The third question Q3 is basically the subject of a very recent work by N.P. Landsman.
In quantum theory, spontaneous symmetry breaking requires the system to be infinite dimensional. When the number of degrees of freedom is finite, spontaneous symmetry breaking does not take place. Consider for example a particle in one dimension moving in the potential of a double well, tunneling takes place between the two degenerate states, corresponding to the minima of the potential, resulting a unique linear superposition ground state. In the infinite number of degrees of freedom limit. The transition probabilities between the degenerate states vanish, thus partitioning the Hilbert space into mutually inaccessible sectors built up over each ground state.
It is well known that in classical systems with finite numbers of
degrees of freedom spontaneous symmetry breaking is possible as
emphasized also in the following review by Narnhofer and Thirring. Given that a (pure) state in a classical system is a point in phase space (a mixed state is a probability distribution over the phase space); then classical spontaneous symmetry breaking means that there are initial conditions leading to time invariant solutions which are not invariant under the symmetry group. For example, in the double well placing the particle in one well without enough energy to get out describes a spontaneously broken state.
There are many other examples of finite classical systems exhibiting
spontaneous symmetry breaking, the most known one is, may be, the
buckling of bars, another example is the Bead, Hoop, and Spring system .
Now, as Landsman emphasizes, Large $N$ quantum systems are analogous to
classical systems in the sense that quantum correlations vanish as
$\frac{1}{N}$, which leads to the posed question that for a finite $N$
no matter how large it is, spontaneous symmetry breaking is not allowed
whereas in the thermodynamic limit the system becomes infinite and the
spontaneous symmetry breaking is allowed. The same question can be asked for $\hbar \rightarrow 0$.
Landsman explanation is that when N becomes very large, the system
becomes exponentially unstable to a symmetry breaking perturbation which drives the system to one of the degenerate states already at a very large but finite $N$. Landsman performs the analysis by means of algebraic quantum mechanics and a full comprehension of the article needs acquaintance with his previous work.
Let me first answer your question "is it wrong to consider topological superconductors (such as certain p-wave superconductors) as SPT states? Aren't they actually SET states?"
(1) Topological superconductors, by definition, are free fermion states that have time-reversal symmetry but no U(1) symmetry (just like topological insulator always have time-reversal and U(1) symmetries by definition). Topological superconductor are not p+ip superconductors in 2+1D. But it can be p-wave superconductors in 1+1D.
(2) 1+1D topological superconductor is a SET state with a Majorana-zero-mode at the chain end. But time reversal symmetry is not important. Even if we break the time reversal symmetry, the Majorana-zero-mode still appear at chain end. In higher dimensions, topological superconductors have no topological order. So they cannot be SET states.
(3) In higher dimensions, topological superconductors are SPT states.
The terminology is very confusing in literature:
(1) Topological insulator has trivial topological order, while topological superconductors have topological order in 1+1D and no topological order in higher dimensions.
(2) 3+1D s-wave superconductors (or text-book s-wave superconductors which do not have dynamical U(1) gauge field) have no topological order, while 3+1D real-life s-wave superconductors with dynamical U(1) gauge field have a Z2 topological order. So 3+1D real-life topological superconductors (with dynamical U(1) gauge field and time reversal symmetry) are SET states.
(3) p+ip BCS superconductor in 2+1D (without dynamical U(1) gauge field) has a non-trivial topological order (ie LRE) as defined by local unitary (LU) transformations. Even nu=1 IQH state has a non-trivial topological order (LRE) as defined by LU transformations. Majorana chain is also LRE (ie topologically ordered). Kitaev does not use LU transformation to define LRE, which leads to different definition of LRE.
Best Answer
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.