# Photon propagator

by praharmitra
Tags: photon-propagator, quantum-field-theory
 P: 311 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?

 Related Discussions Special & General Relativity 2 Quantum Physics 2 Quantum Physics 12 Quantum Physics 5 High Energy, Nuclear, Particle Physics 4