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Operator algebra of chiral quasi-primary fields

  1. Apr 25, 2013 #1
    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.
    Last edited: Apr 25, 2013
  2. jcsd
  3. Apr 25, 2013 #2


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    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.
  4. Apr 26, 2013 #3
    Thanks, fzero :)

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

    1. My result
    [tex]\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)!}.[/tex]

    2. The result in many textbooks
    [tex]\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)!}.[/tex]

    I abbreviated the term includes [itex]\delta _{m,-n}[/itex] which is matche each other.
  5. Apr 27, 2013 #4


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

    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

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


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

    We can therefore write

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


    $$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.
  6. May 10, 2013 #5
    Thank you, fzero

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