
Neutrinos lead to unexpected discovery about eigenvectors and eigenvalues - furcyd
https://www.quantamagazine.org/neutrinos-lead-to-unexpected-discovery-in-basic-math-20191113/
======
garyfirestorm
Can someone explain like I am 5? I work in vibrations and often deal with real
world modes and mode shape vectors which are really eigenvalues and
eigenvectors. I see this might have an impact on the work that I do. But I
can't comprehend what the paper says.

~~~
cshimmin
Particle physicist here... it's a very long article and there's a lot going
on, so I'm not sure what you're seeking clarification on. But I'll take a
whack at it:

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.

~~~
rfhjt
What amazes me is that all this math works. This kind of tricks "lets assume
this clever 3x3 matrix describes how it works" makes sense if you're reverse
engineering a 3D rendering engine, because you know there is some simple math
under the hood. The fact that our world obeys simple mathematical equations is
surprising.

~~~
Redoubts
[https://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness...](https://en.wikipedia.org/wiki/The_Unreasonable_Effectiveness_of_Mathematics_in_the_Natural_Sciences)

~~~
hedvig
If we didn't use mathematics in science, what else could we have used?

~~~
qubex
Narrative. Like we use in history, for instance.

------
earthicus
This collaboration between the three physicists and Tao actually started its
life in a /r/math comment thread! One of the reddit commenters pointed the
physicist (I don't know which of the three was asking the question) to a
mathoverflow answer of Tao's resolving a similar question. Here's the posts
and it's followups:

[https://www.reddit.com/r/math/comments/ci665j/linear_algebra...](https://www.reddit.com/r/math/comments/ci665j/linear_algebra_question_from_a_physicist/)

------
jordigh
Pretty cool. So if the eigenvalues aren't repeated, you get a formula for the
eigenvectors of a matrix in term of its eigenvalues and the eigenvalues of its
submatrices. It doesn't seem computationally useful, but its theoretical value
is up there with Cramer's rule.

As an side, it's kind of funny to hear the physicist accent in this linear
algebra paper.

~~~
dooglius
The article mentions a companion paper about how this was used to speed up
physics computation, so it sounds useful at least in that domain.

~~~
jordigh
I believe "speed up" in this sense is really akin to how Cramer's rule lets
you write down a relatively neat formula for the solution of a linear system,
not that it would actually help with performing the calculation numerically.

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.

------
ntkachov
Link to the paper:
[https://arxiv.org/abs/1908.03795](https://arxiv.org/abs/1908.03795)

------
electricslpnsld
> The identity applies to “Hermitian” matrices, which transform eigenvectors
> by real amounts (as opposed to those that involve imaginary numbers), and
> which thus apply in real-world situations.

This is a weird characterization of Hermitian matrices — this implies that
rotations are not ‘real-world’.

~~~
cshimmin
Rotation matrices are hermitian. It's a popular science article, but the
author is alluding to the fact that Hermitian matrices have real eignenvalues.
In quantum mechanics and quantum field theory, all _physically observable
quantities_ of a system can be expressed as the eigenvalues of a set of a
(matrix) operator associated with that quantity acting on basis states in the
Hilbert space of the system. This makes sense a posteriori since we have no
idea how we would interpret e.g. complex-valued energy or momentum.

Edit: electricslpnsld correctly points out that the first sentence is false,
see below.

~~~
electricslpnsld
> Rotation matrices are hermitian

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.

~~~
contravariant
Yeah rotation matrix are not hermitian (note that you don't need to show the
eigenvalues aren't real, you just need to show it's not self-conjugate). The
OP may have been confused by the fact that you can make a rotation matrix of
eigenvectors of a hermitian matrix, which diagonalises the original matrix
into _two_ conjugate rotation matrices with a diagonal matrix between them.

