Removing theψ†from the left andψfrom the right and dotting intoA, we have the interaction
Hamiltonian.
Hint=ieγ 4 γμAμ
Note the difference between this interaction and the one we used in the non-relativistic case. The
relativistic interaction has just one term, is linear inA, and is naturally proportional to the coupling
e. There isno longer anA^2 termwith a different power ofe. This will make our perturbation
series also a series in powers ofα.
We may still assume thatAistransverseand thatA 0 = 0 by choice of gauge.
Hint=ieγ 4 γkAk36.14Phenomena of Dirac States
36.14.1Velocity Operator and Zitterbewegung
We will work for a while in the Heisenberg representation in which the operators depend on time
and we can see some of the general behavior of electrons. If we work in a state of definite energy,
the time dependence of the operators is very simple, just the usual exponentials.
Theoperator for velocityin the x direction can be computed from the commutator with the
Hamiltonian.
x ̇=i
̄h[H,x] =i
̄hic[γ 4 γjpj,x] =icγ 4 γ 1
vj=icγ 4 γjThevelocity operatorthen isvj=icγ 4 γj.
Its not hard to compute that thevelocity eigenvalues(any component) are±c.
icγ 4 γ 1 =ic
1 0 0 0
0 1 0 0
0 0 −1 0
0 0 0 − 1
0 0 0 −i
0 0 −i 0
0 i 0 0
i 0 0 0
=c
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
c
0 0 0 1
0 0 1 0
0 1 0 0
1 0 0 0
a
b
c
d
=λc
a
b
c
d
∣
∣
∣∣
∣
∣
∣
−λ 0 0 1
0 −λ 1 0
0 1 −λ 0
1 0 0 −λ∣
∣
∣∣
∣
∣
∣
= 0
−λ[−λ(λ^2 )−1(−λ)]−1[λ(λ) + 1(1)] = 0
λ^4 − 2 λ^2 + 1 = 0
(λ^2 −1)^2 = 0