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Peskin and Schroeder mention a few times that the anomalous dimension of a gauge interaction operator is zero. The justification for this is that the charge operator shouldn't get modified under anomalous dimensions. I'm trying to show explicitly for the simplest case, QED, but I'm not getting the right answer so I hoping to get some help. I detail my calculation below, though I suspect my problem is conceptual as opposed to a silly mistake.

We need to consider the diagrams,

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We use $\overline{MS}$ with massless fermions and only keep the divergent pieces. The first diagram is, \begin{align} \require{cancel} i G & = \int d^4 \ell \frac{ \gamma _\mu \cancel{ \ell } \gamma _\nu \cancel{ \ell } \gamma ^\mu }{ \ell ^6 } ( i ) ^2 ( - i ) ^2 ( - i e ) ^3 \\ & = -\frac{ i e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \end{align} The counterterms for the fermion and photon propagators are \begin{equation} - i \frac{ 2e ^2 }{ 16 \pi ^2 \epsilon } \quad \mbox{and} \quad - i \left( \frac{ 8 }{ 3} \frac{ e ^2 }{ 16 \pi ^2 \epsilon } \right) \left( q _\mu q _\nu - q ^2 g _{ \mu \nu } \right) \end{equation} where $q$ is the momentum of the photon. This gives for the last three diagrams, \begin{equation} - i\frac{ e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \quad , \quad - i\frac{ e ^3 }{ 16 \pi ^2 \epsilon } \gamma _\nu \quad , \quad - i \frac{ 8 }{ 3} \frac{ e ^3 }{ 16 \pi ^2 \epsilon } \left( \frac{q _\nu q _\mu}{q^2} - g _{ \nu \mu } \right) \gamma ^\mu \end{equation} , respectively.

Now if I understand correctly, for these not to give the 3-point interaction an anomalous dimension they must cancel, but I don't see how they would so I feel like my understanding of anomalous dimensions is off. Any ideas about what's going on here?

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  • $\begingroup$ Where do they say this? $\endgroup$ – DJBunk Apr 3 '14 at 1:45
  • $\begingroup$ @DJBunk: They mention this below Eq 12.109 on page 430, for example. $\endgroup$ – JeffDror Apr 3 '14 at 12:03
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This is a special case of a more general phenomena. Conserved currents never acquire anomalous dimensions, they are protected by the symmetry. If you have a conserved current, you have a symmetry algebra for the theory \begin{equation} [Q^a,Q^b]=if^{abc}Q_c \end{equation} For this to hold, the charges need to be dimensionless. But the charges are given by $\int d^3x j^0$, so the conserved current must have dimension 3 and cannot receive an anomalous dimension. This is discussed in the book Foundations of Quantum Chromodynamics by Muta.

More heuristically, you know $[Q,\phi]\approx \delta \phi $ and so by dimensional analysis the dimension of the charge cannot change.

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  • $\begingroup$ Thanks for your answer. In terms of trying to calculate the anomalous dimension as I did above, do I have the right approach? In particular, did I include the correct diagrams and am I right that they should cancel to avoid anomalous dimensions? $\endgroup$ – JeffDror Apr 3 '14 at 10:09

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