At a recent talk, I was told by the speaker that it is not possible to adiabatically transfer from one ground state $|\psi_0 \rangle$ to another $|\psi_1 \rangle$ if these states are orthogonal. The reason they gave was that the spectral gap necessarily closes at some point along the adiabatic path. However, I have not been able to find any reference for this claim.


Mathematically, my question is as follows.

Consider two gapped Hamiltonians $H_0$ and $H_1$ which have unique ground states $|\psi_0 \rangle$ and $|\psi_1 \rangle$ respectively. Further, suppose $\langle \psi_0 | \psi_1 \rangle = 0$. Now consider a continuous one-parameter family of Hamiltonians $\{ H(s) \}$ such that $H(0) = H_0$ and $H(1) = H_1$. Does $\langle \psi_0 | \psi_1 \rangle = 0$ imply that there exists an $s_{*}$ such that $H(s_{*})$ is gapless?


To clarify, by "gapped" and "gapless", I mean in the thermodynamic limit. As Emilio's answer shows, one can construct trivial counterexamples for a single qubit.

It would be interesting to know if there are any "reasonable" restrictions on the Hamiltonian for which this statement becomes true (e.g. the Hamiltonian should be k-local for some k).


No, that is obviously wrong. It's trivial to construct a hamiltonian on a two-level system which maintains a constant gap as it swaps the ground state with the excited state.

As an explicit example: $$ H(s) = \begin{pmatrix} 0 & e^{-is} \\ e^{is} & 0 \end{pmatrix} . $$

It may well be possible to sharpen down the statement (considerably!) and to make it talk about e.g. gaps in the thermodynamic limit, so that it becomes restricted enough that it has a chance of being true. But as stated it is simply false.

  • $\begingroup$ Thanks for your answer. Indeed I was thinking more of gaps in the thermodynamic limit (though of course I should have stated this in my question, sorry!). Do you have any idea under which restrictions this statement may be true? I suppose your counterexample shows that interactions are necessary. $\endgroup$ – anon1802 Feb 23 '19 at 13:07
  • $\begingroup$ I don't know of any such statement, but it's not my field. As you've stated it, there's a huge number of things it could mean, and I don't really know enough to point you in any useful direction. $\endgroup$ – Emilio Pisanty Feb 23 '19 at 13:17
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    $\begingroup$ @OliverLunt Obviously, you can tensor Emilio's example with any spin chain or the like with a Hamiltonian which is not changed, then it gives a counterexample to such extensions of the problem. The way you ask it now the answer is simply "No." -- It might help if you give more context -- why said this, in which context? But there is always the possibility that the statement was simply wrong. $\endgroup$ – Norbert Schuch Feb 23 '19 at 23:34

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