Let’s assume we have a Hamiltonian for spin-1/2 particles with two terms, a classical interaction term and a “quantum” (non-diagonal) term. For simplicity, let’s assume that the quantum term is a simple transverse field or an XX interaction, say $-h\sum X_i$. We further assume that the two terms do not commute and there is a quantum phase transition for some $h$.
We set $h=0$. We know (from other means) that the classical part of the hamiltonian has an exact ground state degeneracy, say 8 states minimise the classical energy function. Now, we turn $h>0$ but remaining in the classical phase. The ground state degeneracy is broken, but there is an asymmetry in the spectrum. Not all classical ground states have the same contributions in the ground state vector. We observe (eg. from exact diagonalization of small system sizes) that a specific classical state is selected by the quantum fluctuations as the ground state.
I want to focus on two scenarios:
The state is selected due to fluctuations from classical states outside of the classical ground state manifold. This is what I would call order by disorder. Since we are at zero temperature, this would be quantum order by disorder.
The state is selected due to fluctuations from states from the classical ground state manifold.
As a comparison, for a case where $H_{classical}=-\sum Z_i Z_{i+1}$ in 1D, I would have SSB (in the thermodynamic limit), meaning that the ground state vector for any $h \neq 0$ but small and any finite system size would have two equal contributions from both classical ground states and all other states would have contributions orders of magnitude smaller.
Questions:
- Do you agree on my definition of quantum order-by-disorder?
- How would you call the second mechanism? How would you study it? Are there known results in the literature with this behaviour?
