# I Compton effect: how can it take place?

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1. May 11, 2017

### RS6

English is not my native language. So, I hope to be understood. :-)

The Compton effect is the dynamics in which high-energy incident photons (X or gamma) are scattered by electrons of certain materials, like graphite. The electrons are supposed to be free, as they are only weakly bounded by their atoms, especially considering the high energy of colliding photons.
Well, the implications ot this effect are to me rather confusing.

If I approach the problem, considering the conservation laws for energy and momentum, I have no difficulty in understanding the wavelenght variation wich affects the scattered photon. Of course, I have to consider the quadrivector in that calculation, since I deal with a relativistic process (photons are involved).
Until here, no problem.
But, when I try to interpret the effect I loose my mind.

The conservation laws span over the ends of the timeframe between the arrival of the incident photon and the scattering of the photon. Anyhow, they don't tell me anything about what happens in that interval.
I would like to know it, because I cannot assume that a single photon is directly deflected or bounced like a bowl, even if this is what appears in terms of ending outcome from our point of view. In the matter of fact, a photon is a quantum, so it should not interact partially, i.e. transfer only a certain amount of its energy. A quantum has instead to interact totally or not interact at all. This means, in my opinion, that the incident photon should be absorbed somewhere and that a second photon, with less energy, should be then emitted. I think so to a two-photons process.

There is now a big problem: a free electron can't absorb or emit a photon. So, the photon, in the case, should interact elsewhere. Is then the atom involved? Does the atom (or the internal electrons) 'digest' the incident photon and then release to the free electron the amount of energy needed for the emission of a second photon? Where do I have a misconcept of the process?

Thank you.

2. May 11, 2017

### Staff: Mentor

Why do you think this must be the case? The quantum hypothesis just says a photon's energy depends on its frequency: it doesn't say a photon can't change its energy/frequency.

Have you looked at the actual models used to describe Compton scattering?

First, this is based on classical energy-momentum conservation; you can't just assume it still holds when we are using quantum mechanics.

Second, you just said it was a two-photon process, not a one-photon process.

3. May 11, 2017

### LeandroMdO

Energy and momentum is not "weaker" in quantum mechanics. For example, the Heisenberg equation for $\hat{p}$ reads $$\frac{d\hat{p}}{dt}=i[H,\hat{p}].$$If spatial translations are a symmetry, $[H,\hat{p}]=0$ and $\hat{p}$ is constant for all time. This is an operator equation, so not just the mean value of $\hat{p}$ but all its higher moments as well are conserved.

In QFT in particular, in order to get a nonvanishing contribution to an amplitude for a given process in perturbation theory, energy and momentum must be conserved at every vertex. The diagram with two external electron legs and one external photon leg is decidedly disallowed (except in specific circumstances, e.g. strong external magnetic fields), and the lowest-order contribution to Compton scattering comes from diagrams with topology as given in this figure: https://upload.wikimedia.org/wikipe...eynman_diagram_-_Compton_scattering_1.svg.png

Calling this a "two-photon process" is accurate (in the sense that all such diagrams have two external photon legs).

@OP: To make sense of this process at the level you're interested in, you need to know a bit of quantum field theory. Thinking of it in terms of classical billiard balls is guaranteed to mislead you. In fact, about the only intuition that you can bring from classical mechanics is conservation of energy and momentum, which still hold for calculating amplitudes in QFT. But remember that this is quantum mechanics, and thus you can say very little about what "really happens" between the incident photon arrives and the emitted photon departs.

With a bit of knowledge you can easily draw suggestive pictures that seem to tell stories like "a photon is absorbed and then emitted", or more strangely, "a photon is emitted and then another is absorbed". But beware: suggestive though they are, these "stories" describe the formulas you have to write down to calculate amplitudes rather than a literal process such as a water wave reflecting off the shore of a pond. The "black box" you're uncomfortable with is just the best we can do at this time, though the QFT black box gives you a bit more information than the "conservation law" black box you're using.

Last edited: May 11, 2017
4. May 11, 2017

### Staff: Mentor

As a diagram with two electron external legs and one photon external leg, yes. But not as an internal vertex in a diagram.

Of course it does; that's the definition of Compton scattering.

This was the general point I was trying to make, and is a better way of putting it.

5. May 11, 2017

### LeandroMdO

Yes, I should have made that clearer.

