Anticommuting relation in $bc$ CFT

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I have a question about conformal field theory in Polchinski's string theory vol 1 p. 61. Given anticommuting fields $b$ and $c$ and the Laurent expansions $$b(z) = \sum_{m=-\infty}^{\infty} \frac{b_m}{z^{m+\lambda}}, \,\,\, c(z) = \sum_{m=-\infty}^{\infty} \frac{c_m}{z^{m+1-\lambda}} \,\,\, (2.7.16)$$

Assuming $\lambda$ is an integer.

It is said

The OPE gives the anticommutators $$\{ b_m, c_n \} = \delta_{m,-n} \,\,\, (2.7.17)$$

I tried to vertify (2.7.17), but I think my approach uses too strong condition. Here is my derivation:

Consider Operator Product Expansion (OPE) (similar with (2.5.9)) $$b(z_1) c(z_2) \sim \frac{1}{z_{12}}, \,\,\, c(z_2) b(z_1) \sim \frac{1}{z_{21}}$$ where $z_{ij}=z_i - z_j \,\, (2.1.22)$.

Therefore (it is just the anticommuting properties) $$\{ b(z_1), c(z_2) \} \sim 0 \,\,\, (3)$$

Plug in Laurent expansions for $b$ and $c$, (2.7.16), into (3), we have $$\{ b(z_1), c(z_2) \} = \sum_{m=-\infty}^{\infty} \sum_{n =-\infty}^{\infty} \frac{ 1}{z_1^{m+\lambda}} \frac{1}{z_2^{n+1-\lambda}} \{ b_m, c_n \} \,\,\, (4)$$

Since I aim at vertify (2.7.17), simply plug in (2.7.17) into (4) $$\{ b(z_1), c(z_2) \} = \sum_{m=-\infty}^{\infty} \frac{1}{z_2} \left( \frac{z_2}{z_1} \right)^{m+\lambda} = \sum_{m=0}^{\infty} \frac{1}{z_2} \left( \frac{z_2}{z_1} \right)^{m} +\sum_{m=-\infty}^{0} \frac{1}{z_2} \left( \frac{z_2}{z_1} \right)^{m} -\frac{1}{z_2}$$ $$= \sum_{m=0}^{\infty} \frac{1}{z_2} \left( \frac{z_2}{z_1} \right)^{m} +\sum_{m=0}^{\infty} \frac{1}{z_2} \left( \frac{z_1}{z_2} \right)^{m} -\frac{1}{z_2}$$ Without lose of generality, assuming $z_1>z_2$, we have $$\{ b(z_1), c(z_2) \} = \frac{1}{z_2} \frac{1}{1-\frac{z_2}{z_1}}+ \frac{1}{z_2}\frac{1}{1-\frac{z_1}{z_2}} - \frac{1}{z_2} =0 \,\,\, (5)$$

But if I use $\{ b_m, c_n \} =0$, it could also give Eq. (3). How to derive (2.7.17) properly?

One can derive the commutation relation directly from the OPE. Recall equation (2.6.14) in Polchinski. We first note $$b_m = \oint_C \frac{dz}{2\pi i} z^{m+\lambda-1} b(z),~~c_n = \oint_C \frac{dz}{2\pi i} z^{n-\lambda} c(z)$$ We then have $$\begin{split} \{ b_m, c_n \} &= \oint_{C_2} \frac{dz_2}{2\pi i} \text{Res}_{z_1 \to z_2} z_1^{m+\lambda-1}z_2^{n-\lambda} b(z_1)c(z_2) \\ &= \oint_{C_2} \frac{dz_2}{2\pi i} \text{Res}_{z_1 \to z_2} \frac{ z_1^{m+\lambda-1}z_2^{n-\lambda}}{z_1-z_2} \\ &= \oint_{C_2} \frac{dz_2}{2\pi i} z_2^{m+n-1} \\ &= \delta_{m+n,0} \end{split}$$
• The proof is the same. Try and follow the procedure to rederive the relation for anticommuting operators. You will find that the LHS simply becomes an anticommutator $\{ b_m, c_n \}$ instead of $[b_m, c_n]$ Commented Jul 25, 2013 at 1:21