Why can atoms only absorb/emit certain frequencies of light? I'm aware that the orbitals of atoms have discrete energy spectra. The energies must obey the atom's Hamiltonian eigenvalue equation $H|\psi_n\rangle=E_n|\psi_n\rangle$ and this only holds for certain values of $E$. What confuses me is that while the allowed eigenvalues are discrete, the expectation values of operators often behave continuous. 
An example would be Lamor precession. Spin states take on discrete eigenvalues (up and down) but the expectation value of the spin $\langle\mathbf{\hat S}\rangle(t)$ rotates continuously.
The absorption of a photon is often depicted as follows. We have some electron in the state $|n\rangle$. When it interacts with the photon it can be absorbed and as result the electron jumps up an energy level: $|n\rangle\rightarrow|n+1\rangle$. But this only happens when the energy of the photon matches the energy difference $E_{n+1}-E_n$.
So my question is, why can't atoms absorb photons with a fraction of this energy difference ? As a result of energy conservation the electron would now be in a superposition $\alpha|n\rangle+\beta|n+1\rangle$. The expectation value of the new energy would match that of the absorbed photon. In this case the energy of the electron would increase with $|\beta|^2$.
 A: The transition is 
\begin{align}
\newcommand{\ra}{\rangle}
 |n+1\ra &\,|\text{no photons}\ra
\\
 &\to
 \alpha|n\ra\,|\text{one photon}\ra
 +\beta|n+1\ra\,|\text{no photons}\ra.
\end{align}
Energy is conserved in both terms, but it is distributed differently between atom and photon in the two terms. If you surround the atom with a 100% efficient photon-detector, then the photon-detection measurement will result in the state being either
$$
 |n\ra\,|\text{one photon}\ra
\hskip1cm
 \text{with probability }\propto |\alpha|^2
$$
or
$$
 |n+1\ra\,|\text{no photons}\ra
\hskip1cm
 \text{with probability }\propto |\beta|^2.
$$
With either outcome, the energy is the same as it was originally.
The process envisioned in the question, where the emitted photon's energy matches the expectation value of the atom's energy in the state $\alpha|n\ra+\beta|n+1\ra$, does not conserve energy, because the final state in that (impossible) process would be
\begin{align}
 |n+1\ra &\,|\text{no photons}\ra \\
 &\to
 \alpha|n\ra\,|\text{mini-photon}\ra
 +\beta|n+1\ra\,|\text{mini-photon}\ra
\end{align}
where "mini-photon" means a photon with just part of the energy $E_{n+1}-E_n$. After any measurement that reveals the energy of the atom, the state will be
either
$$
 |n\ra\,|\text{mini-photon}\ra
\hskip1cm
 \text{with probability }\propto |\alpha|^2
$$
or
$$
 |n+1\ra\,|\text{mini-photon}\ra
\hskip1cm
 \text{with probability }\propto |\beta|^2.
$$
Neither of these has the same energy as the original state, so the process suggested in the question does not conserve energy. Expectation values don't tell the whole story. Conservation laws hold in every instance, not just statistically.
