How Do You Derive the Propagator for the Proca Action?

[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.