Contour integral for commutator of fermionic fields

Suppose we have primary fields $$A$$ and $$B$$ which have the OPE,

$$A(z) B(w) = \frac{1}{z-w} = -B(z)A(w), \quad |z| > |w|,\tag{1}$$

so they have fermionic statistics. Now I was curious how this would affect computing their commutation relations, using the usual contour integral approach. If $$A$$ has weight $$h$$ and $$B$$ has weight $$h'$$, I can get the modes using,

$$A_m = \oint \mathrm dz \, z^{m+h-1} A(z), \quad B_n = \oint \mathrm dw\, w^{n+h'-1} B(w),\tag{2}$$

where the contour integrals encircle $$z=0$$ and $$w=0$$ respectively. Then normally we would write,

$$[A_n,B_n] = \oint_{C_1} \mathrm dw \oint_{C_2} \mathrm dz \, z^{m+h-1} w^{n+h'-1}A(z) B(w),\tag{3}$$

where $$C_1$$ is an integral around $$w=0$$ and $$C_2$$ is an integral around $$z=w$$. This much has been explained in previous questions, but I think this wouldn't be valid for fermionic statistics? Since in writing out the commutator, you swap the fields and pick up a minus sign, but then I am not sure which contour integration you would carry out.

I would appreciate any clarification because I cannot find an example where this is carried out for the free fermion, for example.

2. Also it should probably be stressed that the RHS of eq. (3) contains an implicitly written radial operator ordering $${\cal R}$$. Similarly, eq. (1) should properly read \begin{align} {\cal R}[A(z) B(w)] ~=~&\theta(|z|\!-\!|w|)A(z) B(w)\cr &~+~ (-1)^{|A||B|}\theta(|w|\!-\!|z|)B(w)A(z) \cr ~=~& \frac{1}{z-w} + \text{non-singular terms}.\end{align}\tag{1'}
• If I did want to compute the commutator though, the $z=0$ pole contributions would not cancel, as in the bosonic case, so I'd have $2 \mathrm{Res}_{z=0} + \mathrm{Res}_{z=w}$ then? But when I calculate this, the $\mathrm{Res}_{z=0}$ seems to require $n < 0$ for $A_n$ to be non-zero, and seems like an overly complicated result for the commutator. Jan 19, 2023 at 19:17