- #1

- 4

- 0

Studying conformal field theory, I tried to derive general expression for the commutation relations of the modes of two chiral quasi-primary fields.

At first, I expressed the modes [itex] \phi_{(i)m} [/itex] and [itex] \phi_{(j)n} [/itex] as contour integrals over each fields, and took commutation relation. I used ansatz, [tex] \phi _i(z)\phi_j(w)=\sum_{k,n\geqslant 0}C^k_{ij}\frac{a^n_{ijk}}{n!}\frac{1}{(z-w)^{h_i+h_j-h_k-n}}\partial ^n\phi_k(w) [/tex], to calculate the commutation relation, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]. [/tex] [itex] h_i, h_j, [/itex] and [itex] h_k [/itex] are conformal dimension of the fields, [itex] \phi_i(z), \phi_j(z), [/itex] and [itex] \phi_k(z) [/itex], respectively.

Finally, I obtained the result, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}P(m,n;h_i,h_j,h_k)\phi_{(k)m+n}+d_{ij}\delta _{m,-n}\binom{m+h_i-1}{2h_i-1} ,[/tex] where [tex] P(m,n;h_i,h_j,h_k)=\sum_{r=0}^{h_i+h_j-h_k-1}\binom{m+h_i-1}{h_i+h_j-h_k-1-r}\frac{(-1)^r(h_i-h_j+h_k)_{(r)}(m+n+h_k)_{(r)}}{r!(2h_k)_{(r)}} [/tex]. [tex] (x_{(r)}\equiv \Gamma (x+r)/\Gamma (x)). [/tex] I took advantage of two and three point functions to get the result.

I think my calculation is right. In many textbooks on CFT, however, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}P(m,n;h_i,h_j,h_k)\phi_{(k)m+n}+d_{ij}\delta _{m,-n}\binom{m+h_i-1}{2h_i-1} [/tex], where [tex] \sum_{r,s\in \mathbb{Z},r+s=h_i+h_j-h_k-1}\binom{m+h_i-1}{r}\binom{n+h_j-1}{s}\frac{(-1)^r(2h_k-1)!}{(h_i+h_j+h_k-2)!}\frac{(2h_i-2-r)!}{(2h_i-2-r-s)!}\frac{(2h_j-2-s)!}{(2h_j-2-r-s)!} .[/tex] This result looks different from my result, but two result should be the same. I don't know how to obtain the formula in textbooks and how the two results are the same. Please, teach me with explicit calculation procedures.

Thanks.

At first, I expressed the modes [itex] \phi_{(i)m} [/itex] and [itex] \phi_{(j)n} [/itex] as contour integrals over each fields, and took commutation relation. I used ansatz, [tex] \phi _i(z)\phi_j(w)=\sum_{k,n\geqslant 0}C^k_{ij}\frac{a^n_{ijk}}{n!}\frac{1}{(z-w)^{h_i+h_j-h_k-n}}\partial ^n\phi_k(w) [/tex], to calculate the commutation relation, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]. [/tex] [itex] h_i, h_j, [/itex] and [itex] h_k [/itex] are conformal dimension of the fields, [itex] \phi_i(z), \phi_j(z), [/itex] and [itex] \phi_k(z) [/itex], respectively.

Finally, I obtained the result, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}P(m,n;h_i,h_j,h_k)\phi_{(k)m+n}+d_{ij}\delta _{m,-n}\binom{m+h_i-1}{2h_i-1} ,[/tex] where [tex] P(m,n;h_i,h_j,h_k)=\sum_{r=0}^{h_i+h_j-h_k-1}\binom{m+h_i-1}{h_i+h_j-h_k-1-r}\frac{(-1)^r(h_i-h_j+h_k)_{(r)}(m+n+h_k)_{(r)}}{r!(2h_k)_{(r)}} [/tex]. [tex] (x_{(r)}\equiv \Gamma (x+r)/\Gamma (x)). [/tex] I took advantage of two and three point functions to get the result.

I think my calculation is right. In many textbooks on CFT, however, [tex] \left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}P(m,n;h_i,h_j,h_k)\phi_{(k)m+n}+d_{ij}\delta _{m,-n}\binom{m+h_i-1}{2h_i-1} [/tex], where [tex] \sum_{r,s\in \mathbb{Z},r+s=h_i+h_j-h_k-1}\binom{m+h_i-1}{r}\binom{n+h_j-1}{s}\frac{(-1)^r(2h_k-1)!}{(h_i+h_j+h_k-2)!}\frac{(2h_i-2-r)!}{(2h_i-2-r-s)!}\frac{(2h_j-2-s)!}{(2h_j-2-r-s)!} .[/tex] This result looks different from my result, but two result should be the same. I don't know how to obtain the formula in textbooks and how the two results are the same. Please, teach me with explicit calculation procedures.

Thanks.

Last edited: