Swapping the order of operators in the operator product expansion

Question

I'm looking at the $\beta-\gamma$ ghost CFT, although my question holds more generally for any OPE in any conformal field theory.

The OPE of the two defining fields is $$\beta(z) \gamma(w) \sim \frac{1}{z-w}.$$

To do calculations I also need to know the OPE $$\gamma(z) \beta(w)\sim \;?\;.$$ How can I calculate this using the first OPE? I know this OPE needs to be $\pm$ the first one (and I also know that won't hold more generally for any OPE).

Is the point that I can swap the operators around however I like because they're radially ordered, and so $\beta(z) \gamma(w) = \gamma(w)\beta(z)$?

I tried to repurpose this answer to my current question, but I've always found the OPE a bit cryptic and I need it spelt out to me in simple baby steps.

Answer 1

Yes, you can swap them around as long as they are Bosonic fields. Fermions get a minus sign.

Answer 2

Here's a spell-out in toddler steps. From the $\beta\gamma$ path integral you can derive that $$ \begin{array}{rcl} \bar{\partial}\beta(z)\,\gamma(w)&=&-2\pi \delta(z-w)\delta(\bar{z}-\bar{w}) \\ \bar{\partial}\gamma(z)\,\beta(w)&=&2\pi \delta(z-w)\delta(\bar{z}-\bar{w}). \end{array} $$ Together with the normal ordering $:\ :$, the fact that the $\beta\gamma$ system is a free theory, and the identity $$ \bar{\partial}\left(\frac{1}{z}\right) = 2\pi\delta(z)\delta(\bar{z})$$ you get $$ \begin{array}{rcl} :\beta(z)\,\gamma(w):&=& \beta(z)\,\gamma(w) + \dfrac{1}{z-w} \\ :\gamma(z)\,\beta(w):&=& \gamma(z)\,\beta(w) - \dfrac{1}{z-w}, \\ \end{array} $$ which in turn implies that $$ \begin{array}{rcl} \beta(z)\,\gamma(w) &\sim& -\dfrac{1}{z-w} \\ \gamma(z)\,\beta(w) &\sim& \dfrac{1}{z-w}. \\ \end{array} $$ Note that this gives $\beta(z)\gamma(w)=\gamma(w)\beta(z)=-\gamma(z)\beta(w)$.