= Solution
For two <eigenfunctions>, apply the same <Hermitian operator> identity across the two states:
$$
q_n\langle\psi_m,\psi_n\rangle
=\langle\psi_m,\widehat Q\psi_n\rangle
=\langle\widehat Q\psi_m,\psi_n\rangle
=q_m\langle\psi_m,\psi_n\rangle.
$$
Hence $(q_n-q_m)\langle\psi_m,\psi_n\rangle=0$. Distinct <eigenvalues> therefore imply <orthogonal> <eigenfunctions>. Within a degenerate eigenspace, <orthogonality> is a possible basis choice, rather than a consequence for every pair of vectors.
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