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 σ₁, σ₂, σ₃, 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 Cl_2,0(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 2^n-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 ...