= Solution
A <Hermitian operator> satisfies $\langle u,Av\rangle=\langle Au,v\rangle$ for all admissible states in its domain; in finite dimension this is $A^*=A$. For an unbounded physical observable one specifies a self-adjoint realization, rather than only a formal differential expression.
Here $p=-i\hbar\,d/dx$ and $H=-\hbar^2d^2/(2m\,dx^2)+V(x)$. For states with sufficient regularity and decay, integration by parts gives
$$
\langle u,Hv\rangle-\langle Hu,v\rangle
=-\frac{\hbar^2}{2m}[\bar u v'-\bar u'v]_{-\infty}^{\infty}=0.
$$
The multiplication term cancels because $V$ is real. Thus the <Quantum Hamiltonian> is Hermitian on the stated admissible domain. Equivalent boundary conditions must remove this boundary term if the system is on a bounded interval.
The <Schrödinger equation> $i\hbar\dot\psi=H\psi$ and its conjugate imply, for a time-independent observable $A$ and normalized state,
$$
\frac d{dt}\langle A\rangle
=\frac i\hbar\langle HA-AH\rangle.
$$
This follows by differentiating both factors in $\langle\psi,A\psi\rangle$ and moving $H$ using Hermiticity. The canonical <commutator> $[x,p]=i\hbar$ gives
$$
[H,x]=-\frac{i\hbar}{m}p,\qquad [H,p]=[V,p]=i\hbar V'(x).
$$
Consequently the two <Ehrenfest theorem> identities are
$$
\boxed{\frac d{dt}\langle x\rangle=\frac{\langle p\rangle}{m},\qquad
\frac d{dt}\langle p\rangle=-\langle V'(x)\rangle.}
$$
They require the displayed expectations and domain operations to exist; real-valuedness of a potential alone is not a proof about every possible singular operator domain.
Back to article page
