OPE coefficients of primaries in a 2D CFT

Question

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?

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.