# Antiparticles, CPT and leptogenesis

When people are being careful they'll tell you that antiparticles are the CPT conjugates of particles. You can't say that they are C conjugates or CP because these, while they do reverse the charge, are not symmetries of the Hamiltonian and so are not physical (anti)particles.

However, in leptogenesis, one explicitly calculates the CP asymmetry as a measure of the matter-antimatter asymmetry. Since CP violation is part of the theory, then the CP conjugates of physical particles are not physical antiparticles. This means that the resulting asymmetry is not literally the baryon asymmetry of the universe.

Why is this legitimate? Perhaps it's because CP violation is very slight and appears in the decays which destroy the leptons we are counting. Therefore if we imagine them to have long enough lifetimes to be countable real particles, then they are also essentially CP eigenstates?

First, to corroborate the staatement that antiparticles are the CPT conjugates of particles, here are some highlights from Weinberg (1995), The Quantum Theory of Fields, Volume I:

• Equation (2.C.11) on page 103 in chapter 2 says that CPT converts a single-particle state to the corresponding single-anti-particle state

• Page 133 in chapter 3 says, "It is CPT that provides a precise correspondence between particles and antiparticles, and in particular... tells us that stable particles and antiparticles have exactly the same mass."

Now, consider some single-particle state $$|1\rangle$$, and consider the state $$\Theta|1\rangle$$ obtained by applying a CPT transform $$\Theta$$. By definition, the state $$\Theta|1\rangle$$ is a single-antiparticle state. However, even though CPT is a symmetry, the states $$|1\rangle$$ and $$\Theta|1\rangle$$ might not behave symmetrically when the movie is played forward in time. They would behave symmetrically if the theory also had T symmetry (in which case it would also have CP symmetry); but in general, they won't.

Applying a CPT transform to a state-vector doesn't mean that we're obliged to play the movie backward in time. We can compare the state-vectors $$\exp(-iHt)|\psi\rangle$$ and $$\exp(-iHt)\Theta|\psi\rangle$$ instead, and we can do this whether $$|\psi\rangle$$ is a single-particle state or a state with many particles and antiparticles. This comparison gives meaning to concepts like the baryon asymmetry of the universe in the context of relativistic QFT. The point is that we don't need to know what CP means in order to talk about an asymmetry between the numbers (or behaviors) of particles and antiparticles. CPT is sufficient for this.

Sakharov's criteria for baryogenesis don't require the presence of CPT symmetry (they only require the absence of some symmetries, like CP), yet they still refer to "antiparticles." What does "antiparticle" mean if we haven't assumed the presence of any symmetry by which the particle-antiparticle correspondence would be defined? The answer might be simply that we don't need any precise QFT-based definition of "antiparticle" in this context. The criteria for baryogenesis could be expressed without using that language, something like this: If baryons do have partners with the same mass and spin but opposite charge, then they can't behave symmetrically when the movie is played forward in time, not even under a symmetry that reverses a direction of space. (And if baryons don't have such partners, then there would be nothing for baryogenesis to explain.) The idea here is that Sakharov's criteria refer to particles, not to fields. In contrast, the CPT symmetry of relativistic QFT operates on fields, with consequences for the spectrum of particles. We don't need a fields-based definition of "antiparticle" in order to understand Sakharov's criteria. The post

What are the assumptions that $$C$$, $$P$$, and $$T$$ must satisfy?

may be helpful in this context.

• I can accept that Sakharov's conditions can be defined without a precise definition of an antiparticle in QFT. However, in theories of baryogenesis, for instance leptogenesis, one explicitly considers the particle states and their CP conjugates. The difference of their occupations is what determines the baryon asymmetry ultimately. – Kris Jan 31 at 13:38
• @Kris In order to consider a particle's CP conjugate, the transformation being called CP conjugation must be well-defined and must map single-particle states to single-particle states (not using the prefix "anti" here, because that would beg the question). But if that transformation exists, then we can (in principle) check whether or not it maps a single-particle state $|1\rangle$ to any of its CPT-conjugates $\Theta|1\rangle$. (I say "any of" because CPT composed with any proper Poincare transformation is another valid CPT.) If it has that property, then CP maps particles to antiparticles. – Dan Yand Feb 1 at 15:05
• Thanks for the reply. Surely CP will not map single particle states to the CPT conjugate because CP is not a symmetry of the theory. – Kris Feb 1 at 16:09
• @Kris To be a symmetry, the CP transform would have to leave the Hamiltonian invariant. We can construct unitary maps that exchange particles with their CPT-conjugates but that don't leave the Hamiltonian invariant; we can do this even in the model of a free Dirac spinor field that does have CP symmetry. Merely saying that a transformation exchanges particles with their CPT-conjugates isn't enough information to define the transformation; for example, this doesn't indicate whether it's a unitary or antiunitary transformation, and doesn't imply that it leaves the Hamiltonian invariant. – Dan Yand Feb 1 at 17:33
• @Kris Maybe this is a better way to express it. Let $|n\rangle$, with some integer parameter $n$, be a basis for the set of single-particle states with the property that if $|n\rangle$ is one of the basis vectors, then so is $\Theta|n\rangle$ where $\Theta$ is CPT. We can partition these basis vectors into two sets, which are exchanged with each other by CPT. CPT is antilinear, but we can also construct a unitary $U$ that has the same effect on the given basis vectors. In general, this $U$ won't be a symmetry; but we can still say that it exchanges particles with their CPT conjugates. – Dan Yand Feb 1 at 17:46