It is well known that conformal field theories (CFTs) in general have a gapless linear dispersion (up to $E_0$ shift) $E(\vec{k}) = c|\vec{k}|$ for $E\approx 0$ with $c$ a constant velocity.

How can one mathematically infer this relationship from symmetry alone?

What I tried so far: $c$ is known not to be universal, i.e. it depends on the coupling in the Lagrangian and is not defined by symmetries alone, thus it makes sense that the differential equation $\frac{d^2 E}{d^2k} = 0$ would appear somewhere. From Lorentz-invariance we know $\partial_\mu T^{\mu \nu} = 0$ and from scale-invariance we know $T^\mu_{\ \ \ \mu} = 0$, where $T$ is the Energy-Momentum tensor but I am not sure how that can help me.

  • $\begingroup$ Is that really well known? $\endgroup$ Jun 15, 2023 at 23:54
  • $\begingroup$ Well, is there any counterexample? As far as I know CFTs are never discussed in the context of massive field theories. $\endgroup$
    – VoidStar
    Jun 16, 2023 at 8:14

1 Answer 1


The existence of conformal symmetry means that there cannot be massive particles. Having a massive particle means that the mass of that particle is an energy scale, so the theory wouldn't be scale invariant. This means that all excitations must be massless, implying that the spectrum has $E \sim k$.

There's a subtlety, in that it's allowed to have a continuous mass spectrum as well. I am ignoring this possibility in this answer.

  • $\begingroup$ Thanks but I wanted a mathematical derivation of the law $E \sim k$ $\endgroup$
    – VoidStar
    Jun 13, 2023 at 6:19
  • $\begingroup$ I see -- sorry about that. Here is something that might work: $\endgroup$
    – infinity
    Jun 14, 2023 at 7:19
  • $\begingroup$ Notice that $\partial_x^\mu T \langle T_{\mu \nu}(x) \phi(y) \rangle = \langle T (\partial^\mu^x T_{\mu \nu}) \phi(y) + \delta(x^0 - y^0) \langle [T_{\mu 0}(x), \phi(y)] $. Since $\partial^\mu^x T_{\mu \nu} = 0$ within the correlation function, and since $[T_{\mu 0}(x), \phi(y)] \sim \delta^3(x - y) * stuff $, we have (in momentum space) that $p_\mu \langle T T_{\mu \nu}(p) \phi(-p) \rangle \sim \langle ... \rangle$, implying that the correlator must have a pole at $p = 0$. I'm not 100% confident in this at the moment. Also sorry about the formatting -- not sure what happened with it $\endgroup$
    – infinity
    Jun 14, 2023 at 7:28
  • $\begingroup$ Also, for a scalar field, I think $\partial_\mu T^{\mu \nu} = 0$ implies a gapless spectrum. For a massless scalar field, for example, $T^{00} = |\partial_t \phi|^2 + |\nabla \phi|^2 $, and $T_{0i} = \partial_t \phi \partial_i \phi$, so that in momentum space, $\partial_\mu T^{\mu \nu} = 0$ implies that $-E(E^2 + k_i^2) + E k_i = 0$, which is solved to lowest order by $E \sim k_i$. Maybe something similar can be done for other fields, but I'm not 100% sure how it would work. Also, I think my above comment is wrong bc it only works for SSB. $\endgroup$
    – infinity
    Jun 14, 2023 at 7:43
  • $\begingroup$ As far as I understand $\partial_\mu T^{\mu\nu} = 0$ is true for any Lorentz-invariant QFT but not all of them are massless! The solution must somehow involve scale-invariance $\endgroup$
    – VoidStar
    Jun 14, 2023 at 7:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.