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Background:

Let $\Lambda$ be the Lorentz transformation parameterized by the asymmetric real matrix $w_{\mu \nu}$. That is, let $\Lambda = \exp(\frac{w_{\mu \nu}}{2}J^{\mu \nu})$, where $(J^{\mu \nu})_{\alpha \beta} = \delta^\mu_\alpha \delta^\nu_\beta\ - \delta^\mu_\beta \delta^\nu_\alpha$. All indices run from $0$ to $3$, and I am using the metric signature $+---$.

The spin 1/2 representation of the Lorentz group maps $\Lambda$ to $R[\Lambda]\stackrel{\mathrm{def}}{=}\exp(\frac{w_{\mu \nu}}{2}\gamma^\mu \gamma^\nu)$, where $\{ \gamma^\mu \}_{\mu = 0,1,2,3}$ are 4-by-4 complex matrices satisfying $\gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = 2 \eta^{\mu \nu}$.

Question:

If we interpret the $\gamma^\mu$ not as matrices, but as the basis vectors of the geometric algebra $Cl(1,3)$, then we have

$R[\Lambda] x^\mu \gamma_\mu R^{-1}[\Lambda] = \Lambda^{\mu}_\nu x^\nu \gamma_\mu$.

Why is this true? I have no trouble doing the calculation - I am looking for a deeper understanding.

It seems a total coincidence to me that the matrix representation $R[\Lambda]$ happens to be also be the rotor for the Lorentz transformation $\Lambda$. Maybe this indicates that we can dispense with the matrix representation entirely, and somehow formulate the equivalent of a spin 1/2 rep. using $Cl(1,3)$ alone?

Thanks in advance for any help.

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1 Answer 1

I think the problem here is the tail wagging the dog: you've defined the spin 1/2 representation of a rotation and then found that this is indeed also the rotor that performs rotations in a bilinear fashion. I think it's better to look at it the other way around: one can construct a rotor-based transformation that happens to be a rotation, and then prove that those rotors are the spinors of spin 1/2 that you know about.

Here's how that logic goes: if $n$ is a vector normal to a hyperplane, then write $n = n^\mu \gamma_\mu$ and see that a reflection across this hyperplane is $\underline N(a) = -nan^{-1}$.

Then, observe that two such reflections performs a Lorentz rotation. Thus, a rotation can be written $\underline R(a) = mnan^{-1} m^{-1}$ for two vectors $m,n$.

$mn$ is then a Lorentz spinor. The quantity $w_{\mu \nu} \gamma^\mu \gamma^\nu$ then has the orientation of $m \wedge n$ and the magnitude of $\theta/2 = \cos^{-1} (m \cdot n/|m||n|)/2$. The spinor $mn$ then has a scalar part $m \cdot n$ and a bivector part $m \wedge n$.

In this sense, I thinnk the answer to your ultimate question is yes, it's possible to dispense with the matrices, as they almost always simply represent some kind of object. $w_{\mu \nu}$ for instance is just a collection of components of basis bivectors, describing the plane in which the transformation is being done. The components of a transformation as a whole just tell us how basis vectors change. As a matter of course, all rotations can be performed using the rotors and geometric products without interpreting any of the basis vectors or other quantities as matrices.

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I'm just curious. Where does $\cos$ in your definition of $\theta$ come from? The metric is not Euclidean. –  Andrey Sokolov Nov 27 '13 at 4:52
    
You're right; it's impossible to say which trig function goes there without knowing whether $m \wedge n$ squares to $+1$ or $-1$. –  Muphrid Nov 27 '13 at 5:01
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