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From the lessons on QM, I got impression that there are some discrete orbitals that emit light when electron drops from one to another. Specific molecules emit light in very narrow bands, therefore. Similarly, electrons can absorb these bands. But, my impression was that not any electron can absorb the light for the same reason: it must be the same orbital because electrons can absorb light only when transit from one orbital to another. That is very puzzling because hardly you can find exactly the same electron in the whole Universe, not to speak to find it at on the way of the photon.

The bands are very narrow but they are not like single points in the continuum of frequencies because molecules have various extra speeds in all directions in addition to electron orbitals. This means that the recepting molecule must have the same extra speed. I consider all of that impossible. Does uncertainty principle come to rescue to reconcile the energy differences? This picture, discrete out - discrete in, also does not explain how does our planet convert visible high frequency photons into many low frequency ones. If you can only send and receive photons of specific energies then all scattering can do is to change the direction of light but we see that dissipation can split one high freq photon into multiple low energy ones. How is all of that possible?

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  • $\begingroup$ To clarify: are you asking how it is possible that two different electrons can have exactly the same energy, so that a photon emitted by one can be absorbed by the other? The answer to this is indeed a sort of uncertainty principle argument: the photons emitted by atomic transitions do not have a perfectly sharp energy, but rather a range of energies. This is due to the finite lifetime of the atomic states which decay. $\endgroup$ – Mark Mitchison Jul 26 '15 at 21:26
  • $\begingroup$ Regarding your second question: Dissipation is a different process. Electrons that are bound by an atom will oscillate in due to the electromagnetic field of the photon and thus emit radiation. $\endgroup$ – john Jul 26 '15 at 21:41
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  • $\begingroup$ @john This does not explain anything because excitation I am talking about is excitation of these, atom-bound, electrons. The energy they emit has the same, proper specturm of the energy levels, the one is absorbed. Bound electrons absorb and emit one and the same point spectrum. You cannot explain break down of this specturm by restating it. $\endgroup$ – Val Jul 27 '15 at 10:12
  • $\begingroup$ @RecognizeEvilasWaste your question seems generally unfounded by knowledge, I recommend you don't try to understand higher concepts if you don't know stuff like basic field theory. $\endgroup$ – john Jul 27 '15 at 12:32
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The question is not entirely precise but I can at least clarify that photons emitted by one atom can be absorbed by another atom, even if those atoms are moving at different speeds. The likelihood of excitation is simply maximized when the atoms have the same velocity.

In general, the law which states that transitions are only possible between initial and final states of equal energy assumes infinite time between preparation of initial state and observation of final state. In reality there is only some finite time between these two events and transitions between states of unequal energy are possible. The rule of thumb is that energy conservation must hold within a fudge factor of $\Delta E = \frac{h}{t}$, where $h$ is Planck's constant and $t$ is the time interval between preparation and observation, in order for a transition to be possible.

This is separate, however, to the possibility of a single high energy photon turning into many low energy ones. There are many mechanisms that accomplish this without having to consider the velocities of the scatterers. One such mechanism is inelastic scattering (called Raman scattering) where a photon scatters from an atom/molecule, transferring a portion of its energy to the atom/molecule, leaving the atom/molecule in an excited state. A second mechanism is fluorescence, where a photon is absorbed by an atom, and the atom falls back down to the ground state through an intermediate series of states, each transition emitting a separate photon. In this case the sum of outgoing photon energies equals the energy of the ingoing photon.

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  • $\begingroup$ I am asking why one atom can be absorbed by another atom, even if those atoms are moving at different speeds. The likelihood of excitation is simply maximized when the atoms have the same velocity. What probability has to do with concrete energy levels? If you can buy goods only for 1 or 2 dollars, how can you spend 1.33? What probability has to do with that? $\endgroup$ – Val Jul 26 '15 at 21:27
  • $\begingroup$ You can spend 1 dollar on the goods and leave a 33 cent tip. Likewise, an electron can absorb part of a photon's energy as an excitation and the rest as kinetic energy, without which there would be no such thing as thermal conduction. $\endgroup$ – Asher Jul 26 '15 at 22:13
  • $\begingroup$ @Asher That is interesting because I do not understand how can you dispense me the kinetic energy. The excitation already stands for the kinetic energy, as I understand it. Electron does not have any other energy besides its spectral line/orbital. They teach us that bound electrons have their energy quantized and these acceptable energy levels are called "orbitals". In other words, quantum theory says that there cannot be any kinetic energy for the tip, besides the 1 or 2 dollars. THe orbitals is all that we have, QT says. Therefore what you say is every interesting. It defies QM basics. $\endgroup$ – Val Jul 27 '15 at 10:22
  • $\begingroup$ The entire atom (electrons and nucleus) itself has a collective motion, with an associated kinetic energy. $\endgroup$ – creillyucla Jul 27 '15 at 11:50
  • $\begingroup$ @creillyucla The whole body and planet also has a collective motion with total KE. Why should we stop at the atom level when we need to separate 1 dollar from 1.33? Can we slice 1.33 photon into 1$ absorbed + 0.33$ fly through photon? $\endgroup$ – Val Jul 27 '15 at 14:03
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Say an atom just emitted a photon by an electron falling back from an excited state to its grounstate it would make sense this photon can interact readily with another atom of the same element in the groundstate bringing it up to that excited state. The energy lost by the first atom and gained by the second is the same.

