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What is the antiparticle corresponding to $\nu_L$ ?

-According to this thread : What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation?

"The operation that maps particles to antiparticles is just C" (which looks suspicious)

it would be $\bar{\nu}_L$

-according to : Is $CP$ instead of $C$ responsible for changing a particle to its antiparticle?

it would be $\bar{\nu}_R$

So, I'm lost on what is correct.

Bonus question : let's suppose that the answer is : $\bar{\nu}_R$. Then : let's suppose two hypotheses.

*First hypothesis: nature has no CP violation in neutrino sector :

is $CP (\nu_L)$=$\bar{\nu}_R$ ?

*Second hypothesis : nature has CP violation in neutrino sector : is $CP (\nu_L)$=$\bar{\nu}_R$ meaning nothing because the neutrino would is not a CP eigenstate ?

Mathieu Krisztian
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  • I guess since you've linked to one of my answers, saying more on my part would just add to the confusion, but let me just give one piece of advice. – knzhou Jan 26 '20 at 05:34
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    When I was learning QFT, I was incredibly concerned on what words like "antiparticle" really meant, what was the true antiparticle, and so on. But at the end of the day these are just words with arbitrary definitions that differ from person to person. (Some textbooks are so sloppy on issues like antiparticles, the difference between particles and fields, the difference between C and CP, the difference between chirality and helicity, etc. that they're not even consistent with themselves.) But if you focus on concrete physical consequences, you can't get confused. – knzhou Jan 26 '20 at 05:35
  • thank you @knzhou. I'm interested in the concrete case of my example : what is the antiparticle corresponding to $\nu_L$, for which I see two different interpretations on the forum. – Mathieu Krisztian Jan 26 '20 at 05:37
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    Recall that relativistic quantum field theory requires that a field both annihilates and creates particles. If you take the neutrino field in the SM to annihilate a left-helicity neutrino, then it creates a right-helicity antineutrino. If you define "antiparticle" to mean "the thing that a quantum field which annihilates particles also creates", you could say that these are the antiparticles of each other. – knzhou Jan 26 '20 at 05:39
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    Alternatively, you could take "antiparticle" to mean "give me a particle with the same spacetime properties, like momentum and spin, but with the opposite internal quantum numbers". In that case, the left-helicity neutrino in the SM has no antiparticle, because it would be a right-helicity neutrino, which doesn't exist. – knzhou Jan 26 '20 at 05:40
  • @knzhou Thank you. But then, if there is CP violation in neutrino sector, does this makes sense that there are antiparticles to each other since CP does not commute ? – Mathieu Krisztian Jan 26 '20 at 05:41
  • "In that case, the left-helicity neutrino in the SM has no antiparticle, because it would be a right-helicity neutrino, which doesn't exist." : ok, but that is fine : I discuss "nature", not necessary the SM. Maybe right-helicity neutrino would exist in nature. We don't know yet. – Mathieu Krisztian Jan 26 '20 at 05:42
  • Right, I'm taking the SM to just mean the usual SM fields and no more. Sorry, I'm not sure what "does this makes sense that there are antiparticles to each other since CP does not commute" means. – knzhou Jan 26 '20 at 05:44
  • So let's assume that antiparticle for $\nu_L$ is $\bar{nu}_R$. The problem is that the theoretical definition of an antiparticle is (?) the state obtained by the application of the $CP$ operator. But if $CP$ would not be conserved in neutrino sector (if there would be CP violation), then would it makes sense to apply $CP$ operator on $\nu_L$ ? If this would not make sense, then could we really say that antiparticle for $\nu_L$ is $\bar{nu}_R$ ? – Mathieu Krisztian Jan 26 '20 at 05:51
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    I don't know what you mean. $P$ is not conserved in many interactions, but it still "makes sense" to apply $P$. And angular momentum is not conserved in a lot of laboratory settings, but it still makes sense to apply rotation operators. – knzhou Jan 26 '20 at 06:03

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