Does adiabatic quantum computation require the initial and final ground states to be non-orthogonal?

Question

Background

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.

Question

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?

Edit

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).

Answer 1

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.