B Is Z0 Particle Real Without Flavor-Changing Interactions?

AdvaitDhingra
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If there are no flavor changing z0 weak interactions, how do we even know that the particle exists? I thought that we could only tell which particle was exchanged by the particles it decays into. Is this wrong?
 
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How would the absence of flavour changing Z0 interactions preclude the knowledge of its existence? You will still see the decays to particle-antiparticle pairs just the same. The cleanest way of producing Z0 is to collide electrons with positrons at the Z0 resonance.
 
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Orodruin said:
How would the absence of flavour changing Z0 interactions preclude the knowledge of its existence? You will still see the decays to particle-antiparticle pairs just the same. The cleanest way of producing Z0 is to collide electrons with positrons at the Z0 resonance.
Oh ok. So a z0 Boson interaction is one where charge is conserved? (since the z0 carries no charge)
 
Do you see a peak (Z) or not (no Z)?

1618312813823.png
 
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Charge is always conserved.

Flavor-changing Z interactions would allow e.g. electron plus antimuon to Z, or Z to these two particles. We don't observe that decay (experimental upper limit is ~10-6), but of course we see the flavor-conserving electron+positron->Z production (that's what we can actually collide in colliders) and the equally flavor-conserving decays to quark plus matching antiquark.
 
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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}...

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