# Clifford Algebra dot com

1. Jun 6, 2006

### CarlB

I registered the website http://www.CliffordAlgebra.com .

My purpose is to have a website that gives a decent education in the practical uses of Clifford Algebra, as I am interested in their applications to physics.

Is there any desire over here for a more mathematical introduction to the subject?

Please contact me if anyone is interested in contributing or in commenting on what they want.

Carl

By the way, I also registered http://www.measurementalgebra.com which will explain the Schwinger measurement algebra, and http://densitymatrix.com to explain the density matrix formulation for quantum mechanics. But these are more physics oriented subjects.

Last edited by a moderator: Apr 22, 2017
2. Jul 10, 2006

### CarlB

I suppose I should give an example of what it is that I mean by applications to physics.

The usual way one defines a representation of a Clifford algebra in matrices is by listing the canonical basis vectors. For example, a representation of the Pauli algebra in Clifford algebra notation is:

$$\hat{x} = \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$$
$$\hat{y} = \left(\begin{array}{cc}0&-i\\+i&0\end{array}\right)$$
$$\hat{z} = \left(\begin{array}{cc}+1&0\\0&-1\end{array}\right)$$

In the usual spinor formulation of quantum mechanics, one represents a state corresponding to spin +1/2 in the (a,b,c) [i.e. unit vector] direction by picking a 2x1 vector that is an eigenvector of the spin operator in the (a,b,c) direction. That is, one solves for |(a,b,c)> in the eigenvector equation:

$$(a\hat{x} + b\hat{y}+c\hat{z})\; |(a,b,c)> = +|(a,b,c)>$$

In the above matrix representation, this amounts to solving the eigenvector equation:

$$\left(\begin{array}{cc}+c&a-ib\\a+ib&-c\end{array}\right) \;|(a,b,c)> = |(a,b,c)>$$

If one converts the above eigenvector equation to an "eigenmatrix" equation, the solution is obvious:

$$(a\hat{x} + b\hat{y}+c\hat{z})\; (1 + a\hat{x} + b\hat{y}+c\hat{z}) = + (1 + a\hat{x} + b\hat{y}+c\hat{z}),$$

so an eigenvector solution is to take any (nonzero) column of the matrix $$(1 + a\hat{x} + b\hat{y}+c\hat{z}).$$

The solution $$(1 + a\hat{x} + b\hat{y}+c\hat{z})/2$$ is a projection operator, and it has a trace of 1. In physics, such things are called "pure density matrices", and are associated with the quantum states of physical systems in the same manner as the usual ket formalism.

The use of pure density matrices in physics suggests that we should consider a description of a representation of a Clifford algebra by pure density matrices. In fact, this can be done, and in doing so we can obtain new insight into the structure of representations of Clifford algebras as well as the spectral decomposition of Clifford algebras.

A better illustration of the power of density matrices in representations is seen in the Dirac algebra represented over 4x4 complex matrices. Consider the following four projection operators (i.e. pure density matrices):

$$\rho_0 = \left(\begin{array}{cccc}1&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)$$
$$\rho_1 = \left(\begin{array}{cccc}0&0&0&0\\0&1&0&0\\0&0&0&0\\0&0&0&0\end{array}\right)$$
$$\rho_2 = \left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&0\end{array}\right)$$
$$\rho_3 = \left(\begin{array}{cccc}0&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&1\end{array}\right)$$

These correspond to four different states that are "compatible" in the sense that they commute. That is, it is possible to arrange for quantum measurements that cleanly distinguish between these four states. If these were considered as kets, (i.e. the four 4x1 vectors with a single entry 1 and all other entries zero) they would span all possible states. But as density matrices, they clearly do not, nor do they define the representation.

To define the representation completely from the pure density matrices, add one more pure density matrix, the "democratic" one:
$$\rho_D = \frac{1}{4}\left(\begin{array}{cccc}1&1&1&1\\1&1&1&1\\1&1&1&1\\1&1&1&1\end{array}\right)$$
Sums and products of these five density matrices span the 4x4 complex matrices, and therefore completely define the representation. For example, consider products of the form:

$$\rho_m\;\rho_D\;\rho_n$$

For values of m and n between 0 and 3, the above 16 products give the sixteen 4x4 matrices that have a 1 in a single location and all other entries zero. Clearly these define the representation.

