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 $D^{(\frac{1}{2},0)}$)


$$\overline{\sigma}^{\mu}\partial_\mu \xi^\dagger_\dot{b} =0 $$ 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 change of representation does not happen if the hermitian conjugate of a bispinor Dirac-solution is taken since the hermitian conjugate of a bispinor Dirac-solution transforms in a representation which is equivalent to the original (standard bispinor) one.


1 Answer 1


The massless neutrino is described by a two-component spinor $\chi_{a}$. If we take the complex conjugate we get $(\chi_{a})^{*}=\chi^{*}_{\dot{a}}$. The massless neutrino can be described as $\chi_{a}$ or as $\chi^{*}_{\dot{a}}$. There are not two independent undotted and dotted spinors $\chi_{a},\xi^{*}_{\dot{a}}$. Notice that Landau and Lifshitz volume 4 2nd edition express this point just before equation (30.6) on page 112. This should clear up all the problems posed in the question.


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