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It's usually said that SM Lagrangian violates charge conjugation, and should be obvious from the fact that "only left handed fermions are charged under $SU(2)_{L}$ but left and right handed fermions are treated differently under $U(1)_{Y}$".

To me this statement is not very clear (I don't know very much yet about SM); as far as I know I have a symmetry iff there is an operator $C$ which leaves the SM Lagrangian invariant. Which is this operator? Can someone show me how it's done explicitly?

Qmechanic
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Filippo
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  • This might help, or, better yet, the discussion in your book the quote came from. – Cosmas Zachos Aug 10 '23 at 16:47
  • Do you have any trouble appreciating WP? Charged gauge bosons only couple to L fermions and R antifermions. So, since C does not affect their chirality, but only their charge, a C transformation does not lead to a viable/extant SM interaction: it violates C maximally! The hypercharge piece of the statement is overkill... – Cosmas Zachos Aug 10 '23 at 19:53
  • I've never studied SM, so yeah I'm struggling a bit. You say that C does not alter chirality (is it just the R/L nature of the particle right?) but only their charge (is this because of the minus sign?), so why is this not a good SM interaction? – Filippo Aug 11 '23 at 07:59

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You are asking for either an SM tutorial, or the prerequisite tutorial on the Dirac equation and its symmetries, impossible to summarize in a short answer, or the relevant WP articles. People normally take a course in this, and all the pieces of what you are asking about are in standard books, like the one of M Schwartz, Peskin & Schroeder, Li & Cheng, etc...

In any case, veeery crudely, the easiest mnemonic of the point is that the SM contains terms such as the one below, coupling the charged vector bosons to the electron and the positron, where, indeed, L,R are the left and right chiral projectors, $$ ig(W^+_\mu \overline{\nu} \gamma^\mu e^-_L +W^- \overline{\nu^c} \gamma^\mu e^+_R ), $$ but, most significantly, no such terms for the opposite chiralities! So, either a P or a C transformation would take such terms to the non-existent terms, so would take you completely outside the Standard Model; each of these symmetries would thus violate C and P maximally. Forget about the hypercharge, whose intricate action is not easy to master in a crude answer such as this.

$\phantom{137baronsamedi}$ Note it would appear that CP would preserve the term written, but it does not, even though it mostly does: there is a slight difference in the transformed neutrinos and antineutrinos (which I wisely did not discuss, and has to do with the flavor mixing among the 3 generations involved, so don't worry about it for now!), that shifts a few phases of the terms by a small amount... but it's a truly small effect.

Cosmas Zachos
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  • Did you forget a L(R) in the $\nu(\nu^{c})$?Cause I remember that the term should be something like $\bar{\psi}{L}\gamma{\mu}D^{\mu}\psi_{L}$. However thank you now the logic is clear. For the details I'll wait to take the course :) – Filippo Aug 11 '23 at 14:42
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    No, it's automatic. $\overline{\psi_L}\gamma_\mu\psi_L= \overline{\psi}R\gamma_\mu \psi_L = \overline{\psi} \gamma_\mu L \psi_L = \overline{\psi} \gamma_\mu \psi_L$, as you might learn in your course. – Cosmas Zachos Aug 11 '23 at 14:52