Now to get the insight into representations of the Dirac algebra, take your usual definitions of the Dirac gamma matrices, and work backwards to write the representations in the above format. You will find that a typical representation is of the form:

$$\rho_m = (1\pm S_a)(1\pm S_b)/4$$

where $$S_a, S_b$$ are two commuting operators. Commuting operators are easy to choose in the Dirac algebra. Choose the first operator to be some Dirac bilinear, for example, $$\gamma_3\gamma_0$$, and then choose the other operator to be a different Dirac bilinear such that the two Dirac bilinears are different with an even number of terms. For this example, the available choices for the second Dirac bilinear are:
$$\gamma_1, \gamma_2, \gamma_1\gamma_2, \gamma_1\gamma_3\gamma_0, \gamma_2\gamma_1\gamma_3\gamma_0, \gamma_1\gamma_2\gamma_3\gamma_0$$

It remains only to choose $$\rho_D$$, which can be factored as follows:
$$\rho_D = (1\pm S_c)(1\pm S_d)/4$$

where $$S_c, S_d$$ are two commuting Dirac bilinears that do not commute with at least one of $$S_a,S_b$$. For example, if we choose

$$S_a = \gamma_3\gamma_0, S_b = \gamma_1$$

then a choice for $$S_c, S_d$$ could be:

$$S_c = \gamma_1\gamma_2, S_d = \gamma_3$$

From the definitions of $$\rho_\chi$$, one can work out the representations of the canonical basis vectors and then give the representation in the usual form.

Carl

3. Jul 10, 2006

### CarlB

Let $$\hat{x}, \hat{y}, \hat{z}$$ be the canonical basis elements that generate the real Clifford algebra R(3,0). A particular representation in 2x2 complex matrices would be the Pauli spin matrices.

Define three unit vectors, $$\vec{u}, \vec{v}, \vec{w}$$ and assume that they are no pair of them is parallel (or antiparallel). Then these three unit vectors give the corners of a sperical triangle on the unit sphere, a fact we will use in a moment.

Now in physics, the pure density matrix (i.e. state) that correspond to spin in the $$\vec{u}$$ direction is

$$\rho_u = (1 + S_u)/2 = (1 + u_x\hat{x} + u_y\hat{y} + u_z\hat{z})/2$$

where $$S_u$$ is the spin operator in the $$\vec{u}$$ direction, and the right hand side consists of a dot product between the unit vector u, and a vector of the Pauli spin matrices.

Note that $$(S_u)^2 = 1$$, and thus $$S_u \;\rho_u = + \rho_u$$, and therefore the state satisfies the appropriate eigenvector equation. Also note that $$(\rho_u)^2 = \rho_u$$ as is required for a pure density matrix.

Consider the product $$\rho_u\;\rho_w\;\rho_v\;\rho_u$$. It is easy to see that this product must be a complex multiple of $$\rho_u$$. What is the value of this complex number? It turns out to be:

$$\rho_u\;\rho_w\;\rho_v\;\rho_u = \sqrt{\frac{(1+\vec{u}\cdot\vec{v})}{2} \frac{(1+\vec{u}\cdot\vec{w})}{2}\frac{(1+\vec{v}\cdot\vec{w})}{2}} \exp(\hat{x}\hat{y}\hat{z} S_{uvw}/2)\; \rho_u$$

where $$S_{uvw}$$ is the surface area of the (oriented) spherical triangle defined by $$\vec{u}, \vec{v}, \vec{w}.$$

If one wishes to write $$\hat{x}\hat{y}\hat{z} = i,$$ as the physicsts do, then the exponential becomes a complex phase. The three multiplicative factors are just the factors of (1+cos(theta))/2 familiar to physicists computing probabilities of transitions between spin-1/2 states.

Carl

Last edited: Jul 10, 2006