In most textbooks on field theory you read that "spontaneous symmetry breaking implies degeneracy of the ground state". (Like for example in http://www.itp.phys.ethz.ch/research/qftstrings/archive/13FSQFT2/Chapter07.pdf)

I am feeling somewhat uneasy about this. Say you have a system with time-reversal symmetry. Then Kramers Theorem implies that the ground state is at least two-fold degenerate. Now when I break this symmetry spontaneously (or explicitly) I expect that this degeneracy is removed. (Since Kramers Theorem is no longer valid)

Where is the logical error that I am making?

Edit: Maybe I am also giving a reference to a more advanced application of the question. The Wikipedia Article on Symmetry Protected Topological Order states that "If the boundary is a gapped degenerate state, the degeneracy may be caused by spontaneous symmetry breaking and/or (intrinsic) topological order."


1 Answer 1


It is very important to distinguish whether the symmetry is broken explicitly or spontaneously. I think that the sentence "Now when I break this symmetry spontaneously (or explicitly)" indicates that its author isn't quite distinguishing these things.

An explicit symmetry breaking generally lifts the degeneracy because the different parts of the multiplets no longer have the same energy.

However, spontaneous symmetry breaking increases the degeneracy, particularly of the ground state. It's really how spontaneous symmmetry is defined. It's a fate of the symmetry that remains the symmetry of the laws of physics, but in practice, the environment we encounter, starting from the ground state, is no longer invariant under the symmetry.

Its not being invariant means nothing else than the fact that if we act with a generator $G$ of the continuous symmetry on the ground state, we get $$ G|0\rangle \neq 0 $$ We would get zero if the symmetry were not spontaneously broken. If it is broken, we get a nonzero vector that is independent from the original $|0\rangle$, so we get another "copy" of the ground state. Here, $G$ still commutes with the Hamiltonian $H$ so these copies have the same energy – we have degeneracy. For example, if the electroweak symmetry were global, just to simplify things, the ground state of the Higgs field could have vev $(0,246)$ in the units of GeV, but it could have any other vev with the same magnitude, too. So there are infinitely many vacua. (In gauge theory, they're made equivalent, but if we spontaneously break a global symmetry, they are states related by the symmetry and therefore having the same energy, but distinct elements of the Hilbert space.)

So the claim that the OP is dissatisfied with really holds completely generally, by definition of the spontaneous symmetry breaking! Misunderstanding that the ground state of a spontaneously broken symmetric theory is degenerate is the misunderstanding of the basic idea of the spontaneous symmetry breaking.

Note that spontaneous symmetry breaking still effectively means that the "symmetry is broken for most practical purposes" because the "copies" mentioned above may be imagined to be physically identified and we split the Hilbert space into "superselection sectors" each of which is built upon one copy of the ground state. The action of the symmetry generator on any excited state gives us a vector from another superselection sector which can't be identified with zero so from the perspective of a single superselection sector, the symmetry is simply broken.

None of these questions and discussions about explicit vs spontaneous symmetry breaking has anything to do with the time-reversal symmetry (or the absence of it) which is just another symmetry (one given by an antiunitary transformation, so some of the comments above wouldn't apply to this symmetry). Both systems with and without time-reversal symmetry satisfy the claim that the ground state is degenerate if a symmetry is spontaneously broken.

  • $\begingroup$ Say I pick two ground states that aren't eigenstates of the generator of the continuous symmetry. Isn't it (at least in priniciple) possible to find a linear combination that happens to be an eigenstate of the generator of symmetry? Perhaps I am missing another detail of the definition. $\endgroup$
    – CMFT_guy
    Commented Aug 25, 2014 at 12:39
  • $\begingroup$ Dear CMFT, it's unphysical to be looking for eigenstates of the symmetry generator that is spontaneously broken because they're superpositions of states from different superselection sectors. An example is a superposition of a bar magnet magnetized in one direction with the state of the bar magnet magnetized in a different direction. A typical Schrodinger's cat state. But yes, if you don't mind that this is a superposition across superselection sectors, you may find eigenstates of the generators. What's your problem with that? $\endgroup$ Commented Aug 25, 2014 at 15:46
  • $\begingroup$ Is due decoherence that we encountered this superselection sectors? $\endgroup$
    – Nogueira
    Commented Jul 29, 2015 at 19:29
  • $\begingroup$ I would say Yes - but what is the cause and what is the consequence or a description is debatable... But generally Yes, states in different superselection sectors have "very different" macroscopic consequences for the environment, so the information about the superselection sectors is being copied as classical information as these states decohere - so decoherence is why the separation to the sectors becomes visible. $\endgroup$ Commented Jul 31, 2015 at 4:45

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.