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#1
Jan113, 05:16 AM

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When an electron jumps to a low energy level, it emits a photon.
1. Where does this photon come from? 2. Do the constituents of an electron have any role to play in this emission? 


#2
Jan113, 05:56 AM

P: 136

The photon comes from a threeway interaction 'vertex' that allows electrons (or any other electrically charged particles) to emit or absorb them spontaneously.
Electrons have no currently known substructure, so it's impossible to say anything about "constituents" of them. 


#3
Jan113, 06:29 AM

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Keep in mind that electron does not actually "jump". It's just the explanation we give to chemists so that they stop bugging us about the discrete energy levels. The transition is actually a gradual one. Say, you observe emission in transition from 2p to 1s. In that case, during transition, the total state is going to be a superposition a2p> + b1s>, with aČ + bČ = 1, and this state will actually have an oscillating electric dipole. That electric dipole is responsible for producing electromagnetic radiation emitted due to transition. The reason you get precisely one photon out of it is that the energy difference between 2p and 1s is exactly equal to the energy of a photon with frequency equal to frequency of the dipole oscillation of the a2p> + b1s> superposition state.
P.S. This really should be in atomic. There isn't any particle physics going on here. Well, apart from second quantization of electromagnetic field, but we don't seem to be actually discussing that aspect of it. 


#4
Jan113, 07:06 AM

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Emission of a photon
Arrg, I'm shocked to hear you say something like that, K^2! The view that atomic transitions are in any way continuous went out in the 1930's. By the same reasoning a single U238 nucleus slowly evolves into a Th234 nucleus over a period of 4 billion years.
You know as well as I do that the only thing that evolves continuously in a transition is the probabilities. The superposition state, while mathematically correct, has no objective existence. At any time the electron is either in the ground state or the excited state. It does jump, not fade away like a Cheshire cat. And the photon does not come out gradually, it comes out all at once. 


#5
Jan113, 08:23 AM

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Furthermore, <2px1s> ≠ 0. That means the atom in a transition actually has an expectation value for a dipole moment! How the hell do you reconcile this with your primitive notion of a jumping electron? Sure, it's the expectation value, but neither 2p nor 1s has a dipole moment. Yet in transition, it has a nonzero expectation. That's an actual observable, not some obscure phase. And the photon does, most certainly, get emitted over time. The probability of detecting the photon increases over time. The probability density for photon is EČ+BČ. That means electric field increases over time. There are no jumps in quantum mechanical evolution. The only jumps happen due to collapse in measurement. And even that can all together be avoided with the right interpretation. 


#6
Jan113, 08:38 AM

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#7
Jan113, 08:57 AM

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Blargh. Yes, sorry. My brain foobarred there for a bit. While pointing out my error there, you might as well have pointed out that <ψψ> doesn't give me probability density either. It gives me total probability, which is unity.
What I want is ψ^{*}(x)ψ(x). That picks up the 2p1s crossterm which is most certainly nonzero. So anyways, probability density for transition is still completely different than any linear combination of densities from ground state and excited state. 


#8
Jan113, 09:54 AM

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#9
Jan113, 10:32 AM

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If I hand you an electron that passed through a doubleslit, can you tell me which slit it went through? What you should be able to do, however, is set up an experiment which shows that it did not just go through one or the other. It did, in fact, go through both slits. E.g. Delayed Choice Quantum Eraser.
Similarly, it is possible to set up an experiment that shows absolutely definitively that the atom is neither in the state having emitted zero nor one photon. It's somewhere in between. Now, for the photon itself, you can try to interpret it in different ways. You can try to pretend that superposition is just some voodoo. But that doesn't work with the atom. As demonstrated prior, the actual charge density, which is an observable, of atom in transition is distinct from either ground state or excited state. As further demonstrated, the dipole moment is present and can be measured. This, in turn presents an excellent opportunity to cut through the Gordian knot you present above. Say, rather than make measurement that tells me which energy state the atom is in, I do measure the electric dipole of the atom you handed me. Say I get a nonzero result. That would indicate that the state has collapsed into something that is neither the ground nor excited state. Having conducted that measurement, I can tell you exactly what fraction of the photon has been emitted. Naturally, this measurement is still a product of collapse, so I don't reveal information about what that fraction was originally. And nonetheless, I can make a measurement that gives me an answer that is a fraction. 


#10
Jan113, 10:46 PM

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2. Is it the electrical charge that gets converted into a photon or the photon already exists waiting to be emitted? 


#11
Jan113, 10:47 PM

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Jan213, 12:57 AM

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#13
Jan213, 07:56 AM

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#14
Jan213, 12:10 PM

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#15
Jan213, 01:05 PM

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#16
Jan213, 11:53 PM

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#17
Jan313, 02:41 AM

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no because I could annihilate an electron and positron to form 2 photons.
but charge is conserved Or I could just bend an electrons path in a B field and cause it to emit light. 


#18
Jan413, 01:01 AM

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Please reply. 


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