74

Let's give it a try. Of course, the precise mathematical meaning is perhaps absent, so the answers are sort of heuristic. But if I understand correctly, you want to gain intuition ;)
The first observation is that Planck's constant has units, it is not a numerical constant but carries physical dimension. In some sense, this makes all the difference: in your ...

48

I'm can't claim to have studied the relevant history in a lot of detail, but count me a skeptic of Landsman's claim. Let's take this little paper and the companion that it cites as a test case, which I hope we can all agree is "real physics". The authors are clearly well versed in the calculus of variations and the representation theory of Lie groups. ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

35

I've just finished teaching the first semester of a year-long "Quantum Mechanics for Mathematicians" course. Some of the references I found most useful are
A good, clear, physics textbook. Shankar's "Principles of Quantum Mechanics" that many have mentioned fits the bill.
Faddeev and Yakubovskii, "Lectures on Quantum Mechanics for Mathematics Students" is ...

33

To build intuition for the Planck constant $\hbar$, which I understand is the purpose of the OP, I would start by noting that $\hbar$ is not a dimensionless number: it has dimensions of energy $\times$ time, or of momentum $\times$ position, meaning that it represents an action. (Equivalently, it could represent an angular momentum, I will get back to that ...

answered Sep 11 '19 at 9:46

Carlo Beenakker

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This reminds me the following anecdote. K. Friedrichs once met Heisenberg on a conference. He thanked Heisenberg for creation of quantum mechanics which benefited mathematics so much, and added:
"But mathematicians gave much in return."
Heisenberg: "What?"
Friedrichs: "For example, von Neumann explained the difference between symmetric and self-adjoint ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

28

You can understand this philosophy as a generalization of Noether's theorem. Let me only state Noether's theorem in the quantum case because it's actually easier to understand there than in the classical case (for me, anyway).
As setup, we have a Hilbert space of states $V$ and a state vector $\psi \in V$ which evolves according to a Hamiltonian $H$, ...

answered Mar 16 '17 at 19:22

Qiaochu Yuan

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25

I'd like to try to give a more comprehensive answer.
In the elementary formulation of quantum mechanics, pure states are represented by unit vectors in a complex Hilbert space $H$ and observables are represented by unbounded self-adjoint operators on $H$. The expected value of a measurement of the observable $A$ in the state $v$ is $\langle Av,v\rangle$. We ...

24

Typically in math "quantum X" means a deformation of "X" which is in some sense "less commutative." So quantum groups should be deformations of groups which are "less commutative." Interpreting this is slightly tricky since groups are already non-commutative, but nonetheless they do have some "commutativity" built in which you can see either by noting:
...

23

There has been quite a lot of literature on the applications of $q$-numbers, $q$-derivatives, $q$-deformations, etc, of various algebraic models of physics. Such applications range from $q$-deformations of simple harmonic oscillator(s) and angular momentum algebras to the development of quantum groups and their applications in nuclear physics, particle ...

answered Aug 20 '17 at 21:36

Konstantinos Kanakoglou

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In a recent result with Shay Moran, Pavel Hrubes, Amir Shpilka and Amir Yehudayoff, we show that the answer to a basic question in statistical machine learning is determined by the value of the continuum, and is therefore independent of the ZFC set theory. The paper will be available on Arxiv within a few weeks.
Here is a link to that paper:
https://arxiv....

18

A simple bound on the largest dimension of a complex irreducible representation (which is either equal to or half of the largest dimension of a real irreducible representation) is the following: we know that
$|G| = \sum d_i^2$ where $d_i$ are the dimensions of the irreducibles,
the number of (complex) irreducibles is the number $c(G)$ of conjugacy classes, ...

17

As jjcale mentions in a comment, the index of a Fredholm operator is very important in physics. One way to define the Chern number of a topological insulator is in terms of the index of a Fredholm operator, as explained in [1].
There is also the concept of an index of a pair of projections. This is seen a lot recently in physics papers, for example in [2]. ...

fa.functional-analysis mp.mathematical-physics oa.operator-algebras ho.history-overview quantum-mechanics

16

I like the perspective that the set of states is precisely the set of positive trace class operators $M$ of trace one. A state is called pure if $M=pr^{\perp}_{U}$ is the orthogonal projection onto a one-dimensional subspace $U$.
So, every non-zero vector $\psi$ defines a pure state. Since the orthogonal projection onto $\mathbb{C}\psi$ is the same ...

16

In fact, precisely your integral has been computed in closed form, in: Annie Gervois and Henri Navelet, Some integrals involving three Bessel functions when their arguments satisfy the triangle inequalities, J. Math. Phys. 25 (1984), no. 11, 3350–3356. Their result is
$$
\int_0^\infty J_m(ar)J_n(br)J_{m+n}(cr)r\,dr =
\begin{cases}
0&\text{if }c^2 < (a-...

answered Jan 15 '14 at 22:43

Francois Ziegler

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16

Sorry, not an answer, but too long for a comment.
I am the author of the "Pitowsky's Kolmogorovian models and Super-Determinism" paper. I would reject the claim that this paper is a "philosophical discussion".
Rather, I (try to) demonstrate (using simple physics and mathematics, not philosophy) that the whole Pitowsky model business is physically ...

