Dirac spinor and Weyl spinor

Question

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 1

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.)

Answer 2

The answer does come down to the representation theory of the Lorentz group. A good discussion can be found in the first volume of QFT by Weinberg (and in other places as well). One thing to note is that you postulate a 4-dimensional representation of the Lorentz group. This postulate comes in when you assume that your objects have 4 components. Now a 4-dimensional representation of the Lorentz group can either be irreducible, corresponding to the 4-vectors, or be constructed by two two 2-dimensional representations, corresponding to two 2-component spinors. These are the only two options.

There is nothing that can discern left from right. All we know is that there will be two 2-dimensional subspaces that are independent from each other. (This can be seen in the chiral basis for the gamma matrices). We just call the objects in one of the spaces left-moving and the objects on the other right-moving. However, thw two spaces are absolutely identical. For example, if we write down the Dirac equation in two-component form (and there is an equivalent way of doing every possible calculation using only two component spinors instead of 4-component spinors [1]), then we can see that the equations satisfied by the left and the right spinors are absolutely equivalent.

Hope this helps!

[1]http://arxiv.org/abs/0812.1594

Answer 3

The key point in writing an action for spinors is the existence of a Clifford algebra (expanded by the gamma matrices) $$\{\gamma^a,\gamma^b\} = 2\eta^{ab}\mathbf{1},$$ where the index $a$ runs from $0$ to $3$ (or $0$ to $D-1$, where $D$ is the spacetime dimension).

The whole basis for the algebra is given by the $\gamma$'s and all possible products... due to last equation, only antisymmetric products contribute.

It's possible to show that for every even dimensional spacetime the (antisymmetric) product of all $\gamma$'s, i.e., $\gamma^* \propto \gamma^0\cdots \gamma^{D-1}$ allows to define non-trivial projectors $$P_+^2 = P_+,\quad P_-^2 = P_-, \quad P_+ P_- = 0.$$

These projectors serve to split the spinor into pieces, $$\psi_{\pm} = P_{\pm} \Psi,$$ with $\Psi$ the Dirac spinor, and $\psi_\pm$ the Weyl (or chiral) ones.

Conclusion

The fact that Dirac spinors can be split into Weyl ones is due to the dimensionality of spacetime. In odd dimension, the projectors are trivial because they are $0$ and $1$ respectively.

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Extra comment: Weyl spinors are two dimensional only if the chiral representation of gamma matrices in used. Otherwise they have half of the components (in terms of real dimension of the spinors).