You just answered my question on turbulence and the link you provided was quite fascinating although my current knowledge of physics has limited me to fully understand it.
I wanted to know is there a specific tag to browse such questions ?
Questions that people bring from.everyday life but have deep physics behind them.
I asked this questiion yesterday: Does the shape of the Universe refer to the curvature of spacetime in 5-dimensional space?
While I now understand and got good answers, my question was closed as duplicate. However, in my opinion, the linked questions are questions around the same topic, but a di...
The real thing to focus on is the Cartan classification which is around the end of lecture four and all of five, but he doesn't finish the classification in the videos, however he sets up the sections in his book and it's the clearest explanation I've seen in all of them
One of the 'ugly' things in Lie books is how they just define a semi-simple Lie algebra without giving a good reason to care about them and not others (solvable/nilpotent), I think that's off-putting and takes ages to appreciate (ages), and then even worse they focus on Hermitian Lie algebras (as Zee does), but this is also something worth figuring out to get to the magic of $E_8$
@B.Brekke It is a Hilbert space by construction. (Rigorously, the "normal" tensor product of vector spaces does not produce a Hilbert space from two infinite-dimensional Hilbert spaces, you need to "complete" the normal tensor product w.r.t. the metric induced by the inner product, i.e. $A\otimes B$ means something different if you take the tensor product "as vector spaces" or "as Hilbert spaces". The Fock product is "as Hilbert spaces" by definition.)
user434058
@ACuriousMind Hi! I just transferred Emilio's post (with extremely minor changes) to the other post on MathJax visibility of the new "Ask page". Was I right in doing that?
user434058
2:07 PM
@bolbteppa Reading Lubos diss crackpots is quite entertaining :D
user434058
Though I am afraid that there's a slight (extremely slight, slim) chance that I might grow up to become a crackpot...
user434058
That's probably because I expected the Universe to be simpler (like all other crackpots do), but ,then again, that might probably be because I am completely inexperienced as a physicist :-)
@FakeMod I do not understand what you did there, but making major changes to a question that in particular includes "I" wording, thereby putting words in the author's mouth, is not an appropriate use of your edit privileges. If you have something substantial to add, why not provide a new answer to the post? Why copy the content of a different post when you can just link to it?
user434058
4:17 PM
@ACuriousMind AFAIK, I replace all the possessive pronouns, except "we", because it refers to the site as a whole. But I'll surely link the other post, thanks.
@FakeMod I rolled back your edit. As I said, if you have something to add, post an answer, but the value of copying an existing answer elsewhere into the post is unclear to me. There was at least one 'I' in what you edited in (and it's a personal pronoun, not a possessive one), but that's really beside the point - had you tried to suggest this as an edit it would have 100% been rejected as conflicting with author's intent regardless of the 'I'
The idea of unapproved edit privileges is that you should know what edits are and are not appropriate on your own, not that you should submit edits that would otherwise have been rejected.
@ACuriousMind Hold up, I wasn't allowed to add "I'm ACM and I'm a meany head" to the end of all your answers after I got the edit privilege? Oh man, I got a lot of cleaning up to do...
Then I'm with some of the people in that thread. It would be nice to at least have the option to see small edits. Usually the ones I've noticed are fine (like actual small grammar errors); but it would still be nice to know. One thing that I'm (very mildly) concerned about is people switching my Canadian spellings for American, only because I see it all the time in suggested edits.
I'm encountering a bit of difficulty in reconciling the nice Hilbert space state vector formalism in QM with the one typically introduced in books like Griffiths, where wavefunctions are introduced not as vectors but just as functions. In the case of the infinite square well, the $A\sin(kx)$ function that is usually introduced, a little abruptly, as "the wavefunction". In the vector formalism am I correct in saying this function is the components of the Hamiltonian eigenvector in the...
...position basis? Where the components are effectively labelled by the parameter $x$ which in this case is physically the position in 1D.
Perhaps I mean "Hamiltonian eigenvectors", plural.
It might be more helpful if I post this as an actual question on the main site come to think of it
@Charlie The space of square-integrable functions $L^2(\mathbb{R}^n)$ is a Hilbert space.
Choosing wavefunctions in posiiton space is just one concrete choice to represent the Hilbert space of QM
You can think of the wavefunctions $\psi(x)$ as being the components $\langle \psi\vert x\rangle$ in the "$\lvert x\rangle$ basis", but honestly, the less you use the $\lvert x\rangle$, the better, because they're not well-behaved vectors in the space.
they're not actually elements of the Hilbert space, but only of a larger "rigged" space. It's not worth worrying too much about at your stage, but if weird things happen when you try to compute things with them, this is usually the reason
user434058
@ACuriousMind Aight. I'll keep this in mind the next time :-)
In the context of spin measurement, say the Stern-Gerlach experiment, we're working in a 2D Hilbert space so each operator has 2 eigenvectors. However, the $\hat S^2$ operator commutes with all three of the spin observables, so it must have at least 6 eigenfunctions. But, $\hat S^2$ is Hermitian, so it's eigenvectors have to be orthogonal, but you can't have $>2$ distinct orthogonal vectors in a 2D space, what gives?
so just like the identity is completely degenerate $S^2$ is also completely degenerate with it's eigenvalue being whatever multiple of the identity it is