In QED, according to Schwinger-Dyson equation $^{[1]}$,
$$\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)\langle 0|\mathcal{T}A_{\nu}(x)…|0\rangle = e\,\langle 0|\mathcal{T}j^{\mu}(x)…|0\rangle + \text{contact terms}.$$
And the term $\left(\eta^{\mu\nu}(\partial ^2)-(1-\frac{1}{\xi})\partial^{\mu}\partial^{\nu}\right)$ is just the inverse bare photon propagator$^{[5]}$, so if we put the photon on shell, then the l.h.s will yield the complete $n$-point Green function with the complete photon propagator removed and also multiplied by a factor $Z_3$, the vector field renormalization constant.
But the r.h.s gives, according to the Ward Identity associated with global symmetry of the Lagrangian,
$$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)…|0\rangle = \text{contact terms}^{[2]}$$
which is the sum of complete $(n-1)$-point complete Green functions each multiplied by a certain $\delta$ function.
So if we truncate all the $n-1$ external complete propagators, then we are left with the proper vertex Ward identity.
The problem is, now the constant $Z_3$ appeared.
For, example, if we apply this to $j^\mu=\bar\psi\gamma^\mu\psi$,
$$\partial_{\mu}\, \langle 0|\mathcal{T}j^{\mu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle = -ie[\delta(x-x_1)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle-\delta(x-x_2)\langle 0|\psi(x_1)\bar\psi(x_2)|0\rangle]^{[3]},$$
we will get an identity containing $Z_3$ RATHER THAN the well known proper vertex Ward identity (which is derived from the gauge or local symmetry of the Lagrangian), e.g.
$$q_\mu\Gamma^\mu_P(p,q,p+q)=S^{-1}(p+q)-S^{-1}(p)^{[4]}$$
which doesn't contain $Z_3$.
Where went wrong? Please help.
[1]. Peskin&Schroeder, An Introduction to Quantum Field Theory, page308, eq.(9.88)
[2]. Peskin&Schroeder, An Introduction to Quantum Field Theory, page310, eq.(9.97)
[3]. Peskin&Schroeder, An Introduction to Quantum Field Theory, page311, eq.(9.103)
[4]. Lewis H.Ryder, Quantum Field Theory, page 263-266, eq.(7.112)
[5]. Peskin&Schroeder, An Introduction to Quantum Field Theory, page297, eq.(9.58)
Best Answer
We can write the Fourier transform of $\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$ as $$S(p) D_{\nu\alpha}(q) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q)$$ where $S(p)$ is the full fermion propagator, $D_{\nu\alpha}(q)$ is the full photon propagator, $\Gamma^{\alpha}(p,q,p+q)$ is the proper vertex function, and an overall momentum conservation delta function has been dropped. Similarly, we can write the Fourier transform of $\langle 0|\mathcal{T}j^{\mu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle $ as $$S(p)V^{\mu}(p,q,p+q)S(p+q)$$ where $V^{\mu}(p,q,p+q)$ is a vertex function that we want to relate to $\Gamma^{\mu}(p,q,p+q).$ The vertex function $V^{\mu}(p,q,p+q)$ enters into the derivation of the Ward-Takahashi identity in Peskin and Schroeder on page 311, but the Ward-Takahashi identity is normally stated in terms of $\Gamma^{\mu}(p,q,p+q)$. Your conundrum (as I understand it) is that according to your analysis of the Schwinger-Dyson equation, $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ ought to differ by a factor of $Z_3$, but this contradicts the usual statement of the Ward-Takahashi identity where no such factor of $Z_3$ appears. I will argue from the Schwinger-Dyson equation that the longitudinal parts (in $q^{\mu}$) of $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ are equal, but that the transverse parts differ by the factor of $Z_3$ that you have found. Since only the longitudinal part enters into the Ward-Takahashi identity, the factor of $Z_3$ does not inter into that identity. You may want to review page 246 of Peskin and Schroeder. There they show that only the transverse part of the photon propagator is modified by the self-energy, but that in calculating Feynman diagrams we can simplify the analysis by including the self-energy in the longitudinal part as well because the longitudinal part does not contribute to the Feynman diagrams due to the Ward identity. However, the Schwinger-Dyson equation involves an inverse propagator which does not arise in Feynman diagrams and we need to reevaluate where the self-energy does and does not enter.
