Massive vector (Proca) propagator

[CompuChip]

Hi all,
I'm stuck with this following problem:

Homework Statement

Consider the Proca action,
$$S[A_\mu] = \int \, \mathrm d^4x \left[ - \frac14 F_{\mu\nu} F^{\mu\nu} + \frac12 m^2 A_\mu A^\mu \right]$$
where $F_{\mu\nu} = 2 \partial_{[\mu} A_{\nu]}$ is the anti-symmetric electromagnetic
field tensor.

Derive the propagator for the vector field $A_\mu$.

Homework Equations

I did a Fourier transform to get
$$\left[ (- k^2 + m^2) g^{\mu\nu} + k^\mu k^\nu \right] \tilde D_{\nu\lambda}(k) = \delta^\mu_\lambda.$$ (*)

Zee's book on QFT gives the result on page 13, as if it were trivial, but I can't do the calculation (satisfactorily).

The Attempt at a Solution

I tried to follow the hint in the question: "the calculation involves deriving an identity for $k^\nu \tilde D_{\nu\mu}$".
I contracted (*) with $k_\mu$ which got me
$$k^\nu \tilde D_{\nu\lambda} = k_\lambda$$
or (contracting with $k^\lambda$)
$$k^\lambda k^\nu D_{\nu\lambda} = k^2$$
but I still didn't really see how to solve for $\tilde D_{\nu\lambda}$.

 

[George Jones]

I get

$$k^\nu \tilde D_{\nu\lambda} = \frac{k_\lambda}{m^2},$$

and then I think everything works out okay.

 

[CompuChip]

Thanks, I'll check that calculation.
My problem was how to extract the propagator from that contraction, though.

Anyway, let me get some sleep now, as it's 1:30

 

[George Jones]

Thanks, I'll check that calculation.
My problem was how to extract the propagator from that contraction, though.

Anyway, let me get some sleep now, as it's 1:30

Substitute the identity and then contract with the metric.

 

[CompuChip]

I checked my earlier calculation and the 1/m^2 missing was just a typo.
Also, I see what you mean now and it turns out to be quite easy indeed.

Thank you very much George!

 

[Helgi]

I am having the same problem.

Could you elaborate on what you mean by substituting the identity?

Edit: scratch that. I figured it out.