6. May 12, 2017

### RS6

Many thanks for answering and explaining!
Sorry, if the following text is long and qualitative. I try to look first for an easy and much as possible intuitive understanding of the phenomenon (remembering that "phenomenon" is a greek word and means "what it seems", which is not "what it is").

Yesss! It should be one of my fundamental misconceptions.
Effectively, the quantum theory states, that there is a minimum and irreducible amount of action, not of energy. The action is the product energy*time, so it is not said that if a photon loses a certain amount of energy it should then also be divided in some way.
After all, photons can undergo redshifts or blueshifts. I think for instance to a photon in the gravitational field or to the Bremsstrahlung.
So, the incident photon interacts with the free (weakly bounded) electron in the Compton scattering and transfers partially momentum and energy to that particle, but it remains the same photon, carrying the typical Planck amount of action (energy*time). Anyhow, since it loses energy, the "time portion" of the carried action has to increase after the energy exchange.
Is all that right? Leaving for a while Compton, can I say, that in a complete absorption a photon like the oncoming one would interact faster than a photon like the scattered one? Generally speaking, make it sense to state that longer-wave photons take a longer time to be absorbed or emitted?

Uhm... I think the mechanical conservation laws are never violated at quantum scale, exactly as it happens at macroscopic scale. This is the reason why a free electron can't absorb or emit a photon (bounded electrons can). I read some articles which report that a free electron could absorb a photon, but should right after emit another one (we are not speaking about virtual particles). In other words, the conservation laws could be infringed, but only very shortly, i.e. in a time interval compliant with Heisenberg principle. Well, it seems those texts are wrong. The mechanical laws of conservation (energy and momentum) operate in each moment, that is: in each possible vertex of a Feynman diagram, as LeandroMdO writes in his post. This is what I got. Hope to be on the right track.

When I wrote about a "two-photon" process, I was meaning that the incident photon is not the same photon that is scattered. I was in sum thinking to an articulated dynamics in which a photon is absorbed and another one released. But it was not my intention to let understand that the absorption is carried out directly by the free electron (what is forbidden, as I wrote). Instead, I was wondering if the atom as a whole or other in it bounded electrons could take part to the energy exchange process.

Yes! Looking at photon and electrons only as rigid balls is misleading. This is clear to me. I always assume photons and electrons are waves, which in certain cases can show rigid body behaviour. I hope this can be an adequate approach. Anynow, I didn't look at them as field perturbations. I suppose, this is my essential limit.
Now, I see that these perturbations could behave in many different ways in each quantum phenomenon. If I'm not wrong, Feynman describes an infinite amount of coherent paths which could reflect what from our point of view seems to be a single interaction. This is very complex to me. But I have to accept it.
As wave amplitudes relate to (density of) probability and not to deterministic physical magnitudes, there is an additional difficulty in trying to describe what really happens between the arrival and the departure of the photon in the Compton effect.

-----

To summarize:
1) The incident photon can lose energy and become the scattered photon in the Compton effect;
2) The way this happens possibly involves also other particles, not necessarily only the free electron, and is not determined in the microcosmic environment;
3) The incident photon and the scattered photon can also be two different entities;
4) Various (infinite) paths are possible for 3), i.e. for the photon absorption and emission;
5) We can't know what really happens inside the black box, but we can predict how the box behaves.

I'm absolutely not sure of what I wrote.
Does someone tell me where it make sense and where not?
Thanks.

Last edited: May 12, 2017
7. May 12, 2017

### vanhees71

Your points make sense! As was mentioned before, for a full understanding you have to learn some QFT (here QED, which should be the first theory you study anyway). Here are some qualitative remarks:

You cannot consider a photon to be a classical particle or a classical wave. The only adequate description is relativistic QFT, which tells us that a photon does not even have a position observable. It also tells us that indeed all we can interpret in some sense as "particles" (or "quanta" to emphasize that we talk about "particles" in the sense of QFT) are socalled asymptotic free states, i.e., particles which are far from any other particles they can interact with and thus can be considered free. A scattering process is described by QFT in terms of a socalled S-matrix element. It's absolute square provides the transition probability rate that a given asymptotically free state of some particles (usually two particles with quite definite momenta and energies) ends up after a scattering/interaction process in another given asymptotically free state. For these asymptotically free states for each particle the energy-momentum relation $E=\sqrt{\vec{p}^2+m^2}$ (I use units, where $c=\hbar=1$, which is very convenient in high-energy particle physics) and overall energy and momentum conservation holds, i.e., the sum of all four-momentum vectors of the asymptotic free in-particles equals the sum of all four-momentum vectors of the asymptotic free out-particles.

QFT provides a formalism to evaluate such S-matrix elements using perturbation theory, i.e., you start with non-interacting particles and treat the interactions as perturbation, leading to a formal series in powers of the interaction strength (or coupling constant, in the case of QED it's the charge of the electron). Each contribution to this series can be depicted by Feynman diagrams, which are suggestive for "depicting" what's really going on, and in many popular-science books they are sold with such a meaning. However one has to be careful with that. Strictly speaking they are just very clever notations for the mathematical formulae needed to evaluate the S-matrix perturbatively.

Whether in the Compton scattering the outgoing photon is the same as the ingoing is a meaningless question according to QT. You cannot distinguish photons of a given energy/momentum from any other of the same energy/momentum. As already said above, we also cannot follow a photon like a classical particle somehow and know what's going on with it during it's interacting with the electron at all. All we can observe is that we shoot a photon to an electron and then it scatters (with some probability) ending again with an asymptotically free photon and an electron.

It's also impossible to call it "several paths a photon takes". The very notion of "path" (i.e., trajectory in the sense of a classical particle) doesn't make sense. While in non-relativistic quantum theory you can describe the quantum theory in terms of socalled path integrals (also an ingenious discovery of Feynman) which some over all paths of a particle in space-time, but this is equivalent to the formulation of non-relativistic QT of a particle in terms of a Schrödinger-wave function. This is not possible for relativistic particles since usually in scattering processes you always have some probability to destroy some particles in the initial state and create new particles in the final state. It's not possible to describe that with a single-particle wave function, and that's why we use QFT, which is a formalism to describe processes, where the particle numbers are not conserved and creation and destruction of particles can occur. There is a path integral formulation of QFT too (and it's very elegant, particularly for the modern gauge theories which build the foundation of the Standard Model), but here one doesn't some over space-time trajectories of a particle (or some finite conserved number of several particles) but over field configurations, i.e., it's a pretty abstract "sum over paths".

8. May 12, 2017

### RS6

Thank you, vanhees71!
Thank you also for the simplified way you write. But it all remains a bit criptic to me.

I see perhaps the condition: in QFT it's a question of a boundle of relations reflecting integrated states that evolve as a global probability distribution in space and time (better said: in space-time). In other terms, it's not a question of single separated waves (even if intended as probability wave functions).
As far as I understand from your great post, the evolution is caused by perturbations, a kind of parallel to the concept of interaction in the classic world. More than a parallel: perhaps it's the real meaning of what an interaction is. Anyhow, the best we have at that scale.

So, if I got it, one has always to consider first an immaculate state, or at least a stationary state (?). This is the starting point. Then, one has to introduce a perturbation (mathematical series in power) and make a projection for a possible ending state. At that point one can establish the matrix and evaluate the actual probability of that specific evolution of the system.
Not knowing experimentally what the (more frequent) ending states are, a grat amount of calculation is required to formulate predictions, i.e. to detect the most probable transition. Instead, knowing the ending condition, we get a kind of integrated description of what could happen during the process. But, unfortunally, only the matrix has a clear indentity, not what it is made of. Ok, better than nothing.
Hope not to be too wrong with all this.

If each perturbation is depicted by a Feynman diagram, how many perturbations/series should be involved in the matrix calculation? May be it should be convenient to consider only the most relevant diagrams, i.e. the series that influence at most the outcome. Or, may be, with QFT is not as with the Feynman non-relativistic path-integarls. Or what?

My interpretation: This is so, because the matrix, which is the best we have to reflect reality, is an integrated instrument. The matrix reflects relativistic evolution of two or more particles as a whole, not as a sum of classic interacting parts. In the complex whole we cannot distinguish the fate of the single component or what component can be produced and destroyed. Mathematics helps us to depict/predict the transition between its end, but not what depelopes in the intermediate timeframe. It seems to me, that even what happens first and what after is not any more clearly distinguished in the matrix abstraction.

I'm afraid what I wrote is a misconception. I'm here to learn.

Last edited: May 12, 2017
9. May 12, 2017

### Staff: Mentor

There is a physical constant with units of action, yes, but it does not always represent "a minimum and irreducible amount of action".

Photons don't have identifying labels on them, so there is no meaning to the question of whether the scattered photon is "the same photon" as the incident one. All you can say is that a photon goes in and a photon comes out.

