Can electrons absorb photons?

Question

I see two broadly opposite answers to this question. One is that electrons can absorb photons, and one is that electrons can't absorb photons (see How many photons can an electron absorb and why?).

The reason I ask this question is because I want to understand, from the quantum-level view, why metals are more reflective than nonmetals, or why metals have more elastic scattering (reflection). It's clear that protons or nuclei absorb photons (see How does an electron absorb or emit light?). But I want to go further and ask, if there are a lot of electrons moving around the nucleus in disorder (i.e. free mobile electrons), what effects do they have on photons? Obviously, a photon cannot be repelled by an electron, so the idea that a photon will bounce off an electron or be scattered elastically, while easy to understand, contradicts the fact that a photon is not repelled by an electron. In this case, what is the way in which the photons that hit the metal leave the metal?

Answer 1

Bottom line up front: You don't get absorption phenomena with a single free electron. You need to look at a system such as an atom or a crystal, in which there are more complicated dynamics at play than electron-photon scattering.

A free electron interacts with photons by scattering that photon into some direction (with very high probability, at "reasonable" energies; in general there are other processes that could occur such as $e\gamma\rightarrow e \mu\bar\mu$). The electron does not absorb a photon. (I am ignoring the issue of "soft photons" mentioned in one of the answers linked by the OP, which in my opinion not related to the question the OP is asking).

A Hydrogen atom consisting of an electron orbiting a proton can absorb a photon. This will happen if the photon's energy is close to one of a certain discrete set of energy values that the electron can transition from one state of the Hydrogen atom to another. Alternatively, the photon can ionize the Hydrogen atom if it has more than the binding energy of Hydrogen, $13.6\ {\rm eV}$, since then the photon can excite a (ground state) electron to an unbound state. A similar story works for more complicated atoms, although the calculations are much more difficult.

In a solid crystal, you have to remember that the electrons are not just bound to a single nucleus, but that the nuclei share the electrons, forming a kind of sea. How a photon will interact with the crystal will depend on the properties of this sea. In particular, the available energy states of the crystal tend to form bands of allowable energies, separated by gaps of unallowable energies. A metal is a solid where the most energetic electrons in equilibrium are in the bottom or middle of a band. It is easy for photons to scatter off of these in-band electrons, since the electrons can absorb an arbitrarily small amount of energy while still remaining in an allowed energy state. An insulator is a solid where the most energetic electrons are near the top of the band. Then it is hard for a photon to interact with the electrons, because a photon needs to have enough energy to excite the electron above the gap into the next band. Thus insulators tend to be transparent. However, photons with an energy close to the band gap can interact strongly with the material, and so the insulator may be opaque at discrete frequencies. Finally, there are also materials with impurities which can scatter photons in random directions, and then appear to opaque.

Answer 2

Any charged particle can interact with the electromagnetic field through the exchange of photons. So, what applies for electrons in this context applies for all such charged particles.

For a free electron (or charged particle), the basic interactions are where a photon is either absorbed or emitted by an electron, but such basic interactions cannot happen individually, because they violate conservation principles. So two such interactions always combine to give a physical process. Typically an electron would absorb and re-emit a photon to give a process that is called Compton scattering. One can also combine two absorptions, in other words, where an electron absorb two photons. So in principle, there is no limit to the number of photons that a charged particle can absorb. However, it must conserve energy. Therefore, a free charged particle will move faster every time it absorbs a photon.

When the electron is bound in an atom, it is in constant interaction with the nucleus (protons) via the exchange of photons. In such a case, the electron would absorb a single photon to move to a higher energy level provided that the energy of the photon matches that of the difference in the energy level. There are an infinite number of energy levels. So the electron can absorb an infinite number of photons, but each time the energy of the photon much match the difference in energy levels.

For a metal, it may be easier to think in terms of classical electromagnetism. The reason why a metal reflect light (photons) is because it can conduct. This is because there are free electrons in the metal that can flow freely. In terms of the boundary conditions that such conducting material impose on electromagnetism, it would tend to reflect the light that falls on it.

Answer 3

This is for clearing confusions

But I want to go further and ask, if there are a lot of electrons moving around the nucleus in disorder (i.e. free mobile electrons), what effects do they have on photons?

The electrons are not moving around the nucleus, they are quantum mechanically bound. In solid lattices, as are metalic solid, there can be a lot of bound states to the whole lattice that can be modeled as "free electrons" with quantum models,

Obviously, a photon cannot be repelled by an electron, so the idea that a photon will bounce off an electron or be scattered elastically, while easy to understand, contradicts the fact that a photon is not repelled by an electron.

At the quantum level, which is the correct level to model electrons, the attraction and repulsion is modeled by interactions , see how two electrons repel at the quantum level, first order.

In this case, what is the way in which the photons that hit the metal leave the metal

With more complex diagrams/interactions as explained in Andrew's answer

Answer 4

There is a nice online course on quantum optics in the MIT catalogue, and in a section on atomic transitions it has an interaction term

$$ b_m^{\dagger}b_na_k$$

Which means the photon isn’t absorbed, it’s annihilated; moreover, the electron doesn’t jump (as we always say), it is destroyed in state $n$ and another electron (if that means anything) is created in a higher energy state $m$.

It makes absorption seem a bit classical.