Calculating Quantum Massless Photon Propagator

[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?

 

[eljose79]

I can definitely understand your confusion and difficulty in extending the polarization sum for off-shell photons. Let me try to break it down for you and hopefully it will help you in completing your calculation.

Firstly, let's define the polarization vectors \epsilon_\mu^i(p) for off-shell photons. In general, these vectors can be chosen to satisfy the following conditions:

1) Orthogonality: \epsilon_\mu^i(p) \epsilon_\nu^{i*}(p) = -\eta_{\mu\nu}

2) Transversality: p^\mu \epsilon_\mu^i(p) = 0

3) Completeness: \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}) .

Note that the third condition is the same as the one you have mentioned for on-shell photons. This is because for on-shell photons, p^2 = 0 and therefore p \cdot {\tilde p} = 0, leading to the simplification of the completeness condition.

Now, to extend this for off-shell photons, we need to use the transversality condition. Remember that for on-shell photons, we can write the completeness condition as \sum\limits_{i=1}^2 \epsilon_\mu^i(p) {\epsilon_\nu^i}^*(p) = -\eta_{\mu\nu} + \frac{p_\mu p_\nu}{p^2}. This is possible because for on-shell photons, p^\mu is a null vector and therefore p^\mu \epsilon_\mu^i(p) = 0. However, for off-shell photons, p^\mu is no longer a null vector and therefore we cannot simply write p_\mu p_\nu in place of p \cdot {\tilde p}. This is where the transversality condition comes in. Using this condition, we can show that p_\mu p_\nu = p \cdot {\tilde p} - p^2 \eta_{\mu\nu}. Sub