I’ve been working on a derivation of the tribimaximal MNS matrix from first principles. It looks like it will require an assumption that the underlying particles follow a set of mutually unbiased bases (MUB). The calculation involves spin products. These are messy to compute unless you know some tricks I’ve discussed here previously.

So, did I waste my breath? Is teaching how to compute these things like shouting down a well? Let’s have a contest. First person to solve the following problem gets a $50 prize:

“Solve” means to write the above in reduced closed form in the comments section. The solution must be exact, for example, it must use rather than some numeric approximation such as 3.14159. Now I’ve chosen the values so that it will be easy to do with the techniques shown here, but will cause problems with mathematica or other automatic assistants. So go for the glory! Of course the winner can also ask that the prize be donated to a needy physicist such as Marni Sheppeard.

Meanwhile, I’ll share a few lines on what I’m doing with spin sums and the MNS matrix.

The MNS matrix tells how to get from charged leptons to neutral ones. Written in the new form, as a sum of a 1-circulant and a 2-circulant matrix, the MNS is extremely simple. Now my main claim to undying fame is rewriting the Koide equation as an eigenvalue equation using an extremely simple circulant matrix, and then guessing the form for the neutrino matrix (and thereby predicting the neutrino masses). Both the MNS matrix and the mass matrix for the leptons were built with square roots of 2 and are circulant so they seem very similar and surely must arise from similar reasoning.

We first have to convert the MNS matrix from being oriented towards keeping track of what happens to generations, to keeping track of what happens to the snuarks that make up a particle. This is a Fourier transform, Marni Sheppear will tell me I bet. The conversion preserves 1-circularity and 2-circularity, but transforms the MNS matrix along the following line:

where w is the complex cubed root of unity. So the transformation puts powers of the cubed root of unity on the elements. The above transformation arises from moving from a basis set of

(1,0,0), (0,1,0), (0,0,1)

to

(1,1,1), (1,w,w*), (1,w*,w).

However, there are a lot of choices here of complex phase and they give various results. I don’t mean to say that the above transformation causes the above new version of the MNs matrix, but instead to note that this is the sort of effect these kinds of things have. What is important to note is that when this matrix is shifted by one row, it picks up a complex phase of w. And w is the same cubed root of unity that is used in the generation form for the primitive idempotents that gave the Koide mass formulas.

The reasoning behind the Koide mass matrix was based on the algebra of a set of pure density matrices for spin-1/2 taken from a complete set of mutually unbiased bases (MUB). This algebra automatically generates square roots of 2 and gives the phase difference between the charged leptons and the neutrinos. So it’s natural to look for a similar derivation behind the MNS matrix.

The Koide mass matrix had the square root of 2 on the diagonal. The off diagonals took values of 1. The MNS circulant matrices also have a square root of 2 on the diagonal, but the off diagonals are different; one is 1, the other is zero. And there are two MNS matrices, one is a 1-circulant the other is a 2-circulant. And in addition, the Koide matrices had a complex phase which appears to be cancelled in the MNS form.

For the Koide mass matrix, the eigenvectors can be put into matrix form, where they are then called eigenstates instead of eigenvectors. These particular eigenstates are double sided, that is, they are eigenvectors of the mass matrix on either side. This is a cool property that eigenstates (or we might call them eigenoperators) can have that eigenvectors cannot have. It is also possible for an eigenstate to be an eigenvector of different operators on its left and right sides. A good example of this is the operator . It is an eigenstate of on the left, but it is an eigenstate of on the right.

Non Hermitian stuff does not exist in the usual bra-ket formalism of quantum mechanics because bras and kets only have one side. They amount to making an operator from an eigen bra and an eigen ket that do not match, for example, the example in the previous paragraph is an instance of . By “instance” what I mean is that it is equivalent to this ket-bra product but with a particular choice of phase (the natural one from a density operator point of view). These operators are not Hermitian and are seen very little in the literature (search arxiv for non Hermitian states to learn more, or ask me in the comments).

