I failed to find any book or pdf that explains clearly how we can interpret the different components of a Dirac spinor in the chiral representation and I'm starting to get somewhat desperate. This is such a basic/fundamental topic that I'm really unsure why I can't find anything that explains this concretely. Any book tip, reading recommendation or explanation would be greatly appreciated!
A Dirac spinor is a composite object of two Weyl spinors
\begin{equation} \Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} ,\end{equation}
where in general $\chi \neq \xi$. A special case called Majorana spinor is $\chi=\xi$. The charge conjugated spinor is
\begin{equation} \Psi^c = \begin{pmatrix}
\xi_L \\ \chi_R
\end{pmatrix} . \end{equation}
I want to understand how $\xi_L, \xi_R, \chi_L$ and $\chi_R $, can be interpreted in terms of how they describe particles/antiparticles of a given helicity?
Some Background:
The corresponding equations of motion are
\begin{equation} \big ( (\gamma_\mu (i\partial^\mu+ g A^\mu ) – m \big )\Psi^c = 0, \end{equation}
\begin{equation} \big ( (\gamma_\mu (i\partial^\mu- g A^\mu ) – m \big )\Psi = 0, \end{equation}
where we can see where the notion charge conjugation comes from. These equations can be rewritten in terms of the Weyl spinors:
\begin{equation} (i\partial^\mu- g A^\mu ) \begin{pmatrix} \sigma_\mu \xi_R \\ \bar{\sigma}_\mu \chi_L \end{pmatrix} = m \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} \end{equation}
\begin{equation} (i\partial^\mu+ g A^\mu ) \begin{pmatrix} \sigma_\mu \chi_R \\ \bar{\sigma}_\mu \xi_L \end{pmatrix} = m \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} \end{equation}
The charge conjugation transformation, shows that we have in principle $\chi \leftrightarrow \xi$ (as claimed for example here), which we can maybe interpret as $\chi$ and $\xi$ having opposite charge, i.e. describing particle and anti-particle (which I read in some texts without any good arguments). What bothers me about this point of view is that if we have a purely left-handed Dirac spinor
\begin{equation} \Psi_L = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} ,\end{equation}
the charge conjugated spinor is
\begin{equation} \Psi_L^c = i \gamma_2 \Psi_L = \begin{pmatrix} 0 \\ – i \sigma_2 \chi_L \end{pmatrix} = \begin{pmatrix} 0 \\ \chi_R \end{pmatrix} .\end{equation} This tells us that the charge conjugate of a left-handed spinor $\chi_L$, is the right-handed $\chi_R$ and not $\xi_R$.
A different point of view is explained in this Stackexchange answer. I would be interested in how we concretely can identify Electron and Positron states from the solutions of the Dirac equation (,as recited above)? I think an attempt to explain this can be found here, but I'm unable to understand it with all the math missing. It would be awesome if someone would know some text that explains these matters as they are claimed in the post by Flip Tanedo, but with the math added.
Best Answer
As far as I have been able to determine, it is not possible, in general, to interpret different solutions of the Dirac equation as corresponding to 'electron solutions' or 'positron solutions'.
For massive fermions, we can identify four independent spinors that correspond to a particle with 4-momentum $p$. In the Dirac representation, we have two solutions corresponding to the ansatz $u(p) e^{- i p \cdot x}$, which for particles at rest take the form: $$ u_1 = \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix} \qquad \mathrm{and} \qquad u_2 = \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \,. $$ These are said to correspond respectively to the spin-up electron and the spin-down electron. We also have two solutions corresponding to the ansatz $v(p)e^{i p \cdot x}$, which for particles at rest are: $$ v_1 = \begin{pmatrix}0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \qquad \mathrm{and} \qquad v_2 = \begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \,. $$ These are said to correspond respectively to the spin-down positron and the spin-up positron. This seems all well and good, but now let's repeat this prescription with massless fermions. We'll use the Weyl representation, and assume our 3-momentum is directed along the positive $z$-axis. With our initial ansatz, we find the Dirac equation reduces to: $$ (\gamma^0 - \gamma^3)u(p) = 0 \,.$$ However, with our second ansatz, the Dirac equation reduces to the same thing: $$ (\gamma^0 - \gamma^3)v(p) = 0 \,.$$ These equations would not be identical if $m \neq 0$. There are only two independent solutions: $$ u_1 = v_1 = \begin{pmatrix}0 \\ 1 \\ 0 \\ 0 \end{pmatrix} \qquad \mathrm{and} \qquad u_2 = v_2 = \begin{pmatrix}0 \\ 0 \\ 1 \\ 0 \end{pmatrix} \,. $$ We hence cannot simply interpret a particular spinor as corresponding to a 'spin-up electron', say, because the very same spinor would also have to correspond to a 'spin-down positron'.
The expectation that we should have found four solutions --- one each for the four possible choices of up/down and particle/antiparticle --- seems flawed to me, because antiparticles are only a meaningful concept in the quantum theory, and need not correspond to independent spinors in the classical theory. As way of illustration, a theory with two equal mass but otherwise distinct fermions would have eight different quantum states for each choice of momentum, but you certainly wouldn't find eight different spinor solutions to the classical equations. Your spin-up 'electron' and spin-up 'muon' would be described by the same spinor, but that doesn't make them the same state!