k0		k:
And	because	
,	you	have
When	the	scattering	is	elastic,	the	magnitude	of	k	is	equal	to	the	magnitude	of
k0,	and	you	have
q	=	|k0		k|	=	2k	sin(/2)
where		is	the	angle	between	k0	and	k.
In	addition,	if	you	say	that	V(r)	is	spherically	symmetric,	and	you	can	choose	the
z	axis	along	q,	then	q	.	r'	=	qr'	cos',	so
That	equals
You	know	that	
,	so
You've	come	far	in	this	chapter		from	the	Schrdinger	equation	all	the	way
through	the	Born	approximation,	and	now	to	the	preceding	equation	for	weak,
spherically	symmetric	potentials.	How	about	you	put	this	to	work	with	some
concrete	numbers?
Putting	the	Born	approximation	to	work
In	this	section,	you	find	the	differential	cross	section	for	two	electrically
charged	particles	of	charge	Z1e	and	Z2e.	Here,	the	potential	looks	like	this:
So	here's	what	the	differential	cross	section	looks	like	in	the	first	Born
approximation:
And	because	
,	you	know	that
And	because	q	=	2ksin(/2),	the	following	is	true:
where	E	is	the	kinetic	energy	of	the	incoming	particle:	
.
Now	get	more	specific;	say	that	you're	smashing	an	alpha	particle,	Z1	=	4,
against	a	gold	nucleus,	Z2	=	79.	If	the	scattering	angle	in	the	lab	frame	is	60,
what	is	it	in	the	center-of-mass	frame?
The	ratio	of	the	particles'	mass	in	this	case,	m1/m2,	is	0.02,	so	the	scattering
angle	in	the	center-of-mass	frame,	,	is	the	following,	where	lab	=	60:
Solving	that	equation	for		gives	you		=	61.	So	what's	the	cross	section	for
this	scattering	angle?	Take	a	look:
Plugging	in	the	numbers	if	the	incident	alpha	particle's	energy	is	8	MeV	gives
you	the	following:
That's	the	size	of	the	target		the	cross	section		you	have	to	hit	to	create	the
scattering	angle	seen.
Part	VI
The	Part	of	Tens
In	this	part	.	.	.
I	let	quantum	physics	off	the	leash	in	this	part,	and	it	goes	wild.	You	get	to	see
the	ten	best	online	tutorials	here,	as	well	as	ten	major	triumphs	of	quantum
physics.	Researchers	created	quantum	physics	because	of	the	need	to	handle
issues	such	as	the	wave-particle	duality,	the	uncertainty	principle,	and	the
photoelectric	effect,	and	you	relive	those	triumphs	here.
Chapter	13
Ten	Quantum	Physics	Tutorials
In	This	Chapter
	Understanding	basic	concepts	and	equations
	Viewing	illustrations	and	animations
When	scientists	start	mixing	talk	of	dice,	billiard	balls,	and	a	possibly	undead
cat-in-a-box,	you	know	you're	dealing	with	a	challenging	subject.	Luckily,	you
can	find	plenty	of	online	tutorials,	some	of	them	featuring	animation,	to	help
you	wrap	your	brain	around	quantum	physics.	This	chapter	presents	a	good
starter	list.
An	Introduction	to	Quantum
Mechanics
http://legacyweb.chemistry.ohio-state.edu/betha/qm
What	is	a	wave	function?	What	is	an	orbital?:	An	Introduction	to	Quantum
Mechanics	comes	from	Neal	McDonald,	Midori	Kitagawa-DeLeon,	Anna
Timasheva,	Heath	Hanlin,	Zil	Lilas,	and	Sherwin	J.	Singer	at	The	Ohio	State
University.	This	site	includes	tutorials	on	probability,	particles	versus	waves,
wave	functions,	and	more,	including	Shockwave-based	sound	(though	if	you
don't	have	Shockwave	installed,	that's	not	a	problem).
