0
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ The f->i reaction should be the other way around to be CP conjugated... that shown reaction would be the CPT conjugated. $\endgroup$ – Mr Puh Apr 4 at 7:58
  • 1
    $\begingroup$ Sorry, I just made a mistake there, thanks for the catch! $\endgroup$ – knzhou Apr 4 at 8:47
  • $\begingroup$ @knzhou I have another question related to that post so I edited mine. If you could answer it I'd appreciate it $\endgroup$ – Vicky Apr 4 at 14:48
  • $\begingroup$ 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. $\endgroup$ – knzhou Apr 4 at 14:55
  • $\begingroup$ @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 $\endgroup$ – Vicky Apr 4 at 14:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.