(Non-)mixing of K vector bosons

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Discussion Overview

The discussion revolves around the mixing behavior of vector bosons K(0*) and K-bar(0*) compared to the mixing of pseudo-scalar bosons K(0) and K-bar(0). Participants explore the implications of decay processes and selection rules related to these particles.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions why K(0*) and K-bar(0*) do not mix like K(0) and K-bar(0), suggesting a fundamental difference in their properties.
  • Another participant requests a calculation of the lifetime of K* in relation to the Kshort lifetime, implying a connection to the mixing behavior.
  • A different participant expresses uncertainty about performing the calculation and anticipates a selection rule might be relevant.
  • One participant proposes that the rapid decay of K* into Kπ via strong interaction could explain the lack of mixing.
  • A subsequent reply confirms the previous suggestion and asserts that the calculation can be done by looking up the relevant numbers and performing a division.

Areas of Agreement / Disagreement

Participants express differing views on the mixing of K(0*) and K-bar(0*), with some suggesting decay processes as a factor while others focus on the theoretical aspects. The discussion remains unresolved regarding the exact reasons for the observed behaviors.

Contextual Notes

The discussion includes assumptions about decay processes and potential selection rules, but these aspects are not fully explored or resolved within the thread.

Who May Find This Useful

Readers interested in particle physics, particularly those studying meson mixing and decay processes, may find this discussion relevant.

Smithf
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Why don't the vector bosons K(0*) and K-bar(0*) mix the way the pseudo-scalar bosons,
K(0) and K-bar(0) do?
 
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Please calculate the lifetime of the K* in units of the Kshort lifetime. Is the answer obvious now?
 
I don't know how to do the suggested calculation -- I was expecting some sort of selection rule to apply
 
Is the idea that the K* just decays too quickly into Kπ by a strong interaction?
 
Bingo.

And you do know how to do the calculation. Look up both numbers and divide.
 

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