Quantum	Mechanics	Tutorial
www.gilestv.com/tutorials/quantum.html
This	cool	tutorial	is	one	of	the	Flash-animated	Modern	Physics	Tutorials	by	Giles
Hogben.	Extensively	illustrated,	this	tutorial	probes	questions	such	as	wave-
particle	duality	and	offers	a	good	general	introduction	to	quantum	physics.
Grains	of	Mystique:	Quantum
Physics	for	the	Layman
http://www.faqs.org/docs/qp
This	site	provides	good	historical	and	experimental	background	info		and
they've	documented	their	sources	and	made	some	attempts	at	peer	review.
Quantum	Physics	Online	Version	2.0
www.quantum-physics.polytechnique.fr/index.html
This	is	a	cool	set	of	programs	that	run	in	your	browser,	giving	simulations	of
various	quantum	physics	experiments.	It's	by	Manuel	Joffre,	Jean-Louis
Basdevant,	and	Jean	Dalibard	of	the	cole	Polytechnique	in	France.	Look	for
information	on	wave	mechanics,	quantization,	quantum	superposition,	and	spin
1/2.
Todd	K.	Timberlake's	Tutorial
facultyweb.berry.edu/ttimberlake/qchaos/qm.html
This	tutorial	is	by	Todd	K.	Timberlake,	assistant	professor	of	the	Department	of
Physics,	Astronomy,	&	Geology	of	Berry	College	in	Georgia.	It's	a	fairly	brief	but
well-written	introduction	to	the	ideas	of	quantum	mechanics.
Physics	24/7's	Tutorial
www.physics247.com/physics-tutorial/quantum-physics-billiards.shtml
This	is	a	text-based	tutorial	from	Physics	24/7.	It	includes	material	on	quanta,
the	uncertainty	principle,	and	quantum	tunneling	(as	well	as	some	ads).
Stan	Zochowski's	PDF	Tutorials
www.cmmp.ucl.ac.uk/~swz/courses/SM355/SM355.html
Stan	Zochowski,	from	the	department	of	Physics	&	Astronomy	at	University
College	London,	put	together	these	PDF-based	tutorials	on	quantum	physics.
These	are	tutorial	handouts	for	a	Quantum	Mechanics	course	at	the	University
College,	and	they	serve	as	an	excellent	introduction	to	quantum	physics.
Quantum	Atom	Tutorial
www.colorado.edu/physics/2000/quantumzone/index.html
This	is	a	fun,	cartoon-centric	tutorial	on	the	quantum	nature	of	the	atom	from
the	University	of	Colorado	Physics	2000	project.
College	of	St.	Benedict's	Tutorial
www.physics.csbsju.edu/QM/Index.html
This	is	a	comprehensive	quantum	physics	tutorial	from	the	College	of	St.
Benedict.	It's	a	good,	more	serious,	text	and	equations-based	tutorial	with
plenty	of	illustrations.
A	Web-Based	Quantum	Mechanics
Course
electron6.phys.utk.edu/qm1/Modules.htm
This	one's	from	the	University	of	Tennessee,	and	it's	an	extensive	online	course
in	quantum	physics.	It	includes	modules	on	square	potentials,	harmonic
oscillators,	angular	momentum,	spin,	and	so	on.
Chapter	14
Ten	Quantum	Physics	Triumphs
In	This	Chapter
	Explaining	unexpected	results
	Identifying	characteristics	of	the	quantum	world
	Developing	new	models
Quantum	physics	has	been	very	successful	in	explaining	many	physical
phenomena,	such	as	wave-particle	duality.	In	fact,	quantum	physics	was
created	to	explain	physical	measurements	that	classical	physics	couldn't
explain.	This	chapter	is	about	ten	triumphs	of	quantum	physics,	and	it	points
you	to	resources	on	the	Web	that	examine	those	triumphs	for	further
information.
Wave-Particle	Duality
Is	that	particle	a	wave?	Or	is	that	wave	a	particle?	That's	one	of	the	questions
that	quantum	physics	was	created	to	solve,	because	particles	exhibited	wave-
like	properties	in	the	lab,	whereas	waves	exhibited	particle-like	properties.
