(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?


1 Answer 1


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

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

  • $\begingroup$ Thank You for explanation! To sum up, I can just write $-ig^{\mu\nu}\Pi(f^2)/f^2$? if I use Lorentz gauge? $\endgroup$ Feb 28, 2019 at 7:34
  • $\begingroup$ @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.. $\endgroup$ Feb 28, 2019 at 8:34

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