For the MNS matrix, the natural choices for the neutrino and charged lepton eigenstates are not 3-vectors of complex numbers, but instead primitive idempotent 3×3 matrices whose elements are taken from the algebra of the MUBs. The off diagonal elements in these 3×3 matrices are non Hermitian products of MUB projection operators like . For these matrices to be idempotent requires adjustments as compared to complex matrices. The adjustments are the complex phases of and the square roots of 2. This is necessary because it cancels the Berry-Pancharatnam phase (and the spinor transition probabilities) that otherwise would be picked up in products of the form , hence the pop quiz. Of course in addition to this pi/12, there is the phase associated with the generations, .

In the Koide formula, the masses show up as the eigenvalues of the matrix. They are automatically real because the matrix is Hermitian. However, nature is not built from mass matrices, it’s built from the states. For the generations, those states are the 3×3 matrices. One obtains their masses (or square root masses) by taking the real (or scalar) part of the 9 entries in a matrix. One can then reassemble all this into a simple mass matrix because all (complex number) circulant matrices take the same eigenvectors.

The simplest way of multiplying two 3×3 matrices of MUB elements, one taken from +x, +y, +z and the other taken by -x, -y, -z, will give zero. That is, the +x will cancel the -x, and the same for the other pairs. Instead, we have to send +x to -y or send +x to -z, and so on. For an assignment to work, it must send all the particles on valid paths; it must avoid any of the zero products. Thus we have two possible assignments:

+x to -y,

+y to -z,

+z to -x,

or

+x to -z,

+y to -x,

+z to -y.

As is usual in QM, we will have to sum over these two chocies. I’m guessing that one will give the 1-circulant part of the MNS matrix, the other will give the 2-circulant part. To modify a 3×3 matrix so that it chooses one of the above two cases, we have to modify one of the matrices. For example, we could take the right matrix and shift its elements up by one row, taking the top row back to the bottom.

To make the calculation, we have 3 choices for each of the two 3×3 matrices, corresponding to the three generations. Each can multiply the other 2 different ways, as (+x, +y, +z) to (-y, -z, -x) or (-z, -x, -y). So I have 18 calculations to make. These will hopefully give the 18 elements in the 1-circulant and 2-circulant parts of the MNS matrix.

If it weren’t for the non Hermitian parts (i.e. if these were simple 3×3 matrices of complex numbers), the products of these matrices would all be zero or 1 (as they are idempotents). It is in the non commutativity that we can hope to get the correct answer.

Now to get the symmetry seen in the MNS matrix, I have to have that changing the generation by one in one of those 3×3 matrices is equivalent to changing the generation by one (up or down) in the other. This follows immediately as the MNS matrix, when written in RGB basis instead of the generation basis, picks up a factor of w each time a row / column is shifted around.

So overall, I think I’ve got the pieces necessary to derive the MNS matrix and the Koide mass formulas in the same structure. Now I haven’t actually turned the somewhat lengthy crank to see if the square roots of 2 and 1 come out, but I think they will. So can my audience understand the arithmetic? Tell me you can by solving today’s pop quiz.

Hi Carl,

a naive attempt is

(1+x)(1-y)(1+z)(1+x)=2(1-i)(1+x)

(1+x)^k=2^(k-1) (1+x)

so the answer is ((1-i)/4)^(1591/pi)(1+x)/2

If I got it correct, by all means give some cash to Marni! I’ve got to run now, so I can’t check my answer.

(btw mathematica seems to have no problems with this particular question, just use MatrixPower)

Simon

Simon, of course that’s correct. And I’m impressed that you’re using the abbreviated notation I prefer, where x, y, and z stand for the matrices.

I can’t work out the full calculation over the next few days because we’re trying to sell our ethanol plant and we’re entertaining a visiting dignitary. But after that I should have the time to do it.

Thanks, but apart from the notation (which is almost a trivial issue at this level) I didn’t use anything that that you’ve emphasised on this blog.

Good luck with the sale, and thanks for running such an interesting blog. (Though I have to admit that sometimes I don’t read the posts that closely!)

And thanks to you! I’m sure you’ll be happy to know that Kea has graciously accepted your $50 and will be using it as back-up funds for her move.