8

For relativistic quantum field theories, the Coleman-Mandula theorem places very strong restrictions on the possible symmetry groups $G$ of the aforementioned quantum field theory, forcing it to be a direct product of an internal symmetry group and the Poincare group (or the conformal group for massless particles).

Does a similar result hold for non-relativistic quantum field theories, whose symmetry group must contain the (centrally extended) Galilean group? If not, what stops an analogous result from going through?

Qmechanic
  • 220,844
Ishan Deo
  • 2,101

1 Answers1

5

There is no such a theorem. Non-relativistic field theories are allowed to combine external and internal symmetries into a non-trivial group.

The canonical reference is Weinberg, volume 3 (cf. Historical Introduction). He gives an example, the SU(6) symmetry of nuclear physics, which contains the rotation SO(3) group as a subgroup.

The key difference between relativistic and non-relativistic theories is that the semi-simple part of the former (namely, SO($1,d-1$)) is non-compact, while that of the former (namely, SO($d-1$)) is compact. As such, the former does not admit non-trivial, finite-dimensional unitary representations while the latter does. This plays an important role in the proof of the Coleman-Mandula theorem (intuitively, internal symmetries must transform as finite-dimensional reps of the external symmetries; in the relativistic case the only option is the trivial rep, and we therefore find a direct product, but in the non-relativistic case the rep does not have to be the trivial rep, and hence we no longer require a direct product).

Mauricio
  • 6,886
AccidentalFourierTransform
  • 56,647