1000 Solved Problems in Modern Physics

3.3 Solutions 161

3.9 (ψ,Qψ)=(ψ,qψ)=q(ψ,ψ)

(Qψ,ψ)=(qψ,ψ)=q∗(ψ,ψ)

sinceQis hermitian,

(ψ,Qψ)=(Qψ,ψ) and thatq=q∗

That is, the eigen values are real. The converse of this theorem is also true,

namely, an operator whose eigen values are real, is hermitian.

3.10 (a) The normalization condition requires

∫∞

−∞

|ψ|^2 dx=

∫a

− 3 a

|c|^2 dx= 1 = 4 a|c|^2

Thereforec= 1 / 2

√

a

(b)

∫a

0 |ψ|

(^2) dx=∫α
0 c
(^2) dx= 1 / 4
3.11 (a) The expectation values are

−∞

ψ∗xψdx=

∫a

− 3 a

x

dx

4 a

=−a

<x^2 >=

−∞

ψ∗x^2 ψdx=

∫a

− 3 a

(1/ 4 a)x^2 dx=

a^2

xσ^2 =<x^2 >−<x>^2 =

a^2 −(−a)^2 =

a^2

(b) Momentum probability density is|φ(p)|^2

φ(p)=(2π)−^1 /^2

−∞

dxψ(x)e−ipx/

=(2π)−^1 /^2

∫a

− 3 a

dxce−ipx/

ic

p

2 π

⎣e

−

ipa

 −e

3 ipa



ic

p

2 π

eipa/

⎣e

2 ipa

 −e

− 2 ipa



2 c

p

2 π

2 pa



Therefore|φ(p)|^2 = 2 πap 2 sin^2

2 pa