1000 Solved Problems in Modern Physics

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 becauseψ∗ψvanishes 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∗ψdτ.
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 andλ 2 =−1forthematrixPand the eigen vectors√^12

(

1

1

)

and

√^1

2

(

1

− 1

)

. For the matrix Q, the eigen values areλ 1 =5 andλ 2 =1, the eigen

vectors being√^12

(

1

1

)

and√^12

(

1

− 1

)

. Thus the eigen vectors for the commutat-
ing 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]=AA−AA= 0
[A,p]=α[x,p]+iβ[p,p]
=iα+ 0 =iα

2.75 (a) AsAsatisfies a quadratic equation it can be represented by a 2×2 matrix.
Its eigen values are the roots of the quadratic equation
λ^2 − 4 λ+ 3 = 0 ,λ 1 = 1 ,λ 2 = 3
(b)Ais represented by the matrix

A=

(

10

03

)

The eigen value equation is

(

10

03

)(

a

b

)

=λ

(

a

b

)