# Conformal invariance and tracelessness of the energy-momentum tensor: contradictory statements

Before starting my question, let me define a couple terms to avoid the confusion that usually accompanies this topic:

I define the $$c$$-number valued energy-momentum tensor as $$T^{\mu\nu} = \frac{2}{\sqrt{g}}\frac{\delta S}{\delta g_{\mu\nu}}$$. The corresponding quantum operator $$\hat{T}^{\mu\nu}$$ is then defined as usual, by inserting $$T^{\mu\nu}$$ into the path integral (in this question I won't worry about contact terms). Finally I define $$T=T^\mu_\mu$$ and likewise for the corresponding operator.

I define "conformally invariant" to mean that the action $$S$$ is invariant under the $$SO(d,2)$$ group of conformal transformations of flat space. Note: I do not use "conformal invariance" to mean Weyl invariance (i.e. symmetry under $$g_{\mu\nu}\to\Omega^2 g_{\mu\nu}$$ for arbitrary $$\Omega(x)$$), even though many texts sloppily do so.

Now, my question concerns a conformally invariant QFT on flat space. I've seen contradictory statements regarding the implication "if a QFT on flat space has conformal invariance, then its energy-momentum tensor is traceless at the quantum level" (again, here I don't care about contact terms).

Statement A: In this answer, it is stated that all we can say about $$\hat{T}$$ is that it satisfies $$\int \hat{T} = 0$$ and $$\int x^\mu \hat{T} = 0$$. In particular, this is not enough to conclude $$\hat{T}= 0$$.

Statement B: The trace anomaly equation says that for any state, $$\langle \hat{T} \rangle$$ is the sum of curvature invariants. Therefore $$\langle \hat{T} \rangle = 0$$ on flat space. Since this holds for any state, $$\hat{T}=0$$.

Clearly A and B are in contradiction. Which one is correct? Is it true that $$\hat{T}=0$$ on flat space (up to contact terms) - and if so, how does one prove it?

Note: please do not assume Weyl invariance in your answer. It is easy to show that Weyl-invariance implies $$\hat{T}=0$$. My question is about whether conformal invariance alone implies $$\hat{T}=0$$.

• You can have conformal invariance without a stress tensor at all. But does $T = 0$ in local CFTs (those with a stress tensor)? At minimum you would need to impose unitarity as well since the biharmonic scalar is a counter-example. Commented Nov 28, 2022 at 19:36
• A and B are not in contradiction at all. Your definition of conformally invariant is simply not the one that is standard. With your definition (1) is true. With the standard definition (which involves Weyl invariance), (2) is true. (2) implies (1) but not vice versa. Commented Nov 28, 2022 at 20:49
• @Prahar From Di Francesco's textbook: p95: "a conformal transformation of the coordinates is an invertible mapping which leaves the metric tensor invariant up to a scale" p99: "A field theory has conformal symmetry at the classical level if its action is invariant under conformal transformations." So, at least for classical fields, he uses my definition of conformally invariant. For quantum fields he just says (p104) that the action must be conformally invariant, which is ambiguous, but I believe he means the same as for classical fields: the action is invariant under conformal transfs. Commented Nov 29, 2022 at 2:11
• I'll look into what Francesco said and get back to you. However, your point about AdS/CFT is incorrect. The CFT lives on the asymptotic boundary. Being an asymptotic boundary, the induced metric is only defined up to a Weyl factor -- so the dual theory cannot depend on this Weyl factor and hence must be conformally (Weyl) invariant. Commented Nov 29, 2022 at 8:45
• Also, you should understand that Weyl symmetry in the case of CFT is not a symmetry since it acts on the background field (in this case, a metric). Similarly, diffeomorphisms are not a symmetry either. However, there is a diagonal subgroup of diff+Weyl which leaves the metric invariant (namely conformal transformations) and so constitutes a true symmetry. See these answers of mine which are relevant - physics.stackexchange.com/a/613017/8821 and physics.stackexchange.com/a/635336/8821 Commented Nov 29, 2022 at 8:47