# Ward-Takahashi identity for the 2-point 1PI Green function of photons

I am following Sidney Coleman's lectures of Quantum Field Theory (World Scientific).

For the renormalization of QED, he considered the following Lagrangian (Eq 33.54 in the book) $$$$\mathcal{L} = -\frac{1}{4}F_{\mu \nu}^{2} + \bar{\psi}(i \gamma^{\mu}\partial_{\mu} - m - e \gamma^{\mu}A_{\mu})\psi - \frac{1}{2\xi}(\partial_{\mu}A^{\mu})^{2} + \frac{1}{2}\mu^{2}A_{\mu}^{2}.$$$$ He added a photon mass term, but will send it to zero in the end. The goal here is to derive any exact relation that the one-particle-irreducible (1PI) Green function should satisfy. He denote the effective action, namely the generating functional of the 1PI Green functions, as $$\Gamma$$ and in here since I am only concerned with the two point photon 1PI diagram, we can write (Eq 33.59 in the book) $$$$\Gamma = \frac{1}{2!}\int d^{4}xd^{4}y A^{\mu}(x)A^{\nu}(y)\Gamma^{(0,0,2)}_{\mu\nu}(x,y) + \cdots$$$$ He used $$\Gamma^{(n,n,m)}$$ to denote the 1PI with $$n$$ $$\psi$$, $$n$$ $$\bar{\psi}$$ and $$m$$ $$A_{\mu}$$ external legs.

Due to his argument, when we apply gauge transformation $$$$A_{\mu} \to A_{\mu} + \partial_{\mu} \delta \chi,$$$$ because $$\delta A_{\mu} = \partial_{\mu} \delta \chi$$ is independent of $$e$$, the associated change $$\delta \Gamma$$ must be as well, such that $$$$\delta \Gamma = \delta \Gamma |_{e=0} \text{ when A_{\mu} \to A_{\mu} + \partial_{\mu} \delta \chi}.$$$$ From this, we can derive an useful equation (Eq. 33.63) $$$$k^{\mu}\tilde{\Gamma}_{\mu \nu}^{(0,0,2)}(k) = k^{\mu} \tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}.$$$$ $$\tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}$$ can be thought of as the inverse of the free photon propagator, since we set $$e=0$$. The above equation further implies that there is no correction to the longitudinal part of the photon propagator when we sum up all the photon self-energy. Or, in other words, all the corrections to the 1PI photon self-energy are purely transverse, such that $$k^{\mu} \Pi_{\mu \nu}(k^{2})=0$$ where $$\Pi_{\mu \nu}(k^{2})$$ is the photon self-energy.

Now the above conclusions are all good, but here are several confusion that I have:

(1) If we write the gauge transformation as $$A_{\mu} \to A_{\mu} + e^{-1} \partial_{\mu}\Lambda$$ where $$\Lambda$$ is just some functions. This is definitely a legal gauge transformation. But now $$\delta A_{\mu}$$ depends on $$e$$, and his argument seems to breakdown.

(2) Even though we stick to $$A_{\mu} \to A_{\mu} + \partial_{\mu}\delta \chi$$, if we work out $$\delta \Gamma$$ (not at $$e=0$$), it is given by (Eq 33.61 in the book) $$$$\delta \Gamma = - \int d^{4}x d^{4}y A^{\nu}(y) \delta \chi(x) \left(\partial^{\mu}_{x}\Gamma^{(0,0,2)}_{\mu \nu}(x,y) \right).$$$$ Now my intuition is that, there will be a whole bunch of terms that are already collected in $$\Gamma^{(0,0,2)}_{\mu \nu}(x,y)$$, any 2 point 1P with two external lines of photons will go into $$\Gamma^{(0,0,2)}_{\mu \nu}(x,y)$$. It seems to me that there must be some $$e$$-dependence on $$\Gamma^{(0,0,2)}_{\mu \nu}(x,y)$$. In fact, we are wanting to compute the $$\Gamma^{(0,0,2)}_{\mu \nu}(x,y)$$ as a power series expansion of $$e$$ (I think so?). Then from the above equation for $$\delta \Gamma$$, it seems hard to me to believe that there is no $$e$$ ever appear in $$$$\delta \Gamma = - \int d^{4}x d^{4}y A^{\nu}(y) \delta \chi(x) \left(\partial^{\mu}_{x}\Gamma^{(0,0,2)}_{\mu \nu}(x,y) \right).$$$$ But this is what he said: there is no $$e$$-dependence on $$\delta \Gamma$$. He didn't say more or expand the above argument that $$\delta A_{\mu}$$ is independent of $$e$$ that I highlight in boldface. But that argument seems too fragile for me. I am seeking some other more concrete argument.

Can anyone provide any suggestion on the (1) above argument, or (2) derivation of $$$$k^{\mu}\tilde{\Gamma}_{\mu \nu}^{(0,0,2)}(k) = k^{\mu} \tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0},$$$$ or maybe $$$$(3)\ k^{\mu} \Pi_{\mu \nu}(k^{2})=0 \text{ (the relatively useful one?)},$$$$ which according to Coleman is a result due to $$k^{\mu}\tilde{\Gamma}_{\mu \nu}^{(0,0,2)}(k) = k^{\mu} \tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}$$, that can potentially solve my confusion? Please let me know if there are any unclear points. Thanks!