Arcadian Functor

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Marni D. Sheppeard

Monday, June 30, 2008

M Theory Lesson 203

As usual Carl has jumped ahead with a post on mixing matrices as magic squares. For reference, let us collect here some actual figures for the CKM matrix, given in this article from the Particle Data Group. Absolute value signs are omitted.

$M_{ud} = 0.97377 \pm 0.00027$
$M_{us} = 0.2257 \pm 0.0021$
$M_{ub} = 4.31 \pm 0.30 \times 10^{-3}$
$M_{cd} = 0.230 \pm 0.011$
$M_{cs} = 0.957 \pm 0.017 \pm 0.093$
$M_{cb} = 41.6 \pm 0.6 \times 10^{-3}$
$M_{td} = 7.4 \pm 0.8 \times 10^{-3}$
$M_{ts} = 40.6 \pm 2.7 \times 10^{-3}$
$M_{tb} > 0.78$

This is a little different to the values given in the wikipedia article. Standard Model analyses of these quantities can be quite complicated. Following the notation from before, in a very simple ideal double circulant the magic square property demands that $a + b = c + d$. For the CKM values (squared) we see that rows and columns do indeed sum to 1, and $c + d \simeq 1$ because $b$ is so small.


Blogger CarlBrannen said...

The sum ud+us+ub = 0.97377 + 0.2257 + 0.0043 is significantly greater than 1. However, there are arbitrary phases to consider.

The basic problem, defining a 3x3 matrix with rows and columns that sum to 1 and also in squared magnitude, seems to be a problem solved in a 2-d manifold.

That is, a 2x2 square of values define all the others in order to make the square magic. This is 4 complex numbers or 8 real degrees of freedom.

Then requiring the squared magnitudes sum to 1 gives 6 equations which seem to be independent. So I think there are 2 degrees of freedom left.

By the way, the various permutation matrices are trivial examples of matrices which have this property, i.e. ((0,1,0),(0,0,1),(1,0,0)).

June 30, 2008 11:44 PM  

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