TITLE: photon propagator
QUESTION [6 upvotes]: I am reading Zee's book "QFT in a nutshell". I have a question on the photon propagator computation. For a massive photon, consider the Lagrangian
 $L = -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{2}m^2A_\mu A^\mu + A_\mu J^\mu$, then the path integral is $Z = \int dx ~L = \int dx ~\{ \frac{1}{2}A_\mu[(\partial^2 + m^2)g^{\mu \nu} - \partial^\mu \partial^\nu]A_\nu +  A_\mu J^\mu \}$. From this we get that the photon propagator $D_{\mu \nu}$ satisfies $[(\partial^2 + m^2)g^{\mu \nu} -\partial^\mu \partial^\nu ] D_{\nu \lambda}(x) = \delta^\mu_\lambda \delta^{(4)}(x)$, and solving this, 
$$D_{\nu \lambda}(k) = \frac{-g_{\nu \lambda} + k_\nu k_\lambda/m^2}{k^2 - m^2}.$$ 
I can not see why the numerator has a term  $ k_\nu k_\lambda/m^2$. Any ideas?

REPLY [3 votes]: Theo and David are perfectly correct. To add a bit more of a physical explanation which
might help with the why part, a massive spin one particle has 3 physical degrees of freedom
so there must be some condition on the four components $A_\mu$. The equation of motion
for $A_\mu$ is equivalent to saying that each component of $A_\mu $ satisfies the massive
Klein-Gordon equation and that in addition $\partial^\mu A_\mu=0$. This latter condition in
momentum space implies that $k^\mu D_{\mu \nu}(k)=0$. So one can understand the $1/(k^2-m^2)$ from each component obeying the massive KG equation and the factor in the numerator as ensuring that $k^\mu D_{\mu \nu}(k)=0$.