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The annihilation ($e^+ e^-\rightarrow 2\gamma$) diagrams are Annihilation (time goes upwards)

I'm just wondering if the amplitudes should be

$$(-ie)^2\left[\bar v(p_2)\gamma^b\epsilon_b^*\frac{-i(-(\gamma^\mu p_{1\mu}-\gamma^\nu p_{3\nu})+m)}{(p_1-p_3)^2+m^2-i\epsilon}\gamma^a\epsilon_a^* u(p_1)+ \\ \bar v(p_2)\gamma^a\epsilon_a^*\frac{-i(-(\gamma^\mu p_{1\mu}-\gamma^\nu p_{4\nu})+m)}{(p_1-p_4)^2+m^2-i\epsilon}\gamma^b\epsilon_b^* u(p_1)\right]$$

In particular, are the polarization vectors for outgoing photons correct? In general, do external photons always give a polarisation vector and an $\gamma^a$ contracted together? Also, if $\epsilon^a\rightarrow\epsilon^a+\alpha k^a$ for a constant $\alpha$, why is the amplitude unchanged and what does this mean physically?

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    $\begingroup$ Your expression for matrix element is correct. The external photon line always comes with $\epsilon_\mu^*$ and the factor $\gamma^\mu$ comes from vertex factor. Change in polarization by $K^a$ which does not affect amplitude implies that photon are transverse in nature (real photon). $\endgroup$ – user44895 May 16 '14 at 18:50
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For any diagram all the Lorentz indices must be contracted. If there exists a $\epsilon_\mu$ then its index must be contracted with a vertex somewhere in the diagram. In Quantum Electrodynamics the only objects that can carry an explicit Lorentz indices are the $\gamma_\mu$ matrices. Thus every $\epsilon_a$ will correspond to one $\gamma_a$.

The amplitude that you wrote down is almost correct. You have the wrong sign in the denominator for the propagators.

Lastly, the reason the Ward identity (the vanishing amplitude under $\epsilon_{3,4} \rightarrow p_{3,4}$) holds is due to gauge invariance. The QED Lagrangian is invariant under a gauge symmetry and nowhere in the derivation of the Feynman rules was this gauge fixed. Thus any choice of gauge for $A_a$ is still valid and we should have a symmetry under, $$ A_a \rightarrow A_a - i \partial _a \alpha $$ This transformation in momentum space is given by $$ \epsilon_a \rightarrow \epsilon _a + \alpha p_a $$ This implies that the amplitude should be invariant under this transformation.

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  • $\begingroup$ I'm using (-+++) hence my denominator. But thank you anyways! How can I show directly that $\epsilon_a\rightarrow\epsilon_a+\alpha p$ doesn't change the amplitude though? Any tricks after substituting it into the expression? $\endgroup$ – user46348 May 26 '14 at 20:20
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    $\begingroup$ Ahh, sorry about that. I've gotten so used to +--- I sometimes forget there is an alternative... What you want to use is the on-shell condition for the external fermions (since the Ward identity only holds for physical quantities). In particular you want to use (in -+++) identites such as $(p_\mu\gamma^\mu +m ) \gamma^\nu u (p) = 2p^\nu u(p)$ $\endgroup$ – JeffDror May 26 '14 at 21:00

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