What happens if photon propagator goes on shell?

In summary, the conversation discusses the forward Compton scattering process and the issue of the red photon becoming a propagator with zero momentum, leading to a divergence in the cross section. It is suggested to regulate this divergence by considering a cutoff for detecting soft photons. This is discussed in QFT textbooks such as Peskin-Schröder in chapter 6. The red photon's lack of definite momentum is also compared to understanding soft or collinear divergences.
  • #1
Chenkb
41
1
I am thinking about a problem. Consider the forward Compton scattering process e(p)+γ(k) -> e(p)+γ(k), as shown in the following figure.
If we consider the initial red photon is emitted by another electron which then goes to anything, then how can we write down the whole amplitude for this process?
The problem troubles me is that the red photon becomes a propagator with k^2 = p^2 = 0, which is divergent if in the denominator!

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  • #2
The photon will not be on-shell for anything where you can measure a momentum difference between the particles it attach to at the top.

For an actual cross section computation you have to put a cutoff on how soft photons you can detect and the divergence in the cross section due to the exchange of zero-momentum photons will be regulated by the emission of soft photons. This occurs at the cross section level, not at the amplitude level.

Formally, if you consider a series of detectors that each can measure lower and lower energy soft photons, the cross section diverges. This should be discussed in any QFT textbook, e.g., the discussion in Peskin-Schröder is in chapter 6.
 
  • #3
Orodruin said:
The photon will not be on-shell for anything where you can measure a momentum difference between the particles it attach to at the top.

For an actual cross section computation you have to put a cutoff on how soft photons you can detect and the divergence in the cross section due to the exchange of zero-momentum photons will be regulated by the emission of soft photons. This occurs at the cross section level, not at the amplitude level.

Formally, if you consider a series of detectors that each can measure lower and lower energy soft photons, the cross section diverges. This should be discussed in any QFT textbook, e.g., the discussion in Peskin-Schröder is in chapter 6.
Thanks a lot. This reminds me that the red photon no-longer has a definite momentum, and the situation is similar to the case of understanding soft or collinear divergences.
 

Related to What happens if photon propagator goes on shell?

1. What is a photon propagator?

A photon propagator is a mathematical expression that describes the probability amplitude for a photon to travel from one point to another in space and time. It is used in quantum field theory to calculate the interaction between particles and their corresponding fields.

2. What does it mean for a photon propagator to go on shell?

When a photon propagator goes on shell, it means that the energy and momentum of the photon are equal to its mass and velocity, respectively. This is a condition that is necessary for the photon to be a physical particle and to obey the laws of conservation of energy and momentum.

3. What happens if a photon propagator does not go on shell?

If a photon propagator does not go on shell, it means that the energy and momentum of the photon do not satisfy the condition of being equal to its mass and velocity. This can lead to unphysical results in calculations and is not consistent with the laws of physics.

4. Why is it important for a photon propagator to go on shell?

It is important for a photon propagator to go on shell because it ensures that the photon is a physical particle with well-defined properties. This allows for accurate calculations and predictions in quantum field theory, which is crucial for understanding the behavior of particles and their interactions.

5. How is the on-shell condition of a photon propagator related to the speed of light?

The on-shell condition of a photon propagator is related to the speed of light because it is derived from the relativistic energy-momentum relation, E^2 = p^2c^2, where E is the energy, p is the momentum, and c is the speed of light. This means that for a photon to be on shell, its energy and momentum must be consistent with the speed of light, which is a fundamental constant in physics.

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