------
a1pulley
Here's Terence Tao's blog post about the identity:
[https://terrytao.wordpress.com/2019/08/13/eigenvectors-
from-...](https://terrytao.wordpress.com/2019/08/13/eigenvectors-from-
eigenvalues)

------
leni536
I'm a physicist and I hate to say that I'm stuck on the first equation in the
first lemma. Can anybody clarify what the following notation means?

    
    
        \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.

~~~
Recursing
The blog post [https://terrytao.wordpress.com/2019/08/13/eigenvectors-
from-...](https://terrytao.wordpress.com/2019/08/13/eigenvectors-from-
eigenvalues/) is very readable and interesting

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”."

------
jimnotgym
I feel old and decrepit when I read this. I could remember enough to mentally
link eigenvalues with matrices, but that's all. It makes me sad to think of
all the maths I have forgotten.

~~~
lopmotr
Unless you're really having cognitive decline, you can probably pick it up
again if you need it much more easily than the first time. We forget stuff we
don't use, so forgetting maths probably means you've been doing something else
with your life. Maybe that something else was better than repeating the same
maths over and over again? I have repeated some of the same maths for 10-20
years in my work and it's depressing in a different way - I'm not improving
much. I'd be happy to forget the old and learn some new. Knowing about
eigenvalues doesn't make me feel particularly fulfilled. Any 20-year old
physics student can do that too.

------
est
> neutrinos come in one of three possible “flavors” — electron, muon or tau

This reminds me of the Koide’s coincidence

[https://news.ycombinator.com/item?id=17961139](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”

------
raphlinus
I propose the third generation neutrino be renamed to "tao" in commemoration.

~~~
peterburkimsher
I appreciate the pun with the Tau neutrino. I'm sure that someone could make a
joke, or a trick question for a quiz, out of the homophony.

[https://en.wikipedia.org/wiki/Tau_neutrino](https://en.wikipedia.org/wiki/Tau_neutrino)

------
envyac
This could have some implications in the computational expense of
trilateration for GNSS applications. "Optimal trilateration is an eigenvalue
problem" <title of another paper>.

------
lostmsu
TL;DR; anyone? Originally I thought this lets one calculate eigenvectors
directly from matrix and its submatrices's eigenvalues, but looking at the
paper, it is not exactly that good.

~~~
singularity2001
TL;DR: This lets one calculate eigenvectors directly from the eigenvalues of a
matrix and its submatrices,

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

~~~
jhallenworld
So the textbook way is:

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.

------
ncmncm
Is this Fields Medal stuff? If it's not, ought it to be? Or would they be
considered to have come by it too easily?

~~~
auntienomen
It's a nice but ordinary result.

------
hurrdurr2
Math truly is the language of the universe.

------
envyac
This could have some implications in the computational expense of
trilateration for GNSS?

------
Maro
Since Tao's Erdos number is 2, the 3 physicists are now a 3, at most :)

------
voldacar
Imagine writing an article about a new mathematical formula or equation,
without including the actual formula. Peak quanta

~~~
tzs
Here it is, reformatted a bit for HN:

|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.

~~~
panic
Ideally they wouldn't just drop the equation in as-is, but break it down and
explain what all the parts mean. Then people would actually have a path to
engaging with the real work being done instead of a parallel-universe pop-
science version of it.

------
mc3
[https://xkcd.com/1364/](https://xkcd.com/1364/)

~~~
cjsawyer
Cute stab, but the expression isn’t literal and you know that.

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.

~~~
8ytecoder
“if you can’t teach it to someone else, you really don’t know it yourself”

\-- Richard Feynman

~~~
Koshkin
Well, it's a research paper, not a tutorial.

~~~
phkahler
Yes, but the result was simple enough in hindsight that it should probably be
in every introductory linear algebra text.

~~~
Koshkin
Looks like it's not _that_ simple. It does _make sense_ in hindsight - because
the quantity of information is the same, but that does not necessarily mean
that the actual calculation or the proof must be simple.

