Vector rotation in 3D using exponentials of imaginary vectors

July 7, 2009 Leave a comment

If we define \mathbf r' as the vector \mathbf r rotated about the vector \mathbf n by a counterclockwise angle \theta, we write

(1)\qquad \mathbf r'=e^{-i\mathbf n\theta/2}\,\mathbf r\, e^{i\mathbf n\theta/2}.

To simplify this expression, we first express the position vectors \mathbf r and \mathbf r' in terms of their components parallel and perpendicular to \mathbf n:

(2a)\qquad \mathbf r=\mathbf r_\parallel+\mathbf r_\perp,
(2b)\qquad \mathbf r'=\mathbf r'_\parallel+\mathbf r'_\perp.

Substituting these expressions back into Eq. (1) and using the identities,

(3a)\qquad \mathbf r_\parallel e^{i\mathbf n\theta/2}=e^{i\mathbf n\theta/2}\mathbf r_\parallel
(3b)\qquad \mathbf r_\perp e^{i\mathbf n\theta/2}=e^{-i\mathbf n\theta/2}\mathbf r_\perp.

we arrive at

(4)\qquad\mathbf r'_\parallel +\mathbf r'_\perp=\mathbf r_\parallel+\mathbf r_\perp e^{i\mathbf n\theta}.

The second term of the right-hand side of Eq. (4) may be expanded as

(5)\qquad\mathbf r_\perp e^{i\mathbf n\theta}=\mathbf r_\perp(\cos\theta +\mathbf n\sin\theta)=\mathbf r_\perp\cos\theta+\mathbf r_\perp(i\mathbf n)\sin\theta.

Since \mathbf r_\perp\cdot\mathbf n=0, then by the Pauli identity, we have

(6)\qquad\mathbf r_\perp(i\mathbf n)=i(\mathbf r_\perp\mathbf n)=i(i\mathbf r_\perp\times\mathbf n)=-\mathbf r_\perp\times\mathbf n.

Thus, Eq. (5) becomes

(7)\qquad\mathbf r_\perp e^{i\mathbf n\theta}=\mathbf r_\perp\cos\theta-\mathbf r_\perp\times\mathbf n\sin\theta.

Notice that \mathbf r_\perp e^{i\mathbf n\theta} is a vector.

It is easy to show that \mathbf r_\perp e^{i\mathbf n\theta} is perpendicular to \mathbf n. The straightforward way is to take the dot product of these two vectors and show that the result is zero. The other way is to show that these two vectors anticommute. Indeed they do:

(8)\qquad\mathbf r_\parallel\mathbf r_\perp e^{i\mathbf n\theta}=-\mathbf r_\perp\mathbf r_\parallel e^{i\mathbf n\theta}=-\mathbf r_\perp e^{i\mathbf n\theta}\mathbf r_\parallel.

Thus, we may may separate the parallel and perpendicular parts of Eq. (4) to arrive at

(9a)\qquad\mathbf r'_\parallel=\mathbf r_\parallel
(9b)\qquad\mathbf r'_\perp=\mathbf r_\perp e^{i\mathbf n\theta}.

In other words, only the component of \mathbf r perpendicular to the rotation axis \mathbf n is affected the rotation; the parallel component remains unchanged.

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Pauli identity and vector products: geometric, dot, wedge, and cross

July 6, 2009

Let \mathbf a and \mathbf b be two vectors in the Clifford algebra \mathcal Cl_{3,0}:

(1a)\qquad\mathbf a=a_1\mathbf e_1+a_2\mathbf e_2+a_3\mathbf e_3,
(1b)\qquad\mathbf a=b_1\mathbf e_1+b_2\mathbf e_2+b_3\mathbf e_3.

The geometric (direct) product of \mathbf a and \mathbf b is

(2)\qquad\mathbf a\mathbf b = a_1b_1\mathbf e_1\mathbf e_1+a_1b_2\mathbf e_1\mathbf e_2+a_1b_3\mathbf e_1\mathbf e_3
\qquad\qquad+ a_2b_1\mathbf e_2\mathbf e_1+a_2b_2\mathbf e_2\mathbf e_2+a_2b_3\mathbf e_2\mathbf e_3
\qquad\qquad+ a_3b_1\mathbf e_3\mathbf e_1+a_3b_2\mathbf e_3\mathbf e_2+a_3b_3\mathbf e_3\mathbf e_3

Using the orthonormality axiom, we can show that

(3)\qquad\mathbf a\mathbf b=\mathbf a\cdot\mathbf b+\mathbf a\wedge\mathbf b,

where

(4a)\qquad\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3,
(4b)\qquad\mathbf a\wedge\mathbf b=(a_1b_2-a_2b_1)\mathbf e_1\mathbf e_2+(a_2b_3-a_3b_2)\mathbf e_2\mathbf e_3+(a_3b_1-a_1b_3)\mathbf e_3\mathbf e_1.

In terms of the imaginary number (unit trivector) i=\mathbf e_1\mathbf e_2\mathbf e_3, we may rewrite the wedge product \mathbf a\wedge\mathbf b as

(5)\qquad\mathbf a\wedge\mathbf b=i(\mathbf a\times\mathbf b),

where

(6)\qquad \mathbf a\times\mathbf b=(a_2b_3-a_3b_2)\mathbf e_1+(a_3b_1-a_1b_3)\mathbf e_2+(a_1b_2-a_2b_1)\mathbf e_3

is the cross product of \mathbf a and \mathbf b. Substituting Eq. (5) back to Eq. (3), we obtain the familiar Pauli identity:

\mathbf a\mathbf b=\mathbf a\cdot\mathbf b+i(\mathbf a\times\mathbf b).

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