I try to understand a physical interpretation of the four components of the Dirac 4-spinor for a moving electron (in the simplest case, a plane wave). There is a very good question and answer about the interpretations already at SE. Basically it is shown that going to the rest-frame of the electron (i.e. $p^\mu=(E,0,0,0)$), one finds the four different solutions:
$$\psi_1=N_1\left(\begin{array}{c}1\\0\\0\\0\end{array}\right)e^{-iEt}, \psi_2=N_2\left(\begin{array}{c}0\\1\\0\\0\end{array}\right)e^{-iEt}, \psi_3=N_3\left(\begin{array}{c}0\\0\\1\\0\end{array}\right)e^{iEt}\text{ and } \psi_4=N_4\left(\begin{array}{c}0\\0\\0\\1\end{array}\right)e^{iEt},$$
Where the $\psi_1$ and $\psi_3$ have positive helicity (projection of the spin into the direction of the momentum) while $\psi_2$ and $\psi_4$ have negative helicity.
In addition, the phase factor $e^{\pm i E t}$ shows whether the state has positive or negative energy, thus whether it is a particle or antiparticle.
For moving electrons (in the Dirac representation), the solutions get additional contributions. For instance
$$
\psi_{move}(x)=N_1\left(\begin{array}{c}1\\0\\\frac{p_z}{E+m}\\\frac{p_x+ip_y}{E+m}\end{array}\right)\exp(-ip_\mu x^\mu)
$$
It has a nonvanishing 3rd and 4th component. Here Dominique writes:
When the momentum is NOT equal to zero these different states mix up and you can't make such a simple identification. Usually one says that the electron becomes a mixture of an electron with positrons when it starts moving.
However, the time-dependent phase-factor $e^{-i p_{\mu} x^{\mu}}$ still corresponds to positive energy for all of the four components, thus it can not be interpreted as
$$
\psi_{move}(x)\neq N \left(N_1 \psi_1 + \frac{p_z}{E+m} \psi_3 + \frac{p_x+ip_y}{E+m} \psi_4 \right)
$$
(A similar argument is given in these lecture-notes: The fact that the last two components are non-zero does not mean it contains “negative energy'' solutions.)
Thus my question is:
For a moving electron with helicity $+\frac{1}{2}$ (i.e. $\psi_{move}(x)$), what is the interpretation of non-vanishing 3rd and 4th component (in the Dirac representation)?
This question becomes more physical relevant when one considers no plane wave. Then the 3rd and 4th component might have different intensity distribution than the 1st component.
I am interested in both explanations and literature which covers this question.
Best Answer
In order to check the spin projection on the $i$-th axis for the third and fourth component, you just need to compute the quantity $$ \Sigma_{i}\psi, \quad \Sigma_{i} = \text{diag}(\sigma_{i}, \sigma_{i}) $$ with (for simplicity) $\psi = (0,0,1,0)$ and $\psi = (0,0,0,1)$ correspondingly.
In order to check the helicity of these components, you just need to compute the quantity $$ \frac{(\Sigma \cdot \mathbf p)}{|\mathbf p|}\psi $$ again for $\psi = (0,0,1,0)$ and $\psi = (0,0,0,1)$.
This sentence contains two incorrect statements, related to each other.
First, actually the "elementary" irreducible representations $\phi_{L/R}$ are defined as the eigenstates of the chirality matrix $\gamma_{5}$. The latter defines the way under which the spinors are lorentz-transformed, and has nothing to do with the helicity as long as the mass $m$ isn't zero. In the zero mass limit helicity and chirality formally coincide.
Second, $\phi_{L/R}$ are not Weyl spinors. The Weyl spinor is the one satisfying one of the equations $$ \sigma_{\mu}\partial^{\mu}\psi = 0, \quad \tilde{\sigma}_{\mu}\partial^{\mu}\psi = 0, \ \ \text{where }\ \ \sigma_{\mu} = (1,\sigma), \ \tilde{\sigma}_{\mu} = (1,-\sigma) $$ It is defined to describe massless particles with definite helicity, and has notning to do with the spinors $\psi_{L/R}$ inside the Dirac spinor as long as $m\neq 0$.