2.3 Solutions
125
Integration of (1) yields
∂
∂ x
(ψ
∗
ψ)dτ +
ψ
∗ ∂ψ
∂ x
dτ +
ψ
dψ
∗
dx
dτ
(2)
The integral on the LHS vanishes because ψ
∗
ψ vanishes for large values of
|x| if the particle is confined to some finite region. Thus
∂
∂ x
(ψ
∗
ψ)dxdydz =
|ψ
∗
ψ|
∞
−∞ dydz = 0
Therefore (2) becomes
ψ
∗
∂ψ
∂ x
dτ = −
ψ
∂ψ
∗
∂ x
dτ
(3)
Generalizing to all the three coordinates
(ψ, ∇ψ) = −(∇ψ, ψ)
( 4 )
Hence
(ψ, i∇ψ) = (i∇ψ, ψ)
( 5 )
where we have used, i∇ψ, ψ = −
i∇ ψ
∗
ψ dτ .
This completes the proof that the momentum operator is hermitian.
2.73 Using the standard method explained in Chap. 1, define the eigen values
λ 1 = 3 and λ 2 = −1 for the matrix P and the eigen vectors
1
√
2
1
1
and
1
√
2
1
−1
. For the matrix Q, the eigen values are λ 1 = 5 and λ 2 = 1, the eigen
vectors being
1
√
2
1
1
and
1
√
2
1
−1
. Thus the eigen vectors for the commutating matrices are identical.
2.74 (a) A = αx + iβp
A
†
= ax
†
− iβp
†
(b)[A, x] = α[x, x] + iβ[ p, x]
= 0 + iβ(−i) = β
[ A, A] = A A − A A = 0
[ A, p] = α[x, p] + iβ[ p, p]
= iα + 0 = iα
2.75 (a) As A satisfies a quadratic equation it can be represented by a 2 × 2 matrix.
Its eigen values are the roots of the quadratic equation
λ
2
− 4λ + 3 = 0, λ 1 = 1, λ 2 = 3
(b) A is represented by the matrix
A =
1 0
0 3
The eigen value equation is
1 0
0 3
a
b
= λ
a
b