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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) \begin{equation} \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}. \end{equation} 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) \begin{equation} \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 \end{equation} 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 \begin{equation} A_{\mu} \to A_{\mu} + \partial_{\mu} \delta \chi, \end{equation} because $\delta A_{\mu} = \partial_{\mu} \delta \chi$ is independent of $e$, the associated change $\delta \Gamma$ must be as well, such that \begin{equation} \delta \Gamma = \delta \Gamma |_{e=0} \text{ when $A_{\mu} \to A_{\mu} + \partial_{\mu} \delta \chi$}. \end{equation} From this, we can derive an useful equation (Eq. 33.63) \begin{equation} k^{\mu}\tilde{\Gamma}_{\mu \nu}^{(0,0,2)}(k) = k^{\mu} \tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}. \end{equation} $\tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}$ 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) \begin{equation} \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). \end{equation} 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 \begin{equation} \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). \end{equation} 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 \begin{equation} k^{\mu}\tilde{\Gamma}_{\mu \nu}^{(0,0,2)}(k) = k^{\mu} \tilde{\Gamma}^{(0,0,2)}_{\mu \nu}|_{e=0}, \end{equation} or maybe \begin{equation} (3)\ k^{\mu} \Pi_{\mu \nu}(k^{2})=0 \text{ (the relatively useful one?)}, \end{equation} 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!

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