I am recently puzzled by one question in CFT. I want to compute the correlator of $$\langle T(z)J(w)O_1(x_1) O_2(x_2)\rangle$$ where $T$ is the stress tensor, $J$ is the $U(1)$ Kac-Moody current, $O_1,O_2$ have the same weights $h$ but opposite charges $q$. How can I compute the correlator?

I think I can use the Ward identity corresponding to stress tensor and Kac-Moody current. Starting with $$\langle O_1(x_1) O_2(x_2)\rangle=\frac{1}{(x_1 - x_2)^{2 h}} $$ I can compute $$\langle TO_1O_2\rangle=\frac{h \left(x_1-x_2\right){}^{2-2 h}}{\left(x_1-z\right){}^2 \left(x_2-z\right){}^2} $$ using stress Ward identity. Then I use the Ward identity for $J$ to compute

$$F_1=\langle JTO_1O_2\rangle=\frac{h q \left(x_1-x_2\right){}^{3-2 h}}{\left(w-x_1\right) \left(w-x_2\right) \left(x_1-z\right){}^2 \left(x_2-z\right){}^2}$$

On the other hand, I can change the order: I can first compute $$\langle JO_1O_2\rangle=\frac{q \left(x_1-x_2\right){}^{1-2 h}}{\left(w-x_1\right) \left(w-x_2\right)} $$

and then compute

$$F_2=\langle TJO_1O_2\rangle=\frac{q \left(x_1-x_2\right){}^{1-2 h} \left(\frac{h \left(x_1-x_2\right){}^2}{\left(w-x_1\right) \left(w-x_2\right)}+\frac{\left(z-x_1\right) \left(z-x_2\right)}{(w-z)^2}\right)}{\left(x_1-z\right){}^2 \left(x_2-z\right){}^2}$$

The problem is that I find the two results do not agree $F_1\neq F_2$. I am not sure what is wrong? Is the Ward identity correct?

In general I am wondering how to compute the correlators of involving different types of currents, like $J,T$ here...Intuitively I am not sure why taking different orders for Ward identity should give the same result, although I believe this is should be guaranteed by associativity of the algebra...

  • 1
    $\begingroup$ Maybe there's a mistake in the way you've calculated the correlations (you say they don't agree with each other). Maybe edit the question and show us your work? $\endgroup$
    – SuperCiocia
    Aug 10, 2021 at 21:34
  • $\begingroup$ Thanks! I added my computations but I could not figure out what is wrong, whether it is my mistake or my approach... $\endgroup$
    – Light man
    Aug 10, 2021 at 23:52

1 Answer 1


$F_2$ is correct and indeed it is most easily found with \begin{align} \langle TJO_1O_2 \rangle &= \left [ \frac{1}{(z-w)^2} + \frac{h}{(z - x_1)^2} + \frac{h}{(z-x_2)^2} \right ] \langle JO_1O_2 \rangle \\ &+ \left [ \frac{\partial_w}{z - w} + \frac{\partial_{x_1}}{z - x_1} + \frac{\partial_{x_2}}{z - x_2} \right ] \langle JO_1O_2 \rangle \end{align} and inserting the 3pt function you wrote.

If we took the $J(w)T(z)$ OPE to be regular, we would find your $F_1$ but this would only be valid if the stress tensor were a Kac-Moody primary. To determine what the $J(w)T(z)$ OPE really is, we have to at least use commutativity. \begin{align} J(w)T(z) &= T(z)J(w) \\ &= \frac{J(w)}{(z - w)^2} + \frac{\partial J(w)}{z - w} + \dots \\ &= \frac{J(z)}{(z - w)^2} + \dots \end{align} Therefore, the $\frac{1}{z - w}$ term is zero like you wanted but it wasn't consistent to assume the absence of terms that are even more singular.

Using this OPE in the correlator now, we find \begin{align} \langle TJO_1O_2 \rangle = \frac{1}{(z - w)^2} \langle JO_1O_2 \rangle + \left [ \frac{q}{w - x_1} - \frac{q}{w - x_2} \right ] \langle TO_1O_2 \rangle \end{align} which reassuringly gives $F_2$.

  • $\begingroup$ Great! Thanks a lot! I did not realize that the stress tensor is not affine primary... $\endgroup$
    – Light man
    Aug 12, 2021 at 8:50

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