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According to classical electrodynamics any charged particle is accelerated in an electric field, gaining energy.

However, it is said that a free electron cannot absorb a photon. For example, this extract is from a peer-reviewed paper:

As is well known, a free electron, that is an electron not subjected to the field of an atom or an ion, cannot absorb photons from a laser field because of the condition imposed by the law of the conservation of momentum.

Do the classical and quantum mechanical pictures really contradict each other here?

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    $\begingroup$ Where's the contradiction supposed to be? If the charged particle is accelerated in an electric field, it is not "a free electron not subjected to a field". Please be more explicit what you want to know. $\endgroup$
    – ACuriousMind
    Commented Jul 9, 2022 at 11:14
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    $\begingroup$ @ACuriousMind, by free electron here I mean "an electron not subjected to the field of an atom or an ion". We are considering the interaction of electron with electromagnetic radiation, e.g. a laser pulse, so the electron is not "free" in an absolute sense. But excluding/neglecting this interaction the electron is free. $\endgroup$
    – apadana
    Commented Jul 9, 2022 at 13:01
  • $\begingroup$ The extract in the question defines "free electron". $\endgroup$
    – apadana
    Commented Jul 9, 2022 at 13:04
  • $\begingroup$ I still don't see where the contradiction is. $\endgroup$
    – ACuriousMind
    Commented Jul 9, 2022 at 13:09
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    $\begingroup$ Let me rephrase the question: according to the quantum picture, a free electron (as defined above) cannot absorb photons, so cannot absorb energy from electromagnetic radiation. On the other hand, according to the classical picture, a free electron is accelerated in a electromagnetic field, which means it absorbs/loses energy. $\endgroup$
    – apadana
    Commented Jul 10, 2022 at 4:20

2 Answers 2

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This is a limitation of the photon exchange picture of interaction between EM field and matter. This simply cannot always be represented as a real photon exchange. Sometimes (in fact, often) the interaction does not conform to this mode of thinking. Even EM field itself in QED sometimes does not conform to it, such as for coherent states describing classical EM waves where number of photons is not well-defined or determined by the EM field state.

Real photon absorption or emission is a special kind of imagined process, where one eigenstate of material system's Hamiltonian turns into different eigenstate, in presence of resonant EM field mode; or where EM field Hamiltonian eigenstate turns into different eigenstate. Thus we have a description of a resonant process, it happens only if the EM mode frequency matches the transition frequency of the system.

It turns out that in case of free electron, this special kind of process is not possible, because there are no internal states to the electron, and there is no non-zero transition frequency. Any non-zero exchange of energy and momentum between states of definite energy and momentum of the electron would violate simultaneous conservation of energy and momentum of the electron + EM field.

But we can forget about photons and states with well defined energy and momentum, and thus the whole photon exchange picture. Instead, we can write down the Schroedinger equation for charged particle in presence of classical or even quantum EM field represented by the potentials $\mathbf A, \varphi$:

$$ \partial_t \psi = \frac{1}{2mi\hbar} (\hat{\mathbf p} - q\mathbf A(\mathbf r,t))^2\psi + \frac{q}{i\hbar}\varphi(\mathbf r,t) \psi $$

Kinetic energy of the electron will not be defined except in special cases where $\psi$ is eigenfunction of the kinetic energy operator, but we can always calculate expected average value of kinetic energy:

$$ \langle E_k \rangle = \frac{1}{2m} \int \psi^* (\hat{\mathbf p} - q\mathbf A(\mathbf r,t))^2 \psi d^3\mathbf r. $$

Since the evolution equation is linear differential equation in time with finite terms, the expectation value will change continuously, and not by any quanta $\hbar \Omega$. It should be expected that this quantity should behave similarly to the classical kinetic energy of the classical electron. I.e. in field of linearly polarized EM wave of intensity $E_0$ and frequency $\Omega$, at least in some region of the parameter space, it should be close to the classical prediction

$$ \frac{q^2E_0^2}{4m\Omega^2}\sin^2 \Omega t.~~~(*) $$

Deviations can happen in relativistic regime, if the wave frequency becomes high enough, close to frequency associated with the Compton wavelength of the electron $\frac{h}{mc}$. But this should be studied also using relativistic equations (Dirac equation for the electron).

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    – Buzz
    Commented Jul 10, 2022 at 4:46
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We draw a distinction between absorption and scattering. Classically, if you illuminate an isolated electron with an electromagnetic wave, it oscillates. An oscillating electron radiates electromagnetic waves. The process is known as Thomson scattering. Quantum theory also captures this (well-known) phenomenon, and matches Thomson's classical calculation if the photon energy is small compared to the electron's mass energy.

In quantum terms, we get one photon out for each photon in, so it's considered scattering rather than absorption.

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  • $\begingroup$ 1. When an isolated electron is accelerating in an EM field it radiates energy, but the rate at which it gains energy is higher than it loses energy. So either the incoming photons are more energetic than the outgoing photons (different wavelength), or the outgoing photons are emitted at a lower rate. I guess the latter is the case. This means that a photon is absorbed, then after some delay another photon is emitted. In a scattering we expect that the incoming and the outgoing photons occur at the same time. Sorry for the not-so-good English. $\endgroup$
    – apadana
    Commented Jul 9, 2022 at 13:48
  • $\begingroup$ @apadana If the field oscillates, as in the case of the laser, the rate at which the electron loses energy is equal to the rate at which it gains energy. For a static, non-oscillating field we usually don't use quantum models at all. When the photons have zero energy, the quantum viewpoint ceases to yield insight. $\endgroup$
    – John Doty
    Commented Jul 9, 2022 at 14:10
  • $\begingroup$ In a half-cycle the electron is accelerating, and in the other half-cycle is decelerating. I think in each one of these half-cycles the gain and loss rates are not equal. That is why the kinetic energy of the electron changes in each half-cycle. $\endgroup$
    – apadana
    Commented Jul 9, 2022 at 14:17
  • $\begingroup$ According to Wikipedia, in Thomson scattering "the particle's kinetic energy and photon frequency do not change as a result of the scattering." So Thomson scattering doesn't seem to be responsible here. $\endgroup$
    – apadana
    Commented Jul 9, 2022 at 14:45
  • $\begingroup$ @apadana Wikipedia is slightly sloppy. Classically, a free electron illuminated by electromagnetic radiation will have a bit more energy than one not illuminated, but there's no sustained transfer of energy. At the photon level, once the scattered photon has left the vicinity, the electron has no more energy than it started from. $\endgroup$
    – John Doty
    Commented Jul 9, 2022 at 14:57

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