# What do the four components of Dirac Spinors represent in the Standard Model?

I've been trying to get my head around the formalisms used in the Standard Model. From what i've gathered Dirac Spinors are 4 component objects designed to be operated on by Lorentz Transformations much like 4-Vectors are in Special Relativity. However they also incorporate additional information: Spin and "Handedness". Due to the nature of Spin, Spinors also transform differently then vectors.

This leaves me with the impression that the 4 components can be classified as: Left Handed and Spin Up, Left Handed and Spin Down, Right Handed and Spin up, Right Handed and Spin Down.

My question is if this impression is the right general idea or not?

• Your Dirac bi-spinor basis is called the Weyl basis or chiral basis. But, for the Dirac equation, there are different basis, like the Dirac basis. Every basis corresponds to a different representation of the gamma matrices. – Trimok Oct 14 '13 at 8:56

It depends on the representation of the gamma-matrices. It can be, for example, the four combinations of electron/positron and spin up/spin down.

• In any other representation the components are just linear combinations of these ones. So the information contained in them is the same, it just gets packaged differently. – Michael Brown Oct 14 '13 at 8:21
• @Michael Brown: I agree, but that means that the 4 components cannot be "classified as: Left Handed and Spin Up, " etc., as OP suggests, independent of the representation. – akhmeteli Oct 14 '13 at 8:36
• The 4 d.o.f. are complex, could add that extra d.o.f. are eliminated by EoM etc. – innisfree Oct 14 '13 at 8:48
• Perhaps it's worth clarifying that you can classify the 4 d.o.f. in an invariant way using the projection operators $(\gamma^\mu p_\mu \pm m)/2m$ and $(1\pm\gamma^5)/2$, which happen to take nice diagonal forms in a particular basis. The adapted basis is nice if you care about these projectors, though you can use any basis you want. But you can see by this construction that the physical content of the field is independent of the representation. – Michael Brown Oct 14 '13 at 8:58
• @EdwardHughes What? Electrons and positrons correspond to the positive frequency ($u e^{-ipx}$) and negative frequency ($v e^{+ipx}$) solutions of the Dirac equation. Both are contained in the same Dirac spinor field. See, for instance, Peskin & Schroeder eq 3.99. The projection operators I mentioned project onto the left & right handed parts of $u$ and $v$. Whether you classify by left/right particle/antiparticle is equivalent to, but just a reshuffling of, left/right spin up/down. – Michael Brown Oct 15 '13 at 5:31

the 4 components can be classified as: Left Handed and Spin Up, Left Handed and Spin Down, Right Handed and Spin up, Right Handed and Spin Down.

Your classification is representation-dependent in the context you put it.

However, there is a representation-independent way of doing the same with two (orthogonal) projections: \begin{aligned} \text{Chiral Projection}&:\ \frac{1}{2}(1\pm \gamma^5) = \frac{1}{2}(1\pm i\gamma^0\gamma^1\gamma^2\gamma^3), \\ \text{Spin Projection}&:\ \frac{1}{2}(1\pm 2S^3) = \frac{1}{2}(1\pm i\gamma^1\gamma^2). \end{aligned}

The specific representation of yours is to choose the 4 eigenvectors of above projections as the base.

In a case of representations $\left( \frac{m}{2}, \frac{n}{2}\right)$ of the Lorentz group we need to take the direct sum of $\left( \frac{m}{2}, \frac{n}{2}\right) + \left( \frac{n}{2}, \frac{m}{2}\right)$, if we want to make our representations irreducible. It is caused by acting on discrete space (and time) inverse operators on space of the Lorentz group: they transfer $\left( \frac{m}{2}, \frac{n}{2}\right)$ representation to $\left( \frac{m}{2}, \frac{n}{2}\right)$, so $\left( \frac{m}{2}, \frac{n}{2}\right)$ alone is not the representation of the full Lorentz group. Also, if we want to make our field real (not complex), we also must to take the direct sum of reps (by analogical reasoning). But then we must to act on direct-sum-rep field by projection operator, which leave only $n + m + 1$ independent components of a field as it must be for spin $s = \frac{n + m}{2}$ field.

So, let's talk about special case. Dirac bispinor refers to the direct sum of $\left(\frac{1}{2}, 0\right)$ and $\left( 0, \frac{1}{2}\right)$ representations, which correspond to left-handled and right-handled representations (calling chirality). Each of this representations refer to spin $\frac{1}{2}$-particle, and it's projection can be $\pm \frac{1}{2}$. But Dirac equation, which is the projection operator on two-dimensional space of independent components (as it must be for spin $\frac{1}{2}$ be), mixes these components in general. however in a case of $m = 0$ components of different chirality isn't mixed between each other, and Dirac equation leads to two independent equations which are called Weyl's equations. This may be even in Dirac basis.

Also the anticommutation relations between Dirac matrices and the form of the Dirac equation don't change under unitary transformations: $\gamma ' = U\gamma U^{-1}, \Psi ' = U\Psi$, so by taking $U = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 && \sigma_{y} \\ \sigma_{y} && -1\end{pmatrix}$ you make spinor's equations independent. So in this case you also may use your classification.

A two component spinor can be geometrically interpreted as representing a point on the Riemann sphere, defined by the ratio of its two complex components, and its stereographic projection onto the xy-plane. Similarly, a four component spinor can be interpreted, by a more complicated ratio defined by its four complex components, as a point on the Riemann sphere followed by a Lorentz transformation, and its stereographic projection onto the complex projective plane $P^2_C$. My recent work, "Vector Analysis of Spinors", and "Spacetime Algebra of Dirac Spinors", addressing these perplexing issues can be found on my website: http://www.garretstar.com/