The following comes from Di Francesco et al., section 6.6.4. The exercise is number 6.5.

A conformal block is given by $$\mathcal{F}^{21}_{34}(p\mid x) = x^{h_p-h_3-h_4}\sum_{\{k\}}\beta^{p\{k\}}_{34}x^K\frac{\langle h_1|\phi_2(1)L_{-k_1}\cdots L_{-k_N}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}$$ where the denominator is equal to $(C_{12}^{p})^{1/2}$ and $K=\sum_i k_i$. The blocks can be written as a power series: $$\mathcal{F}_{34}^{21}(p\mid x)=x^{h_p-h_3-h_4}\sum_{K=0}^{\infty} \mathcal{F}_K x^K$$

I want to compute $$\mathcal{F}_1=\beta_{34}^{p\{1\}}\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}=\frac12\frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle}{\langle h_1|\phi_2(1)|h_p\rangle}$$ and in particular show that $$ \mathcal{F}_1 = \frac{(h_p+h_2-h_1)(h_p+h_3-h_4)}{2h_p}$$

The authors suggest to compute the block by commuting the Virasoro generators over $\phi_2(1)$.

$$\langle h_1|\phi_2(1)L_{-1}|h_p\rangle=\langle h_1|L_{-1}\phi_2(1)|h_p\rangle-\langle h_1|[L_{-1},\phi_2(1)]|h_p\rangle$$

I think the second term is equal to $\langle h_1|\partial\phi_2(1)|h_p\rangle$ while the first is zero because $L_{-1}$ annihilates primary fields. The term $\langle h_1|\partial\phi_2(1)|h_p\rangle$ is calculated using $\mathcal{L}_{-1}$, so

$$\langle h_1|\partial\phi_2(1)|h_p\rangle = \mathcal{L}_{-1}\langle h_1|\phi_2(1)|h_p\rangle =-\partial\langle h_1|\phi_2(1)|h_p\rangle$$ But doesn't $\partial \phi_2(1)=0$ if $\phi_2$ is defined at $1$?

Where have I gone wrong? How do you compute $\mathcal{F}_{1}$?


1 Answer 1


Firstly, $\beta_{34}^{p\{1\}}=\frac{1}{2}$ only when $h_3=h_4$, this is the case of (6.177). For general $h_3$ and $h_4$, I believe $$ \beta_{34}^{p\{1\}}=\frac{h_p+h_3-h_4}{2h_p}, $$ because this is the only way $h_3$ and $h_4$ appear in $\mathcal{F}_1$ and gives you $\frac{1}{2}$ when $h_3=h_4$. So we only need to prove $$ \frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle} {\langle h_1|\phi_2(1)|h_p\rangle}=h_p+h_2-h_1. $$ Commute $L_{-1}$ over $\phi_2$ we can get $$ \langle h_1|\phi_2(1)L_{-1}|h_p\rangle= -\langle h_1|\partial\phi_2(1)|h_p\rangle $$ where the $\partial\phi_2(1)$ means there is a field $\partial\phi_2(z)$ whose value at $z=1$ is $\partial\phi_2(1)$. Consult the 3-point function in (5.26) $$ \langle h_1|\phi_2(z)|h_p\rangle\propto\frac{1}{z^{h_p+h_2-h_1}}, $$ we immediately know $$ \frac{\langle h_1|\phi_2(1)L_{-1}|h_p\rangle} {\langle h_1|\phi_2(1)|h_p\rangle} =-\left.\frac{\partial_z\langle h_1|\phi_2(z)|h_p\rangle} {\langle h_1|\phi_2(1)|h_p\rangle}\right|_{z=1}=h_p+h_2-h_1. $$

  • $\begingroup$ Thank you, I got it all! So you "assume" $\beta=\frac{h_p+h_3-h_4}{2h_p}$ because it reduces to $1/2$ for $h_3=h_4$. The one thing I still don't understand is that you put $\langle h_1|\phi_2(1)|h_p\rangle=C_{12p}$ while in the book it is equal to $(C^p_{21})^{1/2}$ (see after eq. 6.187) and in fact I would have said like you $\endgroup$
    – Lorenzo B.
    Apr 14, 2022 at 9:01
  • 1
    $\begingroup$ Ah, sorry that should be $\sqrt{C_{12p}}$. However, it was canceled by itself in the denominator. (Remember that conformal blocks are the furthest we can get from conformal invariance, so no $C_{12p}$ should appear in $\mathcal{F}$, i.e. all $C_{12p}$ should cancel finally.) You can reproduce $\beta$ following the argument from (6.169) to (6.177). $\endgroup$
    – Youran
    Apr 14, 2022 at 10:24

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