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It's an old-new question (I found only one similar question with unsatisfactory (for me) answer: Where did Schrödinger solve the radiating problem of Bohr's model?)

It's strange for me how all books simply pass by such an important question, and mentioning strange and mathematically unsupported reasons such as:

  • orbits are stationary (while as I know this is just idealization, there is no stationary orbits in reality even for Hydrogen)

  • electrons are actually not localized due to uncertainty principle, thus they have no acceleration (while obviously in a non-spherically symmetric orbits a kind of "charge acceleration distribution" always exist)

  • vacuum fluctuations play a major role (according to QED).

I'm not interested in how Bohr or Schroedinger explained it, I want to see a rigorous proof with QM, QED or maybe even the standard model as whole. I would like to see how this question was closed.

Community
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TMS
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    This is a duplicate of the question you linked to. Voting to close. –  Jul 06 '13 at 22:31
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    possible duplicate of Where did Schrödinger solve the radiating problem of Bohr's model? –  Jul 06 '13 at 22:31
  • I believe I was clear that the mentioned answer is unsatisfactory because it has no rigor proof. – TMS Jul 06 '13 at 22:34
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    On stackexchange, you don't respond to an unsatisfactory answer by asking the same question again. Appropriate responses would be to downvote the answer, make a comment on the answer explaining why you think it's wrong, and offer a bounty on the question to try to attract better answers. –  Jul 06 '13 at 22:53
  • @BenCrowell It seems that TMS is not asking about Hydrogenous atoms and, apart from this, wants an explanation beyond the non-relativistic quantum mechanics framework. – Diego Mazón Jul 06 '13 at 23:00
  • But electrons in their orbitals can radiate if there is an open, lower energy level. If there is no place lower energy state they can occupy then they can not radiate. This also works in degenerate systems. This was settled in principle by plain quantum mechanics. – dmckee --- ex-moderator kitten Jul 06 '13 at 23:38
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    @Ben Perhaps the answers was satisfactory for the other question, but not for what TMS wants to ask. That would be a case in which it is appropriate to ask a new question. However (TMS), the new question - this one - should explain explicitly how this question goes beyond the previous question. – David Z Jul 06 '13 at 23:44
  • @Ben: The linked question is old and marked as answered, there is really nothing I can change there. – TMS Jul 07 '13 at 06:02
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    Your question makes no sense. You criticize books for their making completely valid and essential observations and statements while your added statements are all incorrect. The orbits in a QM atom are stationary. The lowest energy eigenstate can't radiate because there's no way to take energy from it - no lower-energy state. Electrons are not quite localized due to the uncertainty principle. It is not true that the acceleration always implies radiation - it only does if there's a lower-energy state. The classical formulae linking radiation to acceleration are just approximations. – Luboš Motl Jul 07 '13 at 06:41
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    And the old question seems to be equivalent to yours and gave some perfectly fine (and high-rated) answers. It is therefore very incomprehensible why you asked the question again, @TMS. – Luboš Motl Jul 07 '13 at 06:42
  • @LubošMotl well, why orbits are stationary? EM field of Uranium nuclei is a stationary one? from QFT perspective? "It is not true that the acceleration always implies radiation", well I asked to show how this applies on QM electron in a rigor proof, and that was one of the differences between two questions. – TMS Jul 07 '13 at 06:52
  • since this came up again : I am continuously amazed how people are platonically biasedwhen asking and replying here. They implicitly assume that the mathematics cause observations and not what is true that: " observations are modeled with mathematics". The non radiation from the atomic levels is an experimental observation that needed to be explained. The spectra of the atoms are discontinuous after all. That is the basic observation both of the stability and the need for quantization. – anna v Jan 29 '15 at 11:48
  • Related: Why don't electrons crash into the nuclei they "orbit"? – Emilio Pisanty Jul 05 '18 at 08:58

