 Solution

For a time-independent Quantum Hamiltonian, a stationary state is an energy eigenstate. The Schrödinger equation separates its time dependence asIts probability density $|\psi (x,t){|}^2=|\chi (x){|}^2$ is time independent. Linearity of the Schrödinger equation givesPut $A_{jk}={\int }_{-\infty }^{\infty }{\chi }_j^{\ast }\hat{A}{\chi }_{k}dx$. The operator for a time-independent observable is a Hermitian operator, so $A_{21}=\overline{A_{12}}$. Substituting the evolved wavefunction into the expectation value givesand therefore, for a normalized state,The normalization is automatic for orthogonal ${\chi }_1,{\chi }_2$, in particular when $E_1\ne E_2$: self-adjointness of $\hat{H}$ gives $(E_1-E_2)⟨{\chi }_1,{\chi }_2⟩=0$. If the energies coincide, normalization of the individual eigenstates alone does not ensure normalization of their displayed sum; its squared norm is $1+\mathrm{Re}⟨{\chi }_1,{\chi }_2⟩$. For a nonzero sum not having unit norm, the physical expectation value is the displayed numerator divided by this squared norm. With orthonormal energy states, the stated formula holds in both the distinct and degenerate cases.