= Solution
Use an <orthonormal basis> of <stationary states> and project the <Time-dependent Schrödinger equation> onto state 2 after removing its free-evolution phase. The stationary <Hamiltonian> terms cancel the derivatives of those phases, leaving
$$
i\hbar\dot c_2=V_{21}(t)e^{i(E_2-E_1)t/\hbar}c_1+V_{22}(t)c_2.
$$
Define $d_{21}=\langle\psi_2|ex|\psi_1\rangle$ and $d_{22}=\langle\psi_2|ex|\psi_2\rangle$. With the supplied <time-dependent quantum perturbation>,
$$
\boxed{\omega_0=\frac{E_2-E_1}{\hbar},\qquad
\Omega=\frac{\mathcal E d_{21}}{\hbar},\qquad
V_{22}(t)=\mathcal E d_{22}\cos\omega t,}
$$
and therefore $\dot c_2=-i\Omega\cos(\omega t)e^{i\omega_0t}c_1-iV_{22}c_2/\hbar$. The state phases may be chosen so $d_{21}$ and $\Omega$ are real for this single driven pair; otherwise <probabilities> depend on $|\Omega|^2$ and the conjugate <matrix element> appears in the equation for $c_1$. Our $\Omega$ is the resonant <Rabi angular frequency> in the usual convention whose rotating-wave off-diagonal <Hamiltonian> is $\hbar\Omega/2$.
For a field-free atom with symmetry under <spatial reflection>, each selected nondegenerate stationary orbital can be chosen with definite <parity>. Since $x$ is odd, $\langle2|x|2\rangle=-\langle2|x|2\rangle$, so $V_{22}=0$; the same reasoning gives $V_{11}=0$. \b[Being an <energy eigenstate> alone is not sufficient.] As an explicit mathematical counterexample, a displaced <quantum harmonic oscillator> with potential proportional to $(x-x_c)^2$ has <energy eigenstates> with $\langle x\rangle=x_c$, giving $V_{22}=e\mathcal E x_c\cos\omega t$ when $x_c\ne0$. The vanishing used in the atomic model rests on its <parity> assumption, not merely on stationarity.
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