Continuous spectra of photons I guess this is a basic quantum mechanics problem, but I'm not entirely sure of my answer.
Suppose we have an electron in a hydrogen atom having the state
\begin{equation}
\frac{1}{\sqrt{x^2+(1-x)^2}}\left[x\ |4s\rangle + (1-x)\ |3p\rangle\right]
\end{equation}
Now, we change all of their energy levels according to Hund's rule. Without going into the details of which states the electron goes into, it suffices for us to say that we can write the change in energy in the form of,
\begin{equation}
\Delta E = \frac{x^2}{x^2+(1-x)^2}E_{4s\rightarrow??}+ \frac{(1-x)^2}{x^2+(1-x)^2}E_{3p\rightarrow??}
\end{equation}
Then it seems that by varying $x\in[0,1]$, we can have some kind of a continuous spectrum of energies being emitted. I am not doubting that if we only have the individual states, the change in energy is discrete, nor am I stating that from a state with a fixed $x$, arbitrary values of the change in energy can be obtained. The change in energy will be decided by the value of $x$.
There are two ways in which I can think of exits to this problem, but I am unsure of either.


*

*The change in energy levels each release a photon (or some part of?). So there are two photons (or some part of) released.

*Electrons can't have such a kind of wave-function.
What do you think?
 A: Because the electron's initial state is a superposition of states with different energies, the electron does not have definite energy. Measuring its energy, you would find either $E_{4s}$ (with probability $\frac{x^2}{x^2+(1-x)^2}$) or $E_{3p}$ (with probability one minus that). Since it's energy isn't a definite number, we have to resort to talking about the electron's expected energy, $\langle E \rangle$, which is the average of $E_{4s}$ and $E_{3p}$ weighted by their probabilities.
If the electron jumps to the next lowest level, meaning it goes to $3p$ if it's in $4s$ or to $3s$ if it's in $3p$, then the expected energy changes, and the change $\Delta\langle E \rangle$ (equal to your $\Delta E$) varies over a continuous range depending on the parameter $x$ in your initial state. And, since the electron jumped to a lower level, a photon was emitted. In principle we could catch the photon and measure its energy. Because energy is conserved, the expected energy of the photon is $-\Delta\langle E \rangle$ (which is positive since $\Delta\langle E\rangle < 0$). So when the electron jumps from your initial state to the next lowest level it emits a photon whose expected energy, $-\Delta\langle E\rangle$, is a continuous function of the parameter $x$.
Nevertheless, the spectrum of the photon is discrete and independent of $x$ (for $0<x<1$). By "spectrum" I mean the set of possible values we would find when measuring the photon's energy. We would find either $E_{4s}-E_{3p}$ or $E_{3p}-E_{3s}$ with probabilities that depend on $x$. This is because energy conservation requires that the combined state of the photon and electron be such that when the electron is in $3p$ the photon has energy $E_{4s}-E_{3p}$ and when the electron is in the other state the photon has the other energy. Explicitly, their combined state after the jump but before anybody measures anything is
$$
\frac{x}{\sqrt{x^2-(1-x)^2}}|3p\rangle |E_{4s}-E_{3p}\rangle
+\frac{1-x}{\sqrt{x^2-(1-x)^2}}|3s\rangle |E_{3p}-E_{3s}\rangle
$$
where the first ket is the state of the electron and the second is the state of the photon, labeled by its energy. If we measure the energy of either the electron or the photon then their combined state will collapse into one or the other of the two product states in the superposition.
