Quantum Theory – Does Quantum Theory Allow an Electron to Take a Fraction of Photon Energy?

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In Photoelectric Effect of Theory of Spectral Lines , an electron takes the entire or none of the energy of the Photon ( it absorbs the entire quanta not its fractionS resulting in the disappearance of the Photon).

But in Compton Effect the Electron takes only a fraction of the Photon Energy and the Photon stays alive.

What I specifically want to know is that Quantum theory says that an electron will either take the whole or none of the Photon energy .

Why then in Compton Effect the electron takes a fraction of the Photon energy Is that allowed by Quantum theory ?

Or is not so ? Or the electron takes up 1 photon only when in an energy level . When an electron is free , can it absorb half the energy of a photon ?

Edit:

I still don't get whether the electrons are allowed to take the energy of the whole photon when they are in energy levels and can they absorb a fraction of the photon energy when they are free.

Best Answer

There is a simple reason why a free election can't absorb a photon completely: you can't conserve both energy and momentum for the system of you start off with an electron and a photon and and up with only an electron. You need a final photon as well for conservation of energy and momentum to be satisfied.

For the photoelectric effect things are different. We don't have a free electron ; it is bound to a nucleus. Consequently it can only have certain precise energy values (and won't interact with photons that would move it to a non existent energy level, which is the basic argument for quantum behaviour, but not really comparable to Compton scattering).

In the situation analagous to Compton scattering, the photon has more energy than the binding energy of the electron, and so we wind up with a free electron. But in this case energy and momentum can be conserved without a final photon since the nucleus is also involved in the interaction. The initial state is {photon, bound electron, nucleus } and the final state {free electron, nucleus}.

It is possible to construct Feynman diagrams with a final photon present too, but since they have an extra vertex they happen less often by a factor of roughly the fine structure constant $\alpha\sim\frac{1}{137}.$

Or to put it another way, sometimes in the photoelectric effect where you end up with a free electron you do get a final photon, but it is in less than 1% of cases (unless I've overlooked some reason why it can't happen).

Related Solutions

[Physics] Show that a photon cannot transmit its total energy to a free electron. Contradiction with Photoelectric effect

The original problem can be seen in terms of energy and momentum conservation.

Before scatter, there are two particles in the center of mass and the center of mass has an invariant mass larger than the mass of the electron. For total absorption of the photon there would be only the electron left. As the electron has a fixed mass and at the center of mass it should be at rest, the reaction cannot happen. It can only happen if a third particle is involved to conserve the overall energy and momentum , and this is what is happening with the photoelectric effect.

photoelectric a)

The incoming photon interacts with an electron that is tied to the atom by a virtual photon . The whole system takes up the energy and momentum conservation.

The inverse problem happens with a gamma generating an e+e- pair. The gamma has zero invariant mass, the pair will have at least two electon masses at the center of mass, so a gamma cannot turn into an electron positron pair , a third particle has to be involved. The simplest is a virtual photon from some nucleus too. The Feynman diagram is the same as the one above with a different interpretation ( incoming e- is read as outgoing e+)

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a) diagram copied

Quantum Mechanics – Why One Electron Absorbs Only One Photon in the Photoelectric Effect

More than one photon can be absorbed, but the probability is minute for usual intensities. As a scale for "usual intensities" note that sunlight on earth has an intensity of about $1000\,\mathrm{W/m^2} = 10^{-1}\mathrm{W/cm^2}$.

The intuitive reason is, that the linear process (an electron absorbs one photon) is more or less "unlikely" (as the coupling between the em. field and electrons is rather weak), so a process where two photons interact is "unlikely"$^2$ and thus strongly suppressed. So for small intensities the linear process will dominate distinctly. The question is only, at what intensities the second order effects will become visible.

In the paper by Richard L. Smith, "Two-Photon Photoelectric Effect", Phys. Rev. 128, 2225 (1962) the photocurrent for radiation above half of the cutoff frequency but below the cutoff frequency is discussed. They note that for usual intensities the photocurrent will be minute, but that given strong enough fields such as those observed in a focus spot of a laser (on the order of $10^7\,\mathrm{W/cm^2}$) the effect might be measurable. They also note, that thermal heating by the laser field may make the pure second order effect unobservable.

The more recent paper S. Varró, E. Elotzky, "The multiphoton photo-effect and harmonic generation at metal surfaces", J. Phys. D: Appl. Phys. 30, 3071 (1997) dicusses the case where high intensities (on the scale of $10^{10}\,\mathrm{W/cm^2}$) produce even higher order effects (and unexpectedly high, coherent non-linear effects, that is absorption of more than two photons by one electron). Their calculations explain the experimental observations of sharp features in the emission spectra of metal surfaces.

Historical fun fact: The 1962 paper is so old, that it talks about an "optical ruby maser"; lasers where so new back then, they did not even have their name yet.

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Related Solutions

[Physics] Show that a photon cannot transmit its total energy to a free electron. Contradiction with Photoelectric effect

Quantum Mechanics – Why One Electron Absorbs Only One Photon in the Photoelectric Effect

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