Overblog Suivre ce blog
Editer l'article Administration Créer mon blog

Matrices of seesaw type and the minimax principle

17 Décembre 2015, 15:13pm

Publié par Fabien Besnard

The seesaw mechanism is a popular explanation for the smallness of the observed neutrino masses. There are several kinds of seesaw mechanisms in fact, but here I will stick to the one where heavy (with mass somewhere near the grand unification scale), still unobserved, right-handed neutrinos are postulated (type 1 seesaw). After all every other particle than the neutrino comes in both right-handed and left-handed version, so it would be quite strange if only left-handed neutrinos existed. It is moreover the preferred scenario in the Noncommutative Geometry approach to the Standard Model.

So what is the seesaw mechanism ? I will explain it first, as it is generally done, in the toy-model case where there is only one generation of particles. I refer to this nice place for the physics, here I will talk only about the math (which I understand better anyway). So, suffice it to say that the mass of the neutrinos (here two of them, one left-handed and one right-handed), are the (absolute values of the) eigenvalues of a two-by-two real symmetric matrix

M=[ 0   mD]

   [mD   mR]


with a zero in the upper left corner. Moreover there are some reasons to believe that the so-called Dirac mass mD is of the order of magnitude of the electroweak scale, that is 102-103 GeV whereas m_R is much much larger, around 1016 GeV (GUT scale). Hence mD/mR will be very small. With this hypothesis it is obvious that the smallest eigenvalue of M, call it m_1 will be close to 0 and the largest, call it m2, close to m_R. If you calculate them you can be more precise and find that |m1|/mR =(mD/mR)2 + terms of order four in mD/mR  and that m2/mR=1 + terms of order 2. So you end up with a very light neutrino with mass in the range 0.001-0.1 eV (very roughly) and a very heavy one with a mass of about 1016 GeV. The order of magnitude are good and the heavy neutrino could play a role in dark matter. Hence the name "seesaw": we only observe the lowest end of the seesaw.

All this is very nice, but the mathematical eye is particularly attracted by the fact that if the original matrix M you had a ratio mD/mR between the smallest and the largest entry, and this ratio is squared for the eigenvalues. One cannot help wondering what becomes of this property when it comes to matrices larger that 2x2. In fact this is an important question also from the point of view of physics, since there exist not 1 but 3 (as far as is currently known) families of particles. This means that the entries of M are no longer real numbers but 3x3 matrices. In fact another complication appears: these matrices are complex matrices. In the one-generation case one could do with real numbers, but no longer with three generations. Hence we have have a matrix

M= [0     mD ]

     [ tmD  mR]

where mR is complex symmetric, and mD is a complex 3x3 matrix. This matrix need not be diagonalizable, but it still has a singular value decomposition : M=USV, with U,V unitary and S diagonal. Since M is symmetric we have V=tU here. The diagonal entries of S are the so-called singular values of M: these are the neutrino masses (now 6 of them). If your are not familiar with the singular value decomposition, this is not a problem here: just write A=M*M, where M* is the adjoint of M. Then A is a positive hermitian matrix, and the singular values of M are just the square roots of the eigenvalues of A. We will call such an M a matrix of seesaw type provided "mD is small relative to mR". But what does small mean here ? When we answer this question our matrix M will provide us with an order of magnitude epsilon<1 (which replaces the ration mD/mwe met before) and it would be nice to prove something like this (n=3 in our this discussion but this is of no importance) :

Gap property : the singular values of M come in two families, a first family of n small ones m1<...< mn  and a family of n large ones mn+1<...<m2n, such that the ratio mn/mn+1 is smaller than epsilon squared.

Once we have stated this, the answer to our question above becomes obvious: the order of smallness epsilon would have to be the ratio of the largest singular value of mD over smallest of mR.

When I considered this I thought it would be a great exam problem for my students: I would just have to look at how the physicists do this and write down the questions. Well physicists do it in this way: using a clever ansatz they write an approximate singular decomposition M=U S tU, where S is diagonal up to epsilon squared and U is unitary up to epsilon squared. This means that S=diagonal matrix + O(epsilon2), and UU*=Id + O(epsilon2) where O(epsilon2) is a matrix the entries of which are of order epsilon2. It is perfectly reasonable from a physical point of view, given the order of magnitudes in play, to neglect the entries of a matrix if they are small with respect to the ones of another which are larger. However there can be devilish cancelations: a matrix with large entries can have singular values as small as 0. Hence this procedure could not give a neat answer to the "gap conjecture" above. Moreover with the O(epsilon2) cluttering everywhere you easily forget that they hide multiplicative constants. Since physically the matrices are not very large (6x6) and epsilon is really small, this is of no consequence, but mathematical generality demands the property to be true as soon as epsilon<1, so you would have to take the constants into account.

Luckilly there is a beautiful theorem coming to the rescue: the Cauchy interlacing theorem is just what you need. It gives you majorations and minorations of eigenvalues of a submatrix of larger hermitian matrix. This is a corollary of the so-called minimax principle, also known as the Courant-Fischer-Weyl theorem. Proving the gap property from the Cauchy interlacing theorem is quite easy. Unfortunately my students didn't know about this theorem, so my idea of exam problem turned out to be too hard anyway. But if you want you can try it for yourself ! A solution is below. Have fun !

Commenter cet article