Suppose $^\dagger$ we have absolutely no knowledge of a hydrogen atom. It is well isolated. And we put a photon detector next to it (and pretty sure it has no interaction with the hydrogen atom.)

Now suppose now the detector receives a photon with a specific frequency, which indicates the hydrogen atom was at a specific energy eigenstate {$n'$}, and now, {$n'-1$}.

The question is: could the hydrogen atom be in a superposition of energy eigenstates?

If yes $^{\dagger\dagger}$, then since the specific frequency indicates that the hydrogen atom now is in a specific energy eigenstate; therefore, it must have been experiencing a wavefunction collapse. And since we premised no interaction or whatsoever (there is only a equally well isolated detector next to the hydrogen atom, or rather we could place the detector $3\cdot 10^8m$ far away... ); therefore, there must be a spontaneous wavefunction collapse.

I wonder, what are the problems with my thought experiement? $^{\dagger\dagger\dagger}$

$\dagger :$I am attempting to show that wavefunction collapse spontaneously; however, I am looking for the flaw of it, as I have just learned my undergrad quantum physics last school year, so I probably missed something here.

$\dagger\dagger$: I have asked this question to some people and place, and I was told yes. So, I will just assume yes here.

$\dagger\dagger\dagger$ : I have asked a question here, and have also read that spontaneous wavefunction collapse is nearly impossible. so I wonder what are the flaws in my thought experiment? Here are a few guesses of mine:

1.) I basically assume measurement is about interaction, however, it can be related to entanglement. I wonder if it is exactly the flaw.

2.) On the other hand. I read a article some time ago, it basically says the whole thing about emitting photon is deterministic, and has something to do with field; however, sadly, I do not quite understand the article, but I doubt it also possibly be the problems with my thought experiment.

Would anyone be kind enough to shed some light on this matter?

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    $\begingroup$ I'm not sure what the question is here. The atom interacts with the electromagnetic field, which is then detected by the detector; it's pretty much a totally standard measurement. What is supposed to be "spontaneous"? $\endgroup$ – knzhou Nov 4 '16 at 22:56
  • $\begingroup$ @knzhou There is not necessarily a electromagnetic field in the experiment, and the detector can be far away. Or rather, if a excited state will never emit a photon without electromagnetic field? (I read that the atom will spontaneously emit a photon.) $\endgroup$ – Shing Nov 4 '16 at 23:02
  • $\begingroup$ The point is that spontaneous emission is an interaction; the atom is always interacting with the (quantum) electromagnetic field via an interaction term in the Hamiltonian, whether or not the classical macroscopic field value is nonzero. $\endgroup$ – knzhou Nov 4 '16 at 23:04
  • $\begingroup$ @knzhou I do not understand the spontaneous emission mechanism. I basically assume it does not take any interaction. Would you mind answering it? $\endgroup$ – Shing Nov 4 '16 at 23:09
  • $\begingroup$ The interaction happens through the EM field. Photons are quanta of the EM field. $\endgroup$ – Prof. Legolasov Nov 5 '16 at 19:03

The resolution is that the wavefunction collapse isn't spontaneous at all -- it's due to the interaction of the atom with the electromagnetic field. This interaction is nonzero even though the classical field value could be zero everywhere.

The electromagnetic field is often treated classically during a first exposure to QM. Then the interaction term would be something like $$H_{\text{int}} \sim \mathbf{d} \cdot \mathbf{E}$$ where $\mathbf{d}$ is the operator for the dipole moment of the atom, and $\mathbf{E}$ is the classical electric field. Then indeed there is no interaction when $\mathbf{E} = 0$.

However, a more thorough treatment should also quantize the electromagnetic field. For simplicity, let's suppose we're in a resonant cavity with only one mode of the electromagnetic field; we can then treat its states as those of a harmonic oscillator, where $|n\rangle$ is the state with $n$ photons, with raising and lowering operators $a^\dagger$ and $a$. Similarly, for simplicity, treat the atom as a harmonic oscillator as well, with raising and lowering operators $b^\dagger$ and $b$. Then the interaction term is $$H_{\text{int}} \sim a^\dagger \otimes b + \text{h.c.}$$ That is, the atom can emit a photon, increasing the number of photons in the field, or absorb a photon, increasing its energy level.

This interaction term is always nonzero even when the initial state of the electromagnetic field is the vacuum, $|0 \rangle$. This allows for "spontaneous" emission, where the atom emits even though no photons are around. However, the name's a misnomer -- it's not really spontaneous, it's due to the interaction of the atom with the (quantum) electromagnetic field. (One way to see this is to think of $\mathbf{E}$ as a classical field, but 'smudged' about its zero value by the uncertainty principle.) If you only ever look at the atom's state, it's the emission of this photon that collapses the superposition.

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    $\begingroup$ So you are saying it's an interaction between the dipole moment and the vacuum fluctuations of the electromagnetic field, as a first approximation? Can the spontaneous emission rates be calculated correctly that way? $\endgroup$ – Bob Bee Nov 5 '16 at 5:33
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    $\begingroup$ How "If you only ever look at the atom's state, it's the emission of this photon that collapses the superposition"? The atom should be 1st become state {$n'$}, and then emit a photon to be state {$n'-1$}. if only considering the atom, the atom "collapse" itself to {$n'$} first. Or you mean all those process happen at the same time? Hence it is reasonable to say "emission of photon that collapses the superposition"? $\endgroup$ – Shing Nov 5 '16 at 10:59
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    $\begingroup$ @Shing After the emission, the atom and field states are entangled, but there's still a perfectly well-defined quantum state. However, if you trace out the field state, the resulting atom state must be described by a density matrix, not a single quantum state. So we say the superposition has been destroyed. $\endgroup$ – knzhou Nov 5 '16 at 21:00

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