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).

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