In the one hand, energy can be emitted\absorbed only by a discrete portions.

In the other hand - according to Coulomb law the higher distance - the lower force, because count of force lines per area decreases with distance. This fact always confused me: the problem is that it's difficult(or impossible) to compare classic explanation and quantum.

We have a frame where electron is fixed. In classic physics - it creates an electric field, in theory of fundamental interactions - it emits a virtual photon, plus we can know about it only if will interact with it. And I'm getting trouble, imagining how it occurs. I know that photon(especially virtual) doesn't have size because of 0 zero mass, as understand we can say, that it doesn't exist in the space at all. We can't even calculate possibility of finding the photon, because, as far as I understand $|\psi|^2$ for photon equals zero.

However, can You explain what will, or may happen in the situation described below

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  • $\begingroup$ I am sorry, because you are a new user, but your questions shows that you have not really studied quantum mechanics. quantum mechanics is about probabilities of interactions, not certainties. The energy of virtual photons is under an integral given by a feynman diagram, there is not one hnu, it depends on the boundary conditions of the quantum mechanical equation that has to be solved to give a wavefunction which will give probabilities of interaction. Quantization comes only in bound states. $\endgroup$ – anna v Sep 6 '18 at 18:32
  • $\begingroup$ A single electron take enrgy of a single photon $\endgroup$ – yuvraj singh Aug 29 at 9:07

A key problem with your setup is that "we definitely know... proton emits a single photon per time." This statement is incorrect.

According to the uncertainty principle, energy is only conserved over a given time period that is inverse to the energy. So one can break the conservation of energy for a short period of time. That means there is no limit to the number of photons a proton can release in a short period of time. Which means that...

As long as the overall virtual particle field is ultimately conserved, you can have any number of particles with all sorts of parameters. When you run the math, you end up with a distribution that looks just like a classical field. They are constantly being created and disappearing again.

It's when one of these interacts with some other object that all the magic happens. In your example, consider what happens if one of the infinity of virtual photons interacts with the electron. In that case the electron gains (negative) momentum and starts to travel towards the proton. But now the proton is lacking one of the photons that would have balanced out the other virtual particles, so now it has leftover (positive) momentum and starts moving toward the electron. Mutual attraction.

  • $\begingroup$ Well, it's a good answer, I even almost understood. So it's a good explanation why virtual photon do not caries an energy, but caries an momentum. But I do not understand "energy is only conserved over a given time period". You talking about the energy of proton? But it doesn't lose energy(if doesn't does work). Probably I don't understand correctly $\endgroup$ – user205695 Sep 6 '18 at 18:08
  • $\begingroup$ If you could look through some sort of "quantum time microscope", you would see the proton's energy is changing all the time. It only remains (seemingly) constant over longer "microcopic" periods. When an interaction occurs some energy (and momentum and so on) moves from one place to another, from the proton to the electron in this case. But overall energy is still conserved. But this is true of any interaction, consider the energy inputs and outputs when you toss a shotput to another person. $\endgroup$ – Maury Markowitz Sep 6 '18 at 18:17
  • $\begingroup$ Well, proton doesn't a fundamental particle at all. If we consider an electron, that moves in empty space. It's energy is changing whatever? "Microscopic period" is a Plank's time? $\endgroup$ – user205695 Sep 6 '18 at 18:22

First, protons, because they have EM charge, do interact with the photon field, but that interaction is described as being mediated by virtual photons. These are a mathematical way of describing how the two, the EM and the photon field are interacting.

Now you are asking about real photon absorption. Protons can interact with photons:

  1. scattering, this is Compton scattering, inelastic scattering. The photon gives part of its energy to the proton, and changes angle.

  2. absorption, this is for a composite system, like a hydrogen atom, and when the hydrogen atom can absorb a photon when the photon's energy matches the difference between the energy levels of the electron around the nucleus

Now in your case, protons can absorb photons, but not in the visible wavelength range. A proton is a composite object and it does have energy levels, but those energy levels where the proton can be excited are ing the GeV range, so only gamma rays can excite protons.

Now let's say that in your case a proton is excited by a gamma ray photon, so the photon is absorbed, and the proton is then relaxed, and emits a photon.

You are saying that this photon needs to be absorbed by one of those electrons that you show.

Now free electrons in a vacuum cannot absorb photons. Only electrons bound to nuclei can absorb photons.

So let's say that the electrons you show are bound to nuclei. Now these electrons could absorb the gamma ray if they have two energy levels as per QM that have a difference in energy levels that match the energy of the gamma ray.

You are asking which electron will absorb the photon. It is all probabilities. Yes it is possible that both electrons will be excited by one photon.

Please see here:


  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind Sep 7 '18 at 19:00

Bloody UI.

Either of them, but only one.

Don't think of photons as things. They are the probability distribution of a thing. Meaning it technically hits both, but if the energy is only sufficient to cause one to do something, only one will be hit, because the available energy is drained out of the probability distribution.

  • $\begingroup$ Distance does play a role, but it is complicated. Distance changes the odds of an electron being "activated", favoring closer electrons. Think of your force lines as probability density and you will be close. $\endgroup$ – Adirian Sep 6 '18 at 17:54
  • $\begingroup$ Wait a minute, You are talking about especially virtual photons, right? I read that they lose it's energy, during the propagation, and if do not hit anything dissolves into real particles $\endgroup$ – user205695 Sep 6 '18 at 17:57
  • $\begingroup$ "Real" photons. Virtual particles aren't explicable outside of math; they are more like probabilities about probabilities. $\endgroup$ – Adirian Sep 6 '18 at 18:07

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