# Photon propagator

1. Jun 15, 2012

### praharmitra

So I want to calculate the quantum massless photon propagator. To do this, I write
$$A_\mu(x) = \sum\limits_{i=1}^2 \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \left( \epsilon_\mu^i (p) a_{p,i} e^{-i p \cdot x} + { \epsilon_\mu^i} ^* (p) a_{p,i}^\dagger e^{i p \cdot x} \right)$$
where $\epsilon_\mu^i(p),~ i = 1,2$ are appropriate basis polarizations in the gauge we choose to work in. I then calculate the propagator, which is defined as $\left< 0 \right| T \left\{ A_\mu(x) A_\nu(y) \right\} \left| 0\right>$. Using the above formula, I calculate this. I get
$$\left< 0 \right| T \left\{ A_\mu(x) A_\nu(y) \right\} \left| 0\right> = \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2 + i\epsilon} \left( \sum\limits_{i=1}^2 \epsilon_\mu^i (p) {\epsilon_\nu^i}^*(p)\right) e^{-i p \cdot (x-y)}$$
To complete this calculation, I now have to show (in the general $\xi$ gauge),
$$\sum\limits_{i=1}^2 \epsilon_\mu^i (p) {\epsilon_\nu^i}^*(p) = -\eta_{\mu\nu} + \left( 1 - \xi \right) \frac{p_\mu p_\nu}{p^2}$$
This last step is where I am getting stuck. I know that this is the classical propagator for the photon in the general $\xi$ gauge. But how do I relate that to the polarizations exactly? Also, I know that for on-shell photons, (with $p^2 = 0$), the polarization sums give
$$\sum\limits_{i=1}^2 \epsilon_\mu^i (p) {\epsilon_\nu^i}^*(p) = -\eta_{\mu\nu}+ \frac{p_\mu {\tilde p}_\nu + {\tilde p}_\mu p_\nu}{p \cdot {\tilde p}}$$
where ${\tilde p}^\mu = (p^0, - \vec{p})$. I have no clue how to extend this for off-shell photons.

Any help?