User:Jheald/sandbox/temp

Geometric algebra identifications of the idempotent projectors

 * Pauli matrices - Cl(3,0):

$$\left(\begin{smallmatrix} 1 & 0 \\ 0 & 0 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 + \mathbf{e}_3); \qquad \left(\begin{smallmatrix} 0 & 0 \\ 0 & 1 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 - \mathbf{e}_3)$$


 * Gamma matrices - Cl(1,3) - Dirac basis:

$$ \left(\begin{smallmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 + \gamma_0) \tfrac{1}{2}(1 + i \gamma_1\gamma_2); \qquad \left(\begin{smallmatrix} 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 + \gamma_0) \tfrac{1}{2}(1 - i \gamma_1\gamma_2) $$

$$\left(\begin{smallmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 - \gamma_0) \tfrac{1}{2}(1 + i \gamma_1\gamma_2); \qquad \left(\begin{smallmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{smallmatrix} \right) \leftrightarrow \tfrac{1}{2}(1 - \gamma_0) \tfrac{1}{2}(1 - i \gamma_1\gamma_2) $$

But where does the i come from ?

Could pick (1/2) (1 ± &gamma;1&gamma;2&gamma;3) as the idempotent, but then this doesn't correspond to a column projector.
 * Problem may be that (1 + &gamma;0)(1 + &gamma;123) ≠ (1 + &gamma;123)(1 + &gamma;0) -- &gamma;0 does not commute with &gamma;123


 * Cl(1,3) - Weyl basis:

$$\tfrac{1}{2}(1 + \gamma_{0123}) \tfrac{1}{2}(1 + i \gamma_1\gamma_2)$$


 * Cl(1,3) - Majorana basis:

$$\tfrac{1}{2}(1 + i \gamma_1) \tfrac{1}{2}(1 - \gamma_0 \gamma_2)$$


 * Cl(3,1) - Majorana basis:

$$\tfrac{1}{2}(1 + \gamma_1) \tfrac{1}{2}(1 + \gamma_0 \gamma_2)$$