OPE coefficients identity operator when I have two canonically normalized operators $\phi_{1}$ and $\phi_{2}$ and I want to compute their OPE in terms of the identity operator, is there any way to actually calculate the first levels of coefficients by hand? So far I have only found publications concerning computer codes that are supposed to do this for you... 
Thx!
Jane
 A: This is standard material for any text on 2d CFT's. First, the Virasoro algebra contains holomorphic and anti-holomorphic parts and you can see (exercise!) that the conformal blocks factorize into respective pieces. So lets consider just the holomorphic part. The key is the decomposition of the identity in the conformal module of the intermediate state (the identity operator in your case, $h=0$), or, in other words, projection onto the corresponding module
$$
P_{h}=\sum_{\{k\},\{l\}}L_{-\{k\}}\left\vert h\right\rangle\left\langle h\right\vert L_{\{l\}}Q_{\{k\},\{l\}}.
$$
Here we sum over the multi-indices $k,l$ so that $L_{-\{k\}}\left\vert h\right\rangle$ span the standard basis for descendants of the intermediate operator, and $Q$ (to be determined later) is introduced to make $P_0$ an orthogonal projection onto these descendants.
We now consider the four-point function, where $|z|<1$,
$$
\langle \phi_1(0)\phi_2(z)\phi_3(1)\phi_4(\infty)\rangle=\left\langle \phi_4\right\vert \phi_3(1) \phi_2(z)\left\vert\phi_1\right\rangle,
$$
and insert the projection $P_h$ to isolate the contribution of the identity,
$$
\text{Conformal block}=\left\langle \phi_4\right\vert \phi_3(1) P_h \phi_2(z)\left\vert\phi_1\right\rangle=\sum_{\{k\},\{l\}}\left\langle \phi_4\right\vert \phi_3(1)L_{-\{k\}}\left\vert h\right\rangle\left\langle  h\right\vert L_{\{l\}} \phi_2(z)\left\vert\phi_1\right\rangle Q_{\{k\},\{l\}}=\\
=\sum_{\{k\},\{l\}}z^{|l|+h-h_{1}-h_{2}}\left\langle \phi_4\right\vert \phi_3(1)L_{-\{k\}}\left\vert h\right\rangle\left\langle h\right\vert L_{\{l\}} \phi_2(1)\left\vert\phi_1\right\rangle Q_{\{k\},\{l\}}.
$$
Here one can compute the expectation values by commuting the $L$'s to the left or to the right and recalling their action on primary fields. Thus one can compute the conformal blocks as soon as one determines the matrix $Q$.
We can determine $Q$ by requiring $P_h L_{-\{s\}}|h\rangle =L_{-\{s\}}|h\rangle $ which is the defining property of $P_0$. That gives
$$
\sum_{\{l\}}Q_{\{k\},\{l\}}\left\langle h\right\vert L_{\{l\}}L_{-\{s\}}\left\vert h\right\rangle = \delta_{\{k\},\{s\}}.
$$
Thus $Q$ is just the inverse to the inner product matrix $\left\langle h\right\vert L_{\{l\}}L_{-\{s\}}\left\vert h\right\rangle$. That latter matrix is block-diagonal, since descendants at different levels are orthogonal. Therefore you can compute it block by block, and then take the inverse of each block. I can perhaps include a calculation of a first non-trivial block later.
Note, however, that this is not the most effective way for numerical computation of the conformal block. The state of art is, I believe, Zamolodchikov's 1987 paper on recursion relations. The idea there is that $Q$ has poles in the dimension $h$ of the exchanged operator when some of the descendants become null (in that case the inner product matrix becomes degenerate, and the inverse $Q$ becomes singular), and thus the full conformal block has a pole as well. It is then demonstrated that the residue itself is proportional to a new conformal block. In this way, one is able to efficiently compute conformal blocks via a recursive procedure. This can be orders of magnitude faster than the "by definition" calculation.
