# How do non-transverse photon polarizations cancel in Euclidean QED?

First, recall how to write scattering amplitudes in covariant fashion in Minkowskian QED.

One starts by considering some process with an external photon whose momentum is chosen to be $k^\mu=(k,0,0,k)$ and let the two transverse polarization vectors be $\epsilon^\mu_1=(0,1,0,0)$ and $\epsilon_2^\mu=(0,0,1,0)$. The Ward identity tell us that if the amplitude for the process is $\mathcal M=\mathcal M^\mu \epsilon_\mu(k)$, where we've factored out the polarization vector for the external photon in consideration, then the amplitude obeys $\mathcal M^\mu k_\mu=0$, on shell. With our setup, this simply tells us $\mathcal M^0=\mathcal M^3$. If we then calculate the square of the amplitude and sum over the physical external polarizations we'd find $|\mathcal M|^2=\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}^*\mathcal M^μ\mathcal M^{ν*}=|\mathcal M^1|^2+|\mathcal M^2|^2$. Due to the Ward identity, this is equal to $−\eta_{\mu\nu}\mathcal M^\mu\mathcal M^\nu$, with $\eta_{\mu\nu}={\rm diag}(1,-1,-1,-1)$, and so we can make the replacement $\sum_{i\in\{1,2\}}\epsilon_{i\mu}\epsilon_{i\nu}^*\to-\eta_{\mu\nu}$ and so it's the Ward identity which allows us to write scattering amplitudes covariantly.

How does this generalize to the Euclidean case where the metric is $\delta_{\mu\nu}={\rm diag}(1,1,1,1)$? Naively, we need the non-Euclidean signature in order to reproduce the cancellation between the manifestly positive squares of matrix elements as found above so how does this procedure generalize to the Euclidean case? That is, if the true scattering amplitude in the Euclidean case is still $|\mathcal M^1|^2+|\mathcal M^2|^2$ then it seems that this can't be equivalent to $\delta_{\mu\nu}\mathcal M^\mu\mathcal M^{\nu *}$ for any Ward identity type relations between the $\mathcal M^\mu$'s and so it wouldn't be possible to write the scattering amplitude in a manifestly covariant fashion.

What's going on? Am I missing something obvious? Do the photon's degrees of freedom change when going to Euclidean space or something?

Well, you could have invented a simpler "paradox" of this kind: there are no on-shell photon (i.e. null) momenta in the Euclidean spacetime except for $(0,0,0,0)$. But this doesn't prevent us from defining the path integral and calculating the correlation functions etc. and continue them back to the Minkowski space.
More precisely, there are complex momenta in the Euclidean spacetime for which everything works just fine. They're obtained by the simple $t\to i\tau$ maps. So the null momentum in the Euclidean spacetime may be formally $(ik,0,0,k)$ and one may similarly write the polarization vectors that only differ from the Minkowskian one by an extra $i$ in the time component, and all the identities are obeyed just like in the Minkowski space – because $1^2+(i)^2=0$.