# 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?

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.

• 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? – gj255 Jun 11 '18 at 10:04
• @gj255 Noted, I tried to fix it! – knzhou Jun 11 '18 at 11:26
• @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. – Vicky Apr 4 '19 at 2:13
• @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 $a - b + c - d$. – knzhou Oct 16 '19 at 6:43
• @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 $c \neq d$. – knzhou Oct 16 '19 at 6:43