He knew Goto Dengo

@ACuriousMind Can you verify that the user we knew as obe is gone?

yes, it was obe

@ACuriousMind Does the "radical" of a ring make sense to you?

@0celouvsky one usually talks about the radical of an ideal

what's that?

I mean the intersection of all maximal ideals of the ring

@0celouvsky Given an ideal $\alpha\subset R$, it's the set of all $r\in R$ such that $\r^n\in\alpha$

Hmm. What's the usual name for what I just said?

@0celouvsky Ah, that's the Jacobson radical

ok

Hey

@AccidentalFourierTransform can I contest something?

I couldn't see an option to in the actual page

@ACuriousMind So if I have $x-\lambda e$ in an ideal, and then I pass to the quotient, I basically just "set it equal to zero?"

@Phase What exactly do you want to contest?

what do you mean by contest?

An edit of mine was rejected and there were comments saying it added nothing, but it was in response to a previous edit:

@0celouvsky Yes, as with all quotients.

@ACuriousMind that seems like a very broad statement

Surely "does is the rain falling" not make sense

*does not

Not a big deal just seemed unnecessary, I was confused why it got met with such negative feedback

@0celouvsky It is, but I can't think of a quotient that I wouldn't describe as basically setting the things that you quotient by to zero, except for cases where the notion of "zero" doesn't make sense.

@Phase The feedback is an automatic message if the reviewer chooses a particular reason to reject. Don't read too much "negativity" into its specific formulation

But not improving readability? From "does is the rain falling"?

I have to say that I wouldn't have rejected that edit, though, correcting grammar mistakes is a valid use of the edit feature

@ACuriousMind so is this "Gelfand representation" something you know about?

Ok thanks, was confused about why it was rejected

@0celouvsky Only in so far as I know it exists

@ACuriousMind do you use any reference management software?

@Phase yeah, sorry for that. When I first read the edit, I thought that you changed "is falling" to "does fall"

which are both correct, so the edit was superfluous

but then I read it again and realised that you did fix something

@Danu Currently, my "reference management" is a giant commented bibTeX file :P

Ah, you make it by hand? I see

I'm trying to avoid that :P

so I edited the post myself and fixed it

Oh ok no worries, I guess I can see how the way it's laid out in the page might give that impression, I just wanted to argue against it in case there was a valid reason : )

yeah, the automatic message is kind of harsh

But if you comment it, it has some actual value to it

I just want it for the bibliography

@Danu I couldn't find anything that really looked as if it would make my life easier

Or rather, for clarity in my head :P

and, well, we reviewers are not flawless, we sometimes make mistakes :-P

What I find very useful is having tags

Currently I've got huge trees of folders

but that's like having mutually exclusive tags, and that sucks

Can I ask a question regarding Learning QM?

sure

Yes, and thank you for asking before asking

but the answer is shut up and calculate

@Phase You can ask whatever you want in chat - whether someone answers is a different matter ;)

@Phase Please don't ask to ask

Atm I'm using Shankar's Principles of QM, as I found Feynman too wordy and Dirac too.. uh... Diracc-y, but is Shankar's good enough to learn or should it just be an extra resource?

Shankar is very good

Shankar is pretty good from what I've heard

as in, should it be a supplement or is it fine on it's own

Oh ok great!

I liked Griffiths quite a lot when I learned from it

but nowadays I don't look back on it as such a great thing

because it's really lacking from a math POV

@Danu Eh, I'm not so keen on having some sort of indexed library on my disk

What's the maths / words split? I don't really have a good enough attention span to keep momentum going from word explanations

the best QM book is Cohen-Tannoudji

@Danu is that not the case with every QM book?

Idk if that's a really bad habit

and after teaching some QM last semester I noticed that not explaining the tensor product is a really bad idea

Cohen-Tannoudji? Never heard of it

@0celouvsky idk, honestly

How can you have a QM book that's math heavy but not a functional analysis nightmare

@ACuriousMind Really? You're not?

I 100% am.

I thought so ;)

I've got huge folders with potentially interesting topics etc

And I say that as an analyst

also my textbook collection...

