Are unstable particles ever really external lines in QFT?

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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?
 

Discussion

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.