It is said that a CP violation would mean that the behaviour of the particle is different from the behaviour of antiparticle. Why is C violation not good/enough?


The operation that maps particles to antiparticles is just $C$. (This is somewhat of a simplification. A better thing to say is that in theories with $C$ symmetry, you can pair particle states with the same spacetime quantum numbers but the opposite internal quantum numbers. When $C$ is violated, there may exist no pairing that gets the quantum numbers right. In extreme cases, there may not be any way to define a $C$-like operator at all, no matter how you modify the quantum numbers; an example is a theory with a single Weyl spinor. In such cases you can still define a pairing using $CP$, if it exists, or failing that using $CPT$, which always exists and is conserved, but these pairings don't have the familiar properties you would expect. For much much more, see here and here.)

Why do people focus on $CP$ violation? The issue is that $C$ violation is ubiquitous in the Standard Model; in fact, in a certain sense it is as strong as possible in the charged current weak interactions. However, there are interesting phenomena that require both $C$ violation and $CP$ violation. So since $CP$ violation is the hard part, we talk about it a lot more.

One key example is the creation of a matter/antimatter imbalance in baryogenesis. For simplicity, suppose that $C$ and $CP$ are both defined, though they may not be obeyed. For any particle states $i$ and $f$, there are four related processes: $$i \to f, \quad \bar{i} \to \bar{f}, \quad i_P \to f_P, \quad \bar{i}_P \to \bar{f}_P$$ where a bar denotes the antiparticle, defined by the action of $C$. If these processes have rates $a$, $b$, $c$, and $d$, and the states have different baryon number, then the rate of baryon number violation is proportional to $$a - b + c - d.$$ If $C$ symmetry is obeyed, then $a = b$ and $c = d$, giving a rate of zero. If $CP$ symmetry is obeyed, then $a = d$ and $b = c$, again giving a rate of zero. One needs both $C$ and $CP$ violation to get baryogenesis.

Unfortunately, in popular science these statements are sometimes oversimplified to just "$CP$ distinguishes matter from antimatter", which is confusing.

  • $\begingroup$ Your first paragraph could potentially be taken to suggest the converse: "if a theory is $C$-symmetric, then particles and antiparticles are the same". Perhaps some clarification of the word "distinct" would be helpful? $\endgroup$ – gj255 Jun 11 '18 at 10:04
  • $\begingroup$ @gj255 Noted, I tried to fix it! $\endgroup$ – knzhou Jun 11 '18 at 11:26
  • $\begingroup$ @knzhou I have made a new question, physics.stackexchange.com/questions/470424/…, related to a detail of your answer, if you could take a look I'd appreciated. $\endgroup$ – Vicky Apr 4 '19 at 2:13
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    $\begingroup$ @SRS Let the rates for $i \to f$, $\bar{i} \to \bar{f}$, $i_P \to f_P$ and $\bar{i}_P \to \bar{f}_P$ be $a$, $b$, $c$, and $d$. The net imbalance is $a - b + c - d$. $\endgroup$ – knzhou Oct 16 '19 at 6:43
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    $\begingroup$ @SRS If C symmetry holds, then $a = b$ and $c = d$, so the net imbalance is zero. If CP symmetry holds, then $a = d$ and $b = c$, so the net imbalance is again zero. This remains true even if C symmetry does not hold, $a \neq b$ and $c \neq d$. $\endgroup$ – knzhou Oct 16 '19 at 6:43

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