# Why hydrogen emission spectrum is discrete?

It is generally known that the emission spectrum of hydrogen is discrete, which is usually explained as follows: when a photon is emitted, the atom jumps to a lower energy eigenstate, the energy difference being the energy of the emitted photon.

However, energy eigenstates are not the only quantum states an atom can be in, since any linear combination of those is a pure state as well (provided we take care of normalization). So, my question is: why cannot a hydrogen atom jump from an excited $$2s^1$$ state to a linear combination of $$1s^1$$ and $$2s^1$$, emitting a photon with some random energy, provided that the expectation of total energy is still the same?

I feel that this has something to do with the measurement problem & wavefunction collapse, yet I cannot grasp what is going on.

• What would be the energy of such a mixed state be? – Ezze Nov 7 '19 at 15:59
• @Ezze It is not a mixed state, it is a pure state, i.e. represented as a vector, not as a density operator. As I understand it, such a state doesn't have a definite value of energy; the energy is that of $1s^1$ or $2s^1$ with some probabilities. – lisyarus Nov 7 '19 at 16:04
• So if the energy is really just the energies of orbital energies with different probabilities, then what values would you observe in the spectrum? – Ezze Nov 7 '19 at 16:06
• @Ezze Well, that's what I am asking about. – lisyarus Nov 7 '19 at 16:07
• You would get either the difference between the original eigenstate and $1 s^1$ or the difference between original eigenstate and $2 s^1$, with the probabilities you just described. In other words, you get back the same spectrum what you would get from pure transitions. – Ezze Nov 7 '19 at 16:09