# Spinors: Relativistic vs Non-Relativistic?

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1. Sep 30, 2014

### referframe

Consider the Spinor object for an electron. Are the non-relativistic and relativistic (Dirac equation) Spinor objects, from a mathematical point-of-view, identical?

2. Sep 30, 2014

### Meir Achuz

The nonrelativistic spinor has 2 components, and the relativistic spinor has 4 components.

3. Oct 1, 2014

### aleazk

As mathematical entities, they are more or less the same thing, only some details change. You start with a two dimensional complex vector space (equipped with an antisymmetric tensor of type 2,0;0,0). This is the "spinor space", and its elements the "spinors". Using this (and perhaps some other additional structure), you can build a real vector space (in general, some subspace of tensors over this initial complex space). The antisymmetric tensor gives rise in this new real space a metric. The key thing is that the group of transformations that preserve this metric (the special isometries, i.e., det = 1), is obtained by applying the construction that allowed to build the real vector space to some subgroup of the general linear group of the original two dimensional complex vector space. Even more: it results that this subgroup is the universal covering group of the special isometries and the map we just built the covering map! The action of this covering group in the spinor space is the "spinor action" of the "spinor group". In this sense, at one side we have the spinor space, the spinor group and the spinor action; on the other, the real vector space, a metric and its special orthogonal group. In the middle, a map (in general, not injective) from the spinor group to the special orthogonal group, and a way to build the vectors in the real space from spinors. So, we could say that the theory of the real vector space, with its metric and group, is some kind of 'emergent' thing from the theory of the spinor space. If the action of the special orthogonal group has some physical interpretation (e.g., as a rotation), we can generalize this notion to the spinor space through the covering map (but one must be careful, since the map is not injective, and so a particular rotation can come from two different elements in the spinor group).

Now, if we define a 'hermitian metric' in the spinor space, we can build with this the subgroup SU(2) (special and unitary) of the general linear group of the spinor space. We take SU(2) to be the spinor group in this example. The real vector space is the space of two index symmetric and self-adjoint tensorial spinors. It turns to be three dimensional. The metric is given by eA1A2eB1B2, where eA1A2 is the antisymmetric tensor we chose. This metric is negative definite (the negative sign is conventional), i.e., we recover the ordinary three dimensional vector algebra. The special orthogonal group is simply SO(3), the usual rotation group. The spinors in this example are called "SU(2) spinors" and are the spinors of non-relativistic QM.

If we take the spinor group to be SL(2, C), we obtain a 4-dimensional vector space equipped with a lorentzian metric. The special orthogonal group is of course the restricted part of Lorentz group SO(1,3). The notion we generalize to the spinor space in this case is of course the "Lorentz transformation". These are the "SL(2, C) spinors" we use in relativistic QM. The four-component Dirac spinors are equivalent to them. A Dirac spinor is just a pair composed of a spinor of the type we defined here and a complex conjugate spinor of this same type (notice that if we put a basis in the spinor space, this pair is characterized by four components).

The timelike four-velocity of an inertial observer can be used to define one of these hermitian metrics and in this way to obtain a 3+1 decomposition of the SL(2, C) spinors.

Last edited: Oct 1, 2014
4. Oct 1, 2014

### referframe

OK. Does the relativistic spinor have 4 components because of the 4 dimensions of space-time or because it also includes the matter-antimatter duality?

5. Oct 1, 2014

### referframe

Thank you for the very complete reply. It confirms what I have read so far, that the answer to my question is NOT simple.

6. Oct 1, 2014

### Meir Achuz

It
The 4 components are not because of 4D space-time, and not BECAUSE of matter-antimatter, but they do lead to antimatter.

7. Oct 1, 2014

### referframe

So, for spin 1/2 particles, the relativistic Spinor is a 4-component column vector with HALF of those components having nothing to do with spin. Correct?

8. Oct 1, 2014

### Staff: Mentor

No. Two components correspond to spin-up, and two components to spin-down.

9. Oct 1, 2014

### king vitamin

The confusing thing about Dirac spinors is the way that parity effects spinors in relativity, which is the only reason why we need four components instead of two.

