130_notes.dvi

We have found aconserved current. Some interpretation will be required as we learn more about
the solutions to the Dirac equation and ultimately quantize it. We may choose an overall constant
toset the normalization. The fourth component of the current should beictimes the probability
density so that the derivative with respect tox 4 turns into∂P∂t. Therefore let us set theproperly
normalized conserved 4-vectorto be

jμ=icψγ ̄μψ.

36.5 The Non-relativistic Limit of the Dirac Equation

Oneimportant requirement for the Dirac equation is that it reproduces what we know
from non-relativistic quantum mechanics. Note that we have derived this equation from
something that did give the right answers so we expect the Dirac equation to pass this test. Perhaps
we will learn something new though.

We know that our non-relativistic Quantum Mechanics only needed a two component spinor. We
can show that, in the non-relativistic limit, two components of the Dirac spinor are large and two are
quite small. To do this, we go back to the equations written in terms ofψAandψB, just prior to the
introduction of theγmatrices. We make the substitution to put the couplings to the electromagnetic
field into the Hamiltonian.

36.5.1 The Two Component Dirac Equation

First, we can write thetwo component equation that is equivalent to the Dirac equation.
Assume that the solution has theusual time dependencee−iEt/ ̄h. We start from the equation in
ψAandψB.
(
−i ̄h∂x∂ 0 −i ̄h~σ·∇~
i ̄h~σ·∇~ i ̄h∂x∂ 0

)(

ψA

ψB

)

+mc

(

ψA

ψB

)

= 0

(

−Ec ~σ·~p

−~σ·~p Ec

)(

ψA

ψB

)

+mc

(

ψA

ψB

)

= 0

Turn on the EM field by making the usual substitution~p→~p+ecA~and adding the scalar potential
term.



−^1 c(E+eA 0 −mc^2 ) ~σ·

(

~p+ecA~

)

−~σ·

(

~p+ecA~

)

1

c(E+eA^0 +mc

(^2) )





(

ψA

ψB

)

= 0

1

c

(E+eA 0 −mc^2 )ψA=~σ·

(

~p+

e

c

A~

)

ψB

1

c

(E+eA 0 +mc^2 )ψB=~σ·

(

~p+

e

c

A~

)

ψA

These two equations can be turned into one by eliminatingψB.

1

c

(E+eA 0 −mc^2 )ψA=~σ·

(

~p+

e

c

A~

) c

(E+eA 0 +mc^2 )

~σ·

(

~p+

e

c

A~

)

ψA