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.
Answer
$D_{\mu\nu} = A g_{\mu\nu}+B k_{\mu} k _{\nu}$ 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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