photon propagator 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?
 A: Just multiply it out in Fourier, where $\partial = ik$:
$$ \bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] \frac{ -g_{\nu\lambda} + m^{-2} k_\nu k_\lambda }{k^2 - m^2}  = \frac1{k^2 - m^2} \bigl( - (-k^2 + m^2) \delta^\mu_\lambda + (-k^2 + m^2) m^{-2} k^\mu k_\lambda - k^\mu k_\lambda + k^\mu k^2 k_\lambda m^{-2} \bigr) = \delta^\mu_\lambda$$
which is a function of $k$.  But converting back to position space, $1(k) = \delta(x)$.
This proves that $D$ is a solution.  To be the solution, you usually have to impose boundary conditions, etc.  In this case, there are no solutions to $\bigl[ (-k^2 +m^2) g^{\mu\nu} + k^\mu k^\nu\bigr] f_\nu = 0$: the corresponding equation in Fourier is $(-k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu = 0$, and contracting with $g_{\mu \nu}$ gives $0 = d(-k^2 + m^2) + k^2 = dm^2 - (d-1)k^2$, where $d$ is the dimension of spacetime, so $k^2 = \frac{d}{d-1}m^2$, but $-\frac1{d-1}m^2 g^{\mu\nu} + k^\mu k^\nu$ cannot equal $0$, as $k^\mu k^\nu$ cannot be an invertible matrix.  So $D$ is the only solution.
A: This is just because of the algebraic matrix inversion in the momentum space. It can be seen from the identity:
$(g_{\mu\nu} - \frac{k_\mu  k_\nu}{(k^2-m^2)}) (g_{\nu\lambda} - \frac{k_\nu k_\lambda}{m^2}) = g_{\mu_\lambda}$
This is a special case of a general result in linear algebra: The inverse of a matrix prportional to a weighted sum of a unit matrix and a Hermitian matrix of unit rank is also proportional to a (generally different) weighted sum of the unit matrix and the Hermitian unit rank matrix.
A: 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 $k^\mu D_{\mu \nu}(k)=0$.
