Neutrinos lead to unexpected discovery about eigenvectors and eigenvalues
quantamagazine.org
quantamagazine.org
As far as we know, neutrinos come in three varieties: electron-neutrino, muon-neutrino, and tau-neutrino. These particles interact weakly with the charged leptons (electrons, muons, and taus). This interaction is mediated by the weak force, and specifically by the charged W-bosons. Basically, what this means is an anti-neutrino and it's associated lepton (or a neutrino and anti-lepton) can "annihilate" each other to create a W-boson. However, the W-boson is not stable and will rapidly disintegrate. Sometimes it disintegrates back into the same particles that created it, but not always.
Anyways, the laws of physics (specifically, the quantum field theory [QFT] formulation of the standard model) are written in terms of these three types of neutrinos. Electron neutrinos ALWAYS interact with electrons, and muon neutrinos ALWAYS interact with muons, etc. Specifically, each particle type is associated with a "field", which is kind of like a function that has a value at every point in space and moment in time. So for example there is a single "electron field", which is like a function that encodes information about every electron in the universe for all time. The theory of particle physics is concerned with writing down the mathematical relationship between all the fields for different particles, which is like reverse engineering the firmware of the universe.
For reasons that we needn't get into, in all the formulas we bunch these neutrino fields together into a group that looks like a vector, [v_e, v_mu, v_tau]^T (transposed so it's a column vector). Now, from very basic principles of physics, the part of the equation that describes the mass of particles (specifically, fermions) generally looks something like m(x†)x, where x is the particle field, and x† is a particular adjoint field (kind of like a vector transpose of the field).
Okay, so if you wanted to encode in the "firmware" all the masses of the fields x,y, and z, you simply have to include terms in your "master formula" (called a Lagrangian) which look like m0 (x†)x + m1 (y†)y + m2 (z†)z, and boom now particles associated with fields x, y, and z have masses m0, m1, and m2, respectively. This part you can just take on faith, but if you've studied classical Lagrangian mechanics in undergrad, it's pretty easy to follow the connection to the quantum regime.
Well, someone got clever and realized that in fact, the most general way to write the equation is to take the whole vector of neutrino fields, V = [v_e, v_mu, v_tau]^T, and add a term like V† M V, where V† is now that fancy transpose-adjoint applied to the whole vector, and M is now a 3x3 matrix. The case where each neutrino type has its own mass would correspond the case where M is diagonal with entries [m_e, m_mu, m_tau]. However, a physicist would naturally ask why should all the other entries in this matrix be exactly zero? Of course it has since been proven experimentally that this matrix is in fact not diagonal.
Now, it turns out that in QFT the thing that governs how particle fields change over time (e.g., how they travel through space) is their energy, which depends on their mass. Specifically, a particle in state A will, at some future time, transition to state B, and the formula that describes that transition depends ONLY on the energy configuration of that state. It turns out that if that mass matrix M is not diagonal, then the particles v_e, v_mu, v_tau are not _eigenvectors_ of the mass, and therefore not eigenvectors of energy. That is to say, if you had a neutrino which was observed to have a specific, known value of energy and mass, it could not be purely one type of neutrino but instead a linear combination of v_e, v_mu, and v_tau which diagonalizes the matrix M.
So herein lies the problem: the "flavor" (electron, muon, tau) of a neutrino defines how it interacts with particles. But these flavors are not themselves definite states of energy/mass.
Therefore, if you know that an (unobserved) neutrino was created along with an electron, it must have been an electron-flavor neutrino. But in order to understand how that electron type neutrino will travel through time and space, you need to translate it into a "mass eigenstate", which is to say, the eigenvectors of the mass matrix M.
So, all of this explains why the neutrino physicists care about eigenvectors and eigenvalues. The particle beam at Fermilab produces almost exclusively muon-type neutrinos, and they want to know how many of each type of neutrino to expect when they show up at the DUNE detector 1300km away. The news is that these physicists have discovered a way to write the eigenvectors only in terms of eigenvalues, which are easier to compute. Which is pretty surprising (even to Terrance Tao!), since linear algebra is an extremely mature branch of mathematics.
https://www.reddit.com/r/math/comments/ci665j/linear_algebra...
As an side, it's kind of funny to hear the physicist accent in this linear algebra paper.
> ...one can multiply any eigenvector by a phase e^{iθ}...
and:
> As a consistency check...
There are also other little things like some of the phrases as pointed out in a sibling comment. Physicists really do have their own distinct style.
This seems strange. I didn't realize A† was mainly only used in quantum mechanics, but it seems better to have a distinct notation, since A* could mean a matrix with all the entries conjugated but in the same places.
So, from a physics standpoint it appears to have a bit of a mathematics accent, which is a bit like how real accents are perceived: someone who has lived in two countries will have a foreign accent according to natives of each.
There are other notations for the Hermitian transpose, such as the dagger or the H. The asterisk is particularly physicist, I think.
And you might have studied physicist-accented mathematics; a lot of the accent often carries over into PDEs, functional analysis and related fields.
The \equiv usage is in D \equiv diag(...)
Maybe if you want to know only a few things about the eigen-vectors, or only a few.
For a full solution, I really doubt it.
Computing a full set of eigenvalues of a numerical matrix is pretty slow and the eigenvectors almost come for free; it seems like a very computationally backwards way to get n+1 full sets of eigenvalues only to get slightly incomplete data about the values of the eigenvectors.
I don't know anything about the physics of it, but I doubt that the researchers had numerical matrices that this formula helped with; I expect it was all done symbolically instead.
This is a weird characterization of Hermitian matrices — this implies that rotations are not ‘real-world’.
Edit: electricslpnsld correctly points out that the first sentence is false, see below.
Is that the case? Consider
|a -b|
|b a|
with Eigenvalues lambda = a +/- i b
a = cos theta and b = sin theta gives a rotation, so the Eigenvalues are complex.
\det(B\quad v_n)
For the determinant to make sense its argument needs to be an m×m matrix. B is an n×(n-1) matrix, v_n is an n dimensional vector.There's a comment with your exact question and the answer "It’s not multiplication, it’s concatenation: stick the column vector after the last column of B to “complete the square”."
This reminds me of the Koide’s coincidence
https://news.ycombinator.com/item?id=17961139
So the arithmetic mean(p=1) mass of (e, μ, τ) is twice of square-root-mean(p=1/2) mass of (e, μ, τ).
the p=1 and p=1/2 are defined in “generalized mean”
Element wise:
v(i,j)=√∏δ(k,i)(λi(A) − λk(A))/δ(k,n)(λi(A) − λk(Mj))
δ(x,y)(f):= ±1 if x=y else f
1. Roots of characteristic equation det(A-λI)=0 give the eigenvalues.
2. Solve (A-λI)x = 0 for each eigenvalue to find each eigenvector x.
It seems very plausible that step 2 can be reduced to an equation.
|vi,j|^2 ∏k=1,n;k!=i;(λi(A)−λk(A)) = ∏k=1,n-1;(λi(A)−λk(Mj))
Including it in the article wouldn't really make much difference for most readers. Those for whom it would can easily find it on the first page of the 2.5 page paper on arxiv.org that the article links to.
Explaining concepts without unnecessary jargon is a reasonable request since there are different levels of background knowledge in any audience. The parent comment to your reply is using undergrad-level math terms, how high do you want the barrier to be?
Knowledge is pointless when poorly articulated.
-- Richard Feynman
Only as pointless as giving someone eggs and a pound of flour when they want cookies.