Why are symmetric quantum ground states cat states iff the ground-state manifold is degenerate?

Question

The usual story of symmetry-breaking quantum phase transitions (I won't consider topological transitions here) goes like this: you have a Hamiltonian $H(g)$ describing an infinite system which depends on some coupling constant $g$ and respects some global symmetry $U$. In one regime (say small $g$), the ground state is unique and therefore invariant under $U$. This state is "stable" in the sense that it obeys the cluster decomposition property that for all local (gauge-invariant, in a gauge theory) operators $O_A$ and $O_B$, $$\lim_{|x - y| \to \infty} \left[ \langle O_A(x) O_B(y) \rangle - \langle O_A(x) \rangle \langle O_B(y) \rangle\right] = 0.$$ But once $g$ grows beyond some critical coupling $g_c$, the ground-state manifold becomes degenerate. We can of course still diagonalize $U$ within this degenerate manifold to get a set of multiple orthogonal symmetric ground states. (By "symmetric" I mean an eigenstate of $U$ (i.e. a state of definite symmetry), not necessarily with eigenvalue $1$. So, for example, I'm counting the antisymmetric ground state $\frac{1}{\sqrt{2}} (|\uparrow \uparrow \dots \uparrow \rangle - | \downarrow \downarrow \dots \downarrow\rangle)$ of the ferromagnetic quantum Ising model as a "symmetric" state.) But these symmetric states are all "Schrodinger's cat" states or "macroscopic superpositions" rather than "stable" states, in the sense that there exist local operators $O_A$ and $O_B$ that violate the cluster decomposition property above (e.g. the $S^z$ operator for the quantum Ising model). I won't get into the controversial debate about exactly why such states are unphysical, but simply posit that they are never found in the lab. Instead, we always experimentally find symmetry-breaking ground states that (a) are "stable" in that they respect the cluster decomposition property and (b) are not eigenstates of $U$, such that there exists a local order parameter $m$ that does not commute with $U$ such that $\langle m \rangle \neq 0$.

This story has proven to be incredibly successful at accurately describing an enormously wide range of physical phenomena. It can be rigorously shown to be correct in a few special cases, like the quantum transverse Ising model. But it makes three assumptions that aren't obvious to me:

1. If the ground state is unique, then it is always stable (i.e. respects cluster decomposition).
2. If there are degenerate ground states, then the symmetric ground states are always cat states (i.e. violate cluster decomposition).
3. If there are degenerate ground states, then there always exist (asymmetric) stable ground states.

Without these three assumptions, the story falls apart (e.g. one could imagine a degenerate ground state manifold in which the symmetric state is stable and therefore physically realistic, so the symmetry remains unbroken).

The usual hand-wavy justification for these assumptions comes from Landau-Ginzburg mean-field theory: we imagine an energy density function that depends continuously on the (spatially uniform) expectation value of some local order parameter $m$. If this function is analytic and has multiple degenerate minima, then they must be separated by energy barriers that (when multiplied by $N$ to get the extensive version) are infinitely high in the thermodynamic limit. But can any of the assumptions above be proven at any level of rigor without using Landau-Ginzburg mean field theory? (As usual, feel free to assume translational invariance, finite-dimensional local Hilbert space, etc. if convenient.)

Answer 1

Short answer: Because the stability of the symmetry breaking states implies the instability of the symmetric states. If you take two macroscopically distinct states with small quantum fluctuations then you might expect the superposition of the two to the have large quantum fluctuations due to the macroscopic distinctness. Large quantum fluctuations is your definition of instability. This is expressed with more rigger below.

Even if there are large quantum fluctuations in the symmetry breaking states, it's unlikely they will interfere in just the right way to cancel when interfering in a symmetric state. This is more intuition and might take to long to express.

Long Answer: First we have to assume the degeneracy is due to symmetry breaking. Otherwise you can just come up with a local degenerate ground associated with a local symmetry and don't break that symmetry when you create the many body system. Then you can have a symmetric degenerate macroscopic state which is not a cat state. For example, the hamiltonian $H=-\sum_i \textbf{L}^2_i$ for spin operator $L$ has energy -$Nl^2$ and a $(2l+1)^N$ degenercy. Basically you can come up with boring states if we don't have a reasonable assumption for the cause of the degeneracy.

This gives us a few things:

1. a global symmetry $U=exp(iJ)$ with $J=\sum_i J_i$ (global)
2. an global order paramter operator, $M$ which for the symetry breaking states has: $\left<M\right>=\sum_i \left<M_i\right>=CN$
3. translational symmetry: $\left<M_i\right>=C$
4. $[M,J]=N[M_i,J_i]\neq0$, otherwise the order paramter doesn't break the symmetry

Therefore there are two different basises in the degenerate ground state manifold. One that diagonalizes $M$, $\{\left|m\right\}$ and one that diagonalizes $J$, $\{\left|j\right\}$. (4) implies the basises are not the same and the symmetric state $\left| j \right>$ are non-trivial super positions of $\left| m \right>$.

Now write a j state in terms of the m states:

$\left|j\right>=\sum \psi_{j,m}\left|m\right>$

Before considering the stability of the $\left| j \right>$, lets establish some preliminaries.

• From $\left|m\right>$ being a basis we know $\left< m|m'\right>=0$
• from translation symetry, we know this non orthogonality can't be completely due to 1 or 2 sites so we can assume $\left< m\right|O_x O_y\left|m'\right>\approx\left< m\left|O_x\right|m'\right>\approx 0$
• The symmetry broken state is stable, i.e. fluctuations $\left<M_i^2\right>-\left<M_i\right>^2$ are not dramatically larger then $\left<M_i\right>^2$ and $\left|m\right>$ does not dramatically mix multiple eigen states of $M_i$.

We can now consider the stability in the operators $M_i$. First $\left< j\right|M_x\left|j\right>$. From symmetry this must be 0, since the symmetry operator distinguishes between non zero values. Therefore for stability we need: $\lim_{(x-y)\rightarrow \inf }\left< j\right|M_x M_y\left|j\right>=0$

By non orthogonality we get:

$ \left< j\right|M_x M_y\left|j\right> = \sum_m |\psi_{j,m}|^2 \left< m\right|M_x M_y\left|m\right> $

From the stability of the symmetry broken states we know $M_i$ does not induce large transitions out of $\left|m\right>$ and we can approximate $M_y\left|m\right>=C_m\left|m\right>$ to get

$ \sum_m |\psi_{j,m}|^2 C_m^2 >0 $

which can't be zero since $C_m^2>0$ and $|\psi_{j,m}|^2 $ needs to be greater then 0 for states with non zero $C_m$ otherwise $[M,J]$ must be zero or the m states don't break the symmetry.

Thus the symmetric states must be unstable if symmetry breaking has occurred.

My intuition for superposition of unstable symmetry breaking states is that it is very unlikely for $M_x M_y$ to induce the exact transition to make $\left| m \right>$ orthogonal to its self. The unlikeliness is due to the complexity of the fluctuations in $M_i$ as the quantum fluctuations (entanglement) increases. You might be able to work this out by quantifying the how much this effect of the transition.