# 1-loop correction to photon propagator

(May be it is a duplicate). I do not understand clearly how should I write down 1-loop correction to photon propagator. I know what is $$i\Pi_{\mu\nu}(k^2)$$ (I need only this specific correction) and write down the following expression: $$D_{(1)}^{\mu\nu}=D_{\mu\alpha}(i\Pi^{\alpha\beta})D_{\beta\nu}=-\frac{i}{f^4}\left[f^2g^{\mu\nu}-f^{\mu}f^{\nu}\right]\Pi(f^2).$$

My question: is it true or $$f^{\mu}f^{\nu}$$ term vanishes with help of Ward identity?

Actually, the Ward identity brings out the main complication with the photon propagator.

Our whole goal is to solve the matrix equation

$$(\eta_{\mu\nu} \partial^2 - \partial_{\mu}\partial_\nu)\Delta^{\nu\lambda} = \delta_\mu^\lambda \delta^4(x-y). \tag{a}$$

The usual way to do this is to go to Fourier space and invert the matrix

$$\Omega_{\mu\nu}(k) : = -(k^2 \eta_{\mu\nu} - k_\mu k_\nu)$$

However the ward identity implies that

$$\Omega_{\mu\nu}k^\nu = 0$$

which implies the existence of a nontrivial eigenvector $$k^\nu$$ with eigenvalue $$0$$ which implies that $$\Omega_{\mu\nu}$$ is not invertible in Fourier space.

We can eliminate the second term if we fix the gauge using the Lorentz condition $$\partial_\mu A^\nu = 0 \stackrel{\mathcal{F}}{\longmapsto}k^\mu A_\mu = 0$$. If we plug this into equation $$(a)$$ we catch a term like

$$\sim \partial_\mu \partial_\nu \Delta^{\nu\lambda} = \partial_\mu\langle 0| A^\lambda (\partial_\nu A^\nu)| 0 \rangle = 0$$

leaving behind just

$$\boxed{\eta_{\mu\nu} \partial^2 \Delta^{\nu\lambda} = \delta_\mu^\lambda \delta^4(x-y)}.$$

• Thank You for explanation! To sum up, I can just write $-ig^{\mu\nu}\Pi(f^2)/f^2$? if I use Lorentz gauge? Feb 28, 2019 at 7:34
• @ArtemAlexandrov yeap that’s right.. you can also do it by adding a gauge fixing term in the Lagrangian then it just comes out of the EOM for the gauge field.. Feb 28, 2019 at 8:34