If a massless neutrino or anti-neutrino is considered (in the whole post I consider neutrinos res. anti-neutrinos as mass-less), it is described by the Weyl-equation:
$$\overline{\sigma}^{\mu}\partial_\mu \chi_a =0 $$ if it is left-handed and neutrino (i.e. transforming according to $D^{(\frac{1}{2},0)}$)
or
$$\overline{\sigma}^{\mu}\partial_\mu \xi^\dagger_\dot{b} =0 $$ if it is right-handed and anti-neutrino (i.e. transforming according to $D^{(0,\frac{1}{2})}$)
Instead a Dirac-particle is described by a bispinor (4-spinor) $$\left(\begin{array}{c} \chi_a \\ \xi^{\dagger}_\dot{b} \end{array} \right)$$ which has 4 degrees of freedom: (spin-up particle; spin-down particle; spin-up anti-particle; spin-down anti-particle) a solution of the Weyl-equation apparently has only one degree of freedom, as the helicity is bound to the neutrino-type (neutrino or anti-neutrino).
However, the Weyl-solution has 2-components. Even worse, according to Landau/Lifschitz volume 4 (I also searched in Srednicki for such a development, but could not find it), the corresponding field operator of the free Weyl-equation can be developed in positive and negative frequency solutions:
$$\chi_a = \sum_p (U(p)_a a_p e^{ipx} + V(p)_a b_p^\dagger e^{-ipx})$$
I use capital letters for the 2-spinors $U(p)$ and $V(p)$ in order to distinguish them from the well-known bispinor solutions of the Dirac-equation $u(p)$ and $v(p)$.
Q:How is such a development in positive and above all negative frequency solutions possible? It seems to be that in a solution $\chi_a$, which solely describes neutrinos, anti-neutrinos mix in due to the appearance of the negative frequency solutions. This is actually the point I don't understand at all.
Q: What is the relation of $U(p)_a$ and above all $V(p)_a$ with the Dirac solutions $u(p)$ and $v(p)$ ?
Q: In particular how is guaranteed that $V(p)_a$ remains a left-handed 2-spinor, i.e. doesn't turn into a right-helicity 2-spinor (which seems to be manifest as $V(p)$ is the coefficient of the negative frequency solution)
I consider this detail as important because upon taking the hermitian conjugate the field-operator turns apparently into a right-handed 2-spinor:
$$ \chi^\dagger_\dot{a} = \sum_p (U(p)_\dot{a} a^\dagger_p e^{-ipx} + V(p)_\dot{a} b_p e^{ipx})$$
Such a thing (change of representation) does not happen if the hermitian conjugate of a bispinor Dirac-solution is taken (at least the hermitian conjugate of a bispinor Dirac-solution transforms in a representation which is equivalent to the original (standard bispinor) one).
