= Solution
Hermiticity of the <Hamiltonian operator> gives $\langle\psi_m,H\psi_n\rangle=\langle H\psi_m,\psi_n\rangle$, so
$$
(E_n-E_m)\langle\psi_m,\psi_n\rangle=0.
$$
The distinct <eigenvalues> force <orthogonality> for $m\ne n$, and the given normalization gives one for $m=n$. Therefore $\langle\psi_m,\psi_n\rangle=\delta_{mn}$.
Under the time-independent <Schrödinger equation>, each energy component acquires its own phase:
$$
\boxed{\Psi(x,t)=\sum_{n=0}^\infty2^{-(n+1)/2}e^{-iE_nt/\hbar}\psi_n(x).}
$$
The sum converges in the state <Hilbert space> because the squared coefficients are summable. <Orthogonality> makes its squared <norm> $\sum_{n\geq0}2^{-(n+1)}=1$ at every time.
The <Born rule> for energy measurement gives
$$
\boxed{\mathbb P(E=E_m)=2^{-(m+1)},\qquad
\langle H\rangle=\sum_{n=0}^\infty\frac{E_n}{2^{n+1}}.}
$$
Both 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.
Back to article page
