130_notes.dvi

From the symmetry of the problem, we can guess (and verify) that[H,pz] = [H,Lz] = 0. These
variables will be constants of the motion and we therefore choose

ψ(~r) = umk(ρ)eimφeikz.

Lzψ =

̄h

i

∂

∂φ

ψ=m ̄hψ

pzψ =

̄h

i

∂

∂z

ψ= ̄hkψ

∇^2 ψ = −k^2 ψ−

m^2

ρ^2

ψ+

∂^2 u

∂ρ^2

eimφeikz+

1

ρ

∂u

∂ρ

eimφeikz

d^2 u

dρ^2

+

1

ρ

du

dρ

−

m^2

ρ^2

u−

e^2 B^2

4 ̄h^2 c^2

ρ^2 u+

(

2 meE

̄h^2

−

eBm

̄hc

−k^2

)

u= 0

Letx=

√

eB

2 ̄hcρ(dummy variable, not the coordinate) andλ=

4 mec

eB ̄h

(

E− ̄h

(^2) k 2
2 me

)

− 2 m. Then

d^2 u

dx^2

+

1

x

du

dx

−

m^2

x^2

u−x^2 u+λ= 0

In the limitx→∞,
d^2 u
dx^2

−x^2 u= 0 ⇒ u∼e−x

(^2) / 2
while in the other limitx→0,
d^2 u
dx^2

+

1

x

du

dx

−

m^2

x^2

u= 0

Try a solution of the formxs. Then

s(s−1)xs−^2 +sxs−^2 −m^2 xs−^2 = 0 ⇒ s^2 =m^2

A well behaved function⇒s≥ 0 ⇒s=|m|

u(x) =x|m|e−x

(^2) / 2
G(x)
Plugging this in, we have
d^2 G
dx^2

+

(

2 |m|+ 1

x

− 2 x

)

dG

dx

+ (λ− 2 − 2 |m|)G= 0

We canturn this into the hydrogen equationfor

y=x^2

and hence
dy= 2x dx
d
dy

=

1

2 x

d

dx

.

Transforming the equation we get

d^2 G

dy^2

+

(

|m|+ 1

y

− 1

)

dG

dy

+

λ− 2 − 2 |m|

4 y

G= 0.

Compare this to the equation we had for hydrogen

d^2 H

dρ^2

+

(

2 ℓ+ 2

ρ

− 1

)

dH

dρ

+

λ− 1 −ℓ

ρ

H= 0