Idk, it's only undergrad

I have folders of classic papers by great mathematicians :D

Shankar can absolutely be used for a grad class

@0celouvsky I never said math heavy

the book is massive

Cohen-Tannoudji got a nobel prize you know

I just want it to really basically explain what a Hilbert space is, and how to build other ones out of it

@Danu I call that my "download" folder ;)

so dual spaces, tensor products

@ACuriousMind Oh god, that's terrible. I keep mine very clean :D

Idk if I understand a Hilbert space properly

@Danu Yah but then you need Cauchy sequences and who likes that

No yo udon't

I'm willing to bet I dont

You just brush over completeness

yeah

@ACuriousMind might want to have a look at jabref - I found that to be a "bibtex with additional features" :D

It's totally irrelevant physically AFAIK

Screw completeness for physics man

@Phase A Hilbert space is an inner product space such that it is complete with respect to the induced norm.

@Phase So can you give me a definition (feel free to forget about completeness)

damnit ocelot

lol

@0celouvsky kinda spoiled it : p

Maybe better, @Phase, do you know what the dual space is?

I'm an analyst ;_; what do you want me to do

Or how to practically obtain/work with a tensor product?

@Sanya Hmm, that does look useful. Thanks, I'll give it a try

A dual space, as in [in the context of QM] the vector spaces of kets and bras?

@ACuriousMind yeah, I've got that too

Right now, I'm trying "Zotero" to collect references and make initial bibtex file

Doesn't the tensor product give a new hilbert space where each basis vector is a combination of the two component spaces basis vectors

like

See the problem with bras is that you need completeness for it to make sense

then JabRef to edit the file

@0celouvsky Screw taht

You need the Riesz-Fisher representation theorem!

but zotero has an ugly user interface

The sense is very basic, honestly ocelot.

and even jabref can auto-get reference data once you filled in the DOI

The rigorous proof maybe not so much, but whatever

|0> and |1> state vectors tensored with the same basis state vectors would give a dim(2*2) hilbert space with |00>, |10>... etc

@Sanya I work via MathSciNet

idk if tensored is the right verb, what's it called?

One-click > finding DOI and shit

@Phase As long as you never stray from finite dimensional Hilbert spaces you can forget what I am saying

@Phase Sure, we get what you mean

You can forget what he's saying in any case ;)

Oh come on

I can't now

This functional analysis stuff underlying QM is super unimportant in physics, face it ocelot.

because I'm wondering what you'd use an infinite dimensional Hilbert space for

You wouldn't say that to yugibb, would you?

@Phase everyday QM

decomposing a function into a vector space?

if decomposing is the right word

The wave function for instance lives in one

@Phase Particles

@0celouvsky Hell yes I would!

Can i ask another, completely different question? Prompted by my unhealthy habit of engaging with flat earthers

@Phase, here's a fun brain teaser: Think about the standard Heisenberg relation $[x,p]=i\hbar\operatorname{id}$ or more generally $[A,B]=c \operatorname{id}$ for any complex constant $c$. If these are operators on a finite-dimensional Hilbert space, you know what the trace of the identity is. Take the trace of both sides of the equation. What do you get?

@Phase The classic example is the space of square-integrable functions (actually equivalence classes of them), which is where the wavefunction lives.

@Danu Please slam ACM for bringing in measure theory

It's not physically relevant

What?

Square integrability is probably the most physically relevant thing ever to be formulated in QM

"actually equivalence classes of them" is measure theory

Also WTF chat doesn't allow me to talk at a normal pace

I actually haven't gotten that far yet

but I guess he put it in parentheses

@Danu Is it giving you the spam notification?

Haven't done traces at all really in Physical contexts

@0celouvsky Yush

@Danu if I do that I get non-trivial zeros of $\zeta(s)$ off the critical line

@Danu Yeah it's really random about that

@Phase Do you know how to take the trace of a matrix?

Sum of the diagonal elements right?

Yeah, sure!

I mean

So what is the trace of the identity?

