I'm trying to derive Ward-Takahashi identity $$k_\mu V^\mu(p,q,k)=Z_1 Z_2^{-1}e(S^{-1}(q)-S^{-1}(p))\tag{68.12}$$ using Schwinger-Dyson equation and Ward identity.
In renormalized spinor QED, the Lagrangian is:
$$\mathcal{L}=\bar{\Psi}(iZ_2\gamma^\mu\partial_\mu-Z_m m)\Psi-\frac{1}{4}Z_3F^{\mu\nu}F_{\mu\nu}+Z_1 q\bar{\Psi}\gamma^\mu\Psi A_\mu.$$
Under a global gauge transformation $\Psi\rightarrow e^{-ie\alpha}\Psi,\ \bar{\Psi}\rightarrow\bar{\Psi}e^{ie\alpha}$, the conserved current is:
$$J^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi=Z_2 q\bar{\Psi}\gamma^\mu\Psi\alpha\equiv Z_2 j^\mu \alpha$$
Where $j^\mu$ is defined to be $q\bar{\Psi}\gamma^\mu\Psi$. Plugging that into Ward identity (eq. (22.26) of Srednicki), we can obtain:
$$\begin{aligned} iZ_2 k_\mu & \langle \Omega|\mathrm{T}\{\tilde{j}^\mu(k)\tilde{\Psi}(q_1)...\tilde{\Psi}(q_n)\tilde{\bar{\Psi}}(p_1)...\tilde{\bar{\Psi}}(p_n)\}|\Omega\rangle \\&=e\sum_{i}\langle \Omega|\mathrm{T}\{\tilde{\Psi}(q_1)...\tilde{\Psi}(q_i-k)...\tilde{\Psi}(q_n)\tilde{\bar{\Psi}}(p_1)...\tilde{\bar{\Psi}}(p_n)\}|\Omega\rangle \\&-e\sum_{j}\langle \Omega|\mathrm{T}\{\tilde{\Psi}(q_1)...\tilde{\Psi}(q_n)\tilde{\bar{\Psi}}(p_1)...\tilde{\bar{\Psi}}(p_i+k)...\tilde{\bar{\Psi}}(p_n)\}|\Omega\rangle \end{aligned}$$
Where the Fourier-transformed correlation function was defined to be:
$$\begin{aligned} \langle \Omega&|\mathrm{T}\{\tilde{j}^\mu(k)\tilde{\Psi}(p_1)...\tilde{\Psi}(p_n)\tilde{\bar{\Psi}}(q_1)...\tilde{\bar{\Psi}}(q_n)\}|\Omega\rangle\equiv \\& e\int d^4x\int d^4x_1...\int d^4y_1... e^{ikx}e^{-iq_1x_1}...e^{ip_1y_1}...\langle \Omega|\mathrm{T}\{\bar{\Psi}(x)\gamma^\mu\Psi(x)\Psi(x_1)...\Psi(x_n)\bar{\Psi}(y_1)...\bar{\Psi}(y_n)\}|\Omega\rangle. \end{aligned}$$
Using Schwinger-Dyson equation of $\langle\Omega|\mathrm{T}\{A_\mu\Psi\bar{\Psi}\}|\Omega\rangle$, we have (working in Feynman gauge):
$$-Z_3 \partial^2\langle \Omega|\mathrm{T}\{A_\mu(x)\Psi(y)\bar{\Psi}(z)\}|\Omega\rangle=Z_1 \langle \Omega|\mathrm{T}\{j_\mu(x)\Psi(y)\bar{\Psi}(z)\}|\Omega\rangle.$$
After Fourier transform, that became:
$$Z_3 k^2\langle \Omega|\mathrm{T}\{\tilde{A}_\mu(k)\tilde{\Psi}(q)\tilde{\bar{\Psi}}(p)\}|\Omega\rangle=Z_1 \langle \Omega|\mathrm{T}\{\tilde{j}_\mu(k)\tilde{\Psi}(q)\tilde{\bar{\Psi}}(p)\}|\Omega\rangle.$$
Using Feynman diagrams to evaluate the LHS, we can get:
$$\langle \Omega|\mathrm{T}\{\tilde{j}_\mu(k)\tilde{\Psi}(q)\tilde{\bar{\Psi}}(p)\}|\Omega\rangle=Z_1^{-1}Z_3(2\pi)^4\delta^4(k+p-q)(\delta_{\mu\nu}+\Pi_{\mu\rho}(k)\Delta^{\rho}_{\ \nu}(k)+\cdots)(\frac{1}{i}S(q))(iV^\nu(p,q,k))(\frac{1}{i}S(p))$$
Where $V^\mu$ is the 1PI vertex, and $S(p)$ is the exact fermion propagator. From local gauge invariance, we know that $k_\mu \Pi^{\mu\nu}(k)=0$. So contracting the equation above with $k_\mu$ yields:
$$ik_\mu \langle \Omega|\mathrm{T}\{\tilde{J}^\mu(k)\tilde{\Psi}(q)\tilde{\bar{\Psi}}(p)\}|\Omega\rangle=Z_1^{-1}Z_3(2\pi)^4\delta^4(k+p-q)k_\mu S(q)V^\mu(p,q,k)S(p).$$
Combining that with Ward identity:
$$Z_2 Z_1^{-1}Z_3(2\pi)^4\delta^4(k+p-q)k_\mu S(q)V^\mu(p,q,k)S(p)=(2\pi)^4\delta^4(k+p-q)e [S(p)-S(q)].$$
And we can finally get:
$$Z_3 k_\mu V^\mu(p,q,k)=Z_1 Z_2^{-1} e[S^{-1}(q)-S^{-1}(p)].$$
It turns out that there's an extra factor $Z_3$ in my result, compared to the correct Ward-Takahashi identity $$k_\mu V^\mu(p,q,k)=Z_1 Z_2^{-1}e(S^{-1}(q)-S^{-1}(p)).\tag{68.12}$$ I know that the result I've got must be wrong since LHS is infinite but RHS is finite. But I can't figure out where's wrong in the derivation. Where's my mistake?
