The Photon Propagator

The propagator for a photon is not unique, on account of the freedom in the choice of $A^\mu$. From

$$\Box A^\nu - \partial^\nu(\partial_\lambda A^\lambda) = j^\nu ,$$ (7.113)

we see that the wave equation for a photon can be written in the form

$$(g^{\nu\lambda}\Box - \partial^\nu\partial^\lambda) A_\lambda = j^\nu .$$ (7.114)

In fact, a photon propagator cannot exist until we remove some of the gauge freedom of $A_\lambda$, i.e. the inverse of the ``momentum space operator'' does not exist.

If we chose to work in the Lorentz class of gauges with $\partial_\lambda A^\lambda=0$, the wave equation simplifies to

$$g^{\nu\lambda} \Box A_\lambda = j^\nu .$$ (7.115)

Since $g_{\mu\nu} g^{\nu\lambda} = \delta_\mu^{\ \lambda}$, the propagator (the inverse of the momentum space operator multiplied by $-i$) is

$$\frac{-i g_{\mu\nu}}{k^2} .$$ (7.116)

We are using Heaviside-Lorentz units. If Gaussian units were chosen, the expression would be multiplied by $4\pi$.

The gauge condition $\partial_\lambda A^\lambda = 0$ has been imposed covariantly, and the resulting covariant propagator is ideal for QED calculations.

The condition $\partial_\lambda A^\lambda = 0$ does not fully define the propagator. We are at liberty to rewrite the wave equation as

$$\left[ g^{\nu\lambda} \Box - \left( 1 - \frac{1}{\zeta} \right) \partial^\nu \partial^\lambda \right] A_\lambda = j^\nu .$$ (7.117)

No gauge has yet been chosen in this wave equation. In this case, the propagator is

$$\frac{i}{k^2} \left( -g_{\mu\nu} + (1-\zeta) \frac{k_\mu k_\nu}{k^2} \right) .$$ (7.118)

$\zeta = 1$ is referred to ambiguously as the Feynman gauge. It is really the Lorentz gauge. $\zeta = 0$ is referred to as the Landau gauge. This gauge appears more complicated but the extra term in the propagator vanishes in QED calculations, in which the virtual photon is coupled to conserved currents which satisfy $k_\mu j^\mu = k_\nu j^\nu = 0$.