M Theory Lesson 227
Recall that the two dimensional Fourier operator diagonalises a circulant via conjugation. For the Pauli matrices, again ignoring factors of 2, we have 
so the Fourier operator cycles between the mutually unbiased bases for the two dimensional space. This shows how the non-zero entries of $\sigma_{Z}$ really are identified with the spin eigenvalues of the Fourier generator $\sigma_{X}$. Observe also how these Fourier maps
$\sigma_{X} \rightarrow \sigma_{Z} \rightarrow \sigma_{Y} \rightarrow \sigma_{X}$
cycle through the three directions of the trefoil knot quandle. In other words, this quantum action acts on the embedding space for a trefoil by interchanging the knot crossings. If these crossings are drawn on the squares of an associahedron Riemann sphere, then the action cycles the points $0$, $1$ and $\infty$.
so the Fourier operator cycles between the mutually unbiased bases for the two dimensional space. This shows how the non-zero entries of $\sigma_{Z}$ really are identified with the spin eigenvalues of the Fourier generator $\sigma_{X}$. Observe also how these Fourier maps$\sigma_{X} \rightarrow \sigma_{Z} \rightarrow \sigma_{Y} \rightarrow \sigma_{X}$
cycle through the three directions of the trefoil knot quandle. In other words, this quantum action acts on the embedding space for a trefoil by interchanging the knot crossings. If these crossings are drawn on the squares of an associahedron Riemann sphere, then the action cycles the points $0$, $1$ and $\infty$.
posted by Kea | 3:02 PM
2 Comments:
"so the Fourier operator cycles between the mutually unbiased bases for the two dimensional space." This follows when you write the bases as density operators but it might not be obvious to some of your other readers (who might think it nonsense).
By the way, I'm fairly certain that you can rewrite F so that it doesn't need the minus signs. In any case, since there are two of them, The F map returns a basis state back to where it came from. That is
FoFoF [( 1 + sigma_j)/2]
= (1 + sigma_j)/2
where "F" is the operator map you're talking about.
One thing, I don't think has been properly commented on. The Fourier maps imply that there is a three-fold symmetry present here. It is from this that we eventually get the SU(3) qutrit color symmetry for the mesons. There is also some U(1) mixed in. So it is a way of breaking things apart that is more complicated from the point of view of symmetry, but more natural, from the point of view of Fourier transforms.
Carl, factors of pi, 2, -1, or any combination of these, are highly likely to appear in my posts randomly because (a) I am prone to mistakes and (b) I am too lazy to worry about optimising the exposition.
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