There is no acceptable/viable mechanism for a free electron to absorb or emit energy, without violating energy or momentum conservation. So its wavefunction cannot collapse into becoming a particle, right? How do 2 free electrons repel each other then?
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7$\begingroup$ "There is no known mechanism for free electron to absorb or emit energy"? Who said that? $\endgroup$– knzhouCommented Dec 20, 2018 at 0:24
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$\begingroup$ @knzhou: it will violate simultaneous conservation of momentum and energy $\endgroup$– user6760Commented Dec 20, 2018 at 0:26
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$\begingroup$ Free electron can't absorb a photon and can a free particle absorb emit photons $\endgroup$– user6760Commented Dec 20, 2018 at 2:49
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1$\begingroup$ if your concern is that repelling is cause by the exchange of virtual photons, and a free electron can neither absorb nor emit a photon with violating conversation laws, then don't worry: virtual photons are off shell--they have the energy and momentum required to converse energy and momentum. Unless you are using old fashioned perturbation theory, then energy is violated, but only briefly. $\endgroup$– JEBCommented Dec 20, 2018 at 4:55
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$\begingroup$ imagine a high energy photon hits an electron at rest, my frame is wavelength becomes longer and electron still remains at rest but from zero-momentum frame photon wavelength remains the same except the electron accelerates? what is this zero-momentum frame? too confuse right now!!! $\endgroup$– user6760Commented Dec 20, 2018 at 5:36
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5 Answers
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It is true that the reactions $$e + \gamma \to e, \quad e \to e + \gamma$$ cannot occur without violating energy or momentum conservation. But that doesn't mean that electrons can't interact with anything! For example, scattering $$e + \gamma \to e + \gamma$$ is perfectly allowed. And a classical electromagnetic field is built out of many photons, so the interaction of an electron with such a field can be thought of as an interaction with many photons at once. There are plenty of ways a free electron can interact without violating energy or momentum conservation, so there's no problem here.
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To resolve this paradox requires study of time dependent perturbation theory; solving Schrodinger's equation with a time dependent perturbation corresponding to the interaction time of two particles.
If you do this you arrive at the following conclusions:
A single free electron cannot absorb a free photon ( $e + \gamma \to e$ is not a valid interaction)
A single free electron cannot emit a free photon ( $e \to e + \gamma$ is not a valid interaction)
However, two electrons can scatter by exchange of energy ( $ e + e \to e + e$ is a valid interaction)
In this later case it is common to refer to this process being due to exchange of "a virtual photon" between the two electrons. But this is just a description of the calculation of time dependent perturbation theory.
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Basically, because the process doesn't simultaneously conserve energy and momentum; this is why we say that it's mediated by a virtual photon.
In more technical language, this means that the photons that are exchanged between two interacting electrons are allowed to be "off shell", where the "shell" is the relationship $E^2 = p^2c^2$ (for a massless particle). Real particles are required to be on-shell, but virtual particles are allowed to stray from that condition, at least by some amount.
In the Feynman-diagram description of the scattering between two electrons you often find diagrams where one electron emits a photon which is then absorbed by the other; this is a virtual photon and as such both the 'emission' and 'absorption' processes are exempt from these considerations.
As always, though, it bears repeating that Feynman diagrams are calculational tools, and none of the virtual particles that appear in those diagrams actually physically exist in any definable sense. The fact that the 'emission' and 'absorption' processes, as well as the virtual photon itself, seem to defy energy or momentum conservation is purely a quirk of the way in which we've chosen to interpret limited chunks of our calculation.
answered Dec 20, 2018 at 4:12
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Your first statement is false: energy can indeed be added at will to electrons by accelerating them with electrostatic charge distributions, as for example in the case of rapidly varying radio frequency (electromagnetic) fields. Neither energy nor momentum conservation is violated in this case. Search on SLAC for more details about this.
Your other questions are unclear. I recommend you do the search, read a bit, and return here if you have further questions.
answered Dec 20, 2018 at 1:55
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1$\begingroup$ i know when electron in an atom absorb energy it becomes excited, as it goes from excited to lower state energy is being released as photon and in ur case it is radiowave. However I'm at a loss when I try imagine 2 free electrons (particle/wave) can repell each other? $\endgroup$– user6760Commented Dec 20, 2018 at 2:38
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$\begingroup$ @user6760 The free electron can also gain energy from photon. That also leads to a measurement problem. $\endgroup$ Commented Dec 20, 2018 at 3:20
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1$\begingroup$ @BillAlsept: I'm referring to free electron not valence electron $\endgroup$– user6760Commented Dec 20, 2018 at 3:25
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$\begingroup$ @user6760 if you fire free electrons through a double slit experiment and try to observe them, the photons will physically interfere. $\endgroup$ Commented Dec 20, 2018 at 4:01
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$\begingroup$ @BillAlsept: of course 1 free electron interacts with one atom $\endgroup$– user6760Commented Dec 20, 2018 at 4:08
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Let me start with a simple counter-question. How a free electron in a laser cooling process loses kinetic energy? The photon, hitting the compliant electron, gets absorbed and after is re-emitted with a higher frequency (with a higher energy content).
There is no acceptable/viable mechanism for a free electron to absorb or emit energy,...
There is. Photons are indivisible particles only between their emission and absorption. And the term photons is a summary for a class of particles over all possible frequencies (energy contents). So the re-emission of a photon mostly happens not with the same frequency as the absorbed photon.
So I rewrite the equation from another answer to an interaction between the electron and the photon:
$$e + \gamma \equiv e \leftrightarrow (\gamma_1 + \gamma_2) \to (e + \gamma_1) + \gamma_2 $$
How do 2 free electrons repel each other then?
Beside explanations with virtual photons another explanation is that for equaly charged particles the fields do not exchange energy but work like springs. The electric fields get deformed like springs and get relaxed after by pushing the particles back. But the particles lose meanwhile some amount of their kinetic energy (in relation to each over) by emitting photons. You remember, any acceleration is accompanied by photon emission.
answered Dec 21, 2018 at 5:47