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I understand atomic emission and absorption spectra well - photons of a specific energy can be emitted or absorbed by atoms, if that energy corresponds perfectly to the energy difference between two states of the electron of the atom - but I don't quite understand how the photons are absorbed and emitted during this transition. What process or mechanism underlies this phenomenon?

For Emission: Does it have something to do with the electron being accelerated during the transition, and the accelerating electron radiates a photon? If so, is this process random? What would cause the electron to suddenly drop energy level(s)? Where does the force/impetus for this acceleration come from?

For Absorption: Do the electric and magnetic fields of the photon apply a force to the electron when it interacts with the atom? If so, why do photons of only one energy apply this force, and all others have no effect on the atom?

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    $\begingroup$ The way you're asking the question is the wrong way to think about these things. You have to give up the notion of an electron as a ball and recognize that bound states are extended objects with neither a well defined position nor a well defined momentum. While you are thinking in classical terms with trajectories and positions and so on, you can't get closer than vaguely, kinda, sorta right. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Dec 31, 2015 at 21:16
  • $\begingroup$ Okay, granted. But the question still remains: by what force or interaction does a photon cause a change in the electron's energy state and become absorbed, or by what force or interaction does the electron emit a photon in changing energy states? $\endgroup$
    – D. W.
    Commented Dec 31, 2015 at 21:34
  • $\begingroup$ All answers here are extremely misleading because they mostly go from the direction of classical phenomena. The question is about "where these quanta of light are coming from as particles", like, "are they there around in the probability "field" of all possible permutations of frequencies in vacuum surrounding the electron? Note that the EM field is a FLOW of particles (photons) with different frequencies. $\endgroup$
    – Brian Cannard
    Commented Jun 7, 2019 at 19:15

4 Answers 4

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Neither the photon nor the electron are classical particles and there is no Newtonian picture of the process. Instead you have to imagine relativistic fields that describe the probabilities to detect photons and electrons in different spacetime points. Before the absorption there is a non-zero probability to detect the photon and the probabilities of the electron field are roughly those predicted by the Schroedinger equation for the low energy state of the atom. After the transition the probability to detect the photon is mostly gone and the electron distribution is now in a higher state.

One has to be very careful even with this picture, since one can't do continuous measurements on this system without disturbing it. What these distributions really mean is that we prepare one photon, then perform one measurement on either the photon or the electron. We repeat this experiment many times and then we plot the probability distribution as a function of time. This would have to be a multidimensional plot because of the ways the parts of the quantum system interact. Unfortunately people are not very good at recognizing the finer details of such multi-dimensional phenomena. Whenever we talk about these probability distributions and we show images in books, the problem has already been greatly simplified for our own convenience. From the perspective of human perception it is probably next to impossible to visualize the entire process without some simplification or loss of information.

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  • $\begingroup$ The non-Newtonian picture still leaves room for forces and interactions, no? By what force or interaction does the photon interact with the electron during emission and absorption? What interaction between the photon and the electron accounts for a non-zero probability of detecting the photon before and an infinitesimal probability of detecting the photon after, along with the change in the electrons energy-state? Or are we complacent with just saying "the probability distribution changed, and we don't know what happened in the interim"? $\endgroup$
    – D. W.
    Commented Dec 31, 2015 at 21:31
  • $\begingroup$ @D.W.: No, neither forces not accelerations make any sense in quantum mechanics. Those are macroscopic quantities that can be derived as thermodynamic averages over the dynamics of quantum systems, but if you are talking about atoms and their coupling to electromagnetic fields, then these classical descriptions don't make any sense. There is, of course, a dynamic equation that describes the change of these fields as a function of time and it even has a very deep structural connection to the equations of classical physics, but one does not recover forces from it but so called currents. $\endgroup$
    – CuriousOne
    Commented Dec 31, 2015 at 21:34
  • $\begingroup$ What is that dynamic equation? The Dirac equation? And how does the coupling of the bound electron to the electromagnetic field allow for the system to go from being in a "non-zero photon detection probability and an electron in one state" to a "infinitesimal photon detection probability and electron in a different state"? Or vice versa? The "current" from the electron changing states produces a change in the electromagnetic field? $\endgroup$
    – D. W.
    Commented Dec 31, 2015 at 21:40
  • $\begingroup$ @D.W.: For atomic systems you would use the equations of quantum electro dynamics, the (inhomogeneous) Dirac equation is part of them (for the electrons), the other part is for the description of the electromagnetic field. You also have to chose a particular gauge, I believe, but you better ask a theoretical physicist to explain the details, I have never worked with these equations myself. The currents originate from the charge conservation and the entire form of the equation can be derived from symmetry considerations and he necessity to rediscover EM fields in the classical limit. $\endgroup$
    – CuriousOne
    Commented Dec 31, 2015 at 21:48
  • $\begingroup$ @D.W.: The detection probability for the photon "after" the interaction is finite. The excited state has a finite lifetime, which means that there is (except maybe for infinitesimally short moments) always a finite probability to get an outbound photon distribution. Technically all of this also happens at a finite temperature (the third law of thermodynamics still applies!), so one would actually have to consider the density matrix of the system for finite T, rather than just the T=0 field equations. Moreover, the equations describe stimulated emission, as well. $\endgroup$
    – CuriousOne
    Commented Dec 31, 2015 at 21:52
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The mechanism of interaction is very much the same as a radio wave interacting with an antenna. The "photon" is manifested as oscillations of the electric field, which drive the electron like a mass on a spring. The oscillation frequency is given by the difference between the initial state and the excited state, and you can track the oscillating charge motion by calculating the superposition of the two states.

