CP violation and phase

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  • #1
Utg
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Hello! I have heard many times that CP violation in the SM is due to the presence of phase in the CKM matrix. Can somebody explain how phase is related to CP violation?

Thanks!
 

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  • #2
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If you exchange C and P all the amplitudes of individual processes stay the same. The probability is linked to the squared amplitude. So how do you get an asymmetry at all? The only way to do this is via phases: Some processes keep their phase under CP (EM/strong interaction), some get a different phase (weak interaction). If you have at least two processes that contribute to the overall amplitude, and their phases transform differently under CP, you can get an actual asymmetry.
 
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  • #3
Utg
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If you exchange C and P all the amplitudes of individual processes stay the same. The probability is linked to the squared amplitude. So how do you get an asymmetry at all? The only way to do this is via phases: Some processes keep their phase under CP (EM/strong interaction), some get a different phase (weak interaction). If you have at least two processes that contribute to the overall amplitude, and their phases transform differently under CP, you can get an actual asymmetry.
Thank you for your response. I think my problem is that I don't understand why EM/strong interactions keep phase under CP and weak interaction not. How can I see this? I understand, for example, that P changes left fields to right fields and vice versa. How does C acts on Lagrangian? fields?
 
  • #4
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C changes particles with antiparticles.
 
  • #5
Utg
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C changes particles with antiparticles.
Does it mean that ##\bar{\psi}## changes into ##\psi## and vise versa?
 
  • #6
ChrisVer
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I think, the [itex]\psi[/itex] is generally denoting a Dirac spinor, and as such it contains both the particle and antiparticle states. So, you cannot just say that [itex]\bar{\psi}[/itex] becomes [itex]\psi[/itex]. Basically the [itex]\psi[/itex] will some sort change to [itex]\psi^*[/itex] because you want to flip its charge, which in the local gauge of the U(1) is given by a transformation of the form: [itex]\psi \rightarrow e^{-iqa(x)} \psi[/itex] (so you need to change the sign of [itex]q[/itex]).
Of course it will be a little more complicated and for that you need to look at what operation the charge conjugation does on your field.
 
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