A Are unstable particles ever really external lines in QFT?

bjj
Messages
2
Reaction score
0
TL;DR 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?
 
Physics news on Phys.org
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
\frac{1}{(q^2-m^2)^2+m^2\Gamma^2}\to \frac{\pi}{m\Gamma}\delta(q^2-m^2)

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.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
4
Views
2K
Replies
6
Views
2K
Replies
11
Views
2K
Replies
5
Views
3K
Replies
10
Views
3K
Replies
1
Views
2K
Replies
6
Views
2K
Replies
6
Views
4K
Back
Top