16

Your question touches on many issues in group representation theory, and I can only give a few general remarks which may point you in interesting directions for further reading.
As to your question regarding the maximal real irreducible representation of a finite group, there is an interesting connection with the Frobenius Schur indicator.
If $\chi$ is a (...

14

I think you can find more in Lieb and Seiringer's book "The Stability of Matter in Quantum Mechanics", or see also Freeman Dyson http://www.webofstories.com/play/4415 and the book review http://arxiv.org/abs/1111.0170.

14

To supplement the excellent answers previously given, some more remarks situated in the canonical quantization formalism, which is equivalent (at physicist level) to the path integral formalism referred to by Carlo Beenakker:
When we open our eyes, we see that we're given a playground in which we can organize what we see according to position $x$, time $t$, ...

answered Sep 11 '19 at 17:39

Michael Engelhardt

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Aspects of symplectic topology hold in the infinite dimensional setting. A notable difference is that there are different notions for what a symplectic form should be: there is the notion of a strong symplectic form where $\omega$ is required to induce an isomorphism $\mathbb H\rightarrow \mathbb H^*$, or a weak one, where this map is only injective. In ...

13

It is actually easier to compute the first column of $e^{tM}$ for each $t \in \mathbb{R}$ rather than computing only the first column of $e^M$.
Indeed, let $u(t) = e^{tM}e_1$ for each $t \in \mathbb{R}$, where $e_1$ denotes the first canonical unit vector. Then $u(t)$ is the first column of $e^{tM}$.
Claim: For every $k \in \mathbb{N} := \{1,2,\dots\}$ and ...

12

They turn up quite often in the study of (exceptional) linear algebraic groups. The most famous instance of this is the fact that algebraic groups of type $F_4$ are precisely the automorphism groups of Albert algebras, i.e. $27$-dimensional exceptional Jordan algebras. One possible reference for this fact is the book "Octonions, Jordan Algebras and ...

12

The Algebra of Grand Unified Theories, by John Baez and John Huerta may well be to your liking:
A full-fledged treatment of particle physics requires quantum field
theory, which uses representations of a noncompact Lie group called
the Poincaré group on infinite-dimensional Hilbert spaces. This brings
in a lot of analytical subtleties, which make it ...

answered Mar 16 '17 at 18:55

Carlo Beenakker

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12

It was shown just two years ago that the presence or absence of a spectral gap of certain short-ranged 2D lattice Hamiltonians is independent of the ZFC axioms:
T. S. Cubitt, D. Perez-Garcia, and M. M. Wolf, "Undecidability of the Spectral Gap", Nature 528, 207-211 (2015)
(A 146-page-long full version can be downloaded from this link.)

12

About your Q1: I think that the simplest—and most obvious—way to think mathematically about the physical meaning of Planck's constant $h$ is that it is a kind of quantitative measure of the departure from commutativity:
Although the word "quantization" does not have a uniquely defined mathematical meaning, it is almost always connected with the passage of ...

answered Sep 11 '19 at 22:54

Konstantinos Kanakoglou

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It is probably dangerous to answer without reading Landsman’s whole paper (and the question seems likely to be closed as “opinion-based”), but I’ll record my first reaction as much like lcv’s (a), namely, it sounds a little bit strawman-ish to separate the two and then pit one (FA) against the other (QM).
Footnotes (by Born) on pp. 583, 585 of the famous ...

11

Okay, here's an explanation in terms of quantum mechanics. Let ${\cal A}$ be a family of observables, modeled as self-adjoint operators on some Hilbert space, and let ${\cal U}$ be the group of all unitary transformations that leave every observable in ${\cal A}$ invariant. You can consider ${\cal U}$ to be a kind of symmetry group. Mathematically it is the ...

11

The first thing to say is that ordinary matter is actually not stable. Suppose a baseball-sized rock finds itself in the vacuum of outer space in the very distant future, isolated by the universe's accelerating expansion within its own cosmological horizon. Even within the standard model of particle physics, the rock will eventually decay by quantum-...

11

Well, there is always the trivially enforced solution
$$\tag{1} S[x,\lambda]~=~\int\! dt \sum_{i=1}^3\lambda_i(t) \left(\ddot{x}^i(t)+\alpha \dot{x}^i(t) \right),$$
where $\lambda_i(t)$ are three Lagrange multiplier variables. From now on we assume that we are not allowed to use other variables than $x^i(t)$.
Whether the 3D ODE $\ddot{\bf x}+\alpha\dot{\bf ...

11

I would like to elaborate on the link to "associative" problems (such as the Zelmanov's solution of the restricted Burnside problems mentioned by Tom De Medts) - mainly because by saying that Jordan algebras are nonassociative algebras satisfying a strange list of axioms, you are forgetting that some reverse engineering takes place here.
First of all, if $...

answered Feb 4 '16 at 13:10

Vladimir Dotsenko

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11

As another example of the second category in Kostantinos Kanakoglou's answer I think it is fair to mention quantum-integrable systems: this topic in physics was pivotal in the historical development of the notion of quantum groups by the Leningrad group (Faddeev et al), and the Japanese group (Jimbo and Miwa et al).
In particular, $U_q(\widehat{\mathfrak{sl}...

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