Arcadian Functor

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Marni D. Sheppeard

Sunday, February 15, 2009

M Theory Lesson 263

In summary, this approximate solution for the CKM matrix uses the four parameters $A^2 = 676$, $B^2 = 697$, $C^2 = 29$ and $X^2 = 1$. With an appropriate choice of signs, the magic circulant parameters become

$3I = 26 + 2 \sqrt{697}$
$3J = 26 - \sqrt{697} - \sqrt{3}$
$3K = 26 - \sqrt{697} + \sqrt{3}$
$3R = \sqrt{53} - 2 \sqrt{2}$
$3G = \sqrt{53} + \sqrt{2} + \sqrt{87}$
$3B = \sqrt{53} + \sqrt{2} - \sqrt{87}$

and the resulting (square) matrix values are given by

0.9495, 0.0502, 0.0003
0.0498, 0.9465, 0.0037
0.0008, 0.0033, 0.9960

Observe that the largest disagreement with experiment is in the very small $V_{td}$ and $V_{ub}$ values.


Anonymous Anonymous said...

Note also that the square root integer parameters $B$, $C$ and $X$ are the same parameters that determine the magic square form of the MNS matrix for neutrino mixing.

Remote Kea

February 15, 2009 1:12 PM  
Blogger Kea said...

P.S. Carl, I can't imagine why people think quarks are so complicated. The Fourier transform is really a very simple gadget.

February 15, 2009 1:31 PM  
Anonymous Anonymous said...

Carl, the small values come out better if one replaces the 1-circulant (I,J,K) with the 1-circulant (sqrt(I^2 + J^2 + K^2),0,0), using the same values for the variables.

Remote Kea

February 15, 2009 6:40 PM  
Blogger CarlBrannen said...


I've been in bed with flu the past few days and had time to cogitate in complete boredom (no internet, no tv, etc.). I had this realization that while it is going to be fairly easy to get the MNS matrix to be unitary, what with that elegant 1-circulant and 2-circulant structure, the same could not be said of the CKM matrix.

The most naive way of calculating the CKM / MNS matrices would be to assume complete commutativity, and use (1,1,1), (1,w,x), and (1,x,w) to represent the three generations, where w is the cubed root of unity and x is its complex conjugate.

Now take the 6 permutation matrices, I, J, K, R, G, and B. Think of these as different ways of hooking up the incoming and outgoing preons. The simplest way of summing over all these possibilities is to, well, just sum over them. (This ignores stuff like coupling constants.)

But I+J+K = R+G+B = D, the democratic matrix. And of the three generations, only the 3rd generation (1,1,1) is an eigenvector of D. All the others are annihilated. That means that the bt entry in the CKM matrix could be larger in magnitude than allowed under unitarity.

It seems to me that increasing that CKM entry could cause more weak decay of heavy quarks and that will give the extra muon signal that Tommaso has been exploring recently. That this kind of thing could go on in the CKM matrix entry would be due to the fact that the entries are determined largely by branching ratios rather than absolute measurements (which would detect changes to the weak coupling constant which is what I'm talking about here).

Of course I+J+K = R+G+B = D gets modified when you do non commutativity. For the MNS matrix, things will be very pure and one could end up with a nice clean 1-circ + 2-circ result. For the CKM matrix, the situation is different because the quarks are not composed of three identical preons. Things get uglified and I feel that there will be a tendency to be less pure and more random, hence the violation of unitarity.

Calculations are proceeding.

February 17, 2009 4:33 PM  

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