What does it mean when a particle is an eigenstate of the charge conjugation operator?

Question

I have limited background in Quantum Physics and am trying to understand some Particle Physics material. I was reading about Charge Conjugation and it reads that "Most particles in nature are not eigenstates of Charge Conjugation". I understood that when the wave function of a state is the eigenvector of the operator of a measurable, it is called the eigenstate of the operator and we will obtain a definite value of the measurable the operator is measuring. I am struggling to relate this to the statement in bold.

Answer 1

Operating twice with the charge conjugation operator, $\cal C$, on a physical state must bring you back to the original state, i.e. ${\cal C}^2 = I$. This means that the eigenvalues of $\cal C$ are $\pm 1$, and so ${\cal C} | p \rangle = \pm | p \rangle$. As a consequence the particle state $| p \rangle$ and ${\cal C} | p \rangle$ can differ by at most a sign. The only candidates for such states are particles which are their own antiparticles, and as your statement says, most particles in nature are not of this type. Probably the most common example is the photon, and there are other examples like charge-neutral mesons like the $\pi^0$. But everyday particles like electrons, protons, and neutrons are not eigenstates of $\cal C$.