I Compton effect: how can it take place?

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The Compton effect describes the scattering of high-energy photons, such as X-rays or gamma rays, by free electrons in materials like graphite. The discussion highlights confusion regarding the interaction process, particularly the assumption that a photon must be fully absorbed or emitted, rather than partially transferring energy. It emphasizes the necessity of understanding quantum field theory (QFT) to accurately describe these interactions, as classical mechanics does not apply. The conversation also clarifies that while conservation laws hold, the specifics of photon interactions involve complex dynamics that cannot be simplified to classical analogies. Ultimately, the process remains largely a "black box," where the exact mechanisms of photon behavior during scattering are not fully understood.
  • #31
RS6 said:
When I say "admit" I mean "we can do as if". Is then my statement right?

Not in any practical sense, because, as I said, we can't actually create photons in the lab that are in energy eigenstates. (Or electrons, for that matter.)
 
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  • #32
PeterDonis said:
That's a good general policy. :wink: That particular article looks OK to me for a start, but also note that I said to look at the references as well. Even Wikipedia articles that aren't very accurate by themselves can give useful references.
True.

PeterDonis said:
RS6: When I say "admit" I mean "we can do as if". Is then my statement right?
Not in any practical sense, because, as I said, we can't actually create photons in the lab that are in energy eigenstates. (Or electrons, for that matter.)
Not in eigenstates, but, so far I got it, we can sometimes assume they are in coeherent states.
You wrote:
It so happens that, for cases like Compton scattering, the mathematical analysis ends up giving practically the same answer whether we use coherent states or energy eigenstates; and since energy eigenstates are much simpler to work with, those are the ones that are used in the analysis.
So, if my "we can do as if" is not adequate to the condition in a "practical sense", then I try following inferences:
1) We can address the phenomenon (e.g. scattering) to eigenstates in a mathematical convenient sense;
2) The results are consistent with the lab observations;
3) But we don't know what are the real states of the involved particles (photon and electron);
4) The real energy state of the particles could differ significantly from an eigenstate;
5) The in the lab measured energies are a superposition of different oscillators.
Am I on a better track?
 
  • #33
RS6 said:
We can address the phenomenon (e.g. scattering) to eigenstates in a mathematical convenient sense;
2) The results are consistent with the lab observations

With the particular lab observations we are talking about (Compton scattering), yes. But the results from this model are not consistent with other lab observations we could make using photons from the same sources.

RS6 said:
we don't know what are the real states of the involved particles (photon and electron)

Yes, we do; they're in coherent states (at least the photons are). More precisely, we can verify that the photons from the sources we use are in coherent states by making other observations (besides Compton scattering) on photons from those sources. The electrons, since they are bound in atoms in Compton scattering experiments, are in whatever bound states correspond to the atomic orbitals they occupy. Those are not the same as the electron states used in the mathematical analysis of Compton scattering; the latter states are simpler mathematically, as I said, and happen to give the right answer for Compton scattering, but they are free states, not bound states.

RS6 said:
The real energy state of the particles could differ significantly from an eigenstate

I don't know what you mean by "the real energy state". Either a state is an energy eigenstate, or it isn't. There are no "energy states" in any more general sense.

RS6 said:
The in the lab measured energies are a superposition of different oscillators

No, they aren't; they're just real numbers, like all measurement results. The expectation value of energy--the real number we compute from the theory and compare with the experiment result--would be derived from the quantum field state we used in our model, which in the general case will be a superposition of energy eigenstates (oscillators). But in the simplified model that is used to make predictions for Compton scattering, simple energy eigenstates are used. Actually, what we derive from the model is a distribution of the probability of different measurement results for the photon energy; we then compare that distribution with the distribution of the actual results from lots of runs of the experiment. The probability for each individual measurement result for energy is derived by summing Feynman diagrams whose input and output photon states are energy eigenstates (the known energy expectation value of the photons from the source for input, and the measurement result for energy whose probability we are calculating for output).
 
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  • #34
PeterDonis said:
RS6: We can address the phenomenon (e.g. scattering) to eigenstates in a mathematical convenient sense;
2) The results are consistent with the lab observations
With the particular lab observations we are talking about (Compton scattering), yes. But the results from this model are not consistent with other lab observations we could make using photons from the same sources.
Ok, you wrote this before.

