1000 Solved Problems in Modern Physics

(Grace) #1

3.3 Solutions 165


Integrating by parts twice

d
dt

<Px>=−


ψ∗

[


∂x

(Vψ)−V

∂ψ
∂x

]


=−


ψ∗

∂V

∂x

ψdτ=<

−∂V

∂x

>

These two examples support the correspondence principle as they show
that the wave packet moves like a classical particle provided the expecta-
tion value gives a good representation of the classical variable.

3.15 (a) Using the Laplacian in the time-independent Schrodinger equation


^2

2 m

[

1

r^2


∂r

(

r^2


∂r

)

+

1

r^2 sinθ

(


∂θ

(

sinθ


∂θ

)

+

1

r^2 sin^2 θ

∂^2

∂φ^2

)]

ψ(r,θ,φ)+V(r)ψ(r,θ,φ)=Eψ(r,θ,φ)(1)

We solve this equation by method of separation of variables

Letψ(r,θ,φ)=ψr(r)Y(θ,φ)(2)

Use (2) in (1) and multiply by

(

−^2 m 2 .r^2

)

/ψr(r)Y(θ,φ) and rearrange

1
ψr

(r)

d
dr

(

r^2 dψr(r)/dr

)

+

2 mr^2
^2

[E−V(r)]

=−

1

Y

(θ,φ)

[

1

sinθ


∂θ

(

sinθ


∂θ

(sinθ∂Y(θ,φ)/∂θ)+

1
sin^2 θ

∂^2 Y(θ,φ)/∂φ^2

)]

(3)

It is assumed thatV(r) depends only onr.
L.H.S. is a function of r only and R.H.S is a function ofθandφonly.
Then each side must be equal to a constant, sayλ.

1
sinθ


∂θ

(

sinθ

∂Y

∂θ

(

θ,φ

)

)

+

1

sin^2 θ

∂^2 Y

∂φ^2

(

θ,φ

)

+λY(θ,φ)=0(4)

The radial equation is

d
dr

r^2

dψr(r)
dr

+

2 mr^2
^2

[E−V(r)−λ]ψr(r)=0(5)

(b) The angular equation (4) can be further separated by substituting
Y(θ,φ)=f(θ)g(θ)(6)
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