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In Peskin & Schroeder's book on page 297 in deriving the photon propagator the authors say that

$$\left(-k^2g_{\mu\nu}+(1-\frac{1}{\xi})k_\mu k_\nu\right)D^{\nu\rho}_F(k)=i\delta^\rho_\mu \tag{9.57b}$$

With the solution given in the next line in equation (9.58) as

$$D^{\mu\nu}_F(k)=\frac{-i}{k^2+i\epsilon}\left(g^{\mu\nu}-(1-\xi) \frac{k^\mu k^\nu}{k^2}\right)\tag{9.58}$$

Which is the propagator. I can verify this equation by inserting $D^{\mu\nu}_F(k)$ into the first equation, but I have no idea how to actually solve $D^{\nu\rho}_F(k)$ from $(9.57b)$. If anyone can help, it would be much appreciated.

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2 Answers 2

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$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ with A and B two unknown functions of the scalar k^2. The two tensor after A and B are the only possible Lorentz invariant tensors . Simply plugin and calculate the unknown functions.

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    $\begingroup$ So in general I make an ansatz of all the possible Lorentz invariant terms that appear in the matrix I wish to invert? $\endgroup$
    – Apogee
    Commented Sep 28, 2014 at 13:28
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    $\begingroup$ Yes, that's it. $\endgroup$
    – Jasper
    Commented Sep 28, 2014 at 14:19
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It's just tensor equation reads: $A_{\mu\nu}D^{\nu\rho}=i\delta_\mu{}^{\rho}$, where $A_{\mu\nu}=-k^2g_{\mu\nu}+\left(1-\frac{1}{\xi}\right)k_\mu k_\nu$. What we need to do it to find its Inverse $A^{\mu\nu}$. Certainly you can find it by brute force with the methods given in linear algebra, but it is easier to guess the answer.

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