Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When renormalizing QED, we calculate the 1 loop correction to the fermion-fermion-photon vertex using the diagram,

$\hskip2in$ enter image description here

When doing the calculation we typically let the photon go off-shell but demand that the fermions be on-shell. In other words, \begin{equation} q ^2 >0, \,(q-p)^2 = p ^2 = m_e^2 \end{equation} We then go on to calculate the diagram using its Lorentz structure and then splitting the contribution into a part contributing to the $g$ factor ($F_2(q^2)$) and another which does not ($F_1(q^2)$) giving a vertex factor of, \begin{equation} \Gamma ^\nu ( p , p ' ) = ( - i e ) \left[ \gamma ^\nu F _1 ( q ^2 ) + \frac{ i \sigma ^{ \mu \nu } q _\mu }{ 2 m } F _2 ( q ^2 ) \right] \end{equation}

However, keeping the fermions on-shell seems like a very strange requirement as in general this vertex may appear in the middle of a diagram and so could have off-shell incoming fermions. In fact, we don't make this requirement when renormalizing the propagator, I believe for this exact reason. So what makes it justified when discussing vertex renormalization?

share|cite|improve this question
up vote 2 down vote accepted

Actually, in renormalization of QED, there is no demand to put the fermions of vertex diagram on mass shell. Renormalization procedure is usually performed on the level of Green functions with general four-momenta of outer legs. Note that the off-shell propagator you mention is connected to vertex function via Ward-Takahashi identity $$ (p'-p)_{\mu} \Gamma^{\mu} (p, p') = S(p')^{-1} - S(p)^{-1}$$ where all momenta are off-shell. I guess that in QED textbooks authors put these fermions on-shell to discuss quantum corrections to form factors (such as Schwinger's $\alpha/2\pi$ correction to $g$-factor), and thus connect the renormalization story with effects on actual measurable physical quantities.

When QCD textbooks do vertex diagram they don't put quark momenta on-shell. Actually, they are often put to "deep Euclidean" off-shell limit of large negative $p^2$ to get away from physical singularities so that you can concentrate on some interesting renormalization effects relevant for QCD (like violation of Bjorken scaling).

share|cite|improve this answer
Thanks for you answer! In Peskin and Schroeder (pg 334) the authors calculate the counterterm for vertex renormalization by assuming on-shell fermions (I think, they are not clear here). So is this in general incorrect and should be done for arbitrary $p^2$? – JeffDror Feb 26 '14 at 12:35
I wouldn't say this is incorrect. You have freedom to fix momenta to some specific value if you want. And in some renormalization schemes you actually impose renormalization conditions on diagrams when some lines are on-shell. But there are also other schemes which leave momenta arbitrary. – kkumer Feb 27 '14 at 10:12
But suppose you calculate your counterterms with the fermions on-shell but then do a calculation where ingoing fermions are allowed to be off-shell then are you allowed to use the same counterterms or do you need to go back and recalculate them? – JeffDror Feb 27 '14 at 11:57

Putting the fermions on-shell or off-shell doesn't change the divergent part of counterterms. In the renormalization schemes, the counter terms are determined to cancel the divergencies. You can put $p^2=m^2$ or $p^2=-\mu^2$ or etc in the diagrams to determine the counter terms, but notice that the derived renormalization constants, $Z_1$, $Z_2$, etc, should not depend on $p^2$ and also doesn't violent Ward identity. After summing the diagrams, counters will cancel the divergencies of other diagrams and just leave some finite parts. I think putting the fermions on-shell is related to the renormalization scheme.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.