Can gapped state and gapless state be adiabatically connected to each other?

Question

Is it possible that we can construct a gapped state and a gapless state which are adiabatically connected?

Here adiabatically connected I mean:

there exists a class of Hamiltonians $H(g)$ with ground state $|\phi(g)>$($g\in[0,1]$), such that $|\phi(0)>$ is gapless and $|\phi(1)>$ is gapped. And the ground state average of any local operator $<A>(g)$ doesn't has singularity for all $g\in[0,1]$

1. If it's possible, can some one give me an example?

2. If it's impossible. Does it imply we can always find a topological order or a normal order parameter to distinguish a gapped phase from a gapless phase.

Answer 1

You can e.g. consider the "uncle Hamiltonians" introduced in https://arxiv.org/abs/1111.5817 and https://arxiv.org/abs/1210.6613 (disclaimer: I'm a coauthor). The ground states described there (such as the Toric Code model) both have a gapped "parent Hamiltonian" $H_1$ and a gapless "uncle Hamiltonian" $H_2$. If you interpolate $\lambda H_1 + (1-\lambda) H_2$, the ground state will not change throughout, so that there will be no discontinuity whatsoever, but the gap will finally close.

Answer 2

To give another, more mainstream, example: The Hamiltonian of the 1D XY model with a transverse field, $$ H=-\sum X_i X_{i+1} + Y_i Y_{i+1} + h Z_i $$ maps to a free fermion Hamiltonian $$ H = -\sum a_i a_{i+1}^\dagger + \mathrm{h.c.} - h a_i^\dagger a_i\ . $$ This Hamiltonian is diagonalized as $$ H= \sum (cos(k)-h) a_k^\dagger a_k\ . $$ Thus, for $h<-1$, all the modes are empty, while at $h=-1$, the gap closes at $k=0$ and the system becomes gapless. However, for all $h\le -1$, the ground state is the same, namely the fermionic vaccuum, and no discontinuity appears.

(Note that at $k=0$, there is a second ground state, which is necessarily different. However, since it differs by the other state by only one mode in $k$ space being occupied, it is indistinguishable from the ground state up to terms of order $1/N$.)

(Note: There is likely some subtlety in the mapping to free fermions which I neglected, in that the boundary conditions couple to the fermion parity. This should lead to a shift in $k$ space, which should not have any significant effects, except that it might actually split up the degeneracy at $h=-1$.)

Answer 3

No. It is impossible for gapped state and gapless state to be adiabatically connected to each other.

First, the concepts of gapped state and gapless state are concepts in infinite size limit. (see What does it mean for a Hamiltonian or system to be gapped or gapless? ). In infinite size limit, the ground state average of some local operator $<A>(g)$ will have singularity at a certain $ g \in [0,1]$.

The above statement does not imply that we can always find a topological order or a normal order parameter to distinguish a gapped phase from a gapless phase.

Answer 4

If I understand you correctly, how about a hopping Hamiltonian similar to what one gets in boron nitride close to the corner of the Brillouin zone: $$ H = \sum_k \begin{pmatrix} a_k^\dagger\quad b_k^\dagger \end{pmatrix} \begin{pmatrix} mg&ke^{-i\theta} \\ ke^{i\theta}&-mg \end{pmatrix} \begin{pmatrix} a_k\\b_k \end{pmatrix}\,, $$ where $k$ is the crystal momentum, which makes sense only when the size of the system goes to infinity. The creation/annihilation operators correspond to the two atomic species: boron and nitrogen.

The eigenvalues are $\pm \sqrt{m^2g^2 + k^2}$. By tuning $g$, you open or close the gap. If you set g = 0, you get a linear dispersion, like what you see in graphene. In fact, you could (at least in principle) open a gap in graphene by breaking the sublattice symmetry.

Answer 5

A gapped and gapless state can indeed be connected adiabatically under your specific definition of "adiabatic", as Norbert Schuch points out. However, your definition is not the standard one, and under the standard definition, this cannot be done. The more common definition of "adiabatic" is motivated by the adiabatic theorem, which refers to the conditions under which initial instantaneous eigenstates of a time-dependent Hamiltonian time-evolve to instantaneous eigenstates of the new Hamiltonian. Considering the ground state for simplicity, the condition is that $|\dot{E}_0| \ll (\Delta E)^2$, where $E_0$ is the ground-state energy and $\Delta E$ is the gap to the first excited state. This condition obviously cannot be satisfied if the gap $\Delta E$ closes at any point in the evolution, so under this definition one cannot adiabatically evolve through a gapless Hamiltonian.