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An extension of Dirac notation and its consequences
Spyridon Koutandos
skoutandos@yahoo.com
skoutandos1@gmail.com
In this article we observe first that there is an extension of Dirac notation to include
the so called “external interchangeability” of the hermitian operators when taking
the real part of the bracket in the braket notation. Then we conclude that this,
together with the usual “internal interchangeability” of the hermitian operators
concerning the real part of the bracket leads to the commutability of operators as to
both cross and dot product therefore (by making use of the fact that operators are
linear) all the vector algebra is valid when taking the real part of the bracket and we
have hermitian operators. We deal with one particle situations and bound states.
Finally we find a new way for calculating the square of the RungeLenz vector and
we make speculations about hidden variable. First principles are used for the
deduction of our conclusions,like the independence of the phase angle of ψ from
its absolute value and the commutation relationships which are valid for any
function. Integration as to volume is meant from zero to infinity and units are
omitted for ease of calculations.
I. INTERCHANGEABILITY OF HERMITIAN OPERATORS
Let us begin by giving some definitions. The so called in this article “external
interchangeability” that is to be proved is the following:
(1.1)
[
]
[
]
ψ
ψ
ψ
ψ

ˆ
ˆ

Re
ˆ


ˆ
Re
B
A
A
B
×
=
×
−
The so called, on the other hand “internal interchangeability” is:
(1.2)
ψ
ψ
ψ
ψ
A
B
AB

Re


Re
=
Both are valid when A,B are hermitian.
We are going to start by proving that the operator ΑΒΒΑ is antihermitian when
Α,Β are hermitian. First we notice that :
*






ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
B
A
B
A
A
B
B
A
BA
AB
−
=
=
−
=
−
(1.3)
The last part of the equality containing the two terms is obviously purely imaginary,
thus:
(1.4)
0


Re
=
−
ψ
ψ
BA
AB
The validity of identity (1.4) by taking in mind that A,B are hermitian and can be
interchanged readily proves (1.2) that is the internal interchangeability of the
operators concerning the real part of the bracket.
Next we are going to prove that :
(1.5)
[
]
[
]
)
Re(

ˆ
ˆ

Re
ˆ


ˆ
Re
C
B
A
A
B
v
=
×
=
×
−
ψ
ψ
ψ
ψ
If (1.5) is to be valid obviously then we also have:
(1.6)
[
] 0

ˆ
ˆ

ˆ


ˆ
Re
=
×
−
×
−
ψ
ψ
ψ
ψ
B
A
A
B
For proving (1.5) we decompose into cartesian components. The x component of
(1.5) is written as:
(1.7)
[
]
[
]
[
]
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ


Re




Re


Re
)
Re(
y
z
z
y
y
z
z
y
y
z
z
y
x
A
B
A
B
A
B
A
B
A
B
A
B
C
−
−
=
−
−
=
=
−
−
=
The vector operators A,B are hermitian and so are their components. Therefore the x
component of the last part of (1.7) is written as:
(1.8)
[
]
[
]
ψ
ψ
ψ
ψ


Re


Re
)
Re(
y
z
z
y
y
z
z
y
x
B
A
B
A
A
B
A
B
C
−
=
−
−
=
By subtracting the second part of (1.8) from the second (out of the three ) part of
(1.7) we get:
(1.9)
]

)
(
)
(

Re[
]


Re[
]
Re[
>
−
−
−
<
=
=
>
+
−
+
−
<
=
ψ
ψ
ψ
ψ
y
z
z
y
z
y
y
z
y
z
z
y
y
z
z
y
x
B
A
A
B
B
A
A
B
B
A
B
A
A
B
A
B
C
Both terms in the bracket are of the form ΑΒΒΑ and therefore are antihermitian
with zero real part which concludes the proof of (1.6) since from symmetry the rest
of the components are going to obey the same laws.
II. DECOMPOSING INTO HERMITIAN AND ANTIHERMITIAN PARTS –
THE ANTI COMMUTATIVITY OF OPERATORS
By using the identity (A.1) proved in the appendix and equation (1.6) we obtain the
following identity where we have used the notation that:
ψ
ψ
B
B
A
A
ˆ
,
ˆ
=
=
v
r
(2.1)
∫
∫
×
−
=
×
=
×
+
×
−
=
=
×
+
×
−
=
×
−
×
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ

ˆ
ˆ

Re
2
*
Re
2
)
*
*
(
Re
ˆ


ˆ
Re
ˆ


ˆ
Re

ˆ
ˆ
ˆ
ˆ

Re
A
B
dV
B
A
dV
B
A
A
B
B
A
A
B
A
B
B
A
v
v
v
v
v
v
From equating the first and last part of (2.1) we conclude that:
(2.2)
[
]
[
]
ψ
ψ
ψ
ψ

ˆ
ˆ

Re

ˆ
ˆ

Re
A
B
B
A
×
−
=
×
So far the so called “anti commutability” of hermitian operators as to the cross
product and the real part of the bracket is readily proved. Similarly we find that:
(2.3)
>
<
>=
<
ψ
ψ
ψ
ψ
B
A
AB

Re


Re
And
(2.4)
>
<
>=
<
ψ
ψ
ψ
ψ
A
B
BA

Re


Re
Since from (1.4) we may equate (2.4) and (2.3) we arrive at:
(2.5)
>
<
>=
<
>⇒
<
>=
<
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ


