# Detail on C vs. CP violation

In the answer given by knzhou to the post What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation? it is said that

"but the reaction $$i \rightarrow f$$ will run at the same rate as its $$CP$$ conjugate $$\bar{f}_P \rightarrow \bar{i}_P$$."

My questions is: is not $$\bar{f}_P \rightarrow \bar{i}_P$$ the $$CT$$ of $$i \rightarrow f$$ instead of its $$CP$$ conjugate since you have the anti-$$f$$ particles in the initial state?

EDIT I found something else that I do not understand. Why does knzhou say that C is not enough? If C distingishes particles and antiparticles and our theory violates C (not necessarily CP), the reactions $$i\rightarrow f$$ and its C conjugate $$\bar{i} \rightarrow \bar{f}$$ will have different rate since $$i, f$$ and its counterparts $$\bar{i}, \bar{f}$$ are different, so why do we need CP and not just C?

Moreover, if CPT is preserved but we look for CP violation, does it imply that T is not preserved? But is it not T a symmetry usually?

• The f->i reaction should be the other way around to be CP conjugated... that shown reaction would be the CPT conjugated. – Mr Puh Apr 4 at 7:58
• Sorry, I just made a mistake there, thanks for the catch! – knzhou Apr 4 at 8:47
• @knzhou I have another question related to that post so I edited mine. If you could answer it I'd appreciate it – Vicky Apr 4 at 14:48
• It's simply that if $i \to f$ converts, say, matter to antimatter at some rate, then $\tilde{i}_P \to \tilde{f}_P$ would convert it right back, giving no net change. – knzhou Apr 4 at 14:55
• @knzhou Yeah, but why do you need to include parity? With C you have already that unbalance effect since your theory would not be C-even – Vicky Apr 4 at 14:57