# CFT OPE: Why do only symmetric traceless operators appear in the scalar-scalar OPE?

Why do only symmetric traceless operators appear in the scalar scalar OPE?

The lecture notes: http://arxiv.org/abs/1602.07982, rephrase this question as: Why does $\langle\mathcal{O}^a| \phi(x) |\phi\rangle$ vanish unless $\mathcal{O}$ is a symmetric tensor?

How do we show this? And how do we show that this is an equivalent statement?

It seems to me that this is only equivalent at level of calculating the scalar 4pt function where after performing one OPE we get $\langle\mathcal{O}^a| \phi(x) |\phi\rangle$. For something else, say $\langle\phi \phi J J\rangle$ this does not work.

The same question was asked here OPE in a general $d$-dimensional CFT, as a subquestion with the main question emphasizing a reference request (which probably didn't help the question get answered, because of the higher standards that are required of reference requests.)

• In order to construct $\langle \mathcal{O}^a | \phi(x) |\phi \rangle$ you need to build a tensor with index $a$ out of $x$ (which is the only variable you have here). This is only possible if $a$ is a symmetric tensor index. Symmetric tensors then decompose into traceless-symmetric ones. The relation between OPE and this matrix element is through operator-state correspondence. – Peter Kravchuk Mar 26 '17 at 22:04
• @Peter Could you expand on this? Thinking about the problem again my take on this is as follows: – Kvothe Mar 27 '17 at 8:55
• First off, at the level of the $\phi \times \phi$ OPE. If $\langle\mathcal{O}^a| \phi(x) |\phi\rangle$ vanishes for any anti symmetric operator $\mathcal{O}^a$ it also means that these operators can't show up in the $\phi \times \phi$ OPE. The argument would go something like perform the OPE of $\phi(x)\times\phi(0)$ inside this 3pt function. If an antisymmetric $\mathcal{O}^{a}$ shows up you would get $| C_b|\langle\mathcal{O}^a|\mathcal{O}^b \rangle\neq0$. But we just showed that $\langle\mathcal{O}^a| \phi(x) |\phi\rangle = 0$. – Kvothe Mar 27 '17 at 8:56
• For the other part. Let me see whether I understand correctly. We have only one variable $x$. The other $\phi$ is put at zero. Thus we can construct only something like $\mathcal{O}^{a...c}(x^\mu \frac{d}{dx^\mu},x^2) \frac{d}{dx^a}...\frac{d}{dx^c}f(x)$ where the number of labels $a...c$ is equal to the spin of $\mathcal{O}^a$. Then if there is any part to $\mathcal{O}^a$ that is anti-symmetric in a...c it will vanish because you will get something like, $(\partial_\mu \partial_\nu - \partial_\mu \partial_\nu)f(x) = 0$. Do you agree? Was this what you meant? – Kvothe Mar 27 '17 at 8:56
• Lastly, if we have a symmetric $\mathcal{O}^{a}$ we can write it as $\sum_n\mathcal{O'}^{a}_n$, where $\mathcal{O'}^{a}$ is symmetric and traceless. (I don't immediately see this last statement. Is it easy to show?) – Kvothe Mar 27 '17 at 8:56

An operator $\mathcal{O}^a$ appears in $\phi_1\times\phi_2$ OPE if and only if the matrix element $$\langle \mathcal{O}^a|\phi_1(x)\phi_2(-x)|0\rangle$$ is non-vanishing, for the reason the OP states in the comments. If $\mathcal{O}$ is a primary then this matrix element is invariant under simultaneous shifts of the two $\phi$'s, so this is just a more symmetrical version of the matrix element from the question. (Because $\langle \mathcal{O}|P_\mu=[K_\mu|\mathcal{O}\rangle]^\dagger=0$.)
Now, this matrix element is simply a function of $x$. The most general expession we can construct out of $x$ is $$\langle \mathcal{O}^a|\phi_1(x)\phi_2(-x)|0\rangle=f_1(|x|)x^{\mu_1}\cdots x^{\mu_l}+f_2(|x|)g^{\mu_1\mu_2}x^{\mu_3}\cdots x^{\mu_\ell}+\ldots,$$ where $g^{\mu\nu}$ is the flat Lorentz or Euclidean metric depending on your signature. It shows that $\mathcal{O}$ is a tensor operator (thought not yet necessarily a symmetric one). Any tensor operator is a sum of traceless tensor operators (see below), so we can assume that $\mathcal{O}$ is traceless, and this fixes the terms containing $g^{\mu\nu}$ in terms of the first term (in particular the traceless part of all the terms except the first one is zero). But the first term is symmetric in its indices, which shows that $a$ must be the symmetric traceless tensor index $a=\mu_1\ldots \mu_\ell$. Thus $\mathcal{O}$ must necessarily be a symmetric traceless tensor. Furthermore, we note that if $\phi_1=\phi_2$ then the matrix element should be invariant under $x\to -x$. We then see that only even $\ell$ are allowed in this case.
If we have a tensor operator $\mathcal{O}^{\mu_1\ldots \mu_\ell}$, it can always be written as a sum of traceless symmetric tensor operators. For example, here is the decomposition for spin-3 symmetric tensor $$\mathcal{O}^{\mu_1\mu_2\mu_3}=\mathcal{O}^{\mu_1\mu_2\mu_3}_3+\frac{1}{d+2}(g^{\mu_1\mu_2}\mathcal{O}_1^{\mu_3}+g^{\mu_2\mu_3}\mathcal{O}_1^{\mu_1}+g^{\mu_3\mu_1}\mathcal{O}_1^{\mu_2}),$$ where $\mathcal{O}_1^\mu=\mathcal{O}_\nu{}^{\nu\mu}$, and $\mathcal{O}_3$ traceless and essentially defined by this equation. Such a decomposition happens because symmetric tensors are not irreducible representations of $SO(d)$, and instead they are composed from traceless symmetric tensors, which are irreducible. (Note that for general symmetric spin-$\ell$ there will be operators with spins $\ell,\ell-2,\ell-4,\ldots$ in the right hand side.) In this example $\mathcal{O}_3$ is the "traceless projection" of $\mathcal{O}$. Traceless projection of any tensor proportional to $g^{\mu\nu}$ is zero.
• Thanks! You make a bit of a leap when you go to the most general expression though. Why can't we have derivatives acting on the $x$'s? – Kvothe Apr 3 '17 at 8:02
• @Kvothe, you are right that I made a leap there (fixed now), but it is not related to the derivatives you are talking about. The expression in the right hand side is a concrete function, so if it had derivatives, you could just compute them. For example, $\partial_\mu f(|x|)\propto x_\mu f'(|x|)/|x|^2$, which is of the same form. – Peter Kravchuk Apr 3 '17 at 16:05