# Four-photon polarisation tensor in QED

Let's consider the four-photon polarisation tensor $\Pi^{\mu\nu\lambda\rho}$ in QED. It follows from Ward identity that $$k_1^\mu \Pi_{\mu\nu\lambda\rho}(k_1,k_2,k_3,k_4) = 0.$$ After applying the derivative $\frac{\partial}{\partial k_1^\sigma}$ to both sides of the above equality we obtain $$\Pi_{\sigma\nu\lambda\rho}(k_1,k_2,k_3,k_4) = -k_1^\mu \frac{\partial}{\partial k_1^\sigma}\Pi_{\mu\nu\lambda\rho}(k_1,k_2,k_3,k_4).$$ Thus, $\Pi_{\sigma\nu\lambda\rho}$ vanishes for $k_1=k_2=k_3=k_4=0$.

I cannot reproduce the above result by direct calculaition in one-loop approximation. There are six contributing diagrams in this order, all with one fermionic loop. The three depicted below and another three witch differ only in the arrow direction (they give the same integrals as the diagrams above). I'm interested only in a value of the polarisation tensor for vanishing external momenta of photons. We have $$\Pi^{\mu\nu\lambda\rho} = T^{\mu\nu\lambda\rho} + T^{\mu\nu\rho\lambda} + T^{\mu\lambda\nu\rho},$$ where $$T^{\mu\nu\lambda\rho}=2(-ie)^4 i^4 \int\frac{d^4 p}{(2\pi)^4}\frac{\text{tr}\left((m+\gamma \cdot p)\gamma ^{\mu }(m+\gamma \cdot p)\gamma ^{\nu }(m+\gamma \cdot p)\gamma ^{\lambda }(m+\gamma \cdot p)\gamma ^{\rho }\right)}{(p^2-m^2+i0)^4}.$$ The integral in the definition of $T^{\mu\nu\lambda\rho}$ is logarithmically divergent but the divergent contribution to $\Pi^{\mu\nu\lambda\rho}$ vanishes. Finally I get finite answer for $\Pi_{\sigma\nu\lambda\rho}(k_1=k_2=k_3=k_4=0)$ which is unfortunatelly different from $0$?

Does $\Pi_{\sigma\nu\lambda\rho}(k_1=k_2=k_3=k_4=0)$ really vanish?

Yes, at $k_1 = k_2 = k_3 = k_4 = 0$, the amplitude vanishes. And, I think this can be understood in terms of F. Low's soft photon theorem.
Use LoopIntegrate to initiate the computation of the Feynman integrals:
Then after replacing all external momenta to zero, use LoopRefine to obtain an explicit analytic representation of that integral: