In the Weyl basis we can separate the spinor field into 2 components: the right-chiral spinor and the left-chiral spinor. Each of these fields has again 2 components which are coupled. What is the physical interpretation of these 2 components that make up the left-chiral (or right-chiral) field?
In the Dirac basis the interpretation of the 4 components is:
1. Electron spin-up
2. Electron spin-down
3. Positron spin-up
4. Positron spin-down
So my question is what is the corresponding interpretation in the Weyl basis (in the massless case). Is it like this?
1. Left-chiral electron $\psi_{4}$
2. Left-chiral positron $\psi_{3}$
3. Right-chiral electron $\psi_{2}$
4. Righ-chiral positron $\psi_{1}$
If this is the case than I don't understand why the left-chiral electron $\psi_{4}$ couples to left-chiral positron $\psi_{3}$ as can be seen in the equations:
$$ \partial_{t} \psi_{4} + \partial_{x} \psi_{4} – i\partial_{y} \psi_{4} + \partial_{z} \psi_{3} = 0 $$
$$ \partial_{t} \psi_{3} + \partial_{x} \psi_{3} + i\partial_{y} \psi_{3} – \partial_{z} \psi_{4} = 0 $$
$$ \partial_{t} \psi_{2} – \partial_{x} \psi_{2} + i\partial_{y} \psi_{2} – \partial_{z} \psi_{1} = 0 $$
$$ \partial_{t} \psi_{1} – \partial_{x} \psi_{1} – i\partial_{y} \psi_{1} + \partial_{z} \psi_{2} = 0 $$
Best Answer
The meaning of different components in the chiral representation are
Spin up and down are with respect to some arbitrary axis, which we often set to the $z$-axis .
Electron and positron(or negative-energy electron) states are identified from the solutions of the Dirac equation. It turns out that in the massless limit (for which the chiral representation is most convenient), the left- and right-handed sectors decouple, and electron(positron) states in the left-handed sector are left(right)-handed, viz. the spin is anti-parallel(parallel) to the momentum, and similarly for the right-handed sector.