# When a photon is absorbed by a particle like an electron or a proton, is capacity for further absorption "used up?"

That's basically the whole question -- does absorbing a photon do something to a particle so that the particle no longer can absorb further photons? If there is a limit beyond 1 (maybe protons can absorb 2 and electrons only one), how is the fact that one photon has been absorbed make itself evident? Does it make it more difficult to absorb the second one? If there comes a point where further photons cannot ne absorbed, does this mean the next photon simply bounces off or something?

• An electron cannot absorb a photon. Commented Sep 30, 2022 at 6:17
• @ProfRob To clarify, you mean a free electron, right? Isn't it the case that an electron bounded to an atom can absorb a photon (although maybe you count the atom as one system)? Commented Sep 30, 2022 at 6:40
• @MaximalIdeal The point is that the system that "absorbs the photon" is the atom and not the electron in the atom. An atom will, for instance, also change its momentum when absorbing a photon. We don't talk about that much because it is rarely of interest. Technically even that is a scattering process because eventually the excited atom will de-excite and give the energy back to the field. What these kinds of discussions show is that the teaching process of QM and atomic physics is still very shoddy. One can't really live without the other but we are still teaching them separately. Commented Sep 30, 2022 at 6:58

First of all, momentum conservation implies that a particle cannot just absorb a photon. Expressed in four-momenta and using units where $$c=1$$, i.e. energies, momenta and masses are all measured in the same units, energy-momentum conservation can be expressed as $$p_{\textrm{part}}^i + p_{\textrm{photon}} = p_{\textrm{part}}^f$$ for the initial ($$i$$) and final ($$f$$) states. Squaring both sides yields $$(p_{\textrm{part}}^i)^2 + (p_{\textrm{photon}})^2 + 2 p_{\textrm{part}}^i p_{\textrm{photon}} = (p_{\textrm{part}}^f)^2$$. Now the square of the four-momentum of a particle is its mass squared, so with the massless photon this reduces to $$m_{\textrm{part}}^2 + 2 p_{\textrm{part}}^i p_{\textrm{photon}} = m_{\textrm{part}}^2$$ or equivalently $$p_{\textrm{part}}^i p_{\textrm{photon}} = 0.$$ Let's insert the components of the four-momentum of the particle, simplifying indices, $$(E^i, \mathbf{p}^i)$$ and of the photon $$(|\mathbf{p}|, \mathbf{p})$$ into this equation to obtain

\begin{align} {}0 & = E^i |\mathbf{p}| - \mathbf{p}^i\cdot\mathbf{p} \\ & = E^i |\mathbf{p}| - |\mathbf{p}^i| |\mathbf{p}| \cos\theta\end{align}

with $$\theta$$ the angle between the two momentum vectors. Now $$E^i > |\mathbf{p}^i|$$ for a massive particle and $$\cos\theta$$ is less than or equal to one, so this condition cannot be fulfilled.

So how are photons absorbed?

There are two possibilities. Either the absorbing system (not particle) is excited into a higher state. In the equations above that would mean that the masses on the right-hand and left-hand sides are different, and they don't cancel. This happens for example when an atom absorbs a photon. The atom ends up in an excited state. From there many things can happen, but in the simplest case it will emit an identical photon after some time and return to the original state. The second possibility would be a virtual particle: the absorbing particle goes (speaking loosely) "off-shell" which is jargon for the situation where its energy and momentum no longer fulfills $$E = \sqrt{m^2 + |\mathbf{p}|^2}$$. After an infinitesimal amount of time (given roughly by $$\Delta E \Delta t > \hbar$$ where $$\Delta E$$ is the difference of the two sides of the equation) the off-shell particle returns to the mass-shell $$E=\sqrt{m^2+|\mathbf{p}|^2}$$ by emitting a photon (or another particle). E.g., an electron absorbing and immediately reemitting a photon like this is called Compton scattering.

Now to return to your question. Energy-momentum conservation puts strict limits on the kind of absorption processes you had in mind. While an electron remains an electron no matter how many photons it meets, it cannot absorb a photon and just go along its merry way. Instead, it has to do something else (interacting with a nucleus, radiating a photon, ...) to conserve the energy-momentum balance.

The single quantum mechanically described particles do not absorb impinging photons, they interact, called scattering, with the photon, an elementary particle, and the interaction follows the algebra of relativistic four momentum. Absorption of photons can happen in interactions with atoms, see my answer here .

So there is no meaning to " is capacity for further absorption "used up?". Particles retain their identity.

A photon is a small amount (a quantum) of electromagnetic field energy. There are systems that can absorb many such small amounts of field energy (like a macroscopic body, which then gets hot) and there are systems that can absorb only one (like an atom) of a certain energy level because after the absorption they are in a state in which they can no longer change their energy by the same amount.

And then there are systems like a solar panel, for instance, which are designed to absorb copious numbers of visible and near IR photons, on the order of 10^23 per second, and to convert them into a macroscopic electric voltage and current (which can also be described by photons, if you absolutely have to, but that's a really poor way of talking about a classical voltage and current).

Plants have a photosynthetic system that can convert the photon energy into chemical reactions. Those molecules need actually photons with two different energy values to perform their function.

So as you can see, we have all kinds of types of systems and many different interesting variations on the topic (like the different types of scattering systems in Anna v's answer). In other words... it is very complicated. Indeed, the "complete" story would fill entire libraries.

• An atom can absorb more than one photon: en.wikipedia.org/wiki/Two-photon_absorption Commented Sep 30, 2022 at 6:29
• @TheTiler: That's why I wrote "one of a certain energy level". The entire point of my post was that the phenomenology of light-matter interaction is very complex. There are no easy answers. Commented Sep 30, 2022 at 6:37
• ""there are systems that can absorb only one (like an atom) of a certain energy level because after the absorption it is in a state in which it can no longer change its energy by the same amount."" – FlatterMann I am not very strong in English and to be sure of my comprehension, I translate into French .... Commented Sep 30, 2022 at 7:23

The distribution of photon by an electron in an atom having certain energy value through the absorption or emission can be given by,

$$f=\dfrac{1}{e^{\frac{h\nu}{kT}}-1}=e^{\frac{-h\nu}{kT}}\left(1-e^{\frac{-h\nu}{kT}}\right)^{-1}\tag1$$

For $$\nu\sim10^{15}$$ hz and $$T\sim5,000$$ k, $$h\nu\gt kT$$, so

$$f=\left(e^{\frac{-h\nu}{kT}}+e^{\frac{-2h\nu}{kT}}+e^{\frac{-3h\nu}{kT}}+\ldots\right)\tag2$$

The exponential terms in right side of (2) are probability of photon to have energy level $$nh\nu$$ or to have $$n$$ photons at $$h\nu$$ level, where $$n$$ is integer.

As value of $$n$$ increases, the probabilty of having photon at that level becoming smaller and smaller, or we can say the absorption of a photon by an electron in an atomic system becomes lesser and lesser.

Now question is why this happen. Then answer is, it very common phenomenon of nature. Compare it with spreading of any infection whose rate depends upon number of infected people or population growth or charging of capacitor which absorb current slowly after some time. Population growth and infection are increasing function, but it is decreasing from photon's population view and increasing from atom's energy population view.

From matter's perspective, absorption of a photon increases entropy because it increase number of available states, this makes matter more conductive, it is similar to heating the given matter.