The oscillating electron radiates electromagnetic energy just like any other antenna. Initially, because of the interaction of the in-phase nature of the radiated field with the incident field, there is actually a net absorption of energy; but eventually the incident field goes away and the atomic system then simply continues to oscillate, re-radiating any residual energy until it returns to the ground state.

You can also analyze this system in terms of "photons" using something called "Fermi's Golden Rule", but it all comes to exactly the same result in terms of what you can actually measure...amount of scattered radiation as a function of the incident field.

DISCLAIMER: I am a recognized crackpot whose opinions are routinely and massively downvoted by the experts in this forum who know much more than me.

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  • $\begingroup$ What you say about the antenna is perfectly true. I found a site which nicely illustrates this for any dipole transistion in a hydrogen atom: falstad.com/qmatomrad $\endgroup$
    – A. P.
    Commented Dec 2, 2020 at 11:59
  • $\begingroup$ Thanks, A.P. It's surprising how hard it is to find such applets. I found one years ago but then it went down, and I couldn't find another. The oscillating charge is an obvious consequence of the superposition of wave functions, but no one wants to admit that it gives you all the correct properties of the hydrogen atom simply by applying Maxwell's Equations. $\endgroup$
    – Marty Green
    Commented Dec 3, 2020 at 2:55
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Of course CuriousOne is completely right

"Neither the photon nor the electron are classical particles and there is no Newtonian picture of the process. Instead you have to imagine relativistic fields that describe the probabilities to detect photons and electrons in different spacetime points. Before the absorption there is a non-zero probability to detect the photon and the probabilities of the electron field are roughly those predicted by the Schroedinger equation for the low energy state of the atom. After the transition the probability to detect the photon is mostly gone and the electron distribution is now in a higher state."

Though, I would like to interpret you question a little bit more as of

"For Absorption: Do the electric and magnetic fields of the photon apply a force to the electron when it interacts with the atom? If so, why do photons of only one energy apply this force, and all others have no effect on the atom?"

And let me interpret it in a way as you would ask, if the photon 'can' practically 'move' the electron, or the atom at all.

There is only a few things I would like to add.

  1. Although CuriousOne is completely right, and we have to use QM to understand absorption, but it is a fact though that EM waves do have an effect on the surface they 'hit'.

  2. And the answer to the question is yes they could 'move' (but not the electron), but just the whole atom or object they 'hit'.

  3. The EM wave will 'hit' the surface, we call this 'exert pressure'.

Reference: https://en.wikipedia.org/wiki/Radiation_pressure

  1. You can use QM and say that the total momentum of the system needs to be conserved, so the momentum of the photon will transfer into the higher energy level of the electron.

  2. Please note that as CuriousOne correctly stated you cannot see this as the absorbing electron being accelerated. In fact, on a 'higher' orbit, the electron will 'move' slower. But it's total energy level is raised, and that now will include the photon's converted momentum. (But if the electron's momentum became 'smaller' then how can it's energy level be higher? In a classical view, it's potential energy level is higher on a higher orbit.)

  3. The only thing that you can think of as being accelerated is the whole atom or the whole object that the EM wave exerted pressure on.

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The question is "By what mechanism is a photon emitted or absorbed in atomic electron state transitions?"

The correct answer is that we have no idea about the MECHANISM.

We have statistical equations that describe probablity, these are GROSS measurements. This is like a population cencus identifying that there is a 2% probability that a house in a certain location will have an individual with red hair. It gives you no insight into how an individual got red hair (genetics, hair dye, wig?).

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