This isn't really any better than thinking of them as little billiard balls. Photons and electrons are excitations of quantum fields. They're not like anything in your ordinary experience.

10. May 12, 2017

### RS6

I don't understand. How is it possible to get an energy exchange where less than a Planck quantum is involved? Or, may be, you refer to 2π...

Yes, ok. It was a simplified sentence, pointing to the different energies of the photon before and after the collision. Even the word "collision" could not be adequate in this context.

Ok, I agree here too. My post was before the explanation on the QFT. May be I'm too rough, but if I see a wave on a lake, for instance, I interpet it as a perturbation of the water surface. In reality, the wind or a stone or something else is the actual perturbation and origin of the wave. It was just a way to say. But it's ok to go at the substance... even if the substance is not clearly identified.

11. May 12, 2017

### Staff: Mentor

There is no such thing as "a Planck quantum". Planck's constant is not the size of a little packet of anything. It's just a physical constant that appears in the equations.

The above comments apply just as well to $\hbar = h / 2 \pi$.

12. May 12, 2017

### RS6

@PeterDonis Ok, if I said that h is the smallest energy content that Nature can allow to be trasported in an em field at a given frequency would you accept it? May be, referring to energy is more up to date than to a classic concept like the action.
Anyhow, to me h is what implies that the electrons levels in an atom have to be discrete.

13. May 12, 2017

### Staff: Mentor

No.

It's more complicated than that. Again, $h$ is just a physical constant.

To see what's wrong with your logic here, consider that $h$ "applies" to all electrons, not just electrons in atoms. So if $h$ means energy levels are discrete, why don't free electrons (electrons not bound in atoms) also have discrete energy levels?

You are basically taking a complex domain--quantum mechanics--and trying to simplify it down to a single statement involving a single constant $h$. That won't work.

14. May 12, 2017

### RS6

I did not write that h "means" that energy level in an atom are discrete. I wrote that it implies it. It seems to me it's very different. It implies also other things, like the Heisenberg principle.
I'm afraid, my English is not good enough to express my thoughts.
Of course, h applies to all electrons.

When I refer to h I think to absorption or emission. I have already had the chance here to see very well how some interactions don't need at all a quantum exchange. This was after all my inital question and misconception related to the Compton effect. The users have very well clarified how complex can be the energy transfer from the arriving photon (matrix construction). You are now going back again to the subject. May be I'm wrong.

I still think that h is the smallest energy content that Nature allows to be carried from A to B in an em field at a given frequency, i.e. with a photon. This, regardless of what happens in A or in B. What I have to add to avoid misundertanding, is that the photon doesn't interact during its propagation (for example, just in a scattering effect) between the two space-time coordinates. Describing h only as a constant in quantum equations let one think it has no meaning. I think that something which is mathematically constant in a science, an equation system, a matrix, a constant value, could be a reflex of a physical truth.

I'm not trying at all to reduce quantum mechanics to the constant h. I'm just trying to understand some dynamics in a forum that is supposed to help in that. But it's true that I try to simplify. I'm convinced that one has really got a knowledge when he is able to simplify what he knows in the best way. Fore sure, this is sometimes very hard. To simplify is not easy.
If I'm misplaced or inappropriate I really apologize and leave. Perhaps the section is only for discussion among experts in QFT.

15. May 12, 2017

### Staff: Mentor

This is just quibbling over words. My objection remains the same with the word "implies". What implies that energy levels of electrons in atoms are discrete, while energies of free electrons are not, is not "h". It's quantum mechanics. And there is a lot more to quantum mechanics than just a finite value of $h$.

It's not the "smallest" energy content of a photon at a given frequency, it is the energy content of a photon at a given frequency, via the formula $E = h \nu$. And the qualifier "at a given frequency" destroys your argument, since it shows that all you have to do to transfer an arbitrarily small amount of energy is to use a photon with an arbitrarily small frequency. Yes, there are practical limitations on how low a frequency we can actually produce, but those limitations have nothing to do with the finite value of $h$. So the finite value of $h$ does not put a lower limit on the amount of energy that can be transferred from A to B.

16. May 12, 2017

### RS6

What is true is that electron levels in atoms are quantized, because their energy status is a stationary vibration (a bit like a harmonic vibration on a guitar string, but in the space). Otherwise, the electrons would fall towards the nucleus. This shows the undulatory nature of the electron. On the other hand, the photoelectric effect shows that you have to release the sufficient amount of energy to that vibration to get a jump to ionisation. This illustrates instead the particle behaviour of the radiation. Now, this energy, carried by photons quanta, depends only on the frequency and not on the intensity of the radiation (number of photons). But the energy amount of photons is also quantized, as Planck discovered. Briefly, the comparison of this discrete amount with the emission of electrons particles (current) demonstrates that the inner atoms levels are discrete and also where they are. Note that the comparison involves h, discriminant and mervellous, a value which is always the same, while the energy (h*freq) can differ from case to case.