In non-relativistic QM, you have spatial rotation symmetry, which is described by the group SO(3). You demand that all states are symmetric under this transformation up to a phase, so that no physical predictions depend on how you rotated your coordinate system (and since overall phases have no physical effect in QM). In principle, this allows you to decompose all states in terms of "irreducible representations of projective SO(3)," which is a fancy way of saying that you can have states with spin (projective means up to a phase here). The math tells you all about how you can have integer and half-integer spin (the latter are the ones which are only symmetries up to a phase), and that these need to be described by (2s+1)-dimensional objects. So spin-0 is 1D, spin-1/2 is 2D, spin-1 is 3D (this is just the vector representation!), etc.

In going to relativity, you need your theory to be symmetric under Lorentz boosts too, and the full symmetry group is called SO(3,1). You need to go through the same math as above, but of course the results will be more complicated (it's a larger group). Two big things happen here. The first is that massless states need to be described by helicity instead of spin (let's ignore this large complication for your question, it's actually not too big a deal for spin-1/2). The second is that, for massive states, in addition to spin, for half-integer states you also need to specify the chirality of your representation. Taking spin-1/2 as the simplest example, you can have "left-handed" or "right-handed" spin-1/2 particles, which transform differently under SO(3,1) even though they have the same spin and are both two-dimensional.

However, under parity (taking (x,y,z) -> (-x,-y,-z)), these two states get sent into each other. Now, if you want to describe a theory with parity invariance (like electrodynamics), these two-dimensional states aren't invariant. Instead, you need to combine the left- and right-handed representations into one which is invariant under SO(3,1)+parity. This object, called a Dirac spinor, is now four-dimensional. Under rotations/boosts, the left- and right-handed parts transform separately, but under parity they switch with each other. You can show that in the non-relativistic limit, two of the four components can be ignored.

10. Oct 2, 2014

### dextercioby

I like this type of technical questions. The reason why the Dirac spinor has 4 components has to do with 2 things which coincide to the value 4: 4 space-time dimensions which make the fundamental representation of the Dirac-Pauli Clifford algebra exactly 4 dimensional, while the fundamental irreducible representation of the covering of SO(1,3) is again exactly 4.

11. Oct 2, 2014

### referframe

Thank you all for your valuable input.

12. Oct 3, 2014

### PhilDSP

I'm wondering now if it would be more fair and accurate to say that the 4 components are required not directly because of relativity but because of the factoring or linearization Dirac performed on the relativistic energy equation. He essentially took the square root of the energy equation and from that separated the 3 spatial components and the single time component.

Mathematically, all of that occurs in Euclidean space rather than Minkowski space.

13. Oct 3, 2014

### aleazk

haha, ouch, that means I was not clear at all!

It's a nice topic, but indeed full of twists, and it depends on what are you interested. One way of looking at spinors is that they 'generalize' in some sense the kind of 'fields' you can have in space or spacetime. For example, in pre-relativity physics you have a real 3-dimensional vector space with the usual euclidean metric. With this metric you have its special orthogonal group SO(3), the usual 3-d 'rotations'. In this vector space, you have vectors and you can build tensors, and they all transform under the rotations in the usual way. SU(2) spinors are more 'general' entities, and they allow you to recover all this usual 3-d vector algebra. These spinors are the usual ones in non-relativistic QM.

But in relativity, the (real) vector space is 4-dimensional and with a metric of Lorentzian signature, the group is the (for now, restricted; i.e., without space or time inversions) Lorentz group, SO'(1,3) (the ' indicates that we are taking the connected component of the identity). One may naturally ask, is there a similar spinor construction for this vector space? And the answer is yes, SL(2,C) spinors, i.e., you just take a bigger subgroup of GL(2, C). The underlying space on which these matrices act is still a 2-dimensional complex vector space, i.e., these spinors are always two-component entities. This approach can be found in, e.g., http://home.uchicago.edu/~geroch/Links_to_Notes.html [Broken] on spinors or in chapter 13 of Wald's GR. So, in this approach, they are pretty analogous entities, you only change the underlying groups, real vector space and metric.

