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In his book, the author says that according to enter image description here

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?

PhilosophicalPhysics
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  • Related, possible duplicate http://physics.stackexchange.com/q/70882/ – innisfree Mar 08 '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 08 '15 at 17:47

1 Answers1

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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

Ali Moh
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  • 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 08 '15 at 23:26
  • He is probably emphasizing that ward identity holds for the amplitude, and not necessarily for each of the feynman diagrams whose sum is the amplitude.. – Ali Moh Mar 08 '15 at 23:33
  • By the way, Ali, you said that this Wrd was a consequence of gauge invariance. I am interested to know how is it that this is a consequence of that. If you have any reference you could link me to, I would appreciate it. – PhilosophicalPhysics Mar 09 '15 at 01:04
  • 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 09 '15 at 02:41