These	Web	sites	offer	more	insight:
	www.gilestv.com/tutorials/quantum.html
	www.physics247.com/physics-tutorial/quantum-physics-billiards.shtml
The	Photoelectric	Effect
Another	founding	pillar	of	quantum	physics	was	explaining	the	photoelectric
effect,	in	which	experimenters	shone	light	on	a	metal.	No	matter	how	strong
the	light,	the	energy	of	ejected	electrons	from	the	metal	didn't	rise.	It	turns	out
that	the	energy	of	electrons	goes	up	with	the	frequency	of	the	light,	not	its
intensity		which	gives	support	to	the	light	as	a	stream	of	discrete	photons
theory.
For	more	info	on	the	photoelectric	effect,	check	out
www.gilestv.com/tutorials/quantum.html.
Postulating	Spin
The	Stern-Gerlach	experiment	results	couldn't	be	explained	without	postulating
spin,	another	triumph	of	quantum	physics.	This	experiment	sent	electrons
through	a	magnetic	field,	and	the	classical	prediction	is	that	the	electron
stream	would	create	one	spot	of	electrons	on	a	screen		but	there	were	two
(corresponding	to	the	two	spins,	up	and	down).
This	Web	site	has	more	info:	electron6.phys.utk.edu/qm1/modules/m9/spin.htm.
Differences	between	Newton's	Laws
and	Quantum	Physics
In	classical	physics,	bound	particles	can	have	any	energy	or	speed,	but	that's
not	true	in	quantum	physics.	And	in	classical	physics,	you	can	determine	both
the	position	and	momentum	of	particles	exactly,	which	isn't	true	in	quantum
physics	(thanks	to	the	Heisenberg	uncertainty	principle).	And	in	quantum
physics,	you	can	superimpose	states	on	each	other,	and	have	particles	tunnel
into	areas	that	would	be	classically	impossible.
You	can	find	a	nice	discussion	of	the	differences	between	classical	and	quantum
physics	at	http://facultyweb.berry.edu/ttimberlake/qchaos/qm.html.
Heisenberg	Uncertainty	Principle
One	of	the	triumphs	of	quantum	physics	is	the	Heisenberg	uncertainty
principle:	Heisenberg	theorized	that	you	can't	simultaneously	measure	a
particle's	position	and	momentum	exactly.	This	is	one	of	the	central	theories
that	has	destroyed	classical	physics.
Here's	where	you	can	find	one	of	the	best	Web	discussions	on	this	topic:
www.physics247.com/physics-tutorial/quantum-physics-billiards.shtml.
Quantum	Tunneling
How	can	particles	go	where,	classically,	they	don't	have	enough	energy	to	go?
For	example,	how	can	an	electron	with	energy	E	go	into	an	electrostatic	field
where	you	need	to	have	more	than	energy	E	to	penetrate?	The	answer	was
postulated	with	quantum	tunneling,	and	you	can	find	more	information	about
that	at
www.physics247.com/physics-tutorial/quantum-physics-billiards.shtml.
Discrete	Spectra	of	Atoms
Modeling	the	quantized	nature	of	atoms	and	orbitals	is	another	triumph	of
quantum	physics.	It	turns	out	that	electrons	can't	have	any	old	energy	in	an
atom,	but	are	only	allowed	particular	quantized	energy	levels		and	that	was
one	of	the	foundations	of	quantum	physics.
For	a	lot	more	on	the	topic,	visit
www.colorado.edu/physics/2000/quantumzone/index.html.
Harmonic	Oscillator
Quantizing	harmonic	oscillators	on	the	micro	level	was	another	triumph	of
quantum	physics.	Classically,	harmonic	oscillators	can	have	any	energy		but
not	quantum	mechanically.	And	guess	which	one	was	right?
Read	all	about	it	here:
	www.physics.csbsju.edu/QM/Index.html
	electron6.phys.utk.edu/qm1/modules/module8.htm
Square	Wells
Like	harmonic	oscillators,	quantizing	particles	bound	in	square	wells	at	the
micro	level	was	another	triumph	for	quantum	physics.	Classically,	particles	in
square	wells	can	have	any	energy,	but	quantum	physics	says	you	can	only	have
certain	allowed	energies.