PeterDonis said:
RS6: we don't know what are the real states of the involved particles (photon and electron)
Yes, we do; they're in coherent states (at least the photons are). More precisely, we can verify that the photons from the sources we use are in coherent states by making other observations (besides Compton scattering) on photons from those sources.
Yes, yes. It was my big mistake, sorry.
Coherent states.

PeterDonis said:
The electrons, since they are bound in atoms in Compton scattering experiments, are in whatever bound states correspond to the atomic orbitals they occupy. Those are not the same as the electron states used in the mathematical analysis of Compton scattering; the latter states are simpler mathematically, as I said, and happen to give the right answer for Compton scattering, but they are free states, not bound states.
In the analysis we suppose that they are free, because they are weakly bound, at least compared to the radiation energy (X or gamma, i think). This big difference makes that the end theoretical results do not differ from the observed reality. Right?

PeterDonis said:
RS6: The real energy state of the particles could differ significantly from an eigenstate
I don't know what you mean by "the real energy state". Either a state is an energy eigenstate, or it isn't. There are no "energy states" in any more general sense.
I don't understand what I wrote, either. :-)
The sentence cointains one "energy" too many. This is what I meant: The real state of the particles could differ significantly from an eigenstate.
The answer should be in agreement with my previous point 5): yes, because it's a coherent state and there are many oscillators, not one.
Right?

PeterDonis said:
RS6: The in the lab measured energies are a superposition of different oscillators
No, they aren't; they're just real numbers, like all measurement results.
Ok. I wanted to say that the lab values are the result of the presence of different oscillators.

PeterDonis said:
The expectation value of energy--the real number we compute from the theory and compare with the experiment result--would be derived from the quantum field state we used in our model, which in the general case will be a superposition of energy eigenstates (oscillators).
Ok. This confirms what I just wrote. More precisely, "expectation value" should mean:
1) The energy values we expect to detect, but which are not necessarely detected in each test;
2) As it's about coherent states and not eigenvalues, the energy is note defined.
Right?

PeterDonis said:
But in the simplified model that is used to make predictions for Compton scattering, simple energy eigenstates are used. Actually, what we derive from the model is a distribution of the probability of different measurement results for the photon energy; we then compare that distribution with the distribution of the actual results from lots of runs of the experiment.
Ok. Clear recap of the above. Thanks.

PeterDonis said:
The probability for each individual measurement result for energy is derived by summing Feynman diagrams whose input and output photon states are energy eigenstates (the known energy expectation value of the photons from the source for input, and the measurement result for energy whose probability we are calculating for output).
Hm, I have to consider this better, before I can say I really got it. May be tomorrow...
Anyhow: Eigenstates for both, input and output, in the calculation (S-Matrix?), because we can use simpler eigenstates in place of coherent states, as you explained. Right?
But, at my level, I'm very unsure on "measurement result whose probability we are calculating". Does this mean that we use experimental output data to obtain how the probability distribution of these output data looks? I'm confused.
 
  • #35
RS6 said:
In the analysis we suppose that they are free, because they are weakly bound, at least compared to the radiation energy

That's how I understand the reason for using the free particle approximation, yes.

RS6 said:
This is what I meant: The real state of the particles could differ significantly from an eigenstate.

Not just "could", "does" differ significantly from an energy eigenstate, yes.

RS6 said:
yes, because it's a coherent state and there are many oscillators

There are many oscillators that are excited. All of the oscillators are always "there", they just aren't always excited.

RS6 said:
I wanted to say that the lab values are the result of the presence of different oscillators.

The result of many different oscillators being excited, yes.

RS6 said:
Eigenstates for both, input and output, in the calculation (S-Matrix?)

It's not really an S matrix calculation, but it's similar to one.

RS6 said:
because we can use simpler eigenstates in place of coherent states, as you explained. Right?

Yes. More precisely, when we are using Feynman diagrams, we are using energy eigenstates (more precisely, energy-momentum eigenstates); at least that's how path integrals are usually done, as integrals in momentum space.

RS6 said:
I'm very unsure on "measurement result whose probability we are calculating". Does this mean that we use experimental output data to obtain how the probability distribution of these output data looks?

