Dirac algebra: Spatial inversion of cliffors via the unit time vector

July 9, 2009 1 Comment

A cliffor in Clifford algebra \mathcal Cl_{3,0} is given by

(1)\qquad \hat A=A_0+\mathbf A_1+i\mathbf A_2+iA_3,

which is a sum of a scalar, a vector, an imaginary vector (bivector), and an imaginary scalar (trivector).

Let \mathbf e_4\equiv\mathbf e_0\equiv\,^\circ be the unit time vector in the Clifford algebra \mathcal Cl_{4,0}, also known as the Dirac algebra with the (+,+,+,+) signature. With this vector \mathbf e_0, we can define the spatial inverse \hat A^\dagger of the cliffor \hat A in Eq. (1) as

(3)\qquad \hat A^\circ=\,^\circ\hat A^\dagger.

Distributing the time vector in each component of \hat A on the left side of the equation and using the orthonormality axiom, we arrive at

(4)\qquad\hat A^\dagger = A_0-\mathbf A_1+i\mathbf A_2+iA_3.

Thus, the spatial inversion operation only affects vectors and imaginary numbers; the scalars and imaginary vectors remain unchanged.

The algebraic definition of spatial inversion in Eq. (3) lets us easily obtain two theorems:

(5a)\qquad (\hat A+\hat B)^\dagger=\hat A^\dagger+\hat B^\dagger,
(5b)\qquad (\hat A\hat B)^\dagger=\hat A^\dagger\hat B^\dagger

That is, the spatial inverse of a sum (product) of two cliffors is the sum (product) of their individual spatial inverses. Furthermore, if \hat A^2 is a scalar, then

(6)\qquad \,^\circ e^{\hat A}=e^{\hat A^\dagger}\,^\circ,

where we used the expansion of the exponential of \hat A in terms of circular or hyperbolic cosine and sine functions.

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