# Gauge invariance (QED)

In his book, the author says that according to the Feynman diagrams of this process in QED $$e^+ e^- \rightarrow \gamma \gamma,$$ gauge invariance requires that $$k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0=k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu}).$$ My question is how does gauge invariance set this statement equal to zero?

• Related, possible duplicate physics.stackexchange.com/q/70882 – innisfree Mar 8 '15 at 17:21
• @innisfree if the amplitude in this process is $$M=\epsilon_{1\nu}^*\epsilon_{2\nu}^*(A^{\mu\nu} +\tilde{A}^{\mu\nu})$$ are you telling me that this applies to what drake said in the post you linked me to that$k_\mu M^\mu=0$? If so, can you please elaborate for I can't see the analogy completely. – PhilosophicalPhysics Mar 8 '15 at 17:47

The amplitude for emission of n photons with polarizations $\epsilon_{\mu_j}$ written as $\epsilon_{\mu_1}\ldots \epsilon_{\mu_n}\mathcal{M}^{\mu_1 \ldots \mu_n}$ satisfies $k_{\mu_1} \mathcal{M}^{\mu_1 \ldots \mu_n} = k_{\mu_2} \mathcal{M}^{\mu_1 \ldots \mu_n} = \ldots =0$ due to gauge invariance (do you know this fact? This is a consequence of ward identity & abelian gauge symmetry)
For your process of two photon emission $\mathcal{M}^{\mu_1 \mu_2}\equiv A^{\mu_1 \mu_2}+\tilde{A}^{\mu_1 \mu_2}$. So it should become clear
• Thank you Ali for your reply. But the author insisted that these conditions are met although the quantities in the equation $k_{1\nu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})=0=k_{2\mu}(A^{\mu\nu} + \tilde{A}^{\mu\nu})$ each separately are all different from zero. Why would he say that if it is already a consequence of ward identity? – PhilosophicalPhysics Mar 8 '15 at 23:26
• intuitively the amplitude should be invariant under $\epsilon_\mu (p)\rightarrow \epsilon_\mu (p) + \alpha p_\mu$ because this represents a change in the longitudinal mode of the photon. See Srednicki Ch 67, or Zee II.7 or Weinberg 10.5 – Ali Moh Mar 9 '15 at 2:41