130_notes.dvi

withλ=nr+ℓ+ 1. The equations are the same if WE set ourλ 4 =nr+1+ 2 |m| wherenr=

0 , 1 , 2 ,.... Recall that ourλ=^4 eBme ̄hc

(

E− ̄h

(^2) k 2
2 me

)

− 2 m. This gives us theenergy eigenvalues

⇒ E−

̄h^2 k^2

2 me

=

eB ̄h

mec

(

nr+

1 +|m|+m

2

)

.

As in Hydrogen, theeigenfunctionsare

G(y) =L|nmr|(y).

We can localize electrons in classical orbits for large E andnr≈0. This is the classical limit.

nr= 0 ⇒ L 0 = const ⇒ |ψ|^2 ∼e−x

2

x^2 |m|

Max when
d|ψ|^2
dx

= 0 =

(

− 2 xe−x

2

x^2 |m|+ 2|m|e−x

2

x^2 |m|−^1

)

|m|=x^2 ⇒ ρ=

(

2 c

eB

̄hm

) 1 / 2

Now let’s put in some numbers: LetB≈20 kGauss = 2× 104 Gauss. Then

ρ=

√

2

(

3 × 1010 cmsec

)

(1. 05 × 10 −^27 erg sec)

(4. 8 × 10 −^10 esu) (2× 104 g)

m≈ 2. 5 × 10 −^6

√

mcm

This can be compared to the purely classical calculation for an electron with angular momentum

m ̄hwhich givesρ=

√

m ̄hc
Be. This simple calculation neglects to count the angular momentum stored
in the field.

20.5.6 Energy States of Electrons in a Plasma II

We are going to solve the same plasma in a constant B field in a different gauge. IfA~= (0,Bx,0),
then

B~=∇×~ A~=∂Ay

∂x

zˆ=Bˆz.

ThisAgives us thesame B field. We can then compute H for a constant B field in the z direction.

H =

1

2 me

(

~p+

e

c

A~

) 2

=

1

2 me

(

p^2 x+

(

py+

eB

c

x

) 2

+p^2 z

)

=

1

2 me

(

p^2 x+p^2 y+

2 eB

c

xpy+

(

eB

c

) 2

x^2 +p^2 z

)

With this version of the same problem, we have

[H,py] = [H,pz] = 0.

We can treatpzandpyas constants of the motion and solve the problem inCartesian coordinates!
The terms inxandpyare actually a perfect square.