I've been working through the chapters in Schwartz on the renormalisation of QED, and I have some confusion to do with the form of the Vertex correction. By my understanding, the correlation function can be expressed \begin{align*} \left<\Omega\right\vert T\{\hat{\psi}(x_1)\hat{A}_\nu(x)\bar{\psi}(x_2)\}\left\vert\Omega\right> = \int\frac{d^4p}{(2\pi)^4}&\int \frac{d^4q_1}{(2\pi)^4}\int\frac{d^4q_2}{(2\pi)^4}e^{iq_1\cdot x_1}e^{ip\cdot x}e^{-iq_2\cdot x_2}\\ &\times (2\pi)^4\delta^{(4)}(q_2-p-q_1)iG(q_1)(-ie\Gamma^\mu)iG(q_1)iG_{\mu\nu}(p). \end{align*} In this expression $G$ stands for the corrected electron or photon propagators. So the vertex function $\Gamma^\mu$ is the 1PI contribution to the vertex.

$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$enter image description here $~~$

The claim is then made the form of the vertex must be $$\Gamma^\mu = F_1(p^2)\gamma^\mu + F_2(p^2)\frac{i\sigma^{\mu\nu}}{2m}p_\nu.$$ In order to derive this expression he uses

  1. 4-momentum conservation.

  2. The Ward identity, which holds for off-shell photons.

  3. The "Gordon identity", which is where I am confused $$\bar{u}(q_2)(q_1^\mu + q_2^\mu)u(q_1) = 2m\bar{u}(q_2)\gamma^\mu u(q_1) + i\bar{u}(q_2)\sigma^{\mu\nu}(q_{1\nu} - q_{2\nu})u(q_1).$$

I thought that the point of renormalising correlation functions rather than S-matrix elements was because they could be embedded in larger Feynman diagrams. In order to use the Gordon identity, we have to take the electrons to be on-shell. Why is it okay to assume the electrons are on-shell, or is there another way to justify this form of the correction for off-shell electrons?

edit: For clarity, when I say off-shell electrons, I mean to say that the entire vertex correction is embedded in a larger Feynman diagram, for instance: enter image description here

In this case the vertex correction is not sandwiched between 2 off-shell spinors but between 2 propagators.

  • $\begingroup$ Are you sure you need to assume the electrons are on-shell? Don't you just need to assume that the functions $u(p)$ satisfy the appropriate equations? $\endgroup$ Commented May 14, 2020 at 12:08
  • $\begingroup$ @Oбжорoв Your comment should be an answer - because it is the answer... $\endgroup$
    – mike stone
    Commented May 14, 2020 at 12:31
  • $\begingroup$ @Oбжорoв So the corrected propagator satisfies ($\gamma^\mu p_\mu - m$)G(p) = 0? $\endgroup$
    – Edbroad123
    Commented May 14, 2020 at 13:40

2 Answers 2


The second part of the question is about the Feynman propagators $G(p)$, and whether they are also annihilated by $(\gamma^\mu p_\mu - m)$.

This is not the case; however applying this operator to $G$ does cancel the pole associated with that propagator, yielding a simple constant. So in your example, this eliminates $q_1$ and $q_2$ from the expression of the vertex function, since for any $\gamma^\mu q_{1\mu}$ appearing in $\Gamma^\mu$, we can rewrite: $$\gamma^\mu q_{1\mu} \rightarrow m$$ And the difference will be suppressed in the final amplitude, since it lacks a $(q_1^2-m^2)^{-1}$ term.

This would be a very roundabout way of computing the vertex correction though. It is easier to consider a simple scattering off a very heavy particle, like Schwartz seems to do (I have not read it). Once we find $\Gamma^\mu$, then we know we must have the same vertex correction on any other vertex.

I hope this helped a little.

Note: I am still assuming that $q_1$ and $q_2$ are on shell. Indeed, if they are not then the vertex might look very different. Consider for example the $e^+e^-\rightarrow \gamma\gamma$ scattering. Here we have a vertex with one virtual electron, and real photon and electron. This looks like:

$$\mathcal{M} = \epsilon_\nu(q_1)^*(-ie\Gamma^\nu)\frac{1}{\gamma^\mu (q_1-q_2)_\mu - m}u(q_2)$$ Which I'm betting does not lead to the same radiative corrections.


You don't need to assume the electrons are on-shell. The relation you need is $(\gamma^\mu p_\mu -m) u(p)=0$, which holds by definition.

  • $\begingroup$ I've just edited the question for clarity, but if the vertex correction is embedded in a larger Feynman diagram, it wouldn't be sandiwched between off-shell spinors, it would be sandwiched between propagators. $\endgroup$
    – Edbroad123
    Commented May 15, 2020 at 13:29

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