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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?

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    $\begingroup$ 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? $\endgroup$
    – ACuriousMind
    Commented May 29, 2015 at 13:44

2 Answers 2

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  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 $$\begin{align} {\cal R} [\beta(z)\gamma(w)]~=~&\frac{1}{z-w} ~+~ :\beta(z)\gamma(w):,\cr{\cal R} [\beta(z)\beta(w)]~=~& :\beta(z)\beta(w):, \cr {\cal R} [\gamma(z)\gamma(w)]~=~&:\gamma(z)\gamma(w):, \end{align}\tag{A}$$ and (ii) Wick's theorem, it follows that $$ {\cal R} [\beta(z)f(\gamma(w))]~=~\frac{f^{\prime}(\gamma(w))}{z-w} ~+~ :\beta(z)f(\gamma(w)):, \tag{B}$$ where $f$ is a differentiable function$^1$, and ${\cal R}$ denotes radial order.

  3. Now put $f(z)=z^{-1}$ in eq. (B) to calculate the sought-for expression (1).

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$^1$ Here it is implicitly assumed that the operator $f(\gamma)$ is a well-defined in the given physical context.

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Constructing a free CFT means you take the Verma modules of your fundamental fields, in this case \begin{equation} V \left ( h = -\frac{1}{2}, c = 11 \right ), V \left ( h = \frac{3}{2}, c = 11 \right ), \end{equation} and then add all the representations formed by taking products of those fields. I.e. where they are raised to non-negative integer powers. This procedure (possibly after summing over spin structures to get modular invariance) is the minimal one needed to get a consistent theory so it's what we mean whenever we say "the $\beta\gamma$ system".

If you're doing something to go beyond this minimal procedure, such that there actually is a local operator which can act on $\gamma(0) \left |0 \right>$ to give $\left |0 \right>$, then you can talk about $\gamma^{-1}(z)$. But otherwise, the best answer to this question is: no it's not possible. In particular, taking a form of Wick's theorem derived for analytic functions and then plugging in the non-analytic $f(z) = z^{-1}$ is not meaningful.

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