Operator algebra of chiral quasi-primary fields

[J.Hong]

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 \phi_{(i)m} and \phi_{(j)n} as contour integrals over each fields, and took commutation relation. I used ansatz, \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), to calculate the commutation relation, \left [ \phi_{(i)m},\phi_{(j)n} \right ]. h_i, h_j, and h_k are conformal dimension of the fields, \phi_i(z), \phi_j(z), and \phi_k(z), respectively.
Finally, I obtained the result, \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} , where 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)}}. (x_{(r)}\equiv \Gamma (x+r)/\Gamma (x)). 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, \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}, where \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)!} . 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.

 

[fzero]

\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)!} .

I don't have the patience to go through this step by step, but you should first note that this isn't really a double sum because of the constraint that $r+s=h_i+h_j-h_k-1$. So the first step would be to eliminate $s$ using this constraint. It looks like both expressions are then sums over the same index $r$ and same range. So you can compare the expressions term-wise, for fixed $r$.

Convert all short-cut notation like the binomial coefficients, the $x_{(r)}$ notation, and the factorials into $\Gamma$ functions. Some common coefficients are already obvious in what you've wrote down so far, if you go a bit further, I expect things to be a bit more clearer. Simplify as much as you can and then post back if you still have questions.

 

[J.Hong]

I don't have the patience to go through this step by step, but you should first note that this isn't really a double sum because of the constraint that $r+s=h_i+h_j-h_k-1$. So the first step would be to eliminate $s$ using this constraint. It looks like both expressions are then sums over the same index $r$ and same range. So you can compare the expressions term-wise, for fixed $r$.

Convert all short-cut notation like the binomial coefficients, the $x_{(r)}$ notation, and the factorials into $\Gamma$ functions. Some common coefficients are already obvious in what you've wrote down so far, if you go a bit further, I expect things to be a bit more clearer. Simplify as much as you can and then post back if you still have questions.

Thanks, fzero :)

Even though eliminate one variable by using the constraint you said, it still remains different term. See below.

1. My result
\left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}\sum_{r=0}^{h_i+h_j-h_k-1}\frac{(-1)^{h_i+h_j-h_k-1-r}(m+n+h_k+r-1)!(m+h_i-1)!(2h_k-1)!(2h_i-2-r)!\phi_{(k)m+n}}{(m+n+h_k-1)!(h_i+h_j-h_k-1-r)!(m+h_i-1-r)!r!(h_i-h_j+h_k-1)!}.

2. The result in many textbooks
\left [ \phi_{(i)m},\phi_{(j)n} \right ]=\sum_{k\geq 0}C^k_{ij}\sum_{r=0}^{h_i+h_j-h_k-1}\frac{(-1)^{r}(n+h_j-1)!(m+h_i-1)!(2h_k-1)!(2h_i-2-r)!(h_j-h_i+h_k-1-r)!\phi_{(k)m+n}}{(n-h_i+h_k+r)!(h_i+h_j-h_k-1-r)!(m+h_i-1-r)!r!(h_i+h_j+h_k-2)!(h_j-h_i+h_k-1)!(h_i-h_j+h_k-1)!}.

I abbreviated the term includes \delta _{m,-n} which is matche each other.

 

[fzero]

I had a chance to look at this a bit more. I was able to reproduce your result

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)}}

but I haven't been able to show that this is equal to the textbook result. I had a few ideas about how to manipulate this, but I'm still left with some strange factors. Maybe you'll be able to straighten things out further.


We need the identities

$$
\binom{n}{k} = \binom{n}{n-k} = \frac{\binom{n}{h} \binom{n-h}{k}}{\binom{n-k}{h}} = \sum_{j=0}^k \binom{m}{j} \binom{n-m}{k-j},$$
where in the 4th term, we can choose any $m$ that we want. We can use the 4th identity to write

$$
\begin{split} (m+n+h_k)_{(r)} & = r! \binom{m+n+h_k+r-1}{r} \\
& = \sum_{t=0}^r r! \binom{m+h_k-h_j+r}{t} \binom{m+h_j-1}{r-t} .\end{split}
$$

Under the sum over $r$, we can shift the index to $s=r-t$, so that

$$ (m+n+h_k)_{(r)} \longrightarrow \sum_{s} r!\binom{m+h_k-h_j+r}{r-s} \binom{m+h_j-1}{s}$$


Set $r+s= h_i+h_j-h_k-1$, we can use the 3rd identity to write

$$\binom{m+h_i-1}{h_i+h_j-h_k-1-r} = \binom{m+h_i-1}{s} = \frac{\binom{m+h_i-1}{s} \binom{m+h_i-r-1}{s}}{\binom{m+h_i-s-1}{r}} . $$

Furthermore,

$$(h_i-h_j+h_k)_{(r)} = \frac{(2h_i-r-2)!}{(2h_i-r-s-2)!} \frac{r!\binom{2h_i-s-2}{r}}{\binom{2h_i-r-2}{s}}.$$

We can therefore write

$$P(m,n;h_i,h_j,h_k) = \sum_{r,s} (-1)^r \binom{m+h_i-1}{s} \binom{n+h_j-1}{s} \frac{(2h_i-r-2)! }{(2h_i-r-s-2)! } \frac{(2h_j-s-2)! }{(2h_j-r-s-2)! } p_{r,s} $$

where

$$p_{r,s} = \frac{\binom{m+h_k-h_j+r}{r-s} \binom{m+h_i-r-1}{s}}{\binom{m+h_i-s-1}{r}}
\frac{\binom{2h_i-s-2}{r}}{\binom{2h_i-r-2}{s}\binom{ 2h_j-s-2}{r}} .
$$

We can use the 3rd identity a couple of times to write this as

$$p_{r,s} = \frac{\binom{m+h_k-h_j+r}{r-s} }{\binom{ 2h_j-s-2}{r}} \frac{\binom{m+h_i-1}{r}}{\binom{m+h_i-1}{s}}
\frac{\binom{2h_i-2}{s}}{\binom{2h_i-2}{r}} .
$$

There are some obvious cancellations, but I haven't been able to get all of the factors to cancel in order to recover the textbook answer. Maybe you'll have better luck and/or turn up some mistake that I made.

 

[J.Hong]

We need the identities

$$
\binom{n}{k} = \binom{n}{n-k} = \frac{\binom{n}{h} \binom{n-h}{k}}{\binom{n-k}{h}} = \sum_{j=0}^k \binom{m}{j} \binom{n-m}{k-j},$$
where in the 4th term, we can choose any $m$ that we want.

Thank you, fzero

I proved two formulas are the same each other using general version of the identities you introduced. Thank you again, fzero.