No. It means that we calculate, from the theory, the probability of measuring a whole range of different output energies for the photon (in Compton scattering), for a given input energy. Then we run the experiment many, many times with the same input energy for the photon (more precisely, with the source set up the same way, to provide photons with the same expectation value of energy), and record the output energy each time, and then compare the probability distribution of the actual output energies with the distribution we calculated from the theory.
 
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  • #36
PeterDonis said:
Yes. More precisely, when we are using Feynman diagrams, we are using energy eigenstates (more precisely, energy-momentum eigenstates); at least that's how path integrals are usually done, as integrals in momentum space.
Ok.
Bearing in mind, that in QFT path integrals (which relates to the minimum action principle, as far as I remember) are not exactly the same as in non-relativistic contexts. Am I wrong?

PeterDonis said:
No. It means that we calculate, from the theory, the probability of measuring a whole range of different output energies for the photon (in Compton scattering), for a given input energy. Then we run the experiment many, many times with the same input energy for the photon (more precisely, with the source set up the same way, to provide photons with the same expectation value of energy), and record the output energy each time, and then compare the probability distribution of the actual output energies with the distribution we calculated from the theory.
Aha. So it looks coherent for my English understanding.
Thanks a lot.

I think that I'm reaching to a much better understanding of the scattering. I feel I am close to the end of my newbie's questions set.
Unfortunally, I'm still convinced of the meaning of h, that I described some posts ago.
 
  • #37
RS6 said:
Of course. Anyhow, my so colled "intuitions" are just trials, which should be an underlaying premise, but which are possibly misunderstood. In this case: my fault.

There's no need to assign "faults" is my point. Just ask questions and we'll be happy to help. Broken English is the language of science and it is spoken by all scientists, except possibly for the British and American ones ;)

RS6 said:
I'm not convinced about this. But, I'm not sure I got correctly what you wrote. What have the units of measure to do in this context? How do they relate to "intuition"?
Anyway, the "the minimum energy that can be carried at a given frequency" does not mean that a photon at that frequency could carry also other energy amounts. The "by a photon" you wrote was not in my original sentence.

The point here is that physicists like to use what are called 'natural units'. For example, if you measure distance in light years, the speed of light is 1 light year per year. When you look at it this way, it becomes almost silly to give a different name to the unit of distance, so you can just as well measure distance in "years", and say that the speed of light is 1 year/year, also known as "1". This is because the speed of light is just a conversion factor between meters and seconds, units we invented because they are convenient on a human scale, and the scales that are actually relevant for natural phenomena.

There are other constants like that, such as Planck's constant, the gravitational constant G, and the Boltzmann constant. You can just as well work in units where Planck's constant is 1, and particle physicists do. In the end you're left with only one unit to keep track of, which we typically choose to be energy. So for example we can measure masses in joules (related to kg via E=mc²), distances in 1/joules, electric fields in (joules)², and so on. For convenience we typically use some other unit more relevant for particle physics, such as mega electron-volts. Then the "size" of a proton, 10^(-15) m, can be written as 1/(180 MeV). The particle that is responsible for binding protons together is called the pion, and it has a mass of ~140 MeV. Interesting, huh?

Anyway, this type of system of units gets called "natural units", because it leaves most human prejudices aside. In natural units, hbar is 1, and the energy of the photon is E=f. Does it then make sense to attribute the meaning you gave to hbar, as a "unit of energy"? We can see that it doesn't: it's role is to translate between the units that nature likes and common units such as joules and seconds, that humans like.
 
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  • #38
RS6 said:
Ok.
Bearing in mind, that in QFT path integrals (which relates to the minimum action principle, as far as I remember) are not exactly the same as in non-relativistic contexts. Am I wrong?

Yes and no. The relationship between quantum mechanical "path integrals" and the functional integrals used in field theory is a bit complicated. Superficially, they look quite different: in quantum mechanics you are integrating over literal particle trajectories and calculating how they interfere. In quantum field theory, you are integrating over entire field configurations. It seems very different, but it turns out that the functional integral for a quantum field theory with 0 dimensions of space and 1 of time is mathematically identical to an ordinary quantum mechanical path integral. Furthermore, there is a device (commonly attributed to Schwinger but it has been used earlier) that allows one to express a quantum field theory functional integral as an ordinary quantum mechanical path integral in a space with one extra dimension.

I advise you not to think too hard about this at this moment. You can safely think of the quantum field theory functional integrals as a generalization of the path integrals of quantum mechanics. Same principle, adapted to a slightly different problem.
 
