@JohnRennie

Good Afternoon sir.

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.

@MadameAkira hi :-)

@MadameAkira there is an everyday-life tag. Have a look at those questions to see if they look interesting.

Thank you very much sir.

@Slereah just read Zee's group book, e.g. the Dynkin part to construct the classical Lie algebras with diagrams, it's worth it

So a Fock space is a tensor product of single particle Hilbert spaces, but that means that a Fock space is a Hilbert space as well?

@B.Brekke Well it's not trivial to show, but yes

Also it's not a tensor product, it's the direct sum of tensor products

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 $E_8$

(Perfect excuse to post): motls.blogspot.com/2007/11/exceptionally-simple-theory-of.html

@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.)

At least make it entertaining if you veer off the deep end of physics

Does motl still practice physics professionally?

no

This guy however

The guy on the right look like he's wearing a toupee

I can't unsee it

He's also a heavy hitter with his own unified theory too

theguardian.com/science/2013/may/23/… (lol motls.blogspot.com/2013/05/… )

Is this a different weinstein to the one who wrote that GR textbook?

I think that was Einstein

I feel like I've seen interviews he's given before and I remember him being way older than that guy

Oh I'm thinking of steve weinberg

lmao

Really astounding stuff in that clip, string theory is hard, needs to bow to my whims, therefore it's bad

@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?

@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...

Better get to work then, before I see what you've done!

It's always been kinda weird to me that you don't seem to get any notices when your post is edited (like you do for suggested edits).

Don't you?

Do you? I thought it just automatically changed it if it was a privileged edit? I don't recall getting pinged for those.

According to meta.stackexchange.com/q/202359/263383, there's supposed to be nofications, but only for "substantive" edits

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.

Ok thank you I think I'm starting to get it

When you say "not well behaved" is this because it can only be "normalised" to the delta function?

essentially yes

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

@ACuriousMind Ahh I see, I've seen you use the phrase rigged Hilbert space before

@Charlie no word as has ever been better chosen as “rigged” in “rigged Hilbert space.”

the damn system is rigged!

much like some Hilbert spaces...

rigged hilbert space is basically distributions for Hilbert spaces

@bolbteppa Lee Smolin several years ago expounded the same thesis in “The trouble with Physics”.

Of course Weinstein identifies theoretical physics with high energy theory, which in itself this is already debatable.

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?

@Charlie "However, the $S^2$ operator commutes with all three of the spin observables, so it must have at least 6 eigenfunctions." [citation needed]

Or, well, you're not wrong, depending on what you mean by "6 eigenfunctions"

My reasoning is that it commutes with each of these operators, which don't commute with eachother, so it must share eigenfunctions with each of them

@Charlie Yes, it does.

but they don't commute with eachother, so all 6 of their eigenfunctions must be distinct no?

Yes. But no one says they must be orthogonal

Is $\hat S^2$ Hermitian?

Sure

Once again, consider that the identity operator commutes with everything and is Hermitian

isn't one of the properties of Hermitian operators that their eigenfunctions are orthogonal

hmm

Come to think of it $S^2$ must commute with arbitrary axis spin measurements, which actually makes my confusion quite literally infinitely larger

Have you tried to compute the actual form of $S^2$ on the 2d space?

um

I think I've seen it written before when I did the perturbation theory stuff but I've never tried to do it myself

It's literally just a 2-by-2 matrix, not some mysterious operator

I'll just tell you: It's a multiple of the identity

oh

I have to go afk for about 10m sry

i will have a think though

Oh this is to do with its degeneracy, eigenfunctions with distinct eigenvalues have to be orthogonal

Yup :)

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

ahhh

(those were short 10 mins :P)

Matchmaking found me a game but it was on a map I don't want to play :P

what game?

Overwatch

ah, never got into that

I just play with friends, no one wanted to play starcraft with me so I just joined them on their game :C

and now I'm invested

Heh, I know the feeling

I played Dota 2 with friends for a while pretty excessively, but we sort of just gradually stopped after far too many hours :P

I got into Dota for a while, my friends apparently don't like RTS or moba games so that dream died fairly quickly

Funny since I find it debatable that Overwatch might be a MOBA, too! :P

But, well, the learning curve in Dota sucks regardless. It's not a good game to try to get friends into that don't already play it

Yeah that's true, they like the fps part though :P

And dota is definitely a steep learning curve, similar objections were made to my suggestion to play starcraft