  Consider M\o{}ller scattering. Electrons with momentum $p$ and $-p$ scatter
by exchange of photon say in $z$ direction to $p+q$ and $-(p+q)$. The
scattering amplitude is well known, given as Feynman propagator $ \M = \frac{(e
\hbar c)^2}{\epsilon_0 V} \frac{\bar{u}(p+q) \gamma^{\mu} u(p) \
\bar{u}(-(p+q)) \gamma_{\mu} u(-p)}{q^2}$, where $V$ is the volume of the
scattering electrons, $e$ elementary charge and $\epsilon_0$ permitivity of
vacuum. But this is not completely correct. Since we exchange photon momentum
in $z$ direction, we have two photon polarization $x,y$ and hence the true
scattering amplitude should be $$ \M_1 = \frac{(e \hbar c)^2}{\epsilon_0 V}
\frac{ \bar{u}(p+q) \gamma^{x} u(p) \ \bar{u}(-(p+q)) \gamma_{x} u(-p)\ \ +
\bar{u}(p+q) \gamma^{y} u(p) \ \bar{u}(-(p+q)) \gamma_{y} u(-p) \ }{q^2}. $$
But when electrons are non-relativistic, $\M_1 \sim 0$. This is disturbing, how
will we ever get the coulomb potential, where $\M \sim \frac{(e \hbar
c)^2}{\epsilon_0 V q^2}$. Where is the problem ? The problem is with the gauge
in Dirac equation.
  For a plane wave along $z$ direction, with electric field $E_x \sin (kz -
\omega t)$, the Lorentz gauge is $$ (A_0, A_x, A_y, A_z) = \frac{E_x}{\omega}
\cos(kz-\omega t)(0, 1, 0, 0)$$. But this gauge is not suited for calculating
optical transitions, because we don't recover the Rabi frequency $q E_x d$ ($d$
electric dipole moment). What we find is something orders of magnitude smaller.
Nor is it suitable for calculating electron electron scattering because we
don't recover Coulomb potential. What we find is something orders of magnitude
smaller. Instead, we work with $\E \cdot x$ gauge $$ (A_0, A_x, A_y, A_z) =
\frac{-E_x}{2} ( x\ \sin(kz-\omega t), -\frac{\cos(kz-\omega t)}{\omega}, 0,
\frac{x}{c} \sin(kz-\omega t) ) $$ ($c$ light velocity) to find everything
correct. What we get is new propagator.
