1000 Solved Problems in Modern Physics

(Tina Meador) #1

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

)


(

a
b

)
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