The existence of degenerated ground states of a many-body quantum system is usually taken as a signature of the quantum order.

I am considering the follow question:

If we have a many-body quantum system with a Hamiltonian $H$, also we know its ground states are degenerated given by $\psi_1, \psi_2,...\psi_N$.

Can the $N$ different degenerated ground states be transformed to each other without phase transition? Or in general these degenerated ground states may belong to different phases?

  • $\begingroup$ What do you mean by "transformed without phase transition"? Which basis do you fix? Otherwise, yes, you can design Hamiltonians which have ground states in different phases -- just couple a symmetry breaking model to two different topological models. $\endgroup$ – Norbert Schuch Nov 3 '17 at 23:21
  • $\begingroup$ @Norbert Schuch Could you please give me an example on 'couple a symmetry breaking model to two different topological models'? Thanks. $\endgroup$ – XXDD Nov 4 '17 at 2:03
  • $\begingroup$ Maybe I have a misunderstanding. But for me 'phase of a quantum state' is a property of the 'state' itself but not the Hamiltonian of the system. So if two states $\psi_1,\psi_2$ of a quantum system belong to the same phase iff they can be connected by a gapped adiabatic evolution $H(g)$ with $\psi_1,\psi_2$ as the ground states of $H(0),H(1)$ respectively(Or equivalently they can be connected by a local unitary evolution). $\endgroup$ – XXDD Nov 4 '17 at 2:33
  • $\begingroup$ So in my understanding, the states $\psi_1,...\psi_N$ belong to the same phase if they can evolve between each other by local unitary evolutions. So my question is : if the states $\psi_1,...\psi_N$ can be written as the degenerated ground states of some $H$, then they can be connected by local unitary evolutions. Or in another way, if there exists two states that can not be connected by a finite length quantum circuit, but they are the degenerated ground states of a gapped system. $\endgroup$ – XXDD Nov 4 '17 at 2:42
  • 2
    $\begingroup$ Take an Ising ferromagnet. Use the Ising variables as a control qubit to either activate a Toric Code Hamiltonian or a Double Semion Hamiltonian. (E.g. choose the local terms to be block-diagonal, with the two blocks the TC and DS Hamiltonian (shifted to be negative), and the block label the Ising spin.) The system will have two sets of ground states, one with TC order and one with DS order. --- If such an example suffices as an answer, I might write one. $\endgroup$ – Norbert Schuch Nov 4 '17 at 17:25

We usually define a Hamiltonian to be in a particular phase, not a single quantum state. Specifically, we consider the space of all possible (local) Hamiltonians $H(\{g\})$ with all possible local coupling constants $g$, and consider the free energy density in the thermodynamic limit $$f(\{g\}, T) := \lim_{N \to \infty} \frac{-T \ln \left( \text{Tr } e^{-H(\{g\})/T} \right)}{N}$$ where $N$ is the system size. The phase boundaries are the points in Hamiltonian parameter space where $f$ depends non-analytically on $T$ or on one of the coupling constants $g$. So you can't talk about transform one state to another with or without a phase transition, only transforming Hamiltonians.

A well-posed (but different) question is whether the ground states of a Hamiltonian must always be connected by local unitary transformations. For a Hamiltonian in the topologically trivial phase, the answer is yes. For a topologically ordered Hamiltonian, the answer is no.

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