130_notes.dvi

Q.E.D.

So we have
Fi=−eEi−

e

c

(

~v×B~

)

i
which is the Lorentz force law. So this is the right Hamiltonian for an electron in a electromagnetic
field. We now need to quantize it.

20.5.3 The Hamiltonian in terms of B

Start with theHamiltonian

H=

1

2 μ

(

~p+

e

c

A~

) 2

−eφ

Now write theSchr ̈odinger equation.

1

2 μ

(

̄h

i

∇~+e

c

A~

)

·

(

̄h

i

∇~ψ+e

c

Aψ~

)

= (E+eφ)ψ

− ̄h^2

2 μ

∇^2 ψ−

ie ̄h

2 μc

∇·~

(

Aψ~

)

−

ie ̄h

2 μc

A~·∇~ψ+ e

2

2 mc^2

A^2 ψ = (E+eφ)ψ

− ̄h^2

2 μ

∇^2 ψ−

ie ̄h

2 μc

(

∇·~ A~

)

ψ−

ie ̄h

μc

A~·∇~ψ+ e

2

2 mc^2

A^2 ψ = (E+eφ)ψ

The second term vanishes in theCoulomb gaugei.e.,∇·~ A~= 0, so

−

̄h^2

2 μ

∇^2 ψ−

ie ̄h

μc

A~·∇~ψ+ e

2

2 mc^2

A^2 ψ= (E+eφ)ψ

Now for constantBz, wechoose the vector potential

A~=−^1

2

~r×B~

since
(
∇×~ A~

)

k

=

∂

∂xi

Ajεijk=−

1

2

∂

∂xi

(xmBnεmnj)εijk

= −

1

2

δimBnεmnjεijk=−

1

2

Bnεinjεijk

=

1

2

Bn





∑

i

∑

j

εijnεijk



=^1

2

Bk





∑

i

∑

j

ε^2 ijk



=Bk

it gives the right field and satisfies the Coulomb gauge condition.

Substituting back, we obtain

− ̄h^2

2 μ

∇^2 ψ+

ie ̄h

2 μc

~r×B~·∇~ψ+

e^2

8 mc^2

(

~r×B~

) 2

ψ= (E+eφ)ψ

Now let’s work on thevector arithmetic.
(
~r×B~·∇~ψ

)

=riBjεijk

∂ψ

∂xk

=−Bj

(

ri

∂ψ

∂xk

εikj

)

=−B~·~r×∇~ψ=−

i

̄h

B~·~Lψ