In the context of Conformal Field Theory (CFT), I have a primary field $\phi_{(r,s)}$ with a level 1 null-descendant, i.e. $(r,s)=(1,1)$ and $h_{(1,1)}=0$. My goal is to understand how this condition constraints the operator algebra, especially regarding the OPE (Operator Product Expansion).

Looking at the 3-point function, I see the following:

$$\langle L_{-1} \phi_{(1,1)} \phi_1 \phi_2 \rangle = \partial_z \langle \phi_{(1,1)} \phi_1 \phi_2 \rangle, \tag{1}$$

but we know that:

$$L_{-1} \phi_{(1,1)} = 0, \tag{2}$$

which in turn means

$$\langle L_{-1} \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0. \tag{3}$$

We also know that in CFT, the general form of a 3-pt function is (with $h_{(1,1)}=0$):

$$\langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = c_{12}^h (z-z_1)^{-h_1+h_2} (z_1-z_2)^{-h_1-h_2} (z_2-z)^{-h_2+h_1}. \tag{4}$$

Thus equating $(1)$ with $(3)$, and using $(4)$, we find:

$$\left( \frac{-h_1+h_2}{z-z_1} - \frac{-h_2+h_1}{z_2-z} \right) \langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0, \tag{5}$$

and therefore that $h_1 = h_2$ or $\langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = 0$.

So far so good. Now the claim is that, as a consequence, we can write the following OPE:

$$\phi_{(1,1)} \phi_1 = \sum c_{12}^h \phi_2 + \text{descendants} \tag{6}$$

when the condition $h_1 = h_2$ is satisfied. I don't see that. Can someone explain how to obtain eq. $(6)$ from eq. $(5)$?

Thanks in advance!


Here is a picture of the script of my professor, maybe his intentions are clearer then?

enter image description here

  • $\begingroup$ What are you summing over in eq $(6)$? $\endgroup$
    – MannyC
    Jul 8, 2019 at 21:46
  • $\begingroup$ @MannyC I guess that I understand my stupid confusion now thanks to your comment. Actually, the notation of the script of my professor is $\phi_{(1,1)} \phi_i = \sum c_{ij}^h \phi_j \text{descendants}$, but because he used a $j$ instead of a $2$ all along during the derivation, and did not write over what he was summing over, I assumed that this field $\phi_j$ was the same as the field that I called $\phi_2$ in the OP, which is obviously wrong. So I guess that the condition below eq. ($6$) would also change to $h_1 = h_j$ for all $j$? $\endgroup$
    – Pxx
    Jul 8, 2019 at 22:52
  • $\begingroup$ @MannyC Well, he wrote explicitly $h_i = h_j$ next to the equation (although he did not write "for all $j$"). I've added a screenshot of the script in the OP. Eq. (223) is the one that I don't understand. $\endgroup$
    – Pxx
    Jul 9, 2019 at 8:39

1 Answer 1


The note is simply saying this: the three-point function $$ \langle \phi_{(1,1)} \phi_1 \phi_2 \rangle = c^{h = 0}_{12}\, (z-z_1)^{-h_1+h_2} (z_1-z_2)^{-h_1-h_2} (z_2-z)^{-h_2+h_1} \,, $$ satisfies a differential equation that implies $$ c_{12}^0 \,(h_1 - h_2) = 0\,. $$ So this means either $c_{12}^h = 0$ or $h_1 = h_2$. The first trivial consequence is that in the OPE $\phi_1 \times \phi_2$ $$ \phi_1 \times \phi_2 = \sum_j c^{h_j}_{12} \,\phi_j $$ you can find the operator $\phi_{(1,1)}$ (i.e. $c^{0}_{12} \neq 0$) only if $h_1 = h_2$.

But you can say other things: using the associativity of the OPE you can flip this around and look at the OPE $\phi_{(1,1)} \times \phi_1$ $$ \phi_{(1,1)} \times \phi_1 = \sum_j c_{0 1}^j \phi_j \underset{\mathrm{Associativity}}{=}\sum_j c_{1j}^0 \,\phi_j \,. $$ If $h_j = h_1$ (for example if $\phi_j$ is $\phi_2$ or $\phi_1$ itself), then the operator appears in the OPE, otherwise it does not.

Indeed $\phi_{(1,1)}$ can be thought of as an "identity element" in the OPE algebra: $\phi_j \times \phi_{(1,1)} \sim \phi_j$, which makes this thing very intuitive.

  • 1
    $\begingroup$ I see. Thank you very much for this very detailed and instructive remark! $\endgroup$
    – Pxx
    Jul 9, 2019 at 17:52

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