OPE coefficients of primaries in a 2d CFT
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. 402
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? 1021