Let

Q ψ_{n} = q_{n} ψ_{n}

with

ψ_{n} \neq = 0

. The Hermitian operator identity and conjugate linearity of the first slot of the inner product imply

q_{n} ⟨ ψ_{n}, ψ_{n} ⟩ = ⟨ ψ_{n}, Q ψ_{n} ⟩ = ⟨ Q ψ_{n}, ψ_{n} ⟩ = q_{n}^{*} ⟨ ψ_{n}, ψ_{n} ⟩ .

The squared norm is positive, so

q_{n} = q_{n}^{*}

: every such eigenvalue is real.

2

Solution

For two eigenfunctions, apply the same Hermitian operator identity across the two states:

q_{n} ⟨ ψ_{m}, ψ_{n} ⟩ = ⟨ ψ_{m}, Q ψ_{n} ⟩ = ⟨ Q ψ_{m}, ψ_{n} ⟩ = q_{m} ⟨ ψ_{m}, ψ_{n} ⟩ .

Hence

(q_{n} - q_{m}) ⟨ ψ_{m}, ψ_{n} ⟩ = 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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