# 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?

There is no natural pairing between particles and anti-particles at the level of individual states (any such pairing is purely a matter of convention), but there is a natural pairing between particle species — that is, between species and their anti-species. That's because Poincaré symmetries don't permute species, even though 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 natural 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. That's because in a theory with both CP and C symmetry, P is also a symmetry, so particles related by P are the same species, by definition.

• To accommodate theories that don't have C (or P) 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...

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

explains why this is so. It is also nicely explained in the comments following knzhou's answer to

What distinguishes the behaviour of particle from its antiparticle: C violation or CP violation?

To relate this to part 1 of the present question, note 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).

• 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
Commented Nov 3, 2018 at 17:42
• 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?
– SRS
Commented Nov 4, 2018 at 6:23
• @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.) Commented Sep 28, 2019 at 1:56