Polarization vectors in scattering amplitudes The annihilation ($e^+ e^-\rightarrow 2\gamma$) diagrams are

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?
 A: 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.
