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CarlB
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Apr24-06, 09:03 AM
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The above LaTex came out fairly well. The only problem is that the \rangles that should appear after each |n got deleted, likely because I use a macro for these in the original.

Since the above, I've got a new result to add to the pile. There were some mysterious 12s showing up in the above, particularly with the charged / neutral mass ratio. It turns out that the MNS mixing matrix in tribimaximal form (which is generally assumed to be the correct one in most recent reviews of neutrinos) can be put into a form where it is a 24th root of unity. To get it into this form, one multiplies it on the left by nothing other than a 3x3 matrix of eigenvectors of the circulant matrix. Here's the details:

[tex]
\frac{1}{\sqrt{3}}
\left(\begin{array}{ccc}1&e^{+2i\pi/3}&e^{-2i\pi/3}\\
1&1&1\\
1&e^{-2i\pi/3}&e^{+2i\pi/3}\end{array}\right)
\left(\begin{array}{ccc}\sqrt{2/3}&\sqrt{1/3}&0\\
-\sqrt{1/6}&\sqrt{1/3}&-\sqrt{1/2}\\
-\sqrt{1/6}&\sqrt{1/3}&\sqrt{1/2}\end{array}\right)[/tex]

[tex]=\left(\begin{array}{ccc}
\sqrt{1/2}&\;\;\;\;0\;\;\;\;&-i\sqrt{1/2}\\
0&1&0\\
\sqrt{1/2}&0&i\sqrt{1/2}\end{array}\right) =
\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)^{1/24}[/tex]

In the above, the MNS matrix is on the top right, the eigenvectors are top left. The MNS matrix transforms vectors of [tex](\nu_e, \nu_\mu, \nu_\tau)[/tex] into vectors of [tex](\nu_1,\nu_2,\nu_3)[/tex] where the first are what you get when you arrange a weak interaction with a charged lepton, and the second are the mass eigenstates of the neutrinos.

Some notes. First, there is one column of [tex]\sqrt{1/3}[/tex] in the MNS matrix and one row of the same in the array of eigenvectors. This implies that the (1,1,1) eigenvectors of the operator must correspond to the muon neutrino and the 2nd mass eigenstate. This is very useful information because it allows us to remove almost all the redundancy from the values of [tex]\delta[/tex] in the eigenvalue equation. That is, instead of simply finding the smallest value of [tex]\delta[/tex] that gives the three desired eigenvalues, we can instead choose the value that makes the (1,1,1) eigenvalue be the muon in the case of the charged leptons, and the middle neutrino in the case of the uncharged leptons. That is,

[tex]\lambda_{(1,1,1)} = \mu(1 + 2\eta^2\cos(\delta))[/tex]

with no need to guess which value of n to choose in the [tex]2n\pi/3[/tex] that otherwise shows up in the cosine.


As a result of making this change, the new delta value for the electron can be chosen as:

[tex]\delta_1 = 2\pi/3 - .22222204717(48) = \frac{\pi}{180}(120 - 12.73)[/tex]
or
[tex]\delta_1 = 4\pi/3 + .22222204717(48) = \frac{\pi}{180}(240 + 12.73)[/tex]

where the two options correspond to the two solutions for cos(x) = y, and the neutrino delta is:

[tex]\delta_0 = 7\pi/12 - .22222204717(48) =\frac{\pi}{180}(105-12.73)[/tex]
or
[tex]\delta_0 = 17\pi/12 + .22222204717(48) =\frac{\pi}{180}(75+12.73)[/tex]

It turns out that having the angle delta_1 between 60 and 120 degrees is useful in that it implies that we can add a hidden dimension to modify the Pauli algebra's usual relationship between phase and probability so that we can match the above. This is a somewhat mysterious statement that I will expound on later. It's very simple math and not at all difficult. Accordingly, we have to choose the upper of each of the two equations, and for the angle calculation, we can use either 120-12.73 or 105-12.73, with an assumption that the electron is the simple mass relationship particle, or that the neutrino is, respectively.

Carl