User:Jheald/sandbox/GA/3D spinors workpage

Some thoughts towards adding a GA-based translation to Spinors in three dimensions

cf. User:Jheald/sandbox/Spinors in two dimensions

Introduction
Should set out physical significance, most relevantly in the Pauli equation.

+ elsewhere: for finding irreps of SU(2)/SO(3) ?

Representation of the Clifford Algebra
The signature of the algebra requires
 * $$e_1^2 = e_2^2 = e_3^2 = 1\;$$

and the Clifford condition requires that
 * $$e_i e_j + e_j e_i = 0, \; \mbox{for } i \neq j$$

making
 * $$e_{1234}^2 = -1\;$$

These conditions can be fulfilled by using the Pauli matrices &sigma;1, &sigma;2, &sigma;3, so that:

1 \simeq \begin{pmatrix} 1&0\\0&1 \end{pmatrix};\; e_1 \simeq \begin{pmatrix} 0&1\\1&0 \end{pmatrix};\; e_2 \simeq \begin{pmatrix} 0&-i\\i&0 \end{pmatrix};\; e_3 \simeq \begin{pmatrix} 1&0\\0&-1 \end{pmatrix}$$ giving

e_{12} \simeq \begin{pmatrix} i&0\\0&-i \end{pmatrix};\; e_{23} \simeq \begin{pmatrix} 0&i\\i&0 \end{pmatrix};\; e_{31} \simeq \begin{pmatrix} 0&1\\-1&0 \end{pmatrix};\; e_{123} \simeq \begin{pmatrix} i&0\\0&i \end{pmatrix};$$

Spinors, per Lounesto

 * (1+e3)(1+e3) = 2 (1+e3)
 * so ½(1+e3) is an idempotent


 * Lounesto, Clifford Algebras and Spinors (2e, 2001), p. 60, (Google books) gives the example of the projection of Cl3 when right-multiplied by the idempotent ½(1+e3), to give a linear subspace spanned by:
 * f0 = ½(1+e3) = ½(1)(1+e3),
 * f1 = ½(e23+e2) = ½(e2)(1+e3),
 * f2 = ½(e31-e1) = ½(-e13)(1+e3),
 * f3 = ½(e12+e123) = ½(e12)(1+e3),
 * It is clear we could also choose f0= ½(1)(1+e3), -f2= ½(e1)(1+e3), f1= ½(e2)(1+e3), f3 = ½(e12)(1+e3), to give a spinor subspace isomorphic to Cl2,0(R)


 * Evidently, another idempotent we could also have used to project a complex vector would be ½(1-e3)
 * Why the preferred choice? Why doesn't ½(1+e2) do just as well ?
 * Perhaps we should look at the equivalent 4x4 real matrix representation, and see what happens when we bop off columns. Knocking off 1 column of complex numbers is equivalent to knocking off 2 columns of reals, so lets see what the elements are; it should give us 6 ways to choose 2 from 4, enough to take care of the base choice and the ± choice.


 * Using the representation
 * $$ i \simeq \begin{pmatrix} 0&-1\\1&0 \end{pmatrix};\;$$

we get:

1 \simeq \left(\begin{smallmatrix} 1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{smallmatrix}\right);\; e_1 \simeq \left(\begin{smallmatrix} 0&0&1&0\\0&0&0&1\\1&0&0&0\\0&1&0&0\end{smallmatrix}\right);\; e_2 \simeq \left(\begin{smallmatrix} 0&0&0&1\\0&0&-1&0\\0&-1&0&0\\1&0&0&0\end{smallmatrix}\right);\; e_3 \simeq \left(\begin{smallmatrix} 1&0&0&0\\0&1&0&0\\0&0&-1&0\\0&0&0&-1\end{smallmatrix}\right)$$ and

e_{12} \simeq \left(\begin{smallmatrix} 0&-1&0&0\\1&0&0&0\\0&0&0&1\\0&0&-1&0 \end{smallmatrix}\right);\; e_{23} \simeq \left(\begin{smallmatrix} 0&0&0&-1\\0&0&1&0\\0&-1&0&0\\1&0&0&0 \end{smallmatrix}\right);\; e_{31} \simeq \left(\begin{smallmatrix} 0&0&1&0\\0&0&0&1\\-1&0&0&0\\0&-1&0&0 \end{smallmatrix}\right);\; e_{123} \simeq \left(\begin{smallmatrix} 0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0 \end{smallmatrix}\right);$$

Not as helpful as I hoped it was going to be. Can't apparently adjust col.1 independently of col.2; and $$1+e_2$$ is pretty dense -- hard to identify any subspaces that it creates. Besides, we wouldn't be trying to annul one column (would we?) -- we'd be trying to annul all but one column, to produce our real column vector.
 * I guess what we have is that projecting out 1 dimension (a) cuts the size of the Clifford algebra from 2n to 2n-1 elements, and (b) block-zeros one half of the corresponding matrix, by multiplying by something equivalent to the block matrix $$\left(\begin{smallmatrix}1 & 0 \\ 0 & 0\end{smallmatrix}\right)$$ -- similarly cutting the number of live elements by a factor of 2
 * ... curious how all this is discriminating between dimensions with signature +1 and dimensions with signature -1 ...