There's	plenty	on	the	Web	about	it,	including	these	two	good	treatments:
	www.physics.csbsju.edu/QM/Index.html
	electron6.phys.utk.edu/qm1/modules/module2.htm
Schrdinger's	Cat
Schrdinger's	Cat	is	a	thought	experiment	that	details	some	problems	that
arise	in	the	macro	world	from	thinking	of	the	spin	of	electrons	as	completely
non-determined	until	you	measure	them.	For	example,	if	you	know	the	spin	of
one	of	a	pair	of	newly-created	electrons,	you	know	the	other	has	to	have	the
opposite	spin.	So	if	you	separate	two	electrons	by	light	years	and	then	measure
the	spin	of	one	electron,	does	the	other	electron's	spin	suddenly	snap	to	the
opposite	value		even	at	a	distance	that	would	take	a	signal	from	the	first
electron	years	to	cover?	Tricky	stuff!
For	more,	take	a	look	at	www.gilestv.com/tutorials/quantum.html.
Glossary
Here's	a	glossary	of	common	quantum	physics	terms:
amplitude:	The	maximum	amount	of	displacement	of	an	oscillating	particle.
angular	momentum:	The	product	of	the	distance	a	particle	is	from	a	certain
point	and	its	momentum	measured	with	respect	to	the	point.
annihilation	operator:	An	operator	that	lowers	the	energy	level	of	an
eigenstate	by	one	level.
anti-Hermitian:	An	operator	whose	Hermitian	adjoint	is	the	same	as	the
original	operator,	but	with	a	minus	sign;	in	other	words,	an	operator	is	anti-
Hermitian	if	A	=	A.	See	also	Hermitian	operator.
black	body:	A	body	that	absorbs	all	radiation	and	radiates	it	all	away.
Bohr	radius:	The	average	radius	of	an	electron's	orbit	in	a	hydrogen	atom,
about	1010	meters.
bound	state:	A	state	in	which	a	particle	isn't	free	to	travel	to	infinity.
bosons:	Particles	with	integer	spins,	including	photons,	pi	mesons,	and	so	on.
bra-ket	notation:	Abbreviating	the	matrix	form	of	a	state	vector	as	a	ket,	or	|
>,	and	abbreviating	the	ket's	complex	conjugate,	or	bra,	as	<|.
center-of-mass	frame:	In	scattering	theory,	the	frame	in	which	the	center	of
mass	is	stationary	and	the	particles	head	toward	each	other	and	collide.	See
also	lab	frame.
central	potential:	A	spherically	symmetrical	potential.
commute:	Two	operators	commute	with	each	other	if	their	commutator	is
equal	to	zero.	The	commutator	of	operators	A	and	B	is	[A,	B]	=	AB		BA.
complex	conjugate:	The	number	you	get	by	negating	the	imaginary	part	of	a
complex	number.	The	*	symbol	indicates	a	complex	conjugate.
Compton	effect:	An	increase	of	wavelength,	depending	on	the	scattering
angle,	that	occurs	after	incident	light	hits	an	electron	at	rest.
conservation	of	energy:	The	law	of	physics	that	says	the	energy	of	a	closed
system	doesn't	change	unless	external	influences	act	on	the	system.
creation	operator:	An	operator	that	raises	the	energy	level	of	an	eigenstate
by	one	level.
current	density:	See	incident	flux.
electron	volts	(eV):	The	amount	of	energy	one	electron	gains	falling	through
1	volt.
diagonalize:	Writing	a	matrix	so	that	the	only	nonzero	elements	appear	along
the	matrix's	diagonal.
differential	cross	section:	In	scattering	theory,	the	cross	section	for
scattering	a	particle	to	a	specific	solid	angle;	it's	like	a	bull's-eye.
