From the relativistic covariance of the Dirac equation (see Section 2.1.3 in the QFT book of Itzykson and Zuber for a derivation. I also more or less follow their notation.), you know how a Dirac spinor transforms. One has $$\psi'(x')=S(\Lambda)\ \psi(x)$$
under the Lorentz transformation
$$x'^\mu= {\Lambda^\mu}_\nu\ x^\nu= {\exp(\lambda)^\mu}_\nu\ x^\nu=(I + {\lambda^\mu}_\nu+\cdots)\ x^\nu\ .$$
Explicitly, one has $S(\Lambda)=\exp\left(\tfrac{1}8[\gamma_\mu,\gamma_\nu]\ \lambda^{\mu\nu}\right)$.

To show reducibility, all you need is to find a basis for the gamma matrices (as well as Dirac spinors) such that $[\gamma_\mu,\gamma_\nu]$ is block diagonal with two $2\times 2$ blocks. Once this is shown, it proves the reducibility of Dirac spinors under Lorentz transformations since $S(\Lambda)$ is also block diagonal. Such a basis is called the chiral basis. It is also important to note that a mass term in the Dirac term mixes the Weyl spinors in the Dirac equation but that is not an issue for reducibility.

While this derivation does not directly use representation theory of the Lorentz group, it does use the Lorentz covariance of the Dirac equation. I don't know if this is what you wanted.

(*I am not interested in your bounty -- please don't award me anything.*)

This post imported from StackExchange Physics at 2014-03-31 16:04 (UCT), posted by SE-user suresh