Are unstable particles ever really external lines in QFT?

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Discussion Overview

The discussion revolves around the representation of unstable particles, specifically the Z boson, in Feynman diagrams within quantum field theory (QFT). Participants explore whether such particles can be accurately depicted as external lines or if they should always be treated as internal lines due to their unstable nature and decay processes.

Discussion Character

  • Debate/contested

Main Points Raised

  • One participant questions the appropriateness of drawing the Z boson as an external line in Feynman diagrams, arguing that since it cannot be directly detected and is not in a stable eigenstate, it may be more accurate to represent it as an internal line.
  • Another participant asserts that Feynman diagrams depict isolated particles coming from far away, suggesting that short-lived particles like the Z boson do not fit this description.
  • A third participant introduces the narrow-width approximation, explaining how it allows for the simplification of calculations by treating the unstable particle as on-shell, but notes that this approach sacrifices certain details such as spin correlations and finite-width effects.
  • A later reply expresses appreciation for the clarification regarding the use of the narrow-width approximation in breaking down processes involving W or Z bosons, indicating a lack of understanding of why single diagrams are not used.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether unstable particles can be represented as external lines in Feynman diagrams. There are competing views on the implications of using the narrow-width approximation and the nature of unstable particles in QFT.

Contextual Notes

The discussion highlights limitations related to the treatment of unstable particles and the assumptions underlying the narrow-width approximation, including the potential loss of certain physical effects in calculations.

bjj
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TL;DR
Since an unstable particle isn't an eigenstate of the Hamiltonian, and it will eventually decay, should it always be written as an internal line of a Feynman diagram?
Let's for example consider the Z boson. It can't directly be detected; so is it ever really correct to draw it as an external line on a Feynman diagram? I've seen processes involving it before be written as
something -> Z + something, then Z -> ...
but since unstable particles aren't really on their own eigenstates of the Hamiltonian, and they will eventually decay, is it more accurate to draw them always as an internal line?
 
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A Feynman diagram describes individual, isolated, particles coming from far away, meeting at the diagram, and then receding so that they are far away again.

A short-lived particle cannot come from far away.
 
The factorization in the way you described, initial state --> Z and then Z--> final state, is called the narrow-width approximation, in which you replace the Breit-Wigner shaped propagator of the unstable internal particle by a delta function
[tex]\frac{1}{(q^2-m^2)^2+m^2\Gamma^2}\to \frac{\pi}{m\Gamma}\delta(q^2-m^2)[/tex]

In this way the intermediate particle is set on-shell and the producton and the decay are completely factorized in the way mentioned above.
This makes calculations A LOT easier, but you are loosing spin correlations and finite-widht effects of order ##\mathcal{O}\bigl(\frac{\Gamma}{m}\bigr)##.

If this is OK for your calculation, the narrow-width approximation will usually be your method of choice.
If not, you have to do the full calculation, initial state --> final state, with your unstable particle as an internal line in your Feynman diagrams.
 
Thanks! This makes things easier for me to understand. I often see in papers them breaking a process up into scattering into W or Zs, then a W or Z decay, and I didn't understand why they would not just do a single diagram, but I suppose it's probably this approximation.
 

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