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.
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.
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.
thanks very much. i only took introductory linear algebra in university about 11 years ago, so this brought back some of the memories. Very good and interesting to read. Much like gilbert strang himself!
Can you please stop breaking the site guidelines in comments here? Doing this will eventually get your account banned, and we've already had to ask you repeatedly.
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:
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.
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.
You use \equiv for definitions instead of :=, ^* for complex conjugation instead of \overline, you refer to the Hermitian transpose as the adjoint, and you implicitly assume all matrices are over the complex numbers.
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.
Hermitian matrices are always complex matrices, and A^* is a standard notation for the conjugate transpose of a Hermitian matrix. I didn’t find either of those strange, and I studied math, not physics. Nor did I notice any use of \equiv anywhere.
> and A^* is a standard notation for the conjugate transpose
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.
No, Hermitian matrices can be over p-adics too. You just need an inner product so you can define a matrix to be self-adjoint, and you can define that inner product over p-adics. And the result is kind of funny, because although the statement of the result doesn't make any reference to complex numbers nor to adjoints, the proofs do. I do wonder if the result is still valid over other fields.
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.
In my experience (as a physicist), the dagger is much more common in physics literature to represent adjoint; A^* for adjoint is confusing because it could just mean complex conjugation in many contexts. Overlines are practically never used physics notation.
I figured the notational confusion between the Hermitian transpose and the complex conjugate is kind of intentional, since moving from one side to the other of an inner product both conjugates scalars and is the adjoint of matrices, plus AA* is "real squaring" for both matrices and scalars.
> 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’.
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.
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.
D'oh you are correct! It's a bit late here. Anyways the rest of my point stands about the author's intent. To connect that to your original comment: a rotation is not "observable", instead, it's a transformation that modifies a physical system. In quantum mechanics, physical observables (energy, momentum, position, mass, etc) are always associated with Hermitian operators that act on the Hilbert space of states for the system.
It's a bit different. To have real eigenvalues (ie all observables) you need to have hermetian operators. They were working on the hamiltonian operator, which is hermetian since its eigenvalues are energies (ie are observable and are thus real numbers)
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”."
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.
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.
This could have some implications in the computational expense of trilateration for GNSS applications. "Optimal trilateration is an eigenvalue problem" <title of another paper>.
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.
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.
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.
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?
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.
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://news.ycombinator.com/newsguidelines.html
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.
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.
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.
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.
https://en.m.wikipedia.org/wiki/Hermitian_matrix
https://en.wikipedia.org/wiki/Conjugate_transpose
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(...)
> ...one can multiply any eigenvector by a phase e^{iθ}...
and:
> As a consistency check...
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.
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”
https://en.wikipedia.org/wiki/Tau_neutrino
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.
E.g. https://pste.eu/p/3B68.html
|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.
https://arxiv.org/abs/1908.03795
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.