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This question is related to: Physical reason behind one of the selection rules? and links in the comments.

Take a hydrogen atom in an infinite and otherwise empty universe. The hydrogen atom sits in an excited state, $|e\rangle$. Fermi's golden rule says that if we have a perturbation of the form, $\hat V_0 e^{-\omega ti}$, then the rate of decay from our $|e\rangle$ to a state of lower energy $|g\rangle$ is given by: $$R_{if}=\frac{2\pi}{\hbar} | \langle g |\hat V_0|e \rangle|^2 \delta(\omega_f-\omega_i-\omega)$$ In the above linked question, what I interpreted was been said is that the perturbation $\hat V_0 e^{-\omega ti}$ is that due to the emitted photon. However, for surly such a photon to be emitted the atom must already have decayed. So we seem to have a situation that can never occur since for the atom to decay we must have a photon already emitted but for the photon to be emitted the atom must have already decayed. So where is my reasoning wrong and which comes first the decay of the atom or the emission of the photon?

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    $\begingroup$ Why must one come 'first' over the other? Are they not the same event? $\endgroup$ – Jon Custer Mar 29 '16 at 15:09
  • $\begingroup$ @JonCuster since each one seems to be causing the other. $\endgroup$ – Quantum spaghettification Mar 29 '16 at 15:10
  • $\begingroup$ This question has no answer due to quantum indeterminacy. $\endgroup$ – Lewis Miller Mar 29 '16 at 15:11
  • $\begingroup$ Agree with @LewisMiller. This question seems to be beyond the prediction possibilities of QM. The details of microscopic phenomena are irrelevant to QM as far as their macroscopic impact is unchanged. Thus in this framework no final answer can be given, as no answer can be given on wavefunction collapse based on QM. $\endgroup$ – rmhleo Apr 8 '16 at 11:38
  • $\begingroup$ Since the reaction is reversible the only correct answer is "yes" or "no". You can't tell until you make an observation to collapse the wave form. $\endgroup$ – Jim2B Apr 8 '16 at 21:47
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I think I can pinpoint the misunderstanding that lead to your question. It is here:

In the above linked question, what I interpreted was been said is that the perturbation $\hat{V}_0 e^{-\omega t i}$ is that due to the emitted photon.

The mistake is in thinking that the perturbation is due to the emitted photon. The fact is, the perturbation is not due to the presence of any given photon, but rather due to the coupling to the electromagnetic field. Each mode of the quantum electromagnetic field has a quantum mechanical spread (or "noise") in its amplitude, and it is this "noise" that agitates the excited atom leading to its spontaneous emission of a full blown photon.

n.b. Your "Which came first..." title has mislead others about the true nature of your question. The title should have been worded "What is the cause, and what is the effect..."; I suggest you make the appropriate edit.

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  • $\begingroup$ What does explain then the different rates of spontaneous emission vs stimulated emission? Different amplitudes in $\hat{V_0}$ corresponding to 0 or 1 photon? $\endgroup$ – L. Levrel Apr 6 '16 at 19:12
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    $\begingroup$ @L.Levrel In both cases, the form of the interaction $\hat{V}_0$ is fundamentally the full quantum electromagnetic field. Spontaneous emission is the process $|0 \gamma,\text{atom}^*\rangle \rightarrow |1 \gamma,\text{atom}\rangle$. Stimulated emission, on the other hand, is the process $|\text{many} \gamma,\text{atom}^*\rangle \rightarrow |\text{(many+1)} \gamma,\text{atom}\rangle$. The difference comes from the relevant matrix elements $\langle 1\gamma|\hat{V}_0 |0\gamma\rangle$ vs $\langle (\text{many}+1)\gamma|\hat{V}_0 |\text{many}\gamma\rangle$. $\endgroup$ – QuantumDot Apr 7 '16 at 20:30
  • $\begingroup$ Let us note that with zero perturbation field, in the semi-classical view, there would never occur any transition. Spontaneous emission relies on the zero-point fluctuations to start. $\endgroup$ – dominecf Apr 8 '16 at 20:04
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As in all interactions in quantum mechanical terms, one needs a Feynman diagram:

e+e-ann

In this diagram of e+e- annihilation into two photons, time goes up, and both photons are simultaneously emitted . If one reverses the axis its is photon electron scattering (the second e is a e- by the diagram conventions) and the interaction is simultaneous in t1 a virtual electron arises, and it splits into a real photon and an electron at t2 simultaneously.

In Hydrogen deexcitation there is also a before, a during, and an after. In your problem, the "before" the interaction is an excited hydrogen atom, the "during" is the emission of a virtual photon from a virtual electron, and the "final" is a real deexcited atom and a real photon.So the deexcitated atom and the real photon appear on the same time in the time axis of a Feynman diagram.

It is complicated to draw the Feynman diagrams for this,

enter image description here

because of the bulk of the wavefunction of the hydrogen. These are Feynman type diagrams, i.e. there are specific rules how they translated to calculable formulae.

Here is a brave effort: Feynman diagram approach to atomic collisions.

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Assuming ''a hydrogen atom in an infinite and otherwise empty universe'' makes your question meaningless. Your universe does not contain anything that could possibly induce the postulated perturbation $V$.

Thus the state of the hydrogen atom evolves according to the unperturbed Hamiltonian and hence always remains in the excited state without emitting anything.

To have a nontrivial $V$ you need to posit more stuff in the universe, so that it can interact with the hydrogen molecule. But even when you just posit an external electromagentic field coming from nowhere, Fermi's golden rule gives a transition rate for an ensemble of many equally prepared systems, and is meaningless in a universe containing only one system.

https://en.wikipedia.org/wiki/Fermi's_golden_rule: ''Fermi's golden rule is valid when the initial state has not been significantly depleted by scattering into the final states.''

The moral: If one tries to interpret a formula outside the context in which it can be derived one must be prepared for effectively interpreting ''a situation that can never occur''.

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  • $\begingroup$ I don't think I understand your statement "...the state of the hydrogen atom evolves according to the unperturbed Hamiltonian and hence always remains in the excited state without emitting anything." Because the way I interpret it, it sounds false. When the hydrogen atom is coupled to the quantum electrodynamic field, the excited 2s state, say, is no longer a stationary state of the system. And so it will evolve to something else, even when there's nothing around initially. Maybe a clarification is in order? $\endgroup$ – QuantumDot Apr 8 '16 at 19:45
  • $\begingroup$ If the universe contains only the hydrogen atom there is no electromagnetic field to which it couples (the self-field does not count). Putting the hydrogen atom in the context of QED it is a bound state, which does not scatter. $\endgroup$ – Arnold Neumaier Apr 8 '16 at 20:13
  • $\begingroup$ The additional external field must in QED come from other matter in the universe, which was supposed not to be present. $\endgroup$ – Arnold Neumaier Apr 8 '16 at 20:14

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