Rate this Entry

# Koide doublets?

Posted Oct20-09 at 04:30 AM by arivero

From a point of view, there are no Koide doublets. If we define Koide's relationship as coming from three matrix conditions:

$$M^{1/2}= A + B$$ with
1) A multiple of the identity
2) B traceless
3) $$Tr (A^2) = Tr (B^2)$$

Then the 3 in the 3/2 factor is really the dimension of the matrix, and thus the factor is 2/2 for Koide's doublets and 1/2 for Koide "singlets". So in this sense there are no Koide doublets.

If we consider that Koide's is about "45 degrees off from (1,1,1)" the result is the same: the 3 comes from the number of components of (1,1,1). Visually, it is trivial, that 45 degrees off from (1,1) one of the components is going to dissapear.

So the only extand posibility is to consider that a doublet is a triplet with a massless component. If we do the scan in this way, we find two interesting doublets, one composed by eta' and upsilon, and other composed by pion and D. Also, the kaon seem able to contribute to some doublet or triplet, but there are no a good match. Same in the barions, with Lambdas. And that is all: no more matches.

Incidentally, I wonder if there are alternatives to conditions 1,2 above. For instance [A,B]=0, or Tr ((A+B)^2)=Tr(A^2)+Tr(B^2). I am afraid that these alternatives do not fix A beyond being diagonal.
Posted in Uncategorized