Take any 1+1 dimensional conformal field theory on the plane. The Hamiltonian is invariant under the infinite-dimensional Virasoro algebra (with some central charge $c$), generated by $L_i$ ($i\in \mathbb Z$). However, the best one can do is find a ground state $|0\rangle$ which is symmetric w.r.t. $L_{i\geq -1}$ (this is also observed on the Wikipedia page). In other words, the ground state has much less symmetry than the Hamiltonian. This sounds like spontaneous symmetry breaking. Indeed, in footnote 3 of page 3 Maldacena and Stanford say:

In 1 +1 dimensional CFT the conformal symmetry is also spontaneously broken (recall that $L_{-2} |0\rangle \neq |0\rangle$), but it is not explicitly broken.

Can we really think of this as spontaneous symmetry breaking? If so: where is the infinite ground state degeneracy, where are the Goldstone modes, and what about Mermin-Wagner-Coleman even forbidding the breaking of a continuous symmetry? (And what about the slightly obscure Elitzur's 'theorem'/argument that forbids the spontaneous breaking of local symmetries?)

I can imagine some or all of these concerns being alleviated by somehow using the peculiar fact that the local generators $L_{i<-1}$ cannot be extended to global ones, i.e. the way the Virasoro algebra acts in this context cannot be interpreted as the Lie algebra of a Lie group [EDIT: this seems to be a common misconception, see the post by Bruce Bartlett]. Hence although the ground state breaks these local symmetries, there is no corresponding global symmetry being broken (which probably means Goldstone's theorem doesn't apply). Nevertheless I find it a bit hard to wrap my head around this: what is really being broken and what are its consequences? (and how are the aforementioned issues resolved?)

Note 1: This should not be confused with the conformal anomaly, which simply has to do with the fact that we have to replace the de Witt algebra by the Virasoro algebra.

Note 2: A related matter is the following: conformal symmetry (in this dimension) is a local symmetry. We are used to equating local symmetries to gauge symmetries, however at the same time I do not think we would want to call the conformal symmetry a gauge symmetry. Nevertheless, it is an interesting issue, since a gauge symmetry can of course never be broken (but this is perhaps a separate can of worms).

  • 2
    $\begingroup$ The MW theorem only applies to symmetries which commute with the Poincare generators, so there is no contradiction. $\endgroup$ Commented Dec 31, 2016 at 1:43

1 Answer 1


I believe it is an abuse of terminology. In order for you to say there is Symmetry Breaking, you should be able to recognize two phases. (by the way it is not necessary to have an infinite ground state degeneracy, there are models with $\mathbb{Z}_2$ degenerate vacua that feature SB). So there should exist a phase in which your vacuum state has the full symmetry and another in which (either by hand or spontaneously) the phase becomes broken. Up to my knowledge you can't build a vacuum state in your case which is invariant under the full conformal symmetry generators, so the answer is probably it is not symmetry breaking (in the standard sense at least). I was checking in the nice book by Blumenhagen and Plauschin "Introduction to Conformal Field Theory" and there is no mention of symmetry breaking at any point so I presume it is definitely an abuse of the terminology.


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.