# Feynman-Propagator of the gauge-fixed electromagnetic field

I want to find the Feynman propagator for the so called $$R_\zeta$$ gauge fixed electromagnetic field. The lagrangian density is given by: \begin{align} L = -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}-\frac{1}{2\zeta}(\partial_\mu A^\mu)^2. \end{align} The equations of motion are therefore \begin{align} \partial_\mu F^{\mu\nu}+\frac{1}{\zeta}\partial_\mu\partial_\lambda A^\lambda g^{\mu\nu} = 0 \end{align} which can be formulated as the operator equation \begin{align} 0 = [g^{\nu\lambda}\square-(1-\frac{1}{\zeta})\partial^\nu\partial^\lambda]A_\lambda =:O^{\nu\lambda}A_\lambda \end{align}

The feynman propagator $$D_{\nu\rho}(x-y)$$ is defined as the greens function \begin{align} O^{\mu\nu}D_{\nu\rho}(x-y)=-ig^\nu_\rho\delta(x-y) \end{align} I now used a fourier ansatz to obtain \begin{align} -k^2g^{\mu\nu}\tilde{D}_{\mu\rho}+(1-\frac{1}{\zeta})k^\mu k^\nu\tilde{D}_{\nu\rho}+ig^\mu_\rho=0 \end{align} which can be solved for the fourier transformed propagator \begin{align} \tilde{D}_{\nu\rho}(k)=\frac{ig_{\nu\rho}}{k^2(3+\frac{1}{\zeta})} \end{align}

I am slightly confused by a hint at the exercise, whichs says that I should use the ansatz \begin{align} \tilde{D}_{\nu\rho} = A(k^2)k_\nu k_\rho+B(k^2)g_{\nu\rho} \end{align} and I didn't use this ansatz at all. Of course for $$A=0$$, $$B=\frac{i}{k^2(3+\frac{1}{\zeta})}$$ I get exactly what I found for $$\tilde{D}_{\nu\rho}$$ but I cant figure out what the author wants me to do with this information. Furthermore I found this post, where it is mentioned, that the Propagator is \begin{align} \tilde{D}_{\mu\nu} = -\frac{1}{k^2+i\epsilon}[g_{\mu\nu}-(1-\zeta)\frac{k_\mu k_\nu}{k^2}] \end{align} which does not agree with what I got. (They might use a different convention $$\zeta \rightarrow \frac{1}{\zeta}$$ here)

• "which can be solved for the fourier transformed propagator ..." not sure how you did this step, but if you check whether your expression is a solution of the desired equation you'll find that it is not. You can also check whether the other expression is the correct propagator or not. Feb 6 at 15:29

I have studied this problem myself, not too long ago. Here is how I understood it. We are searching for an "inverse" for the equation of motion operator, what you call $$O^{\mu\nu}$$. Writing this in momentum space makes things much easier, where we obtain the equation $$-k^2g^{\mu\nu}D_{\nu\rho} + (1- \frac 1 \xi)k^\mu k^\nu D_{\nu\rho} +ig^\mu_\nu=0$$. What you likely did was try to factor out the gauge boson propagator $$D$$, which cannot be done, as both factors of $$D$$ have different indices. The best thing to do is to find the most general Lorentz invariant ansatz, and narrow it down using the equation of motion. The most general Lorentz invariant ansatz is the one you provided. Plug it into the equation of motion, and match up factors of $$k$$ to find the functions $$A$$ and $$B$$.
• I tried this at first but my problem was, that i couldnt make the term $ig^\mu_\rho$ to vanish. Feb 7 at 9:44
• I end up with this equation: \begin{align} -\frac{1}{\zeta}k^\mu k_\rho k^2A+(1-\frac{1}{\zeta})k^\mu k_\rho B-k^2g^\mu_\rho B+ig^\mu_\rho = 0 \end{align} I would compare orders of $k$ here but its strange since the imaginary term never vanishes for $\mu = \rho$ Feb 8 at 13:18
• Compare the terms with the factor $g^\mu_\rho$. You should find $B = \frac i {k^2}$ Feb 8 at 14:33
• Thanks for you answer this helped me out! My mistake was, that I didn't realize that $B=B(k^2)$ also means, that $B=B(k^{-2})$ and therefore i got confused about the units. But still $A=A(k^{-4})$ and not a function of $k^2$, so the instructions are wrong in my opinion Feb 9 at 8:21