# Is $CP$ instead of $C$ responsible for changing a particle to its antiparticle?

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?

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.

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...

• 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 – SRS Nov 3 '18 at 17:42