6
$\begingroup$

The charge conjugation operator $C$ reverses the charge of a state. But it may or may not convert a particle to its antiparticle. For example, consider a neutrino which is charge-neutral and left-handed while its antiparticle is also charge-neutral but right-handed. Therefore, charge conjugation is not sufficient to produce the antineutrino from a neutrino but CP is. $P$ is responsible for changing the chirality.

  1. So is it not $CP$ instead of $C$ that is responsible for changing a particle to its antiparticle?

  2. Is this fact related to the questions here and here?

$\endgroup$
4
$\begingroup$

My original answer was much too long, it didn't answer the question clearly (if at all), and it made a couple of dubious assertions. For those reasons, I replaced it with this new answer.


Answer to part 1

There is no natural pairing between particles and anti-particles at the level of individual states. There is only a natural pairing between particle species — that is, between species and their anti-species. That's because Poincaré transformations don't permute species, but they do permute states. This statement also applies to discrete Poincaré transformations like a space-reflection transformation P, in theories where P is a symmetry.

Once we agree that the appropriate pairing is between species, not between individual states, the answer to part 1 of the question is easy:

  • In a theory that has both CP and C symmetry, they both permute species the same way. It's not either-or, it's both.

  • To accommodate theories that don't have C symmetry, we can use CP to define the paring between species and their anti-species. This works whether or not the theory has C symmetry, and when it does, C symmetry gives the same pairing.

Even more generally, we could use CPT (instead of CP) to define the pairing between species and their anti-species, as Weinberg does in chapters 2 and 3 of Quantum Theory of Fields, volume 1. This provides some context for part 2 of the question...

Answer to part 2

Part 2 of the question asks whether part 1 is related to a couple of other questions about Sakharov's criteria for baryogenesis. Those criteria assert that to explain baryogenesis, the theory should not have either C or CP symmetry. The answer to

Why are both C and CP violation necessary for baryogenesis?

explains why this is so. To relate that answer to part 1 of the present question, use part 1 to see that Sakharov's criteria can be restated without referring specifically to CP or C, like this: To explain baryogenesis, the theory should not have any symmetries between particle species and their anti-species among all symmetries that preserve the time-orientation (which excludes CPT).

$\endgroup$
  • $\begingroup$ You said the $C$ transformation is not unique. Any $C\Lambda$ is also a valid charge conjugation operation. I believe this is because $\Lambda$ cannot change an electron to something else which is not an electron! I get your point that choosing $C$ to the charge conjugation rather than $C\Lambda$ is just a convention. It will take me some time to go through the full answer. @DanYand $\endgroup$ – SRS Nov 3 '18 at 17:42
  • $\begingroup$ You need CP transformation to convert a left-handed neutrino to a right-handed antineutrino in the Standard Model. SM does not have a left-handed antineutrino. Your answer suggests that the SM is CP-conserving which is not! What am I missing? $\endgroup$ – SRS Nov 4 '18 at 6:23
  • $\begingroup$ @SRS I replaced the answer. (The original answer didn't mean to suggest that the SM is CP-conserving, but it was a rambling and poorly-conceived answer. That's why I replaced it.) $\endgroup$ – Chiral Anomaly Sep 28 at 1:56

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.