This reminds me of one of those facts about the Pauli spin matrices (and more generally, Clifford algebras), that no one ever talks about.
Normally one models the spin of elementary particles with spinors, in this case, those 2x1 vectors, and one treats the matrices as operators. It would be a little cleaner if the matrices were all you needed. It turns out that it is possible to model spin with the matrices alone.
Using the usual basis for the Pauli spin matrices (or any other basis you prefer, or you can refuse to specify a basis and treat it in Clifford algebra form), let \hat{u} = u_x\hat{x} + u_y \hat{y} + u_z\hat{z}, where \hat{u} has length one. For a particle with spin in the +\hat{u} direction, associate the particle with a matrix as follows:
|+1/2_u +1/2 \rangle \equiv (1 + u_x\sigma_x + u_y\sigma_y + u_z\sigma_z)/2.
The matrices so defined are "idempotent", which means that when you square them, you get back the same value. To compute the inner product, \langle A | B \rangle, compute the usual complex conjugate product and compute the trace. The inner product so computed is not quite the same as the inner product computed from spinors, but in certain ways it is an improvement.
From a physical point of view, what we can measure about spin states are probabilities. For example, the probability that a spin-1/2 particle that is oriented in the +z direction will give +1/2 when measured in some other direction.
When computing the transition probability between two spinors, say A and B, one computes:
|\langle A | B \rangle |^2
That is, one computes the spinor inner product, and then takes the magnitude and squares it.
For the idempotent matrices defined above, one uses the matrix inner product and then takes the magnitude. There is no need to square it. The result is the same, but using the idempotents eliminates the extra squaring operation.
Here's the (somewhat trivial) calculation for the inner product between two matrices, one associated with spin+1/2 in the +z direction, the other with spin+1/2 in the +u direction, with \hat{u} = \cos(\theta)\hat{z} + \sin(\theta)\hat{x}:
\textrm{tr}(\left(\begin{array}{cc}1 & 0 \\ 0 & 0 \end{array}\right)<br />
\left(\begin{array}{cc}1+\cos(\theta) & \sin(\theta) \\<br />
\sin(\theta) & 1-\cos(\theta) \end{array}\right)/2)
= (1 + \cos(\theta))/2
While this is obvious in the above case, it generalizes to any two of these idempotent matrices that represent spin in whatever directions you'd like. (Try it if you don't believe me, or better, use \sigma_x notation and the result will be obvious.)
Perhaps the idempotents are a more basic method of modeling elementary particles than spinors.
Carl
