130_notes.dvi

Hamiltonian and will give constants of the motion. We therefore will beable to separate variables
in the usual way.
ψ(~r) =unmk(ρ)eimφeikz

Insolving(see section 20.5.5) the equation inρwe may reuse the Hydrogen solution ultimately
get the energies

E=

eB ̄h

mec

(

n+

1 +m+|m|

2

)

+

̄h^2 k^2

2 m

and associated LaGuerre polynomials (as in Hydrogen) inρ^2 (instead ofr).

The solution turns out to besimpler using the Hamiltonian written in terms of (see section
20.5.6) A~if we choose the right gauge by settingA~=Bxyˆ.

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

)

This Hamiltonian does not depend onyorzand therefore hastranslational symmetry in both
x and yso their conjugate momenta are conserved. We can use this symmetry to write the solution
and reduce to a 1D equation inv(x).

ψ=v(x)eikyyeikzz

Then we actually can use our harmonic oscillator solution instead of hydrogen! The energies come
out to be

En=

eB ̄h

mec

(

n+

1

2

)

+

̄h^2 k^2

2 me

Neglecting the free particle behavior inz, these are called theLandau Levels.This is an example
of the equivalence of the two real problems we know how to solve.

20.5 Derivations and Computations

20.5.1 Deriving Maxwell’s Equations for the Potentials

We take Maxwell’s equations and the fields written in terms of the potentials as input. In the left
column the equations are given in the standard form while the right column gives the equivalent
equation in terms of indexed components. The right column uses thetotally antisymmetric tensor
in 3Dǫijkand assumes summation over repeated indices (Einstein notaton). So in this notation,
dot products can be simply written as~a·~b=aibiand any component of a cross product is written
(~a×~b)k=aibjǫijk.

∇·~ B~= 0 ∂

∂xi

Bi= 0

~∇×E~+^1

c

∂B

∂t

= 0

∂

∂xi

Ejǫijk+

1

c

∂Bk

∂t

= 0

∇·~ E~= 4πρ ∂

∂xk

Ek= 4πρ