I am trying to compute the OPE coefficients in a 2D CFT, and I am convincing myself of something that I know is not true but cant find my mistake.

Given primaries $V_{\Delta_1}$ and $V_{\Delta_2}$ I know that

$$V_{\Delta_1}(z_1)V_{\Delta_2}(z_2) = \displaystyle\sum_\Delta C_{12}^\Delta(z_1,z_2) V_{\Delta}(z_2) $$

Where $\Delta$ runs over all fields in my algebra, primaries and descendants. This far I know, here begins my guesswork. Since this holds as an operator equation, it should hold for expectation values as well. So I know

$\langle V_{\Delta_1}(z_1)V_{\Delta_2}(z_2) \rangle = \displaystyle\sum_\Delta C_{12}^\Delta(z_1,z_2)\langle V_{\Delta}(z_2) \rangle $

Now I know from the global ward identities that

$\langle V_{\Delta_1}(z_1)V_{\Delta_2}(z_2) \rangle = \delta_{\Delta_1 \Delta_2 }|z_{12}|^{-2\Delta_1}$

Where I've normalized my fields so that the overall coefficient in the 2-point function is 1. All this combined tells me that

$\delta_{\Delta_1 \Delta_2 }|z_{12}|^{-2\Delta_1} = \displaystyle\sum_\Delta C_{12}^\Delta(z_1,z_2)\langle V_{\Delta}(z_2) \rangle$

So to compute $C_{12}^\Delta(z_1,z_2)$ I need to know what that $\langle V_{\Delta}(z) \rangle $ can be. Again using global ward identities I know that for primaries $\langle V_{\Delta}(z) \rangle = 0$ unless $\Delta = 0$ in which case it is a constant. So I already know that $C_{12}^\Delta(z_1,z_2) = 0$ for any primary $\Delta \neq 0$. Now we move on to the descendants. I know that

$ \langle L_{-n} V_{\Delta}(z) \rangle = \left[ \frac{\partial_z}{z^{n-1}} +\frac{(n-1\Delta)}{z^n} \right] \langle V_{\Delta}(z) \rangle $

If $\Delta = 0$ then $\langle V_{\Delta}(z) \rangle$ is a constant so $\partial \langle V_{\Delta}(z) \rangle = 0$ and since $\Delta = 0$ I also know that $\Delta \langle V_{\Delta}(z) \rangle = 0$. If $\Delta \neq 0$ then $\langle V_{\Delta}(z) \rangle = 0$, so I get no contributions.

It seems then that the only non zero $C_{12}^\Delta(z_1,z_2)$ is $C_{11}^0(z_1,z_2)$ which I can get by

$\delta_{\Delta_1 \Delta_2 }|z_{12}|^{-2\Delta_1} = C_{11}^0(z_1,z_2)\langle V_{0}(z_2) \rangle$

and so $C_{11}^0(z_1,z_2) = |z_{12}|^{-2\Delta_1}$ (up to a constnat), but this is incorrect, in fact I know that there are contributions from $ \langle L_{-1} V_{\Delta}(z) \rangle$, but where is my mistake?

  • $\begingroup$ OPE actually does not really converge as an operator equation (but this is probably irrelevant to your question). $\endgroup$ Aug 23, 2019 at 5:57
  • 1
    $\begingroup$ "So I already know that $C^\Delta_{12}(z_1,z_2)=0$ for any primary $\Delta\neq 0$" -- this is wrong. It is not $C^\Delta_{12}(z_1,z_2)$ that vanishes, it is $\langle V_\Delta(z)\rangle$. You value for $C^0_{11}(z_1,z_2)$ is, in fact, correct. $\endgroup$ Aug 23, 2019 at 6:01
  • $\begingroup$ @PeterKravchuk Why do you say the OPE doesn't converge? I thought that it does, with radius of convergence given by the closest other operator. $\endgroup$ Aug 24, 2019 at 7:23
  • $\begingroup$ @Wakabaloola it does, but precisely because this depends on where the other operators are, it does but converge as an operator equation. Specifically, it is easy to exhibit states between which OPE does not converge. $\endgroup$ Aug 24, 2019 at 13:44

1 Answer 1


You're almost there, but you are missing some important facts about the $\Delta = 0$ multiplet. You say "in fact I know that there are contributions from $L_{-1} V_\Delta(z)$". This is true in general, that is to say for $\Delta > 0$ (except for some special values of $\Delta$ that depend on $c$). However, in a unitary CFT there can only be one operator with $\Delta = 0$, the unit operator, and this operator obeys $L_{-1} V_{\Delta = 0} = 0$ as an operator equation. (To prove this, you can simply compute that the state $|L_{-1} V_{\Delta = 0} \rangle$ has zero norm, and since the theory is unitary it implies that the state vanishes.) Consequently, your conclusion that

$$C_{11}^0(z_1,z_2) = |z_{12}|^{-2\Delta_1}$$

is completely correct.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.