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 define 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., $\psi_D=\psi_L\oplus\psi_R$ ?

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

Comments

Sorry, by definition in the Chiral/Weyl basis Lorentz transformations is "diagonal" (i.e. it does not mix the two 2-components) because it is just made of two separated representations. So I cannot understand your question.

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

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@Roopam Sinha Thanks, I understand now. Please, I suggest you not to use capital letters, it seems that you are shouting and it does not sound very polite.

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

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@V.Moretti-Sorry. Actually I tried to emphasize. I will edit it.

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

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Try to use something or something instead to emphasize.

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

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Now it looks better!

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

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From the standpoint of representation theory, that's usually taken as the definition of the Dirac spinor representation. Which definition are you working with if not for that one?

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

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I am taking the definition of Dirac spinor as the solution of the dirac equation not the definition that is used in the representation theory of Lorentz group.

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

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.)

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

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

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