2 Answers2

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This question can be answered in the simple framework of non-relativistic quantum mechanics. The electron's electromagnetic charge's density and current — which are the source of the classical electromagnetic field — are given by the electron's probability density and current distributions $$\rho (t,x)=\psi^*(t,x)\,\psi(t,x)\,$$ $$j(t,x)\propto \psi^*(t,x)\,\nabla\psi(t,x)-\psi(t,x)\,\nabla\psi^*(t,x)\,.$$ As in a stationary state $\psi(t,x)=e^{-i\omega\, t}\,\phi(x)$, neither the density nor the current depend on time and therefore they don't emit electromagnetic energy, according to Maxwell equations with $\rho$ and $j$ as sources.

However, when one takes into account the quantum nature of the electromagnetic field, the probability of radiating a photon (quantum of the electromagnetic field) by an atom in a stationary state is different from zero due to the phenomenon of spontaneous emission.

Diego Mazón
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  • Please note that I have mentioned in the question that stationary state is just an Idealization, in reality the nuclei "potential" is actually time dependent and such an explanation is fundamentally not precise. – TMS Jul 06 '13 at 22:44
  • What do you mean by a time dependent potential? What nuclei? In what framework or approximation? Any link or reference? @TMS – Diego Mazón Jul 06 '13 at 22:57
  • (I do know what a time dependent potential is of course, what I ignore is under what conditions the potential is not time independent) @TMS – Diego Mazón Jul 06 '13 at 23:04
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    @TMS "fundamentally not precise"?? Unless you both believe in and have a true Theory of Absolutely Everything, this is not an objection to any argument. A good deal of physics - perhaps even the majority of physics - is knowing what to include and what to leave out in any analysis. If your theory has as many bits of information as the class of systems you are trying to study, then you have duplicated nature but you have not done science. –  Jul 06 '13 at 23:11
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    This is a nice answer and clearly within the scope of the original question http://physics.stackexchange.com/q/68381/ . I would suggest copying it there and then deleting it from this duplicate question. –  Jul 06 '13 at 23:39
  • Wait, I've become unconvinced that this is right. If it were correct, the final (newly added paragraph) would be false. –  Jul 07 '13 at 00:14
  • I meant that this explanation can be considered right only in the approximation of that EM field of protons and neutrons in the nuclei is stationary, only then we can do such a variable separation in Schroedinger equation, unless we can show somehow that those changes (with time) of EM field are too small that making the mentioned approach still valid, and obviously there we are not looking for theory of everything explanation, maybe QEM is enough, especially that I mentioned some reference talks about vacuum fluctuation role. – TMS Jul 07 '13 at 06:11
  • Dear @TMS, be sure that both atoms and nuclei have energy eigenstates - much like any other objects in a quantum universe with a constant Hamiltonian - and an energy eigenstate is always perfectly stationary. Why are you so obsessed with the completely wrong idea that nuclei or atoms aren't stationary? It is one of the key basic results of quantum mechanics that they are stationary. They are stationary because the wave function only changes by the unobservable phase $\exp(Et/i\hbar)$ with time and solutions exist because the Hamiltonian always has eigenstates. – Luboš Motl Jul 07 '13 at 06:44
  • Let me emphasize that the stationarity of a nucleus or atom in its ground state is an absolutely exact fact. On the contrary, be sure that any argument or assumption you are using to derive the converse, namely that the nuclei or atoms or their fields should not be stationary, it inapplicable to nuclei or atoms - it is a classical or another approximation that just breaks down in this situation. – Luboš Motl Jul 07 '13 at 06:46
  • @LubošMotl White "in general, an electron's behavior is not fully described by a single orbital. Electron states are best represented by time-depending "mixtures" (linear combinations) of multiple orbitals." from Atomic orbital and " we can decompose the total eigenvector of a many-electron molecule into separate contributions from individual electron stationary states...use this "single-electron approximation" from Stationary state – TMS Jul 07 '13 at 07:05
