Compton scattering in a Feynman diagram

In summary, the concept of Compton scattering involves the creation and destruction of virtual particles, which exist in a non-measurable time interval and do not have the same properties as macroscopic particles. The distinction between "real" and "virtual" particles can be technical and depends on whether they are on or off mass shell. The diagrams used to represent these interactions are just pictorial representations and do not necessarily correspond to physical processes. The issue of causation and the existence of continuous time also come into play when discussing this topic.
  • #1
nomadreid
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Attempting to understand the following: Compton scattering can happen either
(a) an electron could absorb a photon and later emit a photon, or
(b) an electron could emit a photon and later absorb a photon.
OK, the maths works out, but I am trying to get intuition on track. These two are respectively equivalent to (with apologies to those who don't like the expression "virtual"):
(a*)
(t1) a real electron and a real photon are destroyed and a virtual electron is created
(t2) the virtual electron is destroyed and a real photon and a real electron are created
(b*)
(t1) a real electron is destroyed and a virtual electron and a real photon are created,
(t2) the virtual electron and the real photon are destroyed and a real electron is created.

where t1<t2 (moments in time). The fact that we cannot measure the interval (t1,t2) means that we can't really say what goes on during this time interval, so the paradox that
in (a*) during the interval (t1,t2) nothing real exists, and
in (b*) during the interval (t1,t2) two real photons exist
doesn't bother. So far, so good?
There is the side issue as to how the idea of the existence of continuous time jibes with the idea that small intervals of time don't exist as a real entity. I guess "exists" is taken in two different senses here, right?
 
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  • #2
Feynman diagrams are just pictorial representations of terms in a perturbation series. EM fields interact with the electron field, always. The diagrams for Compton scattering focus on those interactions which contribute to the desired amplitude which is defined by initial and final states. One of those interactions seems out of causal order but none the less exists in the formal series. Perhaps you are reading too much into the pictures? Just a thought.
 
  • #3
You may have a point that it may be too much to ask to have each term in the perturbation series correspond to some physical process. In any case, I lifted my rendition of (a), (b),(a*), (b*) almost verbatim from page 225 of "Student Friendly Quantum Field Theory (Second Edition): Basic Principles and Quantum Electrodynamics" by Robert D. Klauber. But in this formulation using annihilation and creation operators, I do not see where causation might seem to be violated: the sequence of events seems OK. My main question rather concerns the "black box" nature of the non-measurable interval of time. If this part is indeed not measurable, how can one talk about real photons existing during that time, as we are doing in (b*)? (The secondary question about the possible graininess of time is an old one, and so I suppose there is no resolution to that conundrum until a successful quantum theory of gravity comes along.)
 
  • #4
nomadreid said:
how can one talk about real photons existing during that time,

Well, I would suggest we can't. Quantum Mechanics plays havoc with words like "exist" when used in household settings. IMO photons don't exist in the same sense that say bullets or macroscopic particles exist. Personally, I'm happy with this world view. Virtual particles "exist" but only as terms in an operator series expansion. At least, that's how I view it FWIW.
 
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  • #5
Thanks, Paul Coby. That view of virtual particles is sort of what I meant by the conjunction of the two:
nomadreid said:
(with apologies to those who don't like the expression "virtual")
nomadreid said:
"exists" is taken in two different senses
I am not sure, however, where the boundary is for the distinction between real and virtual for:
(a) photons, electrons, quarks etc. and
(b) collections of these
 
  • #6
"real" and "virtual" I think are technical terms for on and off mass shell. What I was getting at with "exist" is that macroscopic particles have properties that photons don't. One can quickly be lead to false conclusions and arguments if these things are forgotten. Much head scratching and confusion occurs due to this. For example, a bullet fired at two detectors, A and B, hitting one, either hits A and not B or hits B and not A. For photons from the vast majority of light sources, this is easily shown to be false by looking at the correlation between A and B counts. So, photons don't exist in the same way as bullets.
 
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  • #7
Granted, at least most of the bullet hits A and not B or vice-versa (and we don't notice the parts that don't follow that pattern). The question is whether the boundary is one of quality or of quantity (quantity becoming quality?). That said, I am thus not sure whether one would say that the situation in the time interval (t1,t2) shouldn't as a whole be described as off-shell, so that during this time interval (and, given the continuity of time, the "during" should have a sense) the photons would also be virtual (as part of a virtual situation, i.e., off shell), and only become "real" (in the sense of being part of a real situation obeying the appropriate equations) outside of this time interval.
 
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  • #8
Well, the intent of the example I gave is to underscore that real on mass shell photons don't have a trajectory before the fact. If they did then the either or clause of the bullet argument would be valid for them and the counts A and B would show an anti-correlation. In real measurements with light, these counts are uncorrelated. The reason bullet counts are correlated is because they have a well defined trajectory and only one. So it's unclear what one is talking about on the internal lines of a Compton scattering event. So, in this sense, does your question have meaning?

Now, all external lines in a Feynman diagram are on mass shell by definition while all internal lines are off mass shell. A single isolated vertex which depicts a particle which spontaneously belching a photon, will integrate to 0 when all lines are external or on mass shell. It has to be this way to conserve energy and momentum.
 
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  • #9
OK, thanks for your guidance, Paul Coby. I will try to look at the scattering diagrams with these concepts in mind.
 

1. What is Compton scattering in a Feynman diagram?

Compton scattering is a type of interaction between a photon and a charged particle, where the photon transfers energy to the particle, resulting in a change in its direction and wavelength. In a Feynman diagram, Compton scattering is represented as an interaction between an incoming photon and an outgoing electron or positron.

2. How does Compton scattering occur in a Feynman diagram?

In a Feynman diagram, Compton scattering occurs when a photon (represented by a wavy line) interacts with a charged particle (represented by a solid line), such as an electron or positron. The photon transfers energy to the particle, causing it to change direction and emit a new photon (represented by a wavy line).

3. What is the significance of Compton scattering in a Feynman diagram?

Compton scattering is an important process in particle physics as it helps to explain the interaction between photons and charged particles. It also plays a crucial role in understanding the behavior of electromagnetic radiation, and has practical applications in fields such as medical imaging and materials science.

4. How is Compton scattering in a Feynman diagram related to the Compton effect?

The Compton effect is a physical phenomenon that describes the increase in wavelength of a photon after it interacts with a charged particle. In a Feynman diagram, this effect is represented by the transfer of energy from the photon to the charged particle, resulting in a change in wavelength.

5. Can Compton scattering occur in reverse in a Feynman diagram?

Yes, Compton scattering can occur in reverse in a Feynman diagram, where an incoming electron or positron interacts with an outgoing photon. This process is known as inverse Compton scattering and is commonly observed in astrophysical environments.

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