Dirac	notation:	See	bra-ket	notation.
eigenvalue:	A	complex	constant	that	represents	the	change	in	magnitude	of	a
vector	when	you	act	on	that	vector	with	an	operator.
eigenvector:	A	vector	that	changes	in	magnitude	but	not	direction	after	you
apply	an	operator.
elastic	collision:	A	collision	in	which	kinetic	energy	is	conserved.
electric	field:	The	force	on	a	positive	test	charge	per	Coulomb	due	to	other
electrical	charges.
electron:	A	negatively	charged	particle	with	half-integer	spin.
emissivity:	A	property	of	a	substance	showing	how	well	it	radiates.
energy:	The	ability	of	a	system	to	do	work.
energy	degeneracy:	The	number	of	states	that	have	the	same	energy.
energy	well:	See	potential	well.
expectation	value:	The	average	value	an	operator	will	return.
fermions:	Particles	with	half-integer	spin,	including	electrons,	protons,
neutrons,	quarks,	and	so	on.
frequency:	The	number	of	cycles	of	a	periodic	occurrence	per	second.
Hamiltonian:	An	operator	for	the	total	energy	of	a	particle,	both	kinetic	and
potential.
Heisenberg	uncertainty	principle:	See	uncertainty	principle.
Hermitian	adjoint:	The	complex	conjugate	of	a	number,	the	bra
corresponding	to	a	ket	vector	or	the	ket	corresponding	to	a	bra	vector,	or	the
conjugate	transpose	A	of	an	operator	A.
Hermitian	operator:	Operators	that	are	equal	to	their	Hermitian	adjoints;	in
other	words,	an	operator	is	Hermitian	if	A	=	A.
incident	flux:	The	number	of	incident	particles	per	unit	area	per	unit	time.
inelastic	collision:	A	collision	in	which	kinetic	energy	isn't	conserved.
intensity	(wave):	The	time-averaged	rate	of	energy	transmitted	by	a	wave	per
unit	of	area.
Joule:	The	MKS	unit	of	energy		one	Newton-meter.
ket:	See	bra-ket	notation.
kinetic	energy:	The	energy	of	an	object	due	to	its	motion.
lab	frame:	In	scattering	theory,	the	frame	in	which	one	particle	is	incident	on
a	particle	at	rest	and	hits	it.	See	also	center-of-mass	frame.
Laplacian:	An	operator,	represented	by	2,	that	you	use	to	find	the
Hamiltonian.
magnetic	field:	The	force	on	a	moving	positive	test	charge,	per	Coulomb,	from
magnets	or	moving	charges.
magnitude:	The	size	or	length	associated	with	a	vector	(vectors	are	made	up
of	a	direction	and	a	magnitude).
mass:	The	property	that	makes	matter	resist	being	accelerated.
momentum:	The	product	of	mass	times	velocity,	a	vector.
MKS	system:	The	measurement	system	that	uses	meters,	kilograms,	and	
seconds.
Newton:	The	MKS	unit	of	force		one	kilogram-meter	per	second2.
normalized	function:	A	function	in	which	the	probability	adds	up	to	1.
orbitals:	Different	angular	momentum	states	of	an	electron,	represented	as
subshells	in	atomic	structure.
orthogonal:	Two	kets,	|>	and	|>,	for	which	<|>	=	0.
orthonormal:	Two	kets,	|>	and	|>,	that	meet	the	following	conditions:	<|
>	=	0;	<	|>	=	1;	and	<|>	=	1.
oscillate:	To	move	or	swing	side	to	side	regularly.
pair	annihilation:	The	conversion	of	an	electron	and	positron	into	pure	light.
pair	production:	The	conversion	of	a	high-powered	photon	into	an	electron
and	positron.
particle:	A	discrete	piece	of	matter.
Pauli	exclusion	principle:	The	idea	that	no	two	electrons	can	occupy	the
same	state	in	a	single	atom.
period:	The	time	it	takes	for	one	complete	cycle	of	a	repeating	event.
perturbation:	A	stimulus	mild	enough	that	you	can	calculate	the	resulting
energy	levels	and	wave	functions	as	corrections	to	the	fundamental	energy
levels	and	wave	functions	of	the	unperturbed	system.
photoelectric	effect:	A	result	in	which	the	kinetic	energy	of	electrons	emitted
from	a	piece	of	metal	depends	only	on	the	frequency		not	the	intensity		of
the	incident	light.
photon:	A	quantum	of	electromagnetic	radiation.	An	elementary	particle	that	is
its	own	antiparticle.
pi	meson:	A	subatomic	particle	that	helps	hold	the	nucleus	of	an	atom
together.
