A chiral eigenstate is always a linear combination of a particle and an antiparticle state and a particle or antiparticle state is always a linear combination of chiral eigenstates. Now, how can we then talk about a left-chiral electron or positron, which are said to take part in weak interactions.
Background:
Chiral eigenstates can be identified through the projection operators
$$ P_L = \frac{1 – \gamma_5}{2} \quad P_L = \frac{1 + \gamma_5}{2} $$
The corresponding eigenstates are in the Chiral/Weyl Basis
$$ \Psi_L= \begin{pmatrix} \chi_s \\0 \end{pmatrix} \quad \text{and } \quad \Psi_R = \begin{pmatrix} 0 \\ \xi_s \end{pmatrix} $$
where the index $s$ denotes the different possible spin configuations and with the two component Weyl Spinors $\chi$, $\xi$.
Particle states can be identified through the solutions of the Dirac equation. In the Chiral/Weyl Basis the solutions (in the rest frame) are
$$ u_s= \begin{pmatrix} \eta_s \\ \eta_s \end{pmatrix} \quad \text{and } \quad v_s =\begin{pmatrix}\zeta_s \\ – \zeta_s \end{pmatrix} $$
Therefore, four linearly independent solutions of the Dirac equation are
$$ e^-_\uparrow = u_1= \begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} \quad \quad e^-_\downarrow = u_2 =\begin{pmatrix} 0 \\1 \\0 \\1 \end{pmatrix} \quad e^+_\uparrow = \ v_1= \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} \quad \quad e^+_\downarrow =\begin{pmatrix} 0\\1\\0\\-1 \end{pmatrix} $$
which correspond to, for example electron or positron with spin up or down.
The solutions of the Dirac equation are what we use in QFT computations. The chiral structure becomes important for weak interactions, because only left-handed particles interact weakly. This is included through $P_L$ in the vertex factor, for example for an incoming muon, decaying weakly we have a factor $ \propto P_L u_s$.
What is $P_L u_s$? The computation in the Weyl/Chiral basis shows
$$ (u_s)_L = P_L u_s = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} \eta_s \\ \eta_s \end{pmatrix} = \begin{pmatrix} \eta_s \\ 0 \end{pmatrix} $$
This is no longer a solution of the Dirac equation, so how do we interpret it? We can see that this is a linear combination of a particle and an antiparticle state. For example
$$ (u_1)_L = \frac{1}{2} ( u_1 + u_2^c) = \frac{1}{2} (u_1 + i\gamma_2 u_2^\star) = \frac{1}{2} (\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 & i \sigma_2 \\ -i \sigma_2 & 0 \end{pmatrix} \begin{pmatrix} 0 \\1 \\0 \\1 \end{pmatrix} ) $$
$$ = \frac{1}{2} (\begin{pmatrix} 1 \\ 0 \\ 1 \\ 0 \end{pmatrix} + \begin{pmatrix} 1 \\ 0 \\ -1 \\ 0 \end{pmatrix} ) = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} $$
with the usual charge conjugation transformation $ i\gamma_2 + $ complex conjugation. Anyway, the moral of the story is that a chiral eigenstate is always a linear combination of a particle and an antiparticle state and particle or antiparticle state is always a linear combination of chiral eigenstates. Now, how can we then talk about a left-chiral electron or positron?
PS: Equivalently we can of course see that the solutions of the Dirac equation are always linear combinations of chiral eigenstates, for example $u_s = (u_s)_L + (u_s)_R $
Best Answer
For anyone with similar problems:
The following observation has helped me immensly: We have in fact four particles directly related to an electron:
These four particles are what we describe by the two Weyl spinors inside a Dirac spinor and its charge conjugate respectively.
The Dirac equation tells us that as time passes on a left-chiral electron transforms into a right-chiral and vice versa. Chirality and therefore weak isospin are not conserved quantities. To be able to talk about electrons evolving in time we therefore need to consider left-chiral and right-chiral electrons at the same time, which is why we use Dirac spinors
$$\Psi_{e^-} = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} $$ This is what is commonly called a physical electron or simply electron. Its charge conjugate is what we commonly call positron
$$\Psi_e^c = \Psi_{e^+} = \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} $$ Only the two particles from above with isospin interact weakly, which means $\chi_L$ and $\chi_R$.
An electron that was created by weak interactions and is therefore purely left-chiral, is described by the spinor $$(\Psi_{e^-})_L = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} $$
The Dirac equation tells us that as time passes on this transformes into a mixture of $\chi_L$ and $\xi_R$, but not $\chi_R$, which would violate charge conservation anyway.