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Peskin in his QFT page 334 argued that $Z_1=Z_2$ to all orders in QED perturbation theory, but I couldn't understand his argument:

... With a generalization of the argument given there (section 7.4 for Ward identity), on can show that the diagrammatic identity (7.68) holds for digrams that include counterterm vertices in loops.

Let's assume this is granted true, but I got lost in his following argument:

Thus, if the counterterms $\delta_1$ and $\delta_2$ are determined up to order $\alpha^n$, the unrenormalized vertex diagram at $q^2=0$ equals the derivative of the unrenormalized self-energy diagram on-shell in order $\alpha^{n+1}$. To satisfy the renormalization conditions (10.40), we must then set the counterterms $\delta_1$ and $\delta_2$ equal to order $\alpha^{n+1}$. This recursive argument gives yet another proof that $Z_1=Z_2$ to all orders in QED perturbation theory.

What does he mean by unrenormalized vertex diagram?
Can somebody please explain the connections in his logic?
Thanks!

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  • $\begingroup$ The photon propagator is gauge-dependent; the one you have chosen is in the Feynman gauge. In the most general covariant gauge, there is a term proportional to ($q_{\mu}q_{\nu}/q^2$). Physical results must be independent of the coefficient of that term -- and $Z_1 - Z_2$ is a consequence of that requirement, $\endgroup$ – MikeV Aug 19 '15 at 17:43
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unrenormalized vertex diagram and unrenormalized self-energy diagram are as below: Vertex Diagram Self-energy diagram

Now, let me summarize what Peskin and Schroeder try to express. Feynman propagators for electron $$ S_F=\frac{i}{\require{cancel}\cancel p-m} $$ end photon $$ D_F=\frac{-ig_{\mu\nu}}{q^2} $$ and propagator must hold $$ S_F(p)S_F^{-1}=\mathbb{1} $$ simply differentiate it $$ \frac{dS_F(p)}{dp^{\mu}}S_F^{-1}=-S_F\frac{dS_F^{-1}}{dp^{\mu}} $$ $$ \frac{dS_F(p)}{dp^{\mu}}=-S_F(p)\frac{dS_F^{-1}}{dp^{\mu}}S_F(p) $$ Since $S_F^{-1}$ must hold $S_F(p)S_F^{-1}=\mathbb{1}$ $$ S_F^{-1}(p)=-i(\cancel p-m) $$ and $$ \frac{dS_F^{-1}(p)}{dp^{\mu}}=-i\gamma_{\mu} $$ $$ \Longrightarrow \frac{dS_F(p)}{dp^{\mu}}=iS_F(p)\gamma_{\mu}S_F(p) $$ That's definitely look like this for $q^2=0$

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This is such like if derivate propagator it is simply equal adding photon to it! Let me write unrenormalized electron self-energy diagram as above.(Just reverse photon and electron momentums) $$ \Sigma_2(p)=-ie^2\int \frac{d^4k}{(2\pi)^4}\gamma^{\mu}S_F(p-k)\gamma_{\mu}D_F(k) $$ If you derivate this one supose to get vertex diagram. Remember adding one more photon!So I can write this as follows $$ \Gamma_{2\nu}(p,p)=\frac{d\Sigma_2}{dp^{\nu}}=e^2\int \frac{d^4k}{(2\pi)^4}\gamma^{\mu}S_F(p-k)\gamma_{\nu}S_F(p-k)\gamma_{\mu}D_F(k) $$ where $$ S_F(p-k)\gamma_{\nu}S_F(p-k)=i\frac{\partial S_F}{\partial p^{\nu}} $$ as we calculated above. As you increase the order of derivative, you add more up propagator. Suppose $\Lambda_{\mu}$ all loop corrections of vertex diagram. We can write $$ \frac{d\Sigma(p)}{dp^{\mu}}=\Lambda_{\mu}(p,p)=\Gamma_{\mu}(p,p)-\gamma_{\mu} $$ So the conditions in 10.40 is hold.

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The unrenormalized vertex diagram is the dimensionally regularized Feynman integral corresponding to the following single Feynman diagram:

Vertex diagram

It is called ‘unrenormalized’ because it is not accompanied by a diagram in which the momentum loop is replaced by a counterterm vertex generated by the QED Lagrangian. Adding a second Feynman diagram with the counterterm vertex in place of the momentum loop would precisely cancel out the divergence of the diagram shown.

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    $\begingroup$ You mean, a non-renormalized vertex diagram, don't you ? And you just give an example here. $\endgroup$ – Adam Nov 28 '14 at 7:47
  • $\begingroup$ @Adam: Hi Adam. You’re right. The unrenormalized vertex diagram is actually a dressed vertex function $ {\Gamma^{\mu}}(p,p') $ — an infinite sum of Feynman integrals corresponding to a Feynman diagram resembling the one above, but with the momentum loop replaced by a $ \text{1PI} $ diagram — with $ p = p' $. Even so, this is a function of $ p $ (and of the continuous spacetime dimension $ d $ and a small fictitious mass to cure an infrared divergence). I intend to elaborate on my response soon, so I’ll make the necessary corrections when I do so. Thanks! $\endgroup$ – Berrick Caleb Fillmore Nov 28 '14 at 18:54

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