I have perhaps meaningless question about Dirac spinors, but I'm at a loss.

The transformation laws for for left-handed and right-handed 2-spinors are
$$
\tag 1 \psi_{a} \to \psi_{a}' = N_{a}^{\quad b} \psi_{b} = \left(e^{\frac{1}{2}\omega^{\mu \nu}\sigma_{\mu \nu}}\right)_{a}^{\quad b}\psi_{b}, \quad \psi^{b}{'} = \psi^{a}(N^{-1})_{a}^{\quad b},
$$
$$
\tag 2 \psi_{\dot {a}} \to \psi_{\dot {a}}' = (N^{*})_{\dot {a}}^{\quad \dot {b}} \psi_{\dot {b}} = \left(e^{\frac{1}{2}\omega^{\mu \nu}\tilde {\sigma}_{\mu \nu}}\right)_{\dot {a}}^{\quad \dot {b}}\psi_{\dot {b}}, \quad \psi^{\dot {b}}{'} = \psi^{\dot {a}}(N^{*^{-1}})_{\dot {a}}^{\quad \dot {b}},
$$
where
$$
(\sigma_{\mu \nu})_{a}^{\quad b} = -\frac{1}{4}\left(\sigma_{\mu}\tilde {\sigma}_{\nu}-\sigma_{\nu}\tilde {\sigma}_{\mu}\right), \quad (\tilde {\sigma}_{\mu \nu})_{\quad \dot {a}}^{\dot {b}} = -\frac{1}{4}\left(\tilde {\sigma}_{\mu} \sigma_{\nu}- \tilde {\sigma}_{\nu}\sigma_{\mu}\right),
$$
$$
(\sigma_{\mu})_{b\dot {b}} = (\hat {E}, \sigma_{i}), \quad (\tilde {\sigma}_{\nu})^{\dot {a} a} = -\varepsilon^{\dot {a}\dot {b}}\varepsilon^{b a} \sigma_{\dot {b} b} = (\hat {E}, -\sigma_{i}).
$$
Why do we always take the Dirac spinor as
$$
\Psi = \begin{pmatrix} \varphi_{a} \\ \kappa^{\dot {b}} \end{pmatrix},
$$
not as
$$
\Psi = \begin{pmatrix} \varphi_{a} \\ \kappa_{\dot {b}} \end{pmatrix}?
$$
According to $(1), (2)$ first one transforms as
$$
\delta \Psi ' = \frac{1}{2}\omega^{\mu \nu}\begin{pmatrix}\sigma_{\mu \nu} & 0 \\ 0 & -\tilde {\sigma}_{\mu \nu} \end{pmatrix}\Psi ,
$$
while the second one - as
$$
\delta \Psi ' = \frac{1}{2}\omega^{\mu \nu}\begin{pmatrix}\sigma_{\mu \nu} & 0 \\ 0 & \tilde {\sigma}_{\mu \nu} \end{pmatrix}\Psi ,
$$
so it is more natural than first, because the first one has both covariant and contravariant components, while the second has only covariant (contravariant components).

This post imported from StackExchange Physics at 2014-04-01 16:08 (UCT), posted by SE-user Andrew McAddams