As for free electrons, we have a quiet different situation (in reality, you write here the exact opposite of your post above, but let's say I understand the intention). In this case there is no binding energy. As we have seen with the Compton effect, for instance, an electron can then receive only a % of the photon energy.

???????
...and it's me the one who quibbles over words!! Please. :-)
I don't see how you can derivate your assertion. It's rather bent what you attribute to me. I think we are now really going off-topic.
"At a given frequency" means of course that if you change the wavelenght also the carried energy varies. In fact, the energy raises with the frequency. And this is all.
So, h still corresponds to the smallest energy that can be transferred from A to B in an em field at a given frequency (i repeat: supposing the photon doesn't interact along his "path").

17. May 12, 2017

### Staff: Mentor

No, that's not correct. If it were, energy eigenstates of free electrons, which can also be described by "stationary vibrations" (waves whose wave parameters do not change with time), would also be quantized, but they aren't.

I'm not sure what you mean by "undulatory nature". I suspect you are basing your understand on pop science sources instead of actual textbooks and peer-reviewed papers. I strongly suggest that you spend some time with the latter.

Are you aware that the photoelectric effect can be analyzed using a classical (non-quantized) model of the EM field?

Once again, I suspect you are basing your understanding on pop science instead of actual science.

Yes, and that % can be arbitrarily small, despite the finite value of $h$. So $h$ does not set any lower limit on energy transfer.

The energy increases with increased frequency (shorter wavelength); the energy decreases with decreased frequency (longer wavelength). Make the frequency low enough (wavelength long enough) and the energy can be as small as you like, despite the finite value of $h$. So $h$ does not set any lower limit on energy transfer.

18. May 12, 2017

### RS6

Of course, I mean quantized in the atom.
But, you're right, my sentence was not rigourosly expressed, as it should.

The electron can notoriously show particle behaviour or wave behaviour, according to the condition. In the mentioned effect it behaves like a particle.

Yes, I'm aware. As long as you refer only to the radiation frequency. But, if you know that radiation is the effect of quanta (since Planck, black body), you have to take this into account. At that point you can't avoid comparing h to the electrons level.

I completely agree. Too bad I never stated something else. There is no lower threshold for an energy change, this is true. But, I was referring to an energy transfer at a given frequency without interaction from A to B.

Of course. It's trivial. This is why I specified "At a given frequency". :-)

(Sorry for the mess with the quote. Hope it appears right now)

19. May 12, 2017

### Staff: Mentor

If all you mean by "undulatory nature of the electron" is "the electron can show wavelike behavior", then yes, you can say the electron has an "undulatory nature". But you have to be very careful, because it's very tempting to start saying more than that--for example, to say that somehow the electron "is a particle" sometimes and "is a wave" other times, and talk about wave-particle duality and so on. And doing that will cause you to misunderstand things and make incorrect predictions and inferences. That's why, IMO, it's better to stick with the more neutral "the electron can show wavelike behavior"; it's less likely to lead to misunderstandings and incorrect inferences.

How can any energy at all be transferred if there is no interaction?

20. May 12, 2017

### RS6

After the second or third post in this section, especially with you, I am careful! :-)
Anyhow, I agree with you. An electron is not a wave, nor a particle. It's something different which can show a double nature before our eyes.
I admit that I find myself mentally more comfortable in supposing it's rather a wave that can shrink or dilatate. But this doesn't mean that I think it's really a wave. Besides, one has always to consider the probabilistic aspects. In this sense an electron, after interaction, should be geometrically simply a point, if it makes sense.

The energy is released in B. It is a total release (absorption if possible) or a % release (like in Compton).
Let's take the Compton effect. The incoming photon originated in A. Nothing happens for a certain time. In B it interacts with the free electron of the graphite. We know the energy transfer from the photon to the electron is at that point only a partial amount of the carried energy. But this doesn't matter. In the matter of fact, the photon has transferred his energy from A to B, the smallest possible amount at his (medium) frequency.

I resume tomorrow, if I can. ;-)
Thank you in the mean time...