Planck's	constant:	A	universal	constant,	h,	that	describes	the	relationship
between	the	energy	and	frequency	of	a	photon.	It	equals	6.626		1034	Joule-
seconds.
positron:	A	positively	charged	anti-electron.
potential	barrier:	A	potential	step	of	limited	extent;	an	electron	may	be	able
to	tunnel	through	the	barrier	and	come	out	the	other	side.
potential	energy:	An	object's	energy	because	of	its	position	when	a	force	is
acting	on	it	or	its	internal	configuration.
potential	step:	A	region	in	which	the	energy	potential	forms	a	stair	shape;
a	particle	striking	the	step	may	be	reflected	or	transmitted.
potential	well:	A	region	in	which	there's	a	dip	in	the	energy	potential
threshold;	particles	without	enough	energy	to	overcome	the	barrier	can
become	trapped	in	the	well,	unable	to	convert	the	potential	energy	to	kinetic.
power:	The	rate	of	change	in	a	system's	energy.
probability	amplitude:	The	square	root	of	the	probability	that	a	particle	will
occupy	a	certain	state.
probability	density:	The	likelihood	that	a	particle	will	occupy	a	particular
position	or	have	a	particular	momentum.
quantized:	Coming	in	discrete	values.
quark:	Particles	that	combine	with	antiquarks	to	form	protons,	neutrons,	and
so	on.
radian:	The	MKS	unit	of	angle		2	radians	are	in	a	circle.
radiation:	A	physical	mechanism	that	transports	heat	and	energy	as
electromagnetic	waves.
scalar:	A	simple	number	(without	a	direction,	which	a	vector	has).
Schrdinger	equation:	An	equation	that	tells	us	how	the	wave	function
(which	describes	the	probable	locations	of	particles	like	electrons)	changes
over	time.
simple	harmonic	motion:	Repetitive	motion	where	the	restoring	force	is
proportional	to	the	displacement.
spherical	coordinates:	Coordinates	that	indicate	location	using	two	angles
and	the	length	of	a	radius	vector.
spin:	The	intrinsic	angular	momentum	of	an	electron,	classified	as	up	or	down.
synchrotron:	A	type	of	circular	particle	accelerator.
state	vector:	A	vector	that	gives	the	probability	amplitude	that	particles	will	be
in	their	various	possible	states.
threshold	frequency:	If	you	shine	light	below	this	frequency	on	metal,	no
electrons	are	emitted.
total	cross	section:	In	scattering	theory,	the	cross	section	for	any	kind	of
particle	scattering,	through	any	angle.
tunneling:	The	phenomenon	where	particles	can	get	through	regions	that
they're	classically	forbidden	to	go.
ultraviolet	catastrophe:	The	failure	of	the	Rayleigh-Jeans	Law	to	explain
black-body	radiation	at	high	frequencies.
uncertainty	principle:	A	principle	that	says	it's	impossible	to	know	an	object's
exact	momentum	and	position.
vector:	A	mathematical	construct	that	has	both	a	magnitude	and	a	direction.
velocity:	The	rate	of	change	of	an	object's	position,	expressed	as	a	vector
whose	magnitude	is	speed.
volt:	The	MKS	unit	of	electrostatic	potential		one	Joule	per	Coulomb.
wave:	A	traveling	energy	disturbance.
wavelength:	The	distance	between	crests	or	troughs	of	a	wave.
wave-particle	duality:	The	observation	that	light	has	properties	of	both	waves
and	particles,	depending	on	the	experiment.
wave	packet:	A	collection	of	wave	functions	such	that	the	wave	functions
interfere	constructively	at	one	location	and	interfere	destructively	(go	to	zero)
at	all	other	locations.
work:	Force	multiplied	by	the	distance	over	which	that	force	acts.