Is the following a correct definition of spontaneous symmetry breaking?

Question

I recall a definition of spontaneous symmetry breaking that I am not sure is fully correct as I am unable to find a reference for it.

Given a quantum system with Hamiltonian $H$ at $T = 0$, i.e. considering ground states only, let $H$ be invariant under a symmetry group $G$, meaning $H$ commutes with all elements of $G$. Then $G$ has a representation on the ground state space of $H$. If this representation is trivial the symmetry is not broken. However, if there exist states in the lowest energy eigenspace which are not invariant under action of $G$, i.e. the representation of $G$ on the ground-state space is nontrivial then the state is spontaneously symmetry broken. It would in fact be broken to the kernel of the representation of the ground-state space.

A specific example could be the transverse field Ising model with

$$H = A \sum_i Z_i Z_{i+1} + B \sum_i X_i$$

where $$G = \{ \mathbb{1}, X^{\otimes n} \}$$ forms the symmetry group. Here the lowest energy sector is given by $|+++ \dots \rangle$ at $A = 0$ on which $G$ acts trivially and by $\text{span}(|111\dots\rangle, |000\dots \rangle )$ at $B = 0$ on which $G$ is represented as trivial + alternating representations of $\mathbb{Z}_2$.

Considering much more sophisticated definitions of SSB in quantum systems have been put forward in this question I suspect something must be wrong with the above definition and I ask for somebody to point out what it is and whether this definition is at least close to a correct one.

Answer 1

Consider the Kitaev's toric code Hamiltonian (wiki, some helpful notes) on a 2-torus. This Hamiltonian has symmetry operators that live on non-trivial loops around the torus. This system is 4-fold degenerate and one can show that the symmetry operators have a nontrivial representation in the ground space, satisfying your definition for spontaneous symmetry breaking (SSB). But conventionally, this system is not said to be SSB but an example of topological order.

SSB and topological order are similar ideas as motivated by your definition, but the crucial difference is that states in the ground space of topologically ordered systems are locally indistinguishable (all local operators have the same expectation value), whereas this is not true for SSB systems.

A modern interpretation of topological order is that it can be thought of as SSB of higher-form symmetries (1-form in the case of toric code), and conventional SSB can be thought of as SSB of 0-form symmetries. In this generalized symmetry paradigm, your definition for SSB would be correct. In the toric code example, $G$ would be the set of loop-like symmetries which get spontaneously broken in the topologically ordered phase. Check out this review article for more on that.

Answer 2

No. The problem with your definition is that it does not rely on any notion of 'large system size' (commonly referred to as a thermodynamic limit). Your definition could already be satisfied by a finite number of qubits. Relatedly, your definition applies to thermodynamically large spin chains in a so-called symmetry-protected topological (SPT) state: such a phase of matter has a unique ground state with periodic boundary conditions but it has a ground state degeneracy when considered on a line segment geometry. Moreover, the degeneracy satisfies your definition. However, the usual definitions of spontaneous symmetry breaking do not (and should not) apply to such a scenario (*).

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_{(*) The reason that it is unappealing to declare that 1D SPT states have symmetry-breaking at the edge is that the zero-dimensional edge has no notion of stability in its non-symmetric state. However, 1D gapless SPT edge states can indeed have zero-dimensional symmetry-breaking edges, e.g., see https://arxiv.org/abs/2208.12258 .}