  • I think @TMS that you are wagging the dog by the tail. All your arguments about radiation and time dependence etc etc come by miniaturizing classical physics.i.e. you view the electrons as tiny billiard balls with a charge distribution. We have arrived to the quantum mechanical framework for the microworld exactly because atoms and molecules and nuclei exist stable in time for the vast majority of cases. Minus this stability we would not be here exchanging opinions. Quantum mechanics encapsulated these numerous observations, and physicists realized that the classical world, including – anna v Jul 07 '13 at 08:01
  • Maxwell's electromagnetic theory , emerges from the fundamental quantum mechanical level. In this QM level we talk of orbitals as a probability distributions for finding the electron ( including its charge), not motion distributions. It has no meaning to be thinking of a swinging little billiard ball because that is not what the solutions of QM equations tell us, and those solutions have been accepted because they fit innumerable real ( not thought) experiments. You should be thinking from QM and up, not classical physics and down. – anna v Jul 07 '13 at 08:05
  • @annav Where is that classical thing in my words? Lubos & drake (and books) said orbits are always stationary wile Wikipedia (and all known atoms/molecules models) says that it's just an approximation, if they are really always stationary in sense of QFT then there will be no questions, but how is that? – TMS Jul 07 '13 at 09:32
  • @TMS just the fact that you call them "orbits" shows this confusion between classical and quantum. They are NOT orbits, they are orbitals. The Bohr model was very very useful when physicists had not realized how they could model with much better theories the quantum nature that was appearing in the experiments. We now have these better theoretical models. – anna v Jul 07 '13 at 10:55
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    The electron is not a classical particle in an orbit but a quantum mechanical "entity" characterized by a probability distribution given by the square of the appropriate wave function which shows the space covered as an "orbital" schematically. One does not solve the hydrogen atom using quantum field theory, as one does not use a surgical knife to cut potatoes. – anna v Jul 07 '13 at 10:55
  • @annav Oh didn't know that, in Russian language there is no such distinguish between orbits and orbitals.. anyway that just a matter of notation, and I think the knife sample is wrong, not radiating in QM depends strictly on stationarity, which is just approximation, that was my point in the need of QFT and a rigor proof. – TMS Jul 07 '13 at 11:35
  • @TMS orbitals is a new word in physics use , made necessary because of the distinction needed : up to the Bohr model one can talk of orbits and stationary orbits, but by the Schrodinger equation one says the electrons are in orbitals, exactly to distinguish the probability_wave/particle duality from the naive planetary model with little billiard ball like electrons with constraints. – anna v Jul 07 '13 at 11:39
  • @annav believe me I'm very aware of that there is no classical orbits in the atom, and you clarified why they are orbitals, but the problem in a different place.. – TMS Jul 07 '13 at 11:43
  • Then I cannot see why you do not accept the simple explanation of energy conservation that someone above gave. To radiate from an orbital a photon should take away the energy in a quantized step, and the electron should fall into a lower orbital. If no lower orbital is empty energy conservation forbids radiation. – anna v Jul 07 '13 at 11:53
  • @annav I was asking bout Hamiltonian time dependence, that was my problem, anyway now after some research I got my flaw, In QFT we can do the same thing of time separation as drake showed above (Schroedinger vs. Heisenberg picture), thus stationary orbits exist even for time dependent Hamiltonian as Luboš Motl stated above. – TMS Jul 24 '13 at 23:07
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You should look at the work of G A Schott in the 1930 for example G.A. Schott, "The electromagnetic field of a moving uniformly and rigidly electrified sphere and its radiationless orbits," Phil. Mag. Vol. 15, Ser. 7 (1933), 752-761.

He has other papers about accelerating but non-radiating orbits and the implications for quantum mechanics. It's a pity that his work went un-noticed and apparently ignored as it would have explained the quantisation phenomena, at least of the hydrogen atom and possibly all of quantum EM.

Bruce Hartley
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