# How can one prove that there cannot exist a conformal primary, in the case of free field theory, that doesn't saturate the unitarity bound?

In free field theory, the full list of conformal primaries, is given by the Twist-2 operators. These have $\Delta = l+2$, which is also the saturation condition for the unitarity bound for $l \neq 0$. So, excepting the $l=0$ twist operator, all others saturate the bound.

How can one prove that for $l>0$, one cannot construct a conformal primary with $\Delta > l+2$ ? In other words, how to prove that the twist two operators are the only ones that occur in the $\phi \times \phi$ OPE.

• Since everyone uses different notations, it would be nice if you explained what the symbols actually mean. Apr 15, 2015 at 17:08
• $\Delta$ is scale dimension and $l$ is the spin of the twist-2 fields, which are defined by $\Delta -l=2$. I have found a reference which explains a solution to my question: arxiv.org/pdf/hep-th/0011040v3.pdf In particular, the paragraph following Eq.6.19. Apr 15, 2015 at 18:25

Here are a couple quick and dirty ways to count these operators:

1. Compute the conformal block expansion of the four-point function $$\langle \phi\phi\phi\phi\rangle$$. This will only contain blocks with $$\Delta-\ell=d-2$$. This is done in http://arxiv.org/abs/1009.5985, equation 64.
2. Compute the character of the conformal group acting on operators in the theory. By decomposing this character into characters of irreducible representations of the conformal group, you can read off the conformal primaries. See http://arxiv.org/abs/hep-th/0508031 for an introduction to conformal characters.

However, the free theory is simple enough that we can just do the analysis from scratch.

An operator in the free theory is built from a string of derivatives $$\partial_\mu$$ and $$\phi$$'s. It's easy to see that the only operators appearing in the $$\phi\times\phi$$ OPE have $$\phi$$ number 0 (the unit operator) or 2. The case of $$\phi$$ number 2 is the most interesting. These are operators of the form $$\partial\dots\partial\phi\partial\dots\partial\phi$$.

Recall that any operator of the form $$\partial_\mu \mathcal{O}$$ (where $$\mathcal{O}$$ is any operator) is a descendant. Let us consider the space of all operators modulo descendant operators. Equivalence classes in this space will be in 1 to 1 correspondence with conformal primaries.

In this quotient space, we have $$\partial(A B) = \partial A B + A\partial B \sim 0$$ (where $$\sim$$ means "is equivalent to"). Using this relation, we can move derivatives from one $$\phi$$ to the other, modulo descendants. Let us put all the derivatives on the right-hand $$\phi$$. We now have operators of the form

$$\phi \partial_{\mu_1}\cdots\partial_{\mu_\ell}\partial^{2n}\phi$$

However, the equation of motion says that $$\partial^2 \phi=0$$, so we're left with

$$\phi\partial_{\mu_1}\cdots\partial_{\mu_\ell}\phi$$

We're not quite done. The above operator could be equivalent to a primary modulo descendants, or it could be a descendant itself. It turns out that when $$\ell$$ is odd, it is a pure descendant (homework exercise!). When $$\ell$$ is even, there is a primary in the same equivalence class. To find it, we must solve the equation $$K_\mu \mathcal{O}=0$$, where $$K_\mu$$ is the special conformal generator. The solution will involve a particular linear combination of derivatives acting on the two $$\phi$$'s.

The primary will be traceless by the equations of motion, so it transforms in a spin-$$\ell$$ representation of the rotation group. It has $$\Delta-\ell=2\Delta_\phi=d-2$$ (where $$d$$ is the spacetime dimension), so it saturates the unitarity bound for $$\ell>0$$.

• When writing the solution key for a problem set with this as one of the problems, I ended up getting a different answer for the form of the two field primaries when doing a direct calculation with $K_\mu$. In particular, I don't see how your formula recovers the improved stress tensor when $s = 2$. Also, I'm a bit confused: what primary is $:\phi\phi\phi:$ supposed to be a descendant of? Nov 22, 2023 at 0:53
• You're right -- that expression is incorrect. I removed it from my answer. The operator :𝜙𝜙𝜙: is a primary on its own. (But it has phi-number 3, which means it cannot appear in the OPE of phi with phi.) Nov 23, 2023 at 21:25