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The scalar product of two quantum states gives the probability of transition between those two states. In particular, for two stationary (eigen) states, the orthogonality implies that the probabillity of transition is zero:
$$(\Psi,\Phi)=\int \Psi^* \Phi dv=0 \tag 1\\$$
That being said, I have a problem with the so-called "spontaneous emission" in atoms, which occurs when an electron undergoes a transition between two stationary states without any perturbation or external agent (that's why it's called "spontaneous").

How is that possible?! According to $(1)$, there would be no transition at all because its probability is zero. But we know it's not and yet this probability is given in terms of the Einstein coefficient $A$.

Question : Is this a contradiction between theory and experiment? Or is it a flaw in the frame work of the Schrodinger picture of QM? If so, how do we fix it ?!

I really need a satisfactory explanation. Thanks!

Samà
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    Have you read the "theory" section of the corresponding Wikipedia article? I think it explains the theoretical QM explanation of the effect rather well and gives the relevant references such as Dirac original paper for the actual calculations. – ACuriousMind Jul 18 '17 at 12:24
  • @ACuriousMind: Thank you, but I'm not familiar with QFT, is there any other explanation in the frame work of QM?! – Samà Jul 18 '17 at 13:05
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    I read the article link provided by ACuriousMind, and it does seem to strongly suggest that our first approximation to quantum phenomena (that is QM) is not adequate to explain the emission. But I don't think the math behind the level of QFT needed is any more of an effort to study than any complicated, ad hoc set of assumptions needed to explain the effect in QM. By the time you have tried to hammer QM enough to explain it, you will likely have ended up with QFT anyway. Of course, as soon as I have written this, somebody may give you a 4 line QM based answer...... –  Jul 18 '17 at 13:28
  • "The scalar product of two quantum states gives the probability of transition between those two states. That is not true. – DanielSank Jul 18 '17 at 15:37
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    As far as I remember spontaneous emission is associated with vacuum fluctuations and with the name Weisskopf. – Semoi Jul 18 '17 at 18:39
  • @JohnKennedy: For me, and according to my current level, an ad hoc quantum mechanical explanation (if it exists) of the emission is all what I need. A QFT explanation would be not helpful 'cause I'm not familiar with the theory at all. – Samà Jul 18 '17 at 19:44
  • @DanielSank: As I've studied, the scalar (or dot) product of a state (a wavefunction) with itself gives the probability of occupation of that state. And the scalar product of two quantum states gives the probability of transition between those two states. It makes sense to me, where is the mistake here? – Samà Jul 18 '17 at 19:47
  • @Semoi: By "vaccum fluctuations" do you mean particle-antiparticle creation/annihilation processes? I know that these processes are described by the Dirac equation. Is that mean spontaneous emission are relativistic effects?! – Samà Jul 18 '17 at 19:51
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    Yes, I mean particle-antiparticle creation/annihilation. I'm not an expert on this field, but for me the "Dirac eq." is the relativistic equation for fermions. Dirac has done many things, so maybe there are other Dirac equations. However, vacuum fluctuations are a general concept of QED / QFT and not specific to any particular equation. I'm quite sure that you need these to describe spontaneous emission. So I agree with Lewis Miller, because I would consider vacuum fluctuations to be perturbations. However, best you read about Weisskopf's theory by yourself. – Semoi Jul 18 '17 at 20:14
  • @Semoi: I just a nice PDF discusses the Weisskopf-Wigner Theory of spontaneous emission. the theory states that an atom in an excited state is not in a stationary state! and it will eventually decay to the ground state due to the intercation of the atom with the electromagnetic vacuum field. The most important thing here is excited states are not stationary states, this clarifies my confusion a lot. – Samà Jul 18 '17 at 22:23
  • @Semoi; thanks a lot! I'm sorry 'cause I can't upvote you comments due to my lack of reputations. – Samà Jul 18 '17 at 22:24

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The flaw in your analysis is the assumption that spontaneous emission occurs between stationary states in the complete absence of perturbations. In the real world there are always perturbations. If these perturbations are weak and not time varying, then we have what we call spontaneous emission. In actuality the upper level state is a mixed state with a small admixture of the lower state and the always present perturbations lead to random decays.

Lewis Miller
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  • Thanks. But if spontaneous emissions occur due some perturbations so it's not "spontaneous ", it's "Stimulated emission" isn't it? – Samà Jul 18 '17 at 19:55
  • "If these perturbations are weak and not time varying, then we have what we call spontaneous emission." I've studied in perturbation theory that if the perturbation is time-independent, the probability of transition between eigen states is zero. the perturbation must be time-dependent in order to have non-zero transition probabilities. – Samà Jul 18 '17 at 19:58
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    The perturbations I was referring to are actually vacuum fluctuations. I didn't mention this in my answer because you asked for an explanation in terms of QM without invoking QFT. Hence (if we are talking about electrons) the perturbations are virtual photons. It would be stimulated emission if the photons were real. I advise that you take ACuriousMind's advise and read the theory section of the Wiki link in the comment. – Lewis Miller Jul 18 '17 at 21:17
  • I see. Your answer combined with your comments were hepful so I accepted your answer. Thank you so much, and I'm sorry 'cause I can't upvote you answer/comments due to my lack of reputations. – Samà Jul 18 '17 at 22:27