q, p
½ Š ¼ iħ,
ð1:140Þ
where the presence of a unit matrix E is implied. Explicitly writing it, we have,
q, p
½ Š ¼ iħE:
ð1:141Þ
The relations (1.140) and (1.141) are called the canonical commutation relation. On
the basis of a relation p ¼
ħ
i
∂
∂q
, a brief proof for this is as follows:
q, p
½ Š ψi ¼ qp À pq
ð
Þ
j
j ψi ¼ q
ħ
i
∂
∂q
À
ħ
i
∂
∂q
q
ψi ¼ q
ħ
i
∂
∂q
ψi À
ħ
i
∂
∂q
qjψi
ð
Þ
¼ q
ħ
i
∂ j ψi
∂q
À
ħ
i
∂q
∂q
ψi À
ħ
i
q
∂ j ψi
∂q
¼ À
ħ
i
ψi ¼ iħ j ψi:
ð1:142Þ
Since jψi is an arbitrarily chosen vector, we have (1.140).
Using (1.117), we have
A, B
½
Š
{ ¼ AB À BA
ð
Þ
{ ¼ B
{ A
{
À A
{ B
{
:
ð1:143Þ
If in (1.143) A and B are both Hermitian, we have
A, B
½
Š
{ ¼ BA À AB ¼ À A, B
½
Š:
ð1:144Þ
If we have an operator G such that
G
{
¼ ÀG,
ð1:145Þ
G is said to be anti-Hermitian. Therefore, [A, B] is anti-Hermitian, if both A and B are
Hermitian. If an anti-Hermitian operator has an eigenvalue, that eigenvalue is zero or
pure imaginary. To show this, suppose that
G j ψi ¼ λ j ψi,
ð1:146Þ
where G is an anti-Hermitian operator and jψi has been normalized. As in the case of
(1.123) we have
ψj
h G j ψi ¼ λ ψj
h ψi ¼ λ:
ð1:147Þ
Taking a complex conjugate of (1.147), we have
ψjG
h
ψi
à ¼ λ
Ã
:
ð1:148Þ
Using (1.116) and (1.145) again, we have
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