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  • #39
Going through the following references will give you a good idea how QFT works for the basic scattering problems. The math is a bit tedious for my old brain, but it should not be bad for you assuming you are young.:biggrin:

http://rjs.phys.uvic.ca/sites/rjs.phys.uvic.ca/files/lec9.pdf

http://www.personal.soton.ac.uk/ab1.../christmas_problems/2014/XmasProb_DMillar.pdf
http://www.personal.soton.ac.uk/ab1...stmas_problems/2014/xmas_problem_solution.pdf
http://isites.harvard.edu/fs/docs/icb.topic624321.files/12-QED-trees.pdf
 
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  • #40
LeandroMdO said:
The point here is that physicists like to use what are called 'natural units'. For example, if you measure distance in light years, the speed of light is 1 light year per year. When you look at it this way, it becomes almost silly to give a different name to the unit of distance, so you can just as well measure distance in "years", and say that the speed of light is 1 year/year, also known as "1". This is because the speed of light is just a conversion factor between meters and seconds, units we invented because they are convenient on a human scale, and the scales that are actually relevant for natural phenomena.

There are other constants like that, such as Planck's constant, the gravitational constant G, and the Boltzmann constant. You can just as well work in units where Planck's constant is 1, and particle physicists do. In the end you're left with only one unit to keep track of, which we typically choose to be energy. So for example we can measure masses in joules (related to kg via E=mc²), distances in 1/joules, electric fields in (joules)², and so on. For convenience we typically use some other unit more relevant for particle physics, such as mega electron-volts. Then the "size" of a proton, 10^(-15) m, can be written as 1/(180 MeV). The particle that is responsible for binding protons together is called the pion, and it has a mass of ~140 MeV. Interesting, huh?

Anyway, this type of system of units gets called "natural units", because it leaves most human prejudices aside. In natural units, hbar is 1, and the energy of the photon is E=f. Does it then make sense to attribute the meaning you gave to hbar, as a "unit of energy"? We can see that it doesn't: it's role is to translate between the units that nature likes and common units such as joules and seconds, that humans like.

This is a nice post! It's in my opinion a harmonic combination of logic, clarity and elegance. My compliments, for what they are worth.
I see the point, but probably not completely. I should like to explain why...

Fore sure, the unitary magnitude for hbar makes it more manageable in the calculations, but this is a technical consideration, not something about principles, meanings or even intuitions.
Instead, I have to confess that putting hbar=1, i.e adimensional, is to me a bit strange (*).
On the other hand, I have no difficulty to accept E=f, once the convention is established.
This indicates that in the world we are dealing with, every thing which shows a frequency is also expressing an energy and vice versa. And there is no room for errors, I guess, since there are after all only frequencies, i.e. exciteable field oscillators. But this doesn't shift my point of view.

The Planck's constant leads me back to the historical experiment with the blackbody emission and to the regularly avoided "Ultraviolet Catastrophe" (Rayleigh formula, classic electromagnetic eigenoscillations in a cavity). The equation E = n h f was then a ruse invented by Planck to describe the non divergent radiation behaviour, presuming a discrete behaviour. But as for the physical interpretation? Planck didn't give any. The constant remained just an arithmetic device.

The constant h is an action, the same type of physical magnitude that relates to Hamilton law. But what does this mean? I think here at Einstein, who was the first to give an answer, studying the photoelectric effect and taking account of Planck's arithmetic adjustment. The "escape energy" needed by the electrons to leave the metal illustrated better that discrete behaviour already seen but not really understood by Planck. Einstein postulated the existence of light's quanta, photons. The equations for the photon's energy and momentum are then a known consequence of the relativity (c= cost; E = mc2; rest mass = 0; and so on).

So, a photon implies a precise amount of action. A photon can't be divided. It's a physical entity all-or-nothing. In this sense, it's the smallest action that can be carried in space-time from a source to an interacting point. This is my interpretation. Since the word "action" is may be to classic, in this discussion I expressed the above in terms of energy, saying a quantum is the smallest energy amount that can be carried from a source to an interacting point in space-time "at a given frequency".

This doesn't rule out the fact that a photon can then exchange only a portion of the energy transportated to the interaction point, as in the Compton effect. Of course, it doesn't rule out the fact either that a photon can have any frequency value, so that energy itself can vary continuosly.

