 Solution

Hermiticity of the Hamiltonian operator gives $⟨{\psi }_m,H{\psi }_n⟩=⟨H{\psi }_m,{\psi }_n⟩$, soThe distinct eigenvalues force orthogonality for $m\ne n$, and the given normalization gives one for $m=n$. Therefore $⟨{\psi }_m,{\psi }_n⟩={\delta }_{mn}$.

Under the time-independent Schrödinger equation, each energy component acquires its own phase:The sum converges in the state Hilbert space because the squared coefficients are summable. Orthogonality makes its squared norm ${\sum }_{n\ge 0}{2}^{-(n+1)}=1$ at every time.

The Born rule for energy measurement givesBoth depend only on squared coefficient magnitudes, so both are time-independent. The expectation follows either by summing energy times measurement probability or by inserting the orthogonal expansion in the quadratic form of $H$.

Normalization alone does not guarantee a finite mean for an arbitrary spectrum. The displayed energy series is finite when $\sum |E_n|2^{-(n+1)}<\infty$. Since the spectrum is bounded below by $E_0$, a divergent positive tail gives an extended expectation $+\infty$, still independent of time; for example an abstract diagonal Hamiltonian with $E_n=4^n$ has this property. No growth assumption on $E_n$ is printed, so finiteness should not be asserted without this qualification.