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I've often seen in textbooks that authors renormalize diagrams by setting external momentum to zero. Under what conditions is this justified?

An example of this is done in Manohar and Wise's book on Heavy Quark Physics after they renormalize QED and then calculate the operator renormalization, $ Z _S $, of \begin{equation} S = \frac{1}{ Z _S }\bar{\psi} _b \psi _b = \frac{ Z _\psi }{Z _S } \bar{\psi} \psi \end{equation} where $ \psi_b $ is the bare field. They calulate this through the diagram,

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where the cross indicates an operator insertion. The authors then say ``The operator, $ S $, contains no derivatives (and $Z _S$ is mass independent in the $\overline{MS}$ scheme), so $Z _S$ can be determined by evaluating (the diagram) at zero external momentum (and neglecting the (fermion) mass).'' Are these the two conditions necessary,

  1. The operator doesn't have derivatives
  2. The quantity of interest has no mass dependence

and if so how do we know that the quantity you want to calculate (in this case, $ Z _S $) is mass independent, ahead of time?

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In general, derivative couplings lead to momentum-dependencies in scattering amplitudes. This can be seen from the fact that the Fourier transform of a derivative operator corresponds to a multiplication by the relevant momentum. A mass dependence is implicit through by having a momentum, since the momentum of a fermion depends on its mass. In this case, setting momenta to zero would remove information about the coupling.

However, when the coupling does not contain derivatives, there are no momenta resulting from a Fourier transform. Hence, one can simplify the problem by setting them to zero.

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  • $\begingroup$ Thanks, that makes sense. We could get a momentum dependence if the operator included a propagator, but I guess then the operator wouldn't be local? $\endgroup$ – JeffDror Mar 24 '14 at 17:54
  • $\begingroup$ Yes, I think that what you have in mind is correct. $\endgroup$ – Frederic Brünner Mar 24 '14 at 23:00

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