Work in the Pauli algebra, as a complex Clifford algebra C(3,0). The basis vectors are \hat{1}, \hat{x}, \hat{y}, \hat{z}. To find the primitive idempotents (also called primitive projection operators, also called fundamental states, depending on author), one first finds as many commuting square roots of unity as possible. In the Pauli algebra, "as many" means "one". If U is such a square root of unity, then the associated primitive idempotent is (1+U)/2
We now find the square roots of unity in the Pauli algebra.
Let A= a_1\hat{1} + a_x\hat{x} + a_y\hat{y} + a_z\hat{z} be a square root of unity, so A^2 =1. Computing, we have:
A^2 = a_1^2 + a_x^2 + a_y^2 + a_z^2 +<br />
2a_1(a_x\hat{x} + a_y\hat{y} + a_z\hat{z}) = 1.
The cross product terms like \hat{x}\hat{y} all cancel out. To get a non trivial result, we have to have a_1 = 0, and this leaves:
1 = a_x^2 + a_y^2 + a_z^2
If we assume that a_\chi are real, then we get the usual primitive idempotents, which are Hermitian. The non Hermitian primitive idempotents (NHPIs) are complex. Convert to reals by replacing a_\chi = \alpha_\chi + i\beta_\chi. Splitting into real and imaginary parts we get:
1 = (\alpha_x^2 +\alpha_y^2 + \alpha_z^2) - (\beta_x^2 + \beta_y^2 + \beta_z^2)
and
0 = 2(\alpha_x\beta_x + \alpha_y\beta_y + \alpha_z\beta_z)
Thinking of \alpha_\chi, \beta_\chi as vectors, we have:
1 = |\vec{\alpha}|^2 - |\vec{\beta}|^2
0 = \vec{\alpha} \cdot \vec{\beta}
Thus \vec{\alpha} and \vec{\beta} are two perpendicular vectors with lengths that can be characterized by an angle \kappa, with
|\vec{\alpha}| = \sec(\kappa) and |\vec{\beta}|=\tan(\kappa)
Thus non Hermitian primitive idempotents are characterized by these two perpendicular vectors. The cross product of these two vectors define a third vector and these give a complete basis set for 3-space. This is in contrast to the Hermitian case where only one vector is defined, and there is a freedom in how one could choose the other two vectors.
To see the relationship between the Hermitian primitive idempotents and the non Hermitian ones, let us analyze what happens when two Hermitian primitive idempotents are multiplied. The physical model of this situation is two consecutive Stern-Gerlach filters. If they are antiparallel, their product is zero and this is boring, so we exclude this case. Let us label these two Hermitian primitive idempotents R and G, standing for "red" and "green".
One can use spinors to conveniently prove simple things about Pauli operators. Choose spinors so that R = |R><R| and G = |G><G|. Then
RGR = |R><R| |G><G| |R><R| = (<R|G><G|R>) |R><R| = |<R|G>|^2 R,
and RGR must be a real multiple of R. Of course this real multiple is just (1+cos(RG))/2, where cos(RG) is the cosine of the angle between the spin axes of R and G. Primitive idempotents in the Pauli algebra are the solutions to N^2 = N, other than 1 or 0. It is easy to show that
N = 2RG/(1+cos(RG))
is a primitive idempotent.
If R = G, then 2RG/(1+cos(RG)) = R = G, and the product is a Hermitian primitive idempotent. Otherwise, N is a non Hermitian primitive idempotent (NHPI), but in either case, the product RG is a primitive idempotent. What's more, any primitive idempotent of the Pauli algebra can be written as a product of two Hermitian primitive idempotents in a unique way. Physically, this means that two consecutive Stern-Gerlach filters describe all the possible things you can do with Stern-Gerlach filters if you restrict yourself to things that are idempotent and non trivial (i.e. not 0 or 1), and you ignore the overall amplitude scaling (i.e. you normalize them).