It's evident that I have a too little expertise in QM to try to depict drastic frameworks for this discipline (as it was written in relation to some of my posts, what is understandable). No way. I'm only referring to action and energy, not to perturbations, probabilities, paths. So, I'm not at all trying to reduce the QM to Planck's constant, even if it was the value through which a new world was discovered.

(*) Personal consideration: I guess this is only because I pretend to keep to myself a bridge that connects hbar to my day by day world of seconds, meters, kilojoules, kilograms. I'm born in a dimension where time, space and matter are primitive concepts and energy a derivate physical unit.
I know of course that my ordinary world is a consequence of quanta dynamics e not vice versa. But my understanding goes top-down and not bottom-up. This means, I have to reduce my macroscopic experience to the microcosm, before trying to reconstruct back the first on the basis of the latter. I don't see prejudices in this, but it's for sure more tiring than the one-way interpretation bottom-up. Tiring, but more intuitive. Ok, sorry for the detour...
 
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  • #41
LeandroMdO said:
Yes and no. The relationship between quantum mechanical "path integrals" and the functional integrals used in field theory is a bit complicated. Superficially, they look quite different: in quantum mechanics you are integrating over literal particle trajectories and calculating how they interfere. In quantum field theory, you are integrating over entire field configurations. It seems very different, but it turns out that the functional integral for a quantum field theory with 0 dimensions of space and 1 of time is mathematically identical to an ordinary quantum mechanical path integral. Furthermore, there is a device (commonly attributed to Schwinger but it has been used earlier) that allows one to express a quantum field theory functional integral as an ordinary quantum mechanical path integral in a space with one extra dimension.

I advise you not to think too hard about this at this moment. You can safely think of the quantum field theory functional integrals as a generalization of the path integrals of quantum mechanics. Same principle, adapted to a slightly different problem.

Got it.
:-)
 
  • #42
ftr said:
Going through the following references will give you a good idea how QFT works for the basic scattering problems. The math is a bit tedious for my old brain, but it should not be bad for you assuming you are young.:biggrin:

http://rjs.phys.uvic.ca/sites/rjs.phys.uvic.ca/files/lec9.pdf

http://www.personal.soton.ac.uk/ab1.../christmas_problems/2014/XmasProb_DMillar.pdf
http://www.personal.soton.ac.uk/ab1...stmas_problems/2014/xmas_problem_solution.pdf
http://isites.harvard.edu/fs/docs/icb.topic624321.files/12-QED-trees.pdf

Wrong assumption.
Haha.
Anyhow, "old" is a question of relativity.
:-)
 
  • #43
In accordance with what a wrote before as an attempt, I could also say that Planck constant corresponds to the expected time interval requested by a photon (which travels at velocity c) to pass through an ideal surface which is set perpendicular to the quantum path. I always mean before interacting.
But, I don't state this, because I'm not sure about this, nor I know if the concept of "perpendicular to the photon path" or "photon lenght" make sense in QFT.
I don't know either if a probability wave function applies to photons positions. I don't think so.
Hm...
 
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  • #44
RS6 said:
I don't state this, because I'm not sure about this,

That's good, because what you said doesn't make sense.

RS6 said:
nor I know if the concept of "perpendicular to the photon path" or "photon lenght" make sense in QFT

Not really.

RS6 said:
don't know either if a probability wave function applies to photons positions

Photons don't have well-defined positions in QFT.

Please, before making any further speculations, you need to work through a textbook on QFT. PF rules do not allow personal speculations, and you are on the verge of that now.
 
  • #45
You are right and I understand the rules. I will not make any further speculations, like the one in post #43. It was just an attempt to see were I was wrong, knowing I was possibly wrong. Without outcome.
Basing on your feedback, I now only know that it doesn't make sense to think to Planck's constant as to the time interval requested by a photon (which travels at velocity c) to pass through an ideal surface which is set perpendicular to the quantum path.
Ok.
But, as for the rest on Planck's constant, it's not intended as a speculation at all.
Thanks for answering.
 
  • #46
RS6 said:
as for the rest on Planck's constant, it's not intended as a speculation at all

I know you don't intend it as speculation, but I think that you will have better luck improving your understanding if, instead of continuing to try to come up with questions like "could Planck's constant mean this?", you don't even try to figure out "what Planck's constant means", you just study quantum mechanics and QFT in general. The same goes for trying to figure out what there is a "minimum quantum amount" of (energy, action, ??). Trying to answer that question is not, IMO, a good way to improve your understanding. It just leads you down one false path after another.
 
