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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!

LYg
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2 Answers2

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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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Berrick Caleb Fillmore
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