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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.

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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.

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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 $h=-1$.)

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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.

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    $\begingroup$ Would you agree with this perspective on how this answer is compatible with Norbert Schuch's two counterexamples? $\endgroup$ – tparker Nov 11 '18 at 1:59
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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.

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    $\begingroup$ That's not really what people mean by "gapless". "Gapless is a concept defined in the thermodynamic limit (i.e. a many-body system with the system size going to $\infty$). $\endgroup$ – Norbert Schuch Oct 23 '17 at 6:39
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    $\begingroup$ @NorbertSchuch I think that IcyOtter meant to include an integral over $k$ in his/her Hamiltonian, representing a Fourier transform of a spatially infinite translationally-invariant system, which is indeed in the thermodynamic limit. $\endgroup$ – tparker Oct 23 '17 at 6:53
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    $\begingroup$ How do you know that the ground-state expectation values of all local operators remain analytic at $g = 0$? I don't think that's the case. $\endgroup$ – tparker Oct 23 '17 at 6:55
  • $\begingroup$ @tparker At least in the context of a honey-comb lattice, from which the Hamiltonian comes, the $g = 0$ case is not pathological. It's the low-energy graphene Hamiltonian and I'm pretty sure you can compute expectation values of operators since this is a physical system. $\endgroup$ – IcyOtter Oct 23 '17 at 7:44
  • $\begingroup$ @NorbertSchuch Hi Norbert, tparker is indeed right. Despite it looks like first quantised problem, the Hamiltonian IcyOtter gave is just the one of a non-interacting two band model, e.g. it could be a BdG Hamiltonian, or (as IcyOtter said) a model for graphene. I would naively say that the model IcyOtter gave is completely trivial regarding topological effect beyond Fermi surface stability. $\endgroup$ – FraSchelle Oct 23 '17 at 10:31
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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.

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    $\begingroup$ That's not true -- there is no need that for a gapless Hamiltonian the gap actually closes (except in the tdyn limit). You can very well have gaps which scale as $1/N^\alpha$ and a unique ground state -- indeed, this is done all the time in adiabatic quantum computing, and if you go at a speed 1/poly(N) (i.e., in poly(N)) time, you stay in the g.s. w.h.p.. $\endgroup$ – Norbert Schuch Oct 23 '17 at 7:13
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    $\begingroup$ @tparker In fact what happens in the case of a gap opening/closure is exactly what Wilczek and Zee discussed in [doi.dx.org/10.1103/PhysRevLett.52.2111] : you can still apply the adiabatic theorem but separating all crossing energy bands from the rest via a gap. Now if you try to construct a low-energy theory of the crossing energy bands, your model must contain an explicit gauge structure. Eventually, the gauge structure is non trivial, and the low-energy physics is dominated by topological effects (as long as all the hypothesis to construct the low-energy sector are verified) $\endgroup$ – FraSchelle Oct 23 '17 at 10:26
  • $\begingroup$ @NorbertSchuch I suppose there's an ambiguity in the OP's question, whether "gapless state" means "infinite-system state with zero gap" or "finite-size state with a Hamiltonian that would become gapless in the thermodynamic limit, which is not taken". I think it would be very strange if they referred to the latter as a "gapless state" because, well, it's gapped, so I'm pretty sure that they meant to consider an infinite system. $\endgroup$ – tparker Oct 23 '17 at 15:49
  • $\begingroup$ @tparker I'm not sure we mean the same: My point is that a gapless system can either be gapless for finite size already (e.g. because there is a gapless mode with energy E(k=0)=0 which makes the k=0 state 2-fold degenerate), or because there is a continuum of modes above, with no exact degeneracy, such as the critical Ising model. Now there is a problem with adiabatic evolution in both cases, but it is slightly different: In the latter case, one can do adiabatic evolution at a speed 1/poly(N), which of course till means it becomes impossible for $N\rightarrow\infty$. $\endgroup$ – Norbert Schuch Oct 23 '17 at 16:22
  • $\begingroup$ @NorbertSchuch Ah okay, I see what you're getting at. Totally fair point. I didn't realize that you were including exactly degenerate ground states in your definition of "gapless". $\endgroup$ – tparker Oct 23 '17 at 17:06

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