Elpis1

4 ms·

1.
by Elpis1 4y ago
Well, one of course has to make a distinction between the physical thing and the mathematical representation of said thing. The set of states which model a type of elementary particle furnish an irreducible representation of the appropriate
2.
by Elpis1 4y ago
Good question! What's important is that the Hamiltonian or Lagrangian has these symmetries . Particular solutions having rotational invariance may signify some nice properties of that solution, but it's quite irrelevant to Noethe
3.
by Elpis1 4y ago
Ah, sorry about that! You're right, of course; my brain slipped a bit.
4.
by Elpis1 4y ago
Where?
5.
by Elpis1 4y ago
Of course! What you're looking for is Noether's theorem ; this tells us that for every (continuous) symmetry of a system one may construct a conserved quantity. There are subtleties and exceptions, of course, but that's the
6.
by Elpis1 4y ago
Not at all. To the audience the blog is written for, the article is very sensible. When two folks who know computers quite well discuss some esoteric issue, they will use technical language and assume a certain level of competency and backg
7.
by Elpis1 4y ago
The system would be described by a multi-particle state, which would be reducible. Of course, Wigner's classification is just for classifying (most) elementary particles. A hydrogen atom can be considered a particle in some contexts,
8.
by Elpis1 4y ago
It's not quite measurable; the magnetic potential is gauge invariant, which means, among a great deal many other things , that it has no well-defined measurable value. However, it is certainly physical; things like the Aharonov–Bohm e
9.
by Elpis1 4y ago
Dr. Woit's blog is directed towards physicists, for the most part. This sort of thing is covered very early on in a graduate education in physics; it's old hat for that crew, but incomprehensible to anybody else!
10.
by Elpis1 4y ago
This sort of business is covered in any textbook on quantum mechanics worth its salt. The idea goes something like this: physics should certainly not depend on where you are at or the orientation of your measuring device. If we change this