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?
