# Transversality of compton scattering amplitude for off-shell photons

I am having difficulties applying the concept of Ward-identities to the amplitude Compton scattering at tree-level.

To my knowledge, Ward-identity implies that the scattering amplitude of any (abelian) process with an external photon vanishes upon contraction with the photon momentum and all fermion lines amputated, set on-shell and contracted with bi-spinors $$u,\bar{u}, v, \bar{v}$$. So diagrammatically Ward identity implies:

In particular the photons do not have to be on-shell (but I don't know if that also includes $$q$$). However, if I want to show explicitly that the amplitude of Compton scattering at tree-level is transverse, $$$$k_\mu i \mathcal{M}^{\mu \nu} = k_\mu \left[ \bar{u} (p^\prime) (-ie\gamma^\mu ) \frac{i(\not{p} + \not{k} + m)}{(p + k)^2 - m^2} (-ie \gamma^\nu) u(p) + \bar{u} (p^\prime) (-ie\gamma^\nu ) \frac{i(\not{p} - \not{k}^\prime + m)}{(p + k^\prime)^2 - m^2} (-ie \gamma^\mu) u(p) \right]$$$$ I always need to assume that $$k^2 = k^{\prime 2} = 0$$.

So I would like to know,

1. does the momentum of the photon that is contracted ($$q$$ in the picture) needs to be on-shell?

2. Is the amplitude of Compton scattering transverse without assuming that the photon(s) are(is) on-shell and how can I show it?

EDIT: I have shown that the amplitude is transverse without assuming that $$k_1, k_2$$ are on-shell by now.

• But then, you kind of assumed the photon to be on-shell in the first place, don't you? If you look back on the derivation of the Ward-identity, the momentum enters through a derivative $\partial_\mu$ which becomes $p_\mu$ in momentum space. So I don't really see any reason why the momentum should be regarded on-shell. Jun 21 '21 at 11:27