Interpretation of Dirac Spinor components in Chiral Representation?

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

$$$$\Psi = \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix} ,$$$$

where in general $$\chi \neq \xi$$. A special case called Majorana spinor is $$\chi=\xi$$. The charge conjugated spinor is

$$$$\Psi^c = \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix} .$$$$

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

$$$$\big ( (\gamma_\mu (i\partial^\mu+ g A^\mu ) - m \big )\Psi^c = 0,$$$$

$$$$\big ( (\gamma_\mu (i\partial^\mu- g A^\mu ) - m \big )\Psi = 0,$$$$ where we can see where the notion charge conjugation comes from. These equations can be rewritten in terms of the Weyl spinors:

$$$$(i\partial^\mu- g A^\mu ) \begin{pmatrix} \sigma_\mu \xi_R \\ \bar{\sigma}_\mu \chi_L \end{pmatrix} = m \begin{pmatrix} \chi_L \\ \xi_R \end{pmatrix}$$$$ $$$$(i\partial^\mu+ g A^\mu ) \begin{pmatrix} \sigma_\mu \chi_R \\ \bar{\sigma}_\mu \xi_L \end{pmatrix} = m \begin{pmatrix} \xi_L \\ \chi_R \end{pmatrix}$$$$

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

$$$$\Psi_L = \begin{pmatrix} \chi_L \\ 0 \end{pmatrix} ,$$$$ the charge conjugated spinor is

$$$$\Psi_L^c = i \gamma_2 \Psi_L = \begin{pmatrix} 0 \\ - i \sigma_2 \chi_L \end{pmatrix} = \begin{pmatrix} 0 \\ \chi_R \end{pmatrix} .$$$$ This tells us that the charge conjugate of a left-handed spinor $$\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.

• Neither parity ($P$) nor charge conjugation ($C$) is a symmetry of a Weyl spinor, but $CP$ is. Mar 3, 2017 at 16:12
• I would like to recommend my supervisors book, "ideas and methods of supersymmetry and supergravity, or a walk through super space" S. Kuzenko and I. Buchbinder. Jun 9, 2017 at 1:04

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!

• The position-dependent parts of $u(p) e^{-ip\cdot x}$ and $v(p) e^{ip\cdot x}$ are already orthogonal, so we really should be comparing $u(p) e^{-ip\cdot x}$ with $v(-p) e^{-ip\cdot x}$. Massless limit or not, these two solutions (in particular, their position-independent part) are orthogonal. Mar 3, 2017 at 16:17
• Also, electrons and positrons are nothing other than quantized normal modes of the Dirac field. These normal modes must be present in the classical theory as well; otherwise we have nothing to quantize. Mar 3, 2017 at 16:21
• On the above, $v(-p)e^{-i p\cdot x}$ really should be $v(p^{0}, -\textbf{p}) e^{i(p^{0} t + \textbf{p} \cdot \textbf{x})}$ with the understanding that $p^{0}$ is a function of $\textbf{p}$. Mar 3, 2017 at 21:23

Heristically, it boils down to the fact that the parity transformation (P) reverses the sign of the $$\gamma$$ matrices, while CT turns $$\gamma^\mu\to-\left(\gamma^\mu\right)^T$$. These are again equivalent representations of the Dirac algebra, and the intertwiners are commonly called $$\gamma^5$$ and $$C$$. Note that the C and T transformations themselves are not symmetries of the Dirac algebra.

You seem to use the convention where the $$\gamma$$ matrices are composed of the Pauli matrices,$$\gamma=\begin{pmatrix}0&\sigma^\mu\\\bar\sigma^\mu&0\end{pmatrix}\,.$$ Here, $$\sigma^\mu=\left(\mathbb I,\sigma^i\right)$$ and $$\bar\sigma^\mu=\left(\mathbb I,-\sigma^i\right)$$. ($$\mathbb I$$ is the unit matrix - I can't seem to get the proper double-stroke 1). Then $$\gamma^5=\text{i}\gamma^0\gamma^1\gamma^2\gamma^3 =\begin{pmatrix}\mathbb I&0\\0&-\mathbb I\end{pmatrix}\,,$$ and the chiral projectors $$P_{L/R}=\frac{1}{2}\left(1\pm\gamma^5\right)$$ project onto the upper and lower two components of a four-spinor. Hence, chiral spinors are $$\Psi_L=\begin{pmatrix}\chi_\alpha\\0\end{pmatrix}\quad\text{ and }\quad\Phi_R=\begin{pmatrix}0\\\bar\xi^{\dot\alpha}\end{pmatrix}\,.$$ (The index placements and various signs are matter of convention.) Their "charge conjugates" are $$\left(\Psi_L\right)^c=C\overline{\Psi_L}^T=\begin{pmatrix}0\\\bar\chi^{\dot\alpha}\end{pmatrix}$$ and similar for $$\Phi_R$$. You see that this reverses chirality, but also charge: Charges are determined by the transformation under some $$U(1)$$, and since complex conjugation is involved, the transfromation is reversed. For non-Abelian groups, you end up in the conjugate representation. In both cases, you end up with an object of opposite charge to the original one. Hence, a Majorana spinor, i.e. a Dirac spinor which is eqaul to its charge conjugate, cannot have a $$U(1)$$ charge (or be in a non-real representation). Furthermore, such a spinor can be equally well described by a single left-handed (or right-handed) Weyl spinor. (Note that this is different in six or ten dimensions, where charge conjugation does not reverse chirality.)

So to describe the electron (actually, a toy version with only electric and no weak charges), you can use two left-chiral spinors, $$\chi_\alpha$$ and $$\xi_\alpha$$, of opposite charge or one four-component object $$\Psi=\begin{pmatrix}\chi_\alpha\\\bar\xi^{\dot\alpha}\end{pmatrix}\,.$$