# Special OPE in $\beta\gamma$ system

I would like to find the OPE

$$\beta(z)\gamma(w)^{-1}\tag{1}$$

given

$$\beta(z)\gamma(w)~\sim~\frac{1}{z-w}\tag{2}$$

from the $\beta\gamma$-system in CFT. Is it possible?

• It would make this question far more accessible if you stated what you understand under a "$\beta\gamma$ - system", and why you believe the OPE of $\beta\gamma$ is sufficient to find that of $\beta\gamma^{-1}$. Additionally, is the ${}^{-1}$ the inverse as a field, i.e. $\frac{1}{\gamma}$, or the inverse as an operator? – ACuriousMind May 29 '15 at 13:44

1. Note that radial ordering ${\cal R}$ and normal-ordered/non-singular terms are implicitly implied in OP's eq. (2), see eq. (A) below.
2. From (i) the OPEs $${\cal R} [\beta(z)\gamma(w)]~=~\frac{1}{z-w} ~+~ :\beta(z)\gamma(w):,\tag{A}$$ $${\cal R} [\beta(z)\beta(w)]~=~ :\beta(z)\beta(w):, \qquad{\cal R} [\gamma(z)\gamma(w)]~=~:\gamma(z)\gamma(w):, \tag{B}$$ and (ii) Wick's theorem, it follows that $${\cal R} [\beta(z)f(\gamma(w))]~=~\frac{f^{\prime}(\gamma(w))}{z-w} ~+~ :\beta(z)f(\gamma(w)):, \tag{C}$$ where $f$ is a function, and ${\cal R}$ denotes radial order.
3. Now put $f(z)=z^{-1}$ in eq. (C) to calculate the sought-for expression (1).