Kea said:
BTW, it's the numbering scheme that gets mentioned on slide 38.
Okay, Kea, I see what you mean. The quantum numbers (a,b,c) arise from sums over parts of graphs. That's sort of equivalent to how I combine two primitive idempotents to make a "snuark". The quantum numbers also fall in the following cases:
+1+1 == +1
+1_1 == 0
-1+1 == 0
-1-1 == -1
But they assume that the two center conditions are identical so they only have three cases.
It's probably worthwhile to type up how this works out. It gives a good illustration of how algebraic ways of playing with preons works out differently as compared to topological ways.
My way: First choose a direction, for instance, going in the +z direction. That gives a factor (1-zt)/2, which the rest of the world writes (more or less) as (1-\gamma^3\gamma^0)/2. All the primitive idempotents share the zt quantum number -1. I say "more or less", because I'm using Clifford algebra (4,1), and the gamma matrices are in the Clifford algebra (3,1). The extra guy I call "s", and is sort of like a \gamma^4 that is independent of the four other gammas. The x,y,z,s,t are just like the gamma matrices, but there's an extra one. That is, they anticommute, and I'm using east coast metric so x^2 = y^2 = z^2 = s^2 = 1, t^2 = -1. If you want west coast metric, multiply by a factor of i for each symbol. This notation greatly eases calculation in that it makes stuff shorter. And when one does QFT on qubits, one has no spatial dependence so there's no need to be confused about what "z" means, \gamma^3 or a spatial coordinate.
By the way, as I mentioned above, Smolin etal are making the assumption required by special relativity that the chiral fermions are speed c and frozen. I also make a similar assumption, but at one stage deeper. I'm assuming that the primitive idempotents are "speed c" and therefore frozen and not subject to linear superposition, but I don't assume that the chiral fermions are so restricted.
Choosing two more quantum numbers that commute with zt, we use ixy and s. There are therefore four primitive idempotents to choose between:
\begin{array}{c}<br />
(1-zt)(1-ixy)(1-s)/8,\\<br />
(1-zt)(1-ixy)(1+s)/8,\\<br />
(1-zt)(1+ixy)(1-s)/8,\\<br />
(1-zt)(1+ixy)(1+s)/8,<br />
\end{array}
Writing out an idempotent, and ignoring the common factor (1-z)/8, their components are:
\begin{array}{ccccc}<br />
A = &+1 & -ixy & -s & +ixys\\<br />
B = &+1 & -ixy & +s & -ixys\\<br />
C = &+1 & +ixy & -s & -ixys\\<br />
D = &+1 & +ixy & +s & +ixys\end{array}
Assume that the highest energy is associated with the largest number of reflections (in geometric algebra, each vector is interpreted as a reflection in that direction. For example, x is a reflection in the x direction. That's parly why xx = 1. See http://modelingnts.la.asu.edu/pdf/crystalsymmetry.pdf for a description of how geometric algebra is used in the practical problem of classifying the crystal symmetries.) Therefore, of the above four terms, 1, ixy, s, ixys, the one with the highest energy is ixys.
To cancel off the very high ixys energy, we have to combine the primitive idempotents in pairs. There are four possible pairs, A+B, A+C, B+D, C+D. With the ixys quantum number canceled and the 1 quantum number not distinguishing anything from anything else, there are two quantum numbers left, ixy and s. The sums on these come out as:
\begin{array}{lcc}<br />
\textrm{Snuark}& ixy & s\\<br />
A+B=&-2&0\\<br />
A+C=&0&-2\\<br />
B+D=&0&+2\\<br />
C+D=&+2&0\end{array}
The (+2,-2) are interpreted as a sort of weak isospin doublet, and the (0,0) are two weak isospin singlets. These are the snuarks, roughly equivalent to the ribbons of the helot theory.
So here's the difference between my snuarks and their ribbons. I have four snuarks with quantum charges of +2, 0, 0, -2. They have three ribbons, which carry charges of +1, 0, -1. I've got an extra 0 state, which makes my neutrino be a Dirac particle, and there is an overall factor of two difference, which means nothing.
Carl
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