# 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).

• In your equation (5), a $z^n$ is missing, it should be $L_n\Phi_1(z)|h\rangle = (z^{n+1}\partial_z + (n+1)h z^n )\Phi_1(z)|h\rangle$. – Antoine Mar 15 '17 at 11:26
• Note that this is a misprint in the book as well, but it is mentioned in the errata. – Antoine Mar 15 '17 at 11:27
• @user40085 Thanks for pointing out. Fixed it. You also seem to have come across this particular thing. Have you been able to work it out? – levitt Mar 15 '17 at 12:24
• You have another typo in equation (2), the power of z is missing. – Antoine Mar 15 '17 at 14:57

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 $$\boxed{L_n |N+n,h_p\rangle = (h_p +h(n-1) + N)|N,h_p\rangle }$$