special relativity - Dirac spinor and Weyl spinor

How can it be shown that the Dirac spinor is the direct sum of a right-handed Weyl spinor and a left-handed Weyl spinor?

EDIT: - Let $\psi_L$ and $\psi_R$ be 2 component left-handed and right-handed Weyl spinors. Their transformation properties are known. When I put these two spinors in a column and construct a four-component column which is a direct sum of $\psi_L$ and $\psi_R$ i.e., $\psi_D=\psi_L\oplus\psi_R$. This I defined to be the Dirac spinor. Right? Since it is a direct sum under Lorentz transformation, the corresponding Lorentz transformation matrix is diagonal. Right? Then it is easy to show that, it satisfies the Dirac equation in Chiral basis. Right? This is possible because we started from the definition of left-handed and right-handed Weyl spinors and their transformation properties are known. Right? This is explicitly carried out in the book by Lewis Ryder. But suppose I start the other way around. I solve the Dirac equation in chiral basis. Then no one tells me that the upper two components are really left-handed and lower two are really right-handed. Suppose I take this chiral basis solution of Dirac equation and now take that to be my definition of Dirac spinor. Then how can I show the opposite, that it is made up of two irreps of Lorentz group i.e., $\psi_D=\psi_L\oplus\psi_R$ ?

Answer

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.)