8
$\begingroup$

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?

$\endgroup$
1

1 Answer 1

5
$\begingroup$

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).

$\endgroup$

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.