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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).

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  • $\begingroup$ 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 $. $\endgroup$
    – Antoine
    Mar 15, 2017 at 11:26
  • $\begingroup$ Note that this is a misprint in the book as well, but it is mentioned in the errata. $\endgroup$
    – Antoine
    Mar 15, 2017 at 11:27
  • $\begingroup$ @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? $\endgroup$
    – levitt
    Mar 15, 2017 at 12:24
  • $\begingroup$ You have another typo in equation (2), the power of z is missing. $\endgroup$
    – Antoine
    Mar 15, 2017 at 14:57

1 Answer 1

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

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  • $\begingroup$ Thanks a lot! I feel stupid for not seeing that step. $\endgroup$
    – levitt
    Mar 15, 2017 at 18:56

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