Re


Re

Re

Re
BA
AB
A
B
B
A
Thus we have proved a similar identity for the dot product. By the way we also find
that the operators ˆ
ˆ
ˆ
ˆ
A B B A
× + ×
and ΑΒΒΑ are antihermitian, that is they have no
real part.
The operator ΑΒ+ΒΑ is hermitian if Α,Β are hermitian. Proof:
Due to hermitian property of A,B
(2.6)
>
>=<
<
>
>=<
<
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ



,



BA
A
B
AB
B
A
By observing that:
(2.7)
>
>=<
<
ψ
ψ
ψ
ψ
A
B
B
A


and adding up the terms we get:
(2.8)
real
B
A
B
A
BA
AB
BA
AB
=
>
<
+
>
=<
>=
+
>=<
<
+
>
<
*








ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
ψ
Furthermore the operator
A
B
B
A
C
ˆ
ˆ
ˆ
ˆ
ˆ
×
−
×
=
is hermitian if Α,Β are hermitian .
Proof:
Its x component is written as
Cx=(ΑyBzAzBy)(ByAzBzAy) (2.9)
= (AyBz+BzAy)(AzBy+ByAz)
As can be seen Cx has the form (KL+LK)(MN+NM) where K,L,M,N, are hermitian
operators and thus the operator C is hermitian by making use of what we have
proved so far (see 2.8).
We finally may decompose:
(2.10)
×
+
×
+
×
−
×
=
×
2
ˆ
ˆ
ˆ
ˆ
2
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
A
B
B
A
A
B
B
A
B
A
with the first part being hermitian and the second part being antihermitian.
In a similar fashion we may decompose into hermitian and antihermitian parts the
dot product:
ΑΒ= (ΑΒ+ΒΑ)/2+(ΑΒΒΑ)/2. (2.11)
III. THE EXTENSION OF DIRAC NOTATION
By looking at the form of the decomposition into hermitian and antihermitian
parts of (2.10) we deduce that ˆ
ˆ
A B
×
is hermitian if and only if ˆ
ˆ
ˆ
ˆ
A B
B A
× = − ×
.
Similarly AB is hermitian if and only if [A,B]=0. Now we know that if AB is
hermitian then we have the commutation law. We are going to prove that if ˆ
ˆ
A B
×
is
hermitian, then :
(3.1)
ψ
ψ
ψ
ψ

ˆ
ˆ

ˆ


ˆ
B
A
A
B
×
=
×
−
Proof of (3.1):
We are going to prove that ˆ
ˆ
ˆ
ˆ
A B
B A
× = − ×
by taking its cartesian components. For
the x component we will have to prove(the rest follows by symmetry
considerations):
AyBzAzBy=ByAzBzAy
By taking a look at the first part we may recall that we have proved (1.1) and (1.9):
[
]
(
)
(
) ψ
ψ
ψ
ψ
ψ
ψ



ˆ
ˆ

ˆ


ˆ
z
y
z
y
z
y
y
z
x
A
B
A
B
B
A
A
B
B
A
A
B
−
−
−
=
=
×
−
×
−
The last terms can be rearranged to give –(ByAzBzAy)(AyBzAzBy)=0
This means that in the case
B
A
ˆ
ˆ ×
is hermitian we no longer need to take the real
part of the braket in external interchangeability of operators.
IV. ALGEBRA OF Re(braket)
Since vector calculus is obeyed by operators concerning the real part of the braket
one might be tempted to think that the usual vector identities can be applied.
However according to the author’s investigations these must be applied with
caution. Particularly I have come to the conclusion that each time we transform an
expression every member that is multiplied in the left part of the equation must be
hermitian in order for these identities to hold. In other words one part of the identity
at least must contain hermitian multipliers. For example since the cross product of
two real vectors is again hermitian as is the real vector and so is the momentum
operator p, indeed we could make use of the following identity:
(4.1)
)
(
)
(
)
(
B
A
C
A
C
B
C
B
A
v
v
v
r
v
v
v
v
v
×
⋅
=
×
⋅
=
×
⋅
Using (4.1) we are able to transform:
(4.2)
)
ˆ
(
)
(
ˆ
p
B
A
B
A
p
×
⋅
=
×
⋅
v
v
v
v
V. ORTHOGONALITY OF THE BASIC HERMITIAN OPERATORS –
PROPERTIES OF THE NULL OPERATORS
As a counterexample one could mention that the transformation does not work for
the following expression which is part of the square of the RungeLenz vector:
(5.1) (
)
)
ˆ
ˆ
(
ˆ
ˆ
L
p
p
L
×
⋅
×
This is so because the two parts multiplied are not hermitian although L,p are
hermitian operators. We may not apply a vector identity in that case although one
might argue that in expanding the expression we would find hermitian operators.
However one also gets the operator (pL)
2
which is double the null operator in our
algebra which is problematic for the following reason: Although Re(pLψ >)=0 it
still is Im(pLψ >)≠0 and since (pL)2 is hermitian it splits and one (pL) goes to the
bra and Re(<ψ pL)=0 while Im(<ψ pL)≠0 .
(pLψ >)= Re(pLψ >)+i Im(pLψ >)= +i Im(pLψ >) (5.2)
(<ψ pL)= Re(pLψ >)i Im(pLψ >) = i Im(pLψ >) (5.3)
<ψ pLpLψ >= Re<ψ pLpLψ >=
(
) dV
pL
2

Im
∫
>
ψ
(5.4)
It is also apparent that :
(5.5)
(
)
0

Re
ˆ
ˆ
Re
=
∇
×
⋅
−
=
⋅
ψ
ψ
r
r
i
L
r
r
r
h
Finally we will find that:
Re