Now all this discussion has been in a simplified version of QM in that it is describing only the discrete degrees of freedom. Time and space are not included. In standard quantum mechanics, Hermiticity seems to be related to the breaking of PT symmetry. For example, see
The Physics of non-Hermitian Operators (workshop)
http://academic.sun.ac.za/workshop/
http://academic.sun.ac.za/workshop/documents/abstracts.pdf
So this might be important for the foundations of particle physics. From a physical point of view, when one is dealing with products of two Hermitian PIs, if all one is worried about is amplitudes and phases, then swapping the two PIs will negate the phase, but leave the amplitude unchanged. That is, two Stern-Gerlach filters pass the same amount of stuff no matter which of the two orders they are placed in an unpolarized beam. On the other hand, with non Hermitian PIs, switching the order still negates the phase, but can also change the amplitude. This is another way of saying that time ordering matters more in NHPIs than HPIs.
As you may know, I found a way of describing the masses of the charged leptons using the eigenvalues of a very simple complex circulant 3x3 matrix (see
http://brannenworks.com/MASSES.pdf ). Circulant matrices are already used to decribe the mixing angles of the leptons.
An Hermitian primitive idempotent has a lot of physical interpretations besides just being a projection operator. It is the quantum state picked out by a Stern-Gerlach filter. It is the physical condition in the Stern-Gerlach filter that picks out that quantum state. Finally, and most importantly, when thinking of the situation in terms of Feynman diagrams, the primitive idempotent corresponds to the propagator for that particular type of particle. The requirement of idempotency is therefore a sort of dressing of the propagator.
With this last physical interpretation, the non Hermitian primitive idempotents are interpreted as Feynman diagrams where the particle arrives in one condition and leaves in another. Here, as in the case with a Stern-Gerlach filter, we ignore the mechanism that causes the change in state, and only look at the initial and final fermion states. Physically, we can suppose that what is going on is that a gauge boson is exchanged that is infinitely heavy so all the action happens at a single point in space (as is suitable for the structure inside of a point particle).
Choose three vectors that happen to have the same angle between each pair. Write the (Hermitian) Pauli projection operators in those three directions, say R, G, and B. Now write down the nine possible products of those three projection operators taken in pairs (order matters). Put those nine products of projection operators into a 3x3 matrix:
\left(\begin{array}{ccc}R&RG&RB\\GR&G&GB\\BR&BG&B\end{array}\right)
and solve for idempotency in matrix multiplication. That is, multiply the nine products by arbitary complex numbers, and find what these have to be in order for the 3x3 matrix to be idempotent. Physically, this corresponds to looking for a collection of nine Feynman diagrams that are stable when their final states are fed back into their initial states. For example, one hooks a final G state up to all the initial G states -- this is handled automatically by the matrix multiplication. What this gives is a non perturbational method of dealing with certain types of (very strongly bound) bound states.
To solve this sort of problem, one replaces the non Hermitian PIs with complex constants that happen to satisfy the same multiplication rules as the nine products of projection operators. The result is a circulant 3x3 matrix, for which the primitive idempotent solutions are well known. Then you translate back into Pauli projection operators to get the primitive idempotent projection operator solution.
There are three solutions, corresponding to the three generations. If one adds up all the real (scalar) parts of the nine components, the result satisfies Koide's original formula, 2(m_e+m_\mu+m_\tau)^2 = 3(m_e^2 + m_\mu^2 + m_\tau^2), if and only if one chooses the angle between the vectors as 90 degrees. However, you don't quite get the masses of the charged leptons.
To get the masses of the charged leptons this way, one requires that instead of starting with three circulant symmetric Hermitian PIs, one has to start with three non Hermitian PIs that satisfy the circulant symmetry. So the hope is that there is a connection between the non Hermitian PIs of the Pauli algebra, and the apparently arbitrary masses of the leptons.
The things I know about Clifford algebra primitive idempotents were mostly found by writing computer programs, rather than by just sitting there with a pencil and trying to prove things. I use Java, as I make lots of mistakes and the strong type casting finds saves me from making even more. Anyone who wants a copy of the program can send me an email.
