Consider the QED Lagrangian, \begin{equation} {\cal L} = \bar{\psi} ^{(0)} ( i \partial_\mu \gamma^\mu - m ) \psi ^{(0)} - e A _\mu ^{(0)} \bar{\psi} ^{(0)} \gamma ^\mu \psi ^{(0)} - \frac{1}{4} F ^2 _{ (0) } \end{equation} If we apply the typical renormalization procedure, \begin{align} & A _\mu = \frac{1}{ \sqrt{ Z _A } } A _\mu ^{ ( 0 ) } \\ & \psi = \frac{1}{ \sqrt{ Z _\psi } } \psi ^{ (0) } \\ & e = \frac{1}{ Z _e } e ^{(0)} \\ & m = \frac{1}{ Z _m } m ^{(0)} \end{align} Then we get a mass counterterm, \begin{equation} m ( 1 - Z _m Z _\psi ) \bar{\psi} \psi \end{equation} Up to here, I have no issues.

Now suppose we want to consider the renormalized of some operator, $ {\cal O} ( x ) $. To be explicit, lets take the operator to be $ \bar{\psi} \psi (x) $. The way I would naively think about it is that we should insert this operator into the Lagrangian in terms of the bare fields and then allow its coupling to get renormalized in the same way that we did above. So we would have, \begin{equation} \Delta {\cal L} = g ^{(0)} {\cal O} ^{(0)} ( x ) = g ^{(0)} \bar{\psi} ^{(0)} \psi ^{(0)} \end{equation} then we would say \begin{align} & g = \frac{1}{ Z _g } g ^{(0)} \end{align} and we get a counterterm, \begin{equation} \Delta {\cal L} = g ( 1 - Z _g Z _\psi ) \end{equation} However Wise and Manohar in their book, Heavy Quark Physics, get a slightly different result. They get a counterterm of, \begin{equation} \Delta {\cal L} = g \left( 1 - \frac{ Z _\psi }{ Z _g } \right) \end{equation} What is wrong about my understanding of composite operators?

  • $\begingroup$ Which textbooks are you referring to? $\endgroup$ Mar 24, 2014 at 21:37
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    $\begingroup$ Wise and Manohar in their book, "Heavy Quark Physics", state this explicitly. I was initially under the impression that Peskin and Schroeder also took this viewpoint but now I'm not sure. Maybe I was too quick to say "textbooks". I edited the question. $\endgroup$
    – JeffDror
    Mar 24, 2014 at 21:41
  • $\begingroup$ Actually I think Peskin and Schroeder do agree with Wise and Manohar (see Eq. 12.120, where they add into the Lagrangian directly the renormalized operator instead of the bare one as I would naively expect). I believe this leads to the counterterm as I wrote above. $\endgroup$
    – JeffDror
    Mar 24, 2014 at 21:48
  • $\begingroup$ Are you sure about the convention for how interaction couplings are renormalized? I mean: could they transform as $g \longrightarrow Z_g \; g$ ? $\endgroup$
    – Siva
    Mar 24, 2014 at 22:33
  • $\begingroup$ I could not find the statement in Manohar's book, where did you see it? $\endgroup$ Mar 24, 2014 at 22:39

1 Answer 1


The renormalization factor $1/Z_S$ that Manohar and Wise discuss in their book is not the renormalization of the coupling constant. It is there to remove divergences that come from the fact that we have inserted an operator that is a product of fields. As I understand their line of reasoning, this is before we discuss a coupling constant at all, so if we add one we would get an additional renormalization factor.

  • $\begingroup$ The way I always understood renormalization factors is that they are forced upon you, since when you calculate diagrams to all orders the "effective couplings" change and we account for that by adding in new factors with renormalization conditions. If we are not renormalizing the coupling then why do we even have the factor of $1/Z_S$? $\endgroup$
    – JeffDror
    Mar 25, 2014 at 9:42
  • $\begingroup$ I think my view of renormalization was influenced by this post: physics.stackexchange.com/q/43397 $\endgroup$
    – JeffDror
    Mar 25, 2014 at 10:04
  • $\begingroup$ The statement is that products of local operators themselves are normalized, without any coupling constants. This is explained in chapter 12.4 of Peskin and Schroeder, "Renormalization of Local Operators". $\endgroup$ Mar 25, 2014 at 11:25

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