Why are the energy eigenstates realized in atomic transitions?

Question

I have a question like "Why is it often assumed that particles are found in energy eigenstates?", it is a little different, though.

When one solves the hydrogen atom, one can use a polynomial Ansatz and derive the energy eigenfunctions with that. The energy differences between the eigenstates are exactly the energies that are observed in spectroscopy. So it seems to me that atoms are in a pure eigenstate before and after the transition.

If the state was a superposition of two eigenstates before the transition and a different one after, would the energy difference still be a difference of $E_n$?

Say $$|\text{before}\rangle = \frac1{\sqrt2} \left(|1\rangle + |2\rangle \right)$$ and $$|\text{after}\rangle = \frac1{\sqrt2} \left(|0\rangle + |5\rangle \right).$$

Then before it is $\langle H \rangle = (E_1 + E_2)/\sqrt2 $ and $\langle H \rangle = (E_0 + E_5)/\sqrt2 $ afterwards. The difference would be something which is not simply $E_n - E_m$.

One postulate of quantum mechanics is that each single measurement is an eigenvalue of the operator. So even though $\langle H \rangle$ could be something arbitrary, a single measurement would have to be taken out of the $E_n$? How does this relate to the energy difference (the emited photon) having energies of $E_n - E_m$?

Answer 1

The expectation value of energy is something else than the energy in a particular experiment. With your choice of the initial states, the photons emitted (negative difference) or absorbed (positive difference) will have energies either $$ E_1-E_0 \text{ or } E_2-E_0 \text{ or } E_1-E_5 \text{ or } E_2-E_5 $$ If each of the four transitions were equally likely, with your complex amplitudes, each of the four transitions would be equally likely. But no other energy difference aside from the list of four possibilities above is possible because the energy of that atom is quantized.

The difference of expectation values is just the weighted average of the energy differences – with probabilities' playing the role of the weights. But the actual possibilities are only four, discrete, quantized. The expectation values only change continuously because probabilities are continuous but the energy of the atom is not!

As Danu said, it's also unreasonable to assume that the final state is a nontrivial mixture of different energy eigenstates because by measuring the photon's energy, we measure both the initial and final energy more or less uniquely. If we measure some quantity, we bring the physical system to an eigenstate of that quantity (in this case, and often, energy) corresponding to the measured (eigen)value.

Answer 2

Suppose your initial state is $\lvert 2\rangle$ and that the states $\lvert 0 \rangle$ and $\lvert 1 \rangle$ have lower energies than $\lvert 2 \rangle$. Assuming that there is no so called selection rule that prevents $\lvert 2 \rangle$ from emitting a photon and end up in $\lvert 0 \rangle$ or $\lvert 1 \rangle$, then the final state will be $$ \lvert 2 \rangle_{\mathrm{final}} = a\lvert 0 \rangle\lvert\mathrm{photon}_{20}\rangle + b\lvert 1\rangle\lvert\mathrm{photon}_{21}\rangle $$ It is a linear combination of the two alternatives. One of them is the transition from $\lvert 2 \rangle$ to $\lvert 0 \rangle$ creating a $\mathrm{photon}_{20}$ with energy $E_2-E_0$. That happens with the amplitude $a$ (probability $\lvert a\rvert^2$). The other corresponds to the transition from $\lvert 2 \rangle$ to $\lvert 1 \rangle$. Now you can write down how the final state $\lvert 3 \rangle_{\mathrm{final}}$, assuming $\lvert 3 \rangle$ has somewhat higher energy than $\lvert 2 \rangle$. Finally, if the initial state is $$ \frac{1}{\sqrt{2}}(\lvert 2 \rangle+\lvert 3 \rangle) $$ then the final state will be $$ \frac{1}{\sqrt{2}}(\lvert 2 \rangle_{\mathrm{final}}+\lvert 3 \rangle_{\mathrm{final}}) $$ If photons are measured and their energy is measured, than an individual photon will be one of the photons in the final state e.g. $\mbox{photon}_{31}$. The final state will appear to be collapsed to $\lvert 1\rangle \lvert \mbox{photon}_{31}\rangle$, which says that the atom is in state $\lvert 1 \rangle$.

Answer 3

As you stated already, a measurement of the energy of the hydrogen atom must return an energy eigenvalue. Measuring before and after a transition gives us two energies $E_n$ and $E_m$. This is always true, regardless of the fact that the expectation value of the energy before measuring might not be a difference of $E_p$ and $E_q$ for some $p,q$: The actual transition is always between two of such energy levels!

Now, simple conservation of energy tells us that the energy emitted (or absorped) during the transition has to be the difference of these two energy levels. To avoid confusion, keep in mind that the expectation value of the energy right after the measurement is exactly $E_\text{measured}$ and nothing else.