neutrinos - What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation?

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?

Answer

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.