= Solution
For a time-independent <Quantum Hamiltonian>, a <stationary state> is an energy <eigenstate>. The <Schrödinger equation> separates its time dependence as
$$
\psi(x,t)=e^{-iEt/\hbar}\chi(x),\qquad \widehat H\chi=E\chi.
$$
Its probability density $|\psi(x,t)|^2=|\chi(x)|^2$ is time independent. Linearity of the <Schrödinger equation> gives
$$
\boxed{\psi(x,t)=\frac1{\sqrt2}\left(e^{-iE_1t/\hbar}\chi_1(x)+e^{-iE_2t/\hbar}\chi_2(x)\right).}
$$
Put $A_{jk}=\int_{-\infty}^{\infty}\chi_j^*\widehat A\chi_k\,dx$. 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> gives
$$
\langle\psi,\widehat A\psi\rangle=\frac12\left(A_{11}+A_{22}+e^{i(E_1-E_2)t/\hbar}A_{12}+e^{-i(E_1-E_2)t/\hbar}\overline{A_{12}}\right),
$$
and therefore, for a normalized state,
$$
\boxed{\langle A\rangle_\psi=\frac{A_{11}+A_{22}}2+\operatorname{Re}\!\left(e^{i(E_1-E_2)t/\hbar}A_{12}\right).}
$$
The normalization is automatic for orthogonal $\chi_1,\chi_2$, in particular when $E_1\neq E_2$: self-adjointness of $\widehat H$ gives $(E_1-E_2)\langle\chi_1,\chi_2\rangle=0$. If the energies coincide, normalization of the individual <eigenstates> alone does not ensure normalization of their displayed sum; its squared <norm> is $1+\operatorname{Re}\langle\chi_1,\chi_2\rangle$. 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.
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