where we assumed that jψi is arbitrarily chosen normalized vector. Suppose now
that jψ 0 i is an eigenstate of A that belongs to an eigenvalue a. Then, we have
A j ψ 0 i ¼ a j ψ 0 i:
ð2:137Þ
Taking an adjoint of (2.137), we get
ψ 0 j
h A
{
¼ ψ 0 j
h A ¼ ψ 0 j
h a
Ã
¼ a ψ 0 j
h ,
ð2:138Þ
where the last equality comes from the fact that A is Hermitian. From (2.138), we
would have
ψ 0 j
h AB À BA ψ 0 i ¼ ψ 0 jA
h
Bjψ 0 i À ψ 0 j
h BA
j
j ψ 0 i
¼ ψ 0 ja
h
B ψ 0 i À ψ 0 j
h Ba
j
j ψ 0 i ¼ a ψ 0 j
h B ψ 0 i À a ψ 0 j
h B
j
j ψ 0 i ¼ 0:
This would imply that (2.136) does not hold with jψ 0 i, in contradiction to (2.127),
where ik 6 ¼ 0. Namely, we conclude that any physical state cannot be an eigenstate of
A on condition that (2.127) holds. Equation (2.127) is rewritten as
ψj
h BA À AB j ψi ¼ Àik:
ð2:139Þ
Suppose now that jφ 0 i is an eigenstate of B that belongs to an eigenvalue b. Then, we
can similarly show that any physical state cannot be an eigenstate of B.
Summarizing the above, we restate that once we have a relation [A, B] ¼ ik (k 6 ¼ 0),
their representation matrix does not diagonalize A or B. Or, once we postulate [A,
B] ¼ ik (k 6 ¼ 0), we must abandon an effort to have a representation matrix that
diagonalizes A and B. In the quantum-mechanical formulation of a harmonic oscillator, we have introduced the canonical commutation relation (see Sect. 2.3)
described by [q, p] ¼ iħ (1.140). Indeed, neither q nor p is diagonalized as shown
in (2.69) or (2.70).
Example 2.1 Taking a quantum harmonic oscillator as an example, we consider
variance of q and p in reference to jψ n i (n ¼ 0, 1, Á Á Á). We have
Δq
ð Þ
2
E
D
¼ ψ n j
h q
2
j ψ n i À ψ n j
h q j ψ n i
2 :
ð2:140Þ
Using (2.55) and (2.62) as well as (2.68), we get
ψ n q
j jψ n
h
i¼
ffiffiffiffiffiffiffiffiffi ffi
ħ
2mω
r
ψ n a þ a
{
ψ n
¼
ffiffiffiffiffiffiffiffiffi ffi
ħn
2mω
r
ψ n jψ nÀ1
h
iþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ħ n þ 1
ð
Þ
2mω
r
ψ n jψ nþ1
¼ 0,
ð2:141Þ
where the last equality comes from (2.53). We have
54
2 Quantum-Mechanical Harmonic Oscillator
that jψ 0 i is an eigenstate of A that belongs to an eigenvalue a. Then, we have
A j ψ 0 i ¼ a j ψ 0 i:
ð2:137Þ
Taking an adjoint of (2.137), we get
ψ 0 j
h A
{
¼ ψ 0 j
h A ¼ ψ 0 j
h a
Ã
¼ a ψ 0 j
h ,
ð2:138Þ
where the last equality comes from the fact that A is Hermitian. From (2.138), we
would have
ψ 0 j
h AB À BA ψ 0 i ¼ ψ 0 jA
h
Bjψ 0 i À ψ 0 j
h BA
j
j ψ 0 i
¼ ψ 0 ja
h
B ψ 0 i À ψ 0 j
h Ba
j
j ψ 0 i ¼ a ψ 0 j
h B ψ 0 i À a ψ 0 j
h B
j
j ψ 0 i ¼ 0:
This would imply that (2.136) does not hold with jψ 0 i, in contradiction to (2.127),
where ik 6 ¼ 0. Namely, we conclude that any physical state cannot be an eigenstate of
A on condition that (2.127) holds. Equation (2.127) is rewritten as
ψj
h BA À AB j ψi ¼ Àik:
ð2:139Þ
Suppose now that jφ 0 i is an eigenstate of B that belongs to an eigenvalue b. Then, we
can similarly show that any physical state cannot be an eigenstate of B.
Summarizing the above, we restate that once we have a relation [A, B] ¼ ik (k 6 ¼ 0),
their representation matrix does not diagonalize A or B. Or, once we postulate [A,
B] ¼ ik (k 6 ¼ 0), we must abandon an effort to have a representation matrix that
diagonalizes A and B. In the quantum-mechanical formulation of a harmonic oscillator, we have introduced the canonical commutation relation (see Sect. 2.3)
described by [q, p] ¼ iħ (1.140). Indeed, neither q nor p is diagonalized as shown
in (2.69) or (2.70).
Example 2.1 Taking a quantum harmonic oscillator as an example, we consider
variance of q and p in reference to jψ n i (n ¼ 0, 1, Á Á Á). We have
Δq
ð Þ
2
E
D
¼ ψ n j
h q
2
j ψ n i À ψ n j
h q j ψ n i
2 :
ð2:140Þ
Using (2.55) and (2.62) as well as (2.68), we get
ψ n q
j jψ n
h
i¼
ffiffiffiffiffiffiffiffiffi ffi
ħ
2mω
r
ψ n a þ a
{
ψ n
¼
ffiffiffiffiffiffiffiffiffi ffi
ħn
2mω
r
ψ n jψ nÀ1
h
iþ
ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ħ n þ 1
ð
Þ
2mω
r
ψ n jψ nþ1
¼ 0,
ð2:141Þ
where the last equality comes from (2.53). We have
54
2 Quantum-Mechanical Harmonic Oscillator