QuantumPhysics.dvi

(Wang) #1

The key property is that thegauge covariant derivativeDμ=∂μ+iqAμofψtransforms exactly
asψdoes under gauge transformations. Indeed, we have


Dμψ(x)→D′μψ′(x) = (∂μ+iqA′μ(x))ψ′(x) =e−iqθ(x)Dμψ(x) (22.69)

The modified Dirac equation, in the presence of the electro-magnetic field is obtained by varying
the action with respect toψ ̄, and we find,


(γμ∂μ+iqγμAμ−m)ψ= 0 (22.70)

It is very interesting to work out the non-relativistic limit of this equation. To do so, we multiply
the Dirac equation by the operator (γμ∂μ+iqγμAμ+m), to get a second order equation,
(
γμ∂μ+iqγμAμ+m


)(
γμ∂μ+iqγμAμ−m

)
ψ= 0 (22.71)

or equivalently,
(
γμγν(∂μ+iqAμ)(∂ν+iqAν)−m^2


)
ψ= 0 (22.72)

Using the decomposition of the product into symmetric and anti-symmetric tensors,


γμγν=ημν+ 2Sμν Sμν=

1

4

[γμ,γν] (22.73)

the second order equation becomes,
(
(∂μ+iqAμ)(∂μ+iqAμ) +Sμν[∂μ+iqAμ, ∂ν+iqAν]−m^2


)
ψ= 0 (22.74)

in view of the anti-symmetry inμ,νofSμν. The commutator is easily computed, and we have,


[∂μ+iqAμ, ∂ν+iqAν] =iqFμν (22.75)

Hence, the second order equation becomes,
(
(∂μ+iqAμ)(∂μ+iqAμ) +iqFμνSμν−m^2


)
ψ= 0 (22.76)

In the non-relativistic limit, we havei∂ 0 ψ= (m+E)ψwithE≪m. As a result, we have
(
2 m(E+qA 0 )−(~p+qA~)^2 +iqFμνSμν


)
ψ= 0 (22.77)

The corresponding Schr ̈odinger equation is then,
(
1
2 m


(~p+qA~)^2 −
iq
2 m

FμνSμν−qA 0

)
ψ=Eψ (22.78)

This givesB~·S~couplingμe=e/m.

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