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  • $\begingroup$ I do not see how frequency is changed here. I am asking about specific, high-frequency photons converted into black body spectrum. I do not see how your answer clarifies anything here. $\endgroup$ – Val Jul 27 '15 at 10:16
  • $\begingroup$ Well you asked (if I'm correct) in your first paragraph how a photon emitted by an atom can be absorbed by another. I just stated that should be very probable given that dependent on the situation there could be many other atoms with the same energy levels that can be excited by the just emitted photon. Indeed there is some effect on the frequencies due to the velocities of the atoms called Doppler broadening but I do not understand why you are puzzled how a photon can get absorbed. $\endgroup$ – Jan Bos Jul 27 '15 at 18:50
  • $\begingroup$ Yes, the Doppler boradening seems to be the root the the puzzle. If here were no such fuzzing out of the spectral lines, every atom could absorb your photons. Ok, there are millions of quantum states, which may make your atom not ready to absorb your photon. However, this huge num of electron states is still finite and given millions of atoms (we have per 1 cm of length), it makes it quite likely to find some accepting atoms. Yet, the Doppler broad. is continous! Probability theory says that there is 0 probability to pick specific item in a continous distribution! How can prob. fix it? $\endgroup$ – Val Jul 27 '15 at 21:08
  • $\begingroup$ The fuzzing around the spectral lines doesn't inhibit absorption. The velocities of the emitting and absorbing atom do not need to match. There had to be 4 momentum conservation but part of the energy of the photon can go into exciting the atom and part in changing its velocity. $\endgroup$ – Jan Bos Jul 28 '15 at 5:11
  • $\begingroup$ you basically say that electron slices its energy from the photon, passing the rest, low energy photon through. Otherwise, it is curious to know why remaining energy/low energy photon should be absorbed by the receiving atom (rather than the whole body of the molecule/crystal, for instance). $\endgroup$ – Val Jul 28 '15 at 8:32
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There is something that people used to build in the olden days called a crystal radio. Out on the farms, they didn't have electricity and they couldn't afford batteries, so they found a way of listening to the radio without plugging anything in. Essentially, they were absorbing power from the transmitting station to drive their little earphone speaker.

If people are telling you that such-and-such is how an electron can absorb a photon, you should ask them how a crystal radio can absorb power from a distant transmitter. Because if they can't explain the latter (and most of them can't), I wouldn't trust them to explain the former.

I explain how a crystal radio absorbs power in this blogpost. It's not about the diode or the coil. Those are the tools, but the real mechanism for power absorption is the interaction between two waves: the incoming wave from the transmitter tower, and the outgoing wave from your little receiving antenna. The wave-on-wave calculation shows that there is a net absorption of power by the receiver. And once you understand that mechanism, you can see that it is exactly the same mechanism at work in the atom. I explain all this in the series of blogposts following my initial one about the Crystal Radio.

Schroedinger developed his theory of the wave functions specifically so that he could expalin radiation in just this way. Everything works according to Maxwell's equations, with the charge density calculated as the square of the wave function, and with the absorbption and emission of energy taking place smoothly through the passage of time. Unfortunately, Schroedinger's theory was hijacked by the Copehagen school, who insisted on preserving the trappings of the old Bohr theory. So they re-interpeted the Schroedinger equation in terms of their collapse of the wave function, using their probability interpretation to cobble together a theory.

That is what we are stuck with today. In fact, their is no need for the photon. Everything the atom does in its interaction with thermal radiation can be clearly understood by combining the Schroedinger charge density interpretation with Maxwell's equations for electromagnetism.

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