Hamiltonian and symmetry
I was learning about the symmetry spontaneous breaking (SSB), but confused by this simple minima model$$H=E_0\sum_{i=1}^3|i\rangle\langle i|+J\left(|1\rangle\langle 2|+|2\rangle\langle 3|+|1\rangle\langle 3|+h.c.\right)$$It's matrix form$$H=\begin{pmatrix}E_0 & J & J \\ J & E_0 & J \\ J & J & E_0\end{pmatrix}$$which describes that a particle hops on a lattice consisting of only 3 cells with on-site energy $E_0$ and hopping strength $J$, and the PBC is applied. It's clear that $H$ respects the $Z_3$ symmetry under the PBC, which means all the eigenstates of $H$, $|\psi_{i=1,2,3}\rangle$, will satisfy the symmetry condition$$Z_3|\psi_{i}\rangle=|\psi_i\rangle\ \ i\in1,2,3$$as long as the $Z_3$ symmetry was not spontaneously broken (for simplicity, $Z_3$ denotes both the group and it's unitary representation).
Spectrum of $H$: The degeneracy of eigenvalues and the symmetry of eigenstates
The eigenvalues and eigenstates were very easy to obtain$$\left\{\begin{aligned}&E_e=E_0+2J\\&E^{(2)}_g=E_0-J\end{aligned}\right.\ \ \ \ \left\{\begin{aligned}&|\psi_e\rangle=\frac{1}{\sqrt{3}}(|1\rangle+|2\rangle+|3\rangle)\\&|\psi^{(2)}_g\rangle=|\psi_{a,b}\rangle\ \ \ for \ \ \langle \psi_a|\psi_e\rangle=\langle \psi_b|\psi_e\rangle=\langle \psi_a|\psi_b\rangle=0\end{aligned}\right.$$ where $E_e, E^{(2)}_g$ denotes the excited state and ground state respectively, and the superscript in $E^{(2)}_g$ denotes the two-fold degeneracy of the GS.
The unique ES $|\psi_e\rangle$ stays unchanged under the symmetry transformation $$Z_3|\psi_e\rangle=|\psi_e\rangle.$$
For the two-fold degeneracy GS, one can always choose two referance state $|\psi_a\rangle,|\psi_b\rangle$ that satisfies the orthogonality condition but changes under the action of $Z_3$, for example$$\left\{\begin{aligned}&|\psi_a\rangle=\frac{1}{\sqrt{2}}(|1\rangle-|2\rangle)\ \ \ \ \ \ \ \ \ \ \ \ \ \ Z_3|\psi_a\rangle\ne|\psi_a\rangle\\&|\psi_b\rangle=\frac{1}{\sqrt{6}}(|1\rangle+|2\rangle-2|3\rangle)\ \ \ Z_3|\psi_b\rangle\ne|\psi_b\rangle \end{aligned}\right.$$
Review some important properties of TFI model
Consider the TFI model $$H_{\mathrm{TFI}}=-\sum_{i}{h\sigma^x_i}+\sigma^z_{i}\sigma^{z}_{i+1}$$for $h<1$
1). The true GS keeps the $Z_2$ symmetry, and the two aligned states $|\mathrm{all\ up}\rangle,|\mathrm{all\ down}\rangle$ are neither degenerated nor the eigenstates of TFI model for finite $N$.
2). These two aligned states become both degenerated and the eigenstates of TFI model which breaks $Z_2$ symmetry when $N\to +\infty$.
Questions
- Can I say that this $H$ has SSB due to the spontaneous $Z_3$ symmetry breaking by the two-fold degeneracy GS?
- If the answer of Q1 is Yes. It's obvious that I don't need anything (like $N\to\infty$) to make $|\psi^{(2)}_g\rangle$ become degeneracy and eigen. So my question 2 is: What kinds of quantities make this symmetry breaking become "Spontaneous" in this simple model?