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Just as in the two-dimensional case one can also consider in three dimensions the role of spinors and rotations. To this purpose one first considers a vector transformation of the form

where u is a three-dimensional unit vector: u = u₁σ₁ + u₂σ₂ + u₃σ₃ with . This transformation can be identified as a reflection in a plane orthogonal to u if one decomposes x into a part collinear to u and a part orthogonal to u:

with x = (x · u) u and x = (xu) _Cu = (xu) · u. One can easily check that

This decomposition of x can most easily be obtained if one just star-divides x _Cu = x · u + x u by u, which gives with u^-1 _C=u:

Using (3.28) one sees that the transformation (3.26) turns x into x^′=-u _Cx _Cu=-x +x , so that only the component collinear to u is inverted, which amounts to a reflection at the plane where u is the normal vector. Two successive transformations (3.26) lead to:

where U can be written as:

where the angle between the unit vectors u and v is described by an bivector A = v u = vu = |vu|A₀. Hereby the unit bivector A₀ = vu/|vu| defines the plane in which the angle lies, while the magnitude |vu| gives the angle in radians, furthermore it fulfills A_{0 C}A₀ = −1. If one chooses for example the basis vectors σ_k for u and v, A₀ would be given by one of the bivectors in (3.21). The additional factor 1/2 in (3.31) will become clear if one investigates the action of the transformation (3.30). To this purpose one proceeds analogously to the discussion of the reflection (3.26). One first decomposes the vector x into a part x in the plane defined by A and a part x perpendicular to that plane. This is done analogously to (3.29) by star-dividing x _CA = x · A + x A by A which leads to

with x _C A = −A _Cx and x _CA = A _Cx . We then have for the transformation (3.30):

So the component perpendicular to the plane defined by A is not changed while the component in this plane is rotated with the help of the spinor by an angle of magnitude |A|, just as described in the two-dimensional case above. One sees here why the rotation in the two-dimensional case could be written just by acting with a spinor from the right. This is due to the fact that when the vector lies in the plane of rotation one has

A rotation can be described with the bivector A, but also with the dual vector a defined by A = I_{3 C}a, where the direction of a defines the axis of rotation, while the magnitude gives the radian |a| = |A|. So U can also be written as:

which corresponds to the star exponential (2.24).

The formalism described so far can easily be generalized to the case of d euclidian dimensions. Just as there is a duality in the space spanned by the σ_i there is also the duality between the spaces spanned by the σ_i and the σⁱ. This duality is expressed by the relation . The σⁱ-vectors can be constructed with the help of the pseudoscalar, which is for the d-dimensional euclidian case I_d = σ₁σ₂ σ_d. The hyperplane for which the basis vector σ_j is normal is given for an d-dimensional euclidian space by the (d−1)-blade , where means that this basis vector is missing. The corresponding dual vector is then given by

where is the inverse d-dimensional pseudoscalar.

Note also that the multiple Clifford star product leads to an expansion of Wick type. For example, the Clifford product of four basis vectors is given by

where the contraction of σ_i and σ_j is given by δ_ij. This suggests to use the star product formalism also in the realm of quantum field theory [10].

4. Geometric algebra and classical mechanics

It is now straightforward to use the formalism described so far in classical mechanics as was done in [3]. We will here only give two examples to show where the advantages of geometric algebra lie. Let us first consider the three-dimensional harmonic oscillator, which is defined by the differential equation , where q is now a supernumber: q = q₁σ₁ + q₂σ₂ + q₃σ₃. The ansatz leads to the equation , which is solved by λ = ± iω₀ with . The difference to the conventional formalism is that i is here a bivector with . This gives then the two solutions

In the second term appears the expression a_{± C}i = a_±i + a_± · i, which is the sum of a term of Grassmann grade three and a term of Grassmann grade one. But the result q_± itself is a quantity of Grassmann grade one, so it follows that a_±i = 0, which is the defining equation of the plane in which the oscillatory movement takes place. This plane is defined by the unit bivector i and has to be determined by the initial conditions [3].

As the second example we consider the solution of the Kepler problem by spinors [11]. One uses here the fact that the radial position vector r = r₁σ₁ + r₂σ₂ + r₃σ₃ can be written as a rotated and dilated basis vector:

The components r_i of r can then be expressed in terms of the components u_i of U = u₁ + u₂σ₂σ₃ + u₃σ₃ σ₁ + u₄σ₁σ₂:

which is the well-known Kustaanheimo–Stiefel transformation [12] and [13]. Comparing Figs. (4.2) and (4.3) leads to the notational correspondence

where and are four-dimensional space vectors considered as tupels of numbers as in the conventional formalism. One should note here that the KS-transformation increases the degrees of freedom by one, which means that the bivector U in (4.2) is not unique [11]. This gauge freedom can be reduced by imposing an additional constraint on U as will be shown below. Squaring (4.4) leads to the relations

with . Differentiating (4.4) with respect to t one obtains the KS-transformation for the velocities as

One can then choose for the constraint

which means that the superfluous fourth component r₄ stays zero for all times. With this constraint it is possible to invert the geometric algebra relation (4.6) for U. Implementing Figs. (4.7) and (4.6) gives , which can be solved for , so that the inverse relation to (4.6) is

By introducing a fictitious time s which is defined as

it is then possible to regularize the divergent 1/r-potential so that (4.8) reads or

Substituting now the inverse square force

one obtains:

which is the equation of motion for an harmonic oscillator. This equation can be solved in a straightforward fashion and is much easier than the equation for r. The orbit can then be calculated by (4.2).

The Kepler problem can also be treated in the canonical formalism. For this purpose one first needs the KS-transformation for the momentum. If is the canonical momentum corresponding to the KS-transformation is given by

with W = w₁ + w₂σ₂σ₃ + w₃σ₃σ₁ + w₄σ₁σ₂. For one gets with (4.13)

where and

Equation (4.14) allows to transform the Hamiltonian into u_i- and w_i-coordinates. This is done in several steps [13]. Starting from the Hamiltonian one first extends the phase space by a q₀- and a p₀-coordinate and forms the homogenous Hamiltonian as H₁ = H + p₀. This leads for the zero component to two additional Hamilton equations

which shows that q₀ corresponds to the time t and p₀ is a constant and corresponds to the negative energy of the system, so that H₁ = H + p₀ = 0 for a conservative force. Since the time is now a coordinate the development of the system has to be described with a different parameter. This development parameter is the fictitious time s that is connected to the time by (4.9). The relation (4.9) can be implemented if one chooses H₂ = rH₁. The Hamilton equations that describe then the development according to s are differential equations with respect to s:

Especially for the zero component one gets

which corresponds to (4.9). After having so far regularized the Hamiltonian one can then go over to KS-coordinates and obtains with (4.14)

Imposing now the constraint p₄ = 0, which for in (4.15) is just (4.7), and considering bound states with E < 0 the Hamiltonian is given by

which describes a four-dimensional harmonic oscillator with fixed energy and frequency ω = (|E|/2m)^1/2.

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