Specializing the Schwinger-Dyson equation to the case of $\langle 0|\mathcal{T}A_{\nu}(x)\psi(x_1)\bar\psi(x_2)|0\rangle$ and Fourier transforming, we have $$\tag{1} (D^{(0)\mu\nu}(q))^{-1} D_{\nu\alpha}(q) S(p) \ e\,\Gamma^{\alpha}(p,q,p+q)S(p+q) = \\ e\, S(p)V^{\mu}(p,q,p+q)S(p+q)$$ where $(D^{(0)\mu\nu}(q))^{-1}$ is the inverse of the non-interacting photon propagator. The Dyson equation for the photon propagator is $$\tag{2} D_{\nu\alpha}(q) = D^{(0)}_{\nu\alpha}(q) + D^{(0)}_{\nu\beta}(q) i\Pi^{\beta\gamma}(q)D_{\gamma\alpha}(q) ,$$ so $$\tag{3} (D^{(0)\mu\nu}(q))^{-1} D_{\nu\alpha}(q) = \delta^{\mu}_{\alpha} + i\Pi^{\mu\gamma}(q)D_{\gamma\alpha}(q).$$ Equation (1) then implies $$\tag{4}\Bigl(\delta^{\mu}_{\alpha} + i \Pi^{\mu\gamma}(q)D_{\gamma\alpha}(q)\Bigr) \Gamma^{\alpha}(p,q,p+q) = V^{\mu}(p,q,p+q).$$ The Ward identity forces the longitudinal part of $\Pi^{\mu\gamma}(q)$ to vanish; that is, $q_{\mu}\Pi^{\mu\gamma}(q) = 0.$ Contracting equation (4) with $q_{\mu}$, we therefore have $$\tag{5} q_{\alpha} \Gamma^{\alpha}(p,q,p+q) = q_{\mu}V^{\mu}(p,q,p+q)$$ so no factor of $Z_3$ appears between the longitudinal parts of $\Gamma^{\alpha}(p,q,p+q)$ and $V^{\mu}(p,q,p+q)$ and therefore no factor of $Z_3$ appears in the Ward-Takahashi identity.
The transverse component does not enter the Ward identity for the vertex function but it is useful to consider the transverse component to illustrate where the factor of $Z_3$ does arise. Define $\Pi(q^2)$ by the equation $\Pi^{\mu\nu}(q) = q^2(g^{\mu\nu} - q^{\mu}q^{\nu}/q^2)\Pi(q^2).$ The quantity $(g^{\mu\nu} - q^{\mu}q^{\nu}/q^2)$ can be described as a projection operator that projects out the transverse part of a vector. Contracting equation (4) with $(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2)$ and using the fact that $\Pi^{\mu\gamma}(q)$ is already transverse, we have $$\tag{7} \Bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2 + i q^2\Pi(q^2)D^T_{\nu\alpha}(q)\Bigr) \Gamma^{\alpha}(p,q,p+q) =\\ \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q),$$ where $$D^T_{\nu\alpha}(q) = \frac{-i}{q^2 (1-\Pi(q^2))} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr)$$ is the transverse part of the photon propagator (see page 246, Peskin and Schroeder). Equation (7) can then be written $$\tag{8} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr) \Bigl(1/\bigl(1-\Pi(q^2)\bigr)\Bigr) \Gamma^{\alpha}(p,q,p+q) =\\ \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q).$$ Now consider $q^2$ small enough that $\Pi(q^2)\approx \Pi(0)$ and use the relation (Peskin and Schroeder, page 246) $$Z_3 = \Bigl(1/\bigl(1-\Pi(0)\bigr)\Bigr).$$ We have $$\tag{9} \bigl(g_{\nu\alpha} - q_{\nu}q_{\alpha}/q^2\bigr) Z_3 \Gamma^{\alpha}(p,q,p+q) = \bigl(g_{\nu\mu} - q_{\nu}q_{\mu}/q^2\bigr)V^{\mu}(p,q,p+q).$$ So we see that the transverse parts of $V^{\mu}(p,q,p+q)$ and $\Gamma^{\mu}(p,q,p+q)$ differ by a factor of $Z_3.$