 Solution

A Hermitian operator satisfies $⟨u,Av⟩=⟨Au,v⟩$ for all admissible states in its domain; in finite dimension this is $A^{\ast }=A$. For an unbounded physical observable one specifies a self-adjoint realization, rather than only a formal differential expression.

Here $p=-i\hslash d/dx$ and $H=-{\hslash }^2{d}^2/(2md{x}^2)+V(x)$. For states with sufficient regularity and decay, integration by parts givesThe 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 $i\hslash \dot{\psi }=H\psi$ and its conjugate imply, for a time-independent observable $A$ and normalized state,This follows by differentiating both factors in $⟨\psi ,A\psi ⟩$ and moving $H$ using Hermiticity. The canonical commutator $[x,p]=i\hslash$ givesConsequently the two Ehrenfest theorem identities areThey 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.