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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?

SRS
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1 Answers1

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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.

knzhou
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  • 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, https://physics.stackexchange.com/questions/470424/detail-on-c-vs-cp-violation?noredirect=1&lq=1, related to a detail of your answer, if you could take a look I'd appreciated. – Vicky Apr 04 '19 at 02:13
  • How come you say "which is incorrect"? Does the left handed electrons behave exactly the same as left handed positrons? – Mr Puh Apr 04 '19 at 07:54
  • @Vicky Sorry, I just made a mistake! Thanks for the catch. – knzhou Apr 04 '19 at 08:47
  • @knzhou I have a doubt regarding the 2nd and 3rd paragraph of your answer. If C is the operator which maps particles into antiparticles, then under C, the process $i\to f$ is mapped to $\bar i\to \bar f$. Now, if C is violated, we expect the rates of $i\to f$ and $\bar i\to \bar f$ to be different. So we don't need CP to create the imbalance. What am I missing? – SRS Oct 16 '19 at 05:10
  • @SRS As I said in the answer, the rates $i \to f$ and $\bar{i} \to \bar{f}$ will not balance, but the rates $i \to f$ and $\bar{i}_P \to \bar{f}_P$ will. The overall effect of these two reactions on the overall charge is zero. – knzhou Oct 16 '19 at 05:12
  • I think, you didn't explicitly mention the process $\bar i\to \bar f$. You talked about $i\to f$ and $\bar i_p\to \bar f_p$. Suppose CP is not violated but C is. Then, the difference in the rates of $i\to f$ and $\bar i\to \bar f$ will produce an imbalance. I fail to see why not. Is it obvious? @knzhou – SRS Oct 16 '19 at 05:21
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    @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 06:43
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    @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 06:43
  • This is great! Honestly, this is for the first time I think I have a convincing answer. Thanks. :-) @knzhou – SRS Oct 16 '19 at 06:49
  • Let us continue this discussion in chat. – SRS Oct 16 '19 at 07:06