Problem in derivation of propagator of vector meson I have been reading A. Zee's book on QFT. In chapter I.5, equation (2) is given as,
$$\left[(\partial^2 + m^2) g^{\mu\nu} - \partial^\mu \partial^\nu\right] D_{\nu\lambda}(x) = \delta^\mu_\lambda \delta^4(x).$$
In order to go to momentum space, we find,
$$D_{\nu\lambda}(x) \equiv \int \frac{d^4 k}{(2\pi)^4}\, D_{\nu\lambda}(k)\, e^{ikx}.$$
Plugging this in above equation we write,
$$\left[-(k^2-m^2) g^{\mu\nu} + k^\mu k^\nu\right] D_{\nu\lambda}(k) = \delta^\mu_\lambda.$$
I am fine withupto this. After this he Zee writes,
$$D_{\nu\lambda}(k) = \frac{-g_{\nu\lambda} + k_\nu k_\lambda/m^2}{k^2 - m^2}.$$
Can someone  explain how this final step is obtained?
 A: A standard method of solving for the propagator (which can be extended to e.g. the photon propagator as well) involves taking the ansatz $D_{\nu\lambda}(k)=Ag_{\nu\lambda}+Bk_\nu k_\lambda$, since these are the only rank-2 tensors you can form from the independent variable $k^\mu$.
You insert this into the propagator equation in momentum space:
$$\left[-(k^2-m^2) g^{\mu\nu} + k^\mu k^\nu\right] (Ag_{\nu\lambda}+Bk_\nu k_\lambda) = \delta^\mu_\lambda$$
From here, it's simply a matter of expanding out the terms and solving for $A$ and $B$, which should come out to be $A = \frac{-1}{k^2-m^2}$ and $B = \frac{1}{m^2(k^2-m^2)}$.

The derivation that you have provided is incorrect since you multiply both sides by $g_{\mu\nu}$, but $\nu$ is already used as a fully contracted/dummy index (you cannot mix free and dummy indices in a tensor equation)
A: I think you went wrong in the second to last line since you're not talking about
an on-shell particle
$$
g_{\mu\nu}k^{\mu}k^{\nu}=k^2\neq m^2
$$
The most straight forward way to solve such equations is using Nihars's 'trick'.