  • #47
Planck's constant is just a conversion factor adapting the arbitrary men-made units of the SI to natural units. The same holds true for ##c##, the speed of light in the vacuum or (the most superfluous of such constants at all) ##\epsilon_0## and ##\mu_0##.
 
  • #48
@vanhees71 Yes, I see.
But this is a consequence of the existence itself of the universal constant h.
If h was zero, the classical physics would apply to sub-nanometric world as well. If it was instead very big, our macroscopic dimension would behave in a discrete way like the quantum systems.
Another consequence of h is that it draws a line upon what we can know about a particle.
And so on.
Consequences suggest something, but before knowing what they suggest, they are only consequences.

I have no problem in realizing in how many situations h plays a crucial role and how it's possible thanks to quantum equations involving it to make quiet good (probabilistic) predictions. It's just that equations and predictions are one thing, physical meanings another one. I don't want to look in a black box. This is not the point. I just would like to grasp a fundamental meaning, let's say a barycentric role.
It seems, there isn't any.

As for natural units, I understand they are an elegant way to connect different aspects and to simplify the handling of problems, minimizing the number of needed physical constants to describe phenomena.
Anyway, when dimensionsful constants are reduced to dimensionsless ones I see a threat for people like me, which couldn't understand to where and how far the reduction works. It's of course my problem, but may be the concern is legitimate also in another sense.

E.g.:
http://file.scirp.org/Html/9-7501694_45033.htm

In relation to this and to the physical meaning of h, I found that the "Planck's lenght", for instance, is the "smallest possible unit of measurement". But the consideration refers to measurement of nature, not only to nature itself.
By the way, investigating on this natural unit, I found this (discrete fractal paradigm, not too long). It's tricky for me:

http://www3.amherst.edu/~rloldershaw/newdevyear/2008/March.htm
 
  • #49
RS6 said:
In relation to this and to the physical meaning of h, I found that the "Planck's lenght", for instance, is the "smallest possible unit of measurement"

This is not known physics, it's just speculation.

RS6 said:
By the way, investigating on this natural unit, I found this (discrete fractal paradigm, not too long).

This is not a valid source, it's just somebody's personal page (and from what I can tell on a quick skim, that person is probably a crackpot).

As I've said many times now: learn QM and QFT from textbooks.
 
  • #50
PeterDonis said:
Photons don't have well-defined positions in QFT.

It seems anyhow, their size can be sometimes calculated.
So, if they move at c, it should be possible to evaluate how long they take to pass to an ideal perpendicular plane.

https://arxiv.org/pdf/1604.03869.pdf
 
  • #51
RS6 said:

This paper is correct that you cannot consistently set all three of ##c##, ##h##, and ##G## to 1; you have to either set ##c = G = 1## (as is done in GR), or ##c = h = 1## (as is done in QFT).

What the paper does not do is support its claim that there is a "considerable body of work in theoretical physics" that does in fact claim to set ##c = h = G = 1##, all three at the same time. I've never seen this done. None of the GR textbooks or papers I've seen set ##h = 1## (most of them don't use any equations where ##h## appears at all; the ones that do don't set it to 1). None of the QFT textbooks or papers I've seen set ##G = 1## (most of them talk explicitly about the fact that you can't set ##G = 1## in the context of QFT, you have to give it its proper units in the "natural" units of QFT, which are inverse mass squared).
 
  • #52
RS6 said:
It seems anyhow, their size can be sometimes calculated.

The paper you link to does not appear to be peer-reviewed, and at least some of the things it says, on a quick skim, appear to contradict established theory. I would be very cautious about trying to draw conclusions from it.

The fact that there is no valid position operator for photons is a standard result in QFT.

Do I have to repeat, again, that you should be looking at textbooks? Particularly before you try to wade into advanced technical papers?
 
  • #53
RS6 said:
I just would like to grasp a fundamental meaning, let's say a barycentric role.
It seems, there isn't any.

If you take the time to learn the standard theory from textbooks, you will be able to answer this for yourself. At this point, there's nothing more that can usefully be said here.

This thread is closed.
 

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