
#1
Jun1512, 08:40 PM

P: 311

So I want to calculate the quantum massless photon propagator. To do this, I write
[tex] 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) [/tex] where [itex]\epsilon_\mu^i(p),~ i = 1,2[/itex] are appropriate basis polarizations in the gauge we choose to work in. I then calculate the propagator, which is defined as [itex] \left< 0 \right T \left\{ A_\mu(x) A_\nu(y) \right\} \left 0\right>[/itex]. Using the above formula, I calculate this. I get [tex] \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 (xy)} [/tex] To complete this calculation, I now have to show (in the general [itex]\xi[/itex] gauge), [tex] \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} [/tex] This last step is where I am getting stuck. I know that this is the classical propagator for the photon in the general [itex]\xi[/itex] gauge. But how do I relate that to the polarizations exactly? Also, I know that for onshell photons, (with [itex]p^2 = 0[/itex]), the polarization sums give [tex] \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}} [/tex] where [itex] {\tilde p}^\mu = (p^0,  \vec{p}) [/itex]. I have no clue how to extend this for offshell photons. Any help? 


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