Now, in relativistic QM, you want a state space equipped with the (for now, restricted) Poincaré group acting as the relativity symmetry group of the system. This implies that you need a true unitary irrepresentation of its universal cover. You can build explicitly all these possible representations by using the 'Mackey machine' method. The inequivalent classes are labeled by mass and spin (helicity, if m=0). For m>0, spin-1/2 and positive energy (the ones with negative energy give you the antimatter when you consider the wavefunctions as classical fields to be quantized), the representation space is the Hilbert space of certain (tempered-distributional) solutions of the following equation:

$\left(\square+m^{2}\right)\varphi^{A}=0$

where $$\varphi^{A}$$ is a two-component SL(2,C) spinor field in spacetime. The representation is given by the natural action of the Poincaré group on SL(2,C) spinor fields. The structure of the equation simply comes from the usual relativistic formula $$E^{2}-p^{2}=m^{2}$$ (in momentum space, we are taking our functions on the positive energy mass shell, and that's how this formula enters here).

But this is a second order equation. If we define the following auxiliary variable, $$\sigma_{A'}=\frac{\sqrt{2}}{m}\partial_{A'A}\varphi^{A}$$ (the derivative is just the usual derivative operator generalized to spinors), the second order equation implies $$\partial^{A'A}\sigma_{A'}=-\frac{m}{\sqrt{2}}\varphi^{A}$$, i.e., the initial second order equation leads to a coupled system of two first order equtions. The implication also works in the other direction, i.e., the system implies the initial equation. If you define the Dirac spinor as $$\psi=(\psi_{0},\psi_{1},\psi_{2},\psi_{3})=(\varphi^{A},\sigma_{A'})$$, you can see that this system is simply the Dirac equation (in the Weyl representation). Also, the transformations laws of SL(2,C) spinors imply the correct transformation law for Dirac spinors. In this sense, SL(2,C) spinors can be seen as the 'atoms' of spinors, from which you can build other 'bigger' spinors in relativity. The 4-components in the Dirac spinor arise when one tries to express the 'wave equation' we got as a first order equation. Notice that the Clifford algebra arises when one tries to write a first order multicomponent relativistic-covariant wave equation (this is what Dirac did). We know that the minimum number of components for that is 4, so everything is self-consistent. Of course, with the two-components SL(2,C) spinors one can see more clearly the connection with the usual spin-1/2 representation of SU(2), which is obscured in the 4-components formalism of Dirac spinors. The usual ways in which you can get real spacetime 4-vectors (the conserved current vector, in particular) from Dirac spinors wavefunctions, can be seen as arising from the fundamental property of SL(2,C) spinors that motivated their introduction, and of course the usual formulas can be obtained in this way.

So, if we are only working with the restricted relativity groups, you can choose second order/two components or first order/four components. But, as @king vitamin said, if you add parity to the picture, since this operation interchanges the pairs in the Dirac spinors, then one has to use the 4-components formalism.

Last edited by a moderator: May 7, 2017
14. Oct 3, 2014

### PhilDSP

Hi aleazk,

It doesn't have to be so complicated. Parity relationships and topological support, i.e. SU(2), fall automatically out of Dirac's linearization of the relativistic energy equation. That's one of the great beauties of that approach and why it is so general (leading to equations for other types of particles such as neutrinos).

15. Oct 3, 2014

### aleazk

Hi, Phil. I must say I'm still learning all this, and I tend to look many references and get quite overwhelmed by all the information. I'm sure it can be greatly simplified, and I agree that my reasoning is a little messy. I hope that with time things will get more simpler in my head. In fact, one of the reasons I post here is to get corrections and new insights because I'm very aware I tend to overcomplicate things!

I will think about what you say.

16. Oct 4, 2014

### vanhees71

A particularly useful book for all that is

Sexl, R., Urbandtke, H., Relativity, Groups, Particles, Springer (2001)

You find my own attempt to explain the unitary representations of the Poincare group in my QFT manuscript (particularly in Appendix B):

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf