QED replacing photon field with current in 3-point function

[handrea2009]

TL;DR

QED: replacing photon field with current in the definition of the renormalized 3-point function vertex

I am self-studying QFT in the Schwartz book "Quantum Field Theory and the Standard Model", currently I am struggling to understand the all-orders proof that $Z_1=Z_2$ using Ward-Takahashi identity (page 352).

He states that $-ie_R\Gamma^\mu$, which is the sum of the 1PI contributions to matrix elements for the 3-point function $\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle$ with externa legs amputated , can be formally defined as follow:

$$-ie_R\Gamma^\mu(p,q_1,q_2)(2\pi)^4\delta^4(p+q_1-q_2) \\
\equiv -ie_R\int d^4x d^4x_1 d^4x_2 e^{ipx}e^{iq_1x_1}e^{-iq_2x_2}\\(iG)^{-1}(\not q_1) \langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle (iG)^{-1}(\not p+\not q_1) \tag {19.78}
$$

I don't understand how he gets to replace in $\langle \psi(x_1)A_\nu(x)\bar\psi(x_2)\rangle$ the $A_v$ photon field with the current $j^\mu = \bar\psi\gamma^\mu \psi$.

The only clue I could find is on page 281 where we have the following formula which comes from Schwinger-Dyson equation:

$$
\square_{\alpha\beta}^k \square_{\mu\nu} \langle A_\nu(x)...A_\beta(x_k)...\rangle = \langle j_\mu(x)...j_\alpha(x_k)...\rangle
$$

so basically you can remove $A_\mu$ field and insert current $j_\mu$.

However, even using that result it seems to me that in the formula (19.78) in the right-hand-side we have a wrong extra $ie_R$ factor, since we already have one which comes from exploding the expectation value $\langle ... \rangle$ ( see (7.77) ):

$$
\langle j^\mu(x) \psi(x_1) \bar\psi(x_2)\rangle \equiv \langle \Omega|T\{j^\mu(x)\psi(x_1) \bar\psi(x_2)\}|\Omega\rangle = \langle 0|T\{j^\mu_0(x)\psi_0(x_1) \bar\psi_0(x_2)e^{-ie_R\int \bar\psi_0\gamma^\mu\psi_0 A_\mu}\}|0\rangle_{no bubbles}
$$

where $-ie_R\bar\psi\gamma^\mu\psi A_\mu$ is the interaction term in the QED Lagrangian density

 

[handrea2009]

I believe in the book the formula on page 281:

$$
\square_{\alpha\beta}^k \square_{\mu\nu} \langle A_\nu(x)...A_\beta(x_k)...\rangle = \langle j_\mu(x)...j_\alpha(x_k)...\rangle \tag{14.152}
$$

is missing an $e$ factor on the RHS, this seems to be confirmed by the Schwinger-Dyson equation $(14.117)$ which is used to get the $(14.152)$:

$$
\square^x_{\mu\nu}\langle A^\nu(x)A^\alpha(y)\bar\psi(z_1)\psi(z_2)\rangle = \\
e \langle j_\mu(x)A^\alpha(y)\bar\psi(z_1)\psi(z_2)\rangle -i\delta^4(x-y)\delta^\alpha_\mu\langle\bar\psi(z_1)\psi(z_2)\rangle \tag{14.117}
$$

In that way we have:

$$
\square^{\mu\nu}_x \langle A_\nu(x)\psi(x_1)\bar\psi(x_2)\rangle = e_R\langle j^\mu(x)\psi(x_1)\bar\psi(x_2)\rangle
$$

If I substitute that in the $(19.78)$ and as a check I do the calculation at leading order, I correctly get $\Gamma^\mu = \gamma^\mu$