Question on 2D Conformal bootstrap recursion relation I am trying to derive the following recursion relation for 2D conformal bootstrap from Di Francesco (page182), however I am stuck.
Starting from the operator algebra(note that I have not written the parts in $\bar{z}$ variable)
$$ 
\Phi_1(z)\Phi_2(0) = \sum_p \sum_{\{k\}}C_{12}^{p\{k\}}z^{h_p -h_1-h_2+K}\Phi_p^{\{k\}}(0) \tag{1}
$$
where $h_i$ are the scaling dimensions of the fields $\phi_i(z)$. $K= \sum_i k_i$ and ${\{k\}}$ means the collection of $k_i$ indices.
The claim is that, $C_{12}^{p\{k\}}=C_{12}^p\beta_{12}^{p\{k\}}$ where $C_{12}^p$ is the coefficient of 3 point correlation function. 
As a special case of this, the text considers, when $h_1=h_2=h$ acting on vacuum $|0\rangle$
$$
\Phi_1(z)|h\rangle = \sum_p \sum_{\{k\}}C_{12}^p\phi(z)|h_p\rangle \tag{2}
$$
where $\phi(z)= \sum_{\{k\}}z^K\beta_{12}^{p\{k\}}L_{k_1}...L_{k_N}$.
Next a state is defined,
$$|z,h_p\rangle\equiv \phi(z)|h_p\rangle=\sum_{N=0}^\infty Z^N|N,h_p\rangle \tag{3}$$
where $|N,h_p\rangle$ is a descendant state at level $N$ in the verma module.
$$L_0|N,h_p\rangle = (h_p+N)|N,h_p\rangle \tag{4} $$
Next, it says that $L_n=\frac{1}{2\pi i}\oint dz z^{n+1}T(z)$ is operated on both sides of (1). On LHS, I could see that it was 
$$ L_n\Phi_1(z)|h\rangle = (z^{n+1}\partial_z + (n+1)hz^n)\Phi_1(z)|h\rangle \tag{5}$$. I however could not see the RHS part where the text claims that it is
$$\sum_p \sum_{\{k\}}C_{12}^pL_n|z, h_p\rangle = \sum_p \sum_{\{k\}}C_{12}^p[h_p + h(n-1)z^n +z^{n+1}\partial_z]|z, h_p\rangle \tag{6}$$
And then he finally obtains, 
$$L_n|N+n,h_p\rangle=(h_p + h(n-1) +N)|N,h_p\rangle  \tag{7}$$
I could intuitively see that acting by $L_n$ on $|N+n,h_p\rangle$ will bring it down to $|N,h_p\rangle$ but I am not able to derive the factor in front. I am thinking that there should be a way to derive (7), directly from (4).
 A: Note first that while copying the equations from the book, you forgot several powers of $z$. 
You can substitute the corrected equation (2) into your equation (5), and obtain 
$$L_n \left[ \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} |z,h_p\rangle \right] = (z^{n+1}\partial_z + (n+1)hz^n) \left[ \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} |z,h_p\rangle \right]  $$
On the LHS, the operator $L_n$ goes through the scalars and hits the state $|z,h_p\rangle$. On the RHS, you just compute the derivative, and obtain 
$$ \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} L_n  |z,h_p\rangle  =  \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} (z^n (h_p - 2h) + z^n h (n+1) + z^{n+1} \partial_z) |z,h_p\rangle  $$
Finally, you get 
$$\boxed{ \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} L_n  |z,h_p\rangle  =  \sum_p \sum_{\{k\}}C_{12}^p z^{h_p - 2h} (z^n (h_p +h(n-1)) + z^{n+1} \partial_z) |z,h_p\rangle } \tag{6} $$
as claimed in the book. 
From this, one deduces the relation 
$$ L_n  |z,h_p\rangle  =  (z^n (h_p +h(n-1)) + z^{n+1} \partial_z) |z,h_p\rangle \tag{8} $$
Now to prove equation (7), just substitute 
$$|z,h_p\rangle=\sum_{N=0}^\infty z^N|N,h_p\rangle \tag{3}$$
 into (8) : 
$$ \sum z^N L_n  |N,h_p\rangle  =  (z^n (h_p +h(n-1)) + z^{n+1} \partial_z) \sum z^N|N,h_p\rangle $$
and simplify both sides : 
$$ \sum z^{n+N} L_n  |N+n,h_p\rangle  =   \sum z^{N+n} (h_p +h(n-1) + N)|N,h_p\rangle $$
Identifying equal powers of $z$, you obtain
$$ \boxed{L_n  |N+n,h_p\rangle  =  (h_p +h(n-1) + N)|N,h_p\rangle }$$
