A Are unstable particles ever really external lines in QFT?

bjj

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Summary
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?
 
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.
 

bjj

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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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