2

well

@0celouvsky It's physically meaningful though since it underlines that the value of the wavefunction at a point is actually physically meaningless.

Wait

how many dimensions

$n$, a finite number

all of them

Oh, ok then $n$

Right, easy!

@AccidentalFourierTransform So, which large cardinal axioms are you working with here ;P

@ACuriousMind I don't buy that because we always require continuity when solving the SE

And the SE, being a differential equation, has pointwise meaning

waiting for it to soar out of my range of knowledge

@ACuriousMind $N_\mathrm{Slereah}+1$

Now, do you happen to know some properties of the trace @Phase? Like for instance, how does $\operatorname{tr}(AB)$ relate to $\operatorname{tr}(BA)$?

Unless you're telling me it's only meant in a distributional sense 100% of the time

It's commutative right?

yep

@0celouvsky That we might require the wavefunction to be continuous in order to solve the SE (which is really just because no physicist wants to learn about weak derivatives and whatnot) does not make the value at a point any more physically meaningful

@Phase It has a cyclicity property. So the trace of $ABC$ equals that of $CAB$, etc. For $AB$, you get equality with $BA$ (trace of, that is)

What do you know about trace(A+B)?

trace(A) + trace(B)?

Exactly, it's linear

So now can you tell me trace([A,B])?

commutative and linear so

0?

@ACuriousMind Maybe, but what you're saying is that the value doesn't mean anything on a set of measure zero. I've shown the chat how to construct a pretty horrible set of measure zero.

@Phase YUP! So what do we find by tracing both sides of the Heisenberg relation on a finite-dimensional Hilbert space?

I guess I find it strange that nature should know about Lebesgue measure.

nobody should know about Lebesgue measure

are the operators diagonalised?

^lel

haha

I'm not entirely confident with this sort of stuff yet really sorry > >

well youre almost there!

@Phase Hmm? It doesn't matter. You already computed the trace of both sides. Now just equate them. What do you get?

@Phase It doesn't matter - the trace is basis-independent

@Phase Heisenberg uncertainty principle says you cannot simultaneously diagonalize x and p

(sorry for the slow responses; chat is #@!(+*!@$* and doesn't allow me to answer quicker)

@JohnRennie I went to show the professor my code today, "You cannot use data structures for this."

I think I got it wrong

:^)

Kill me please

No, don't worry. THe answer is weird

I get 0 = nc

correct

exactly

which makes no sense

oh

That's the point!

You just derived a contradiction.

Ohhhh i see

that means there is no finite dimensional representation

unless you take $n$ to be orthogonal to $c$

@0celouvsky Stop being a spoilsport :P

So no finite dimensional Hilbert space has QM-style position and momentum operators

wtf did I do now???

@ACuriousMind Yeah, seriously.

Thanks, that's pretty cool :^)

THIS IS MY LESSON :D

ok now do the same with the Weyl form of the canonical algebra

lmao

its almost 1am

why am I here

I cant take a shower now, its too late

smelly man

@AccidentalFourierTransform You're Fourier transformed, you shouldn't have access to the time domain :P

ugh

@0celouvsky did you just assume my smell?

So anyways @Phase, you'll have to deal with infinite dimensional spaces sometimes in QM (basically the only finite dimensional ones you typically learn about are the ones spanned by spin states). However, it's an amazing fact that you can deal with operators in QM by means of very close analogies with "infinite-dimensional matrices". Many properties still hold.

::twitches::

What are you twitching for? Honestly man, you're just obscuring things.

I'm playing

Is this just a consequence of the fact that the Matrix interpretation must be equivalent to the function interpretation? And that these functions can span continuous domains?

anywayz

@Phase I don't know what you mean by those interpretations

bye peeps

oh sorry, I meant like

heisenberg and schrodinger

@AccidentalFourierTransform cheers

@AccidentalFourierTransform smell ya later

^i'd rather not

@Phase the different "pictures" of quantum mechanics don't really have any effect (they're just pictures, after all :D) on things. Sometimes one of them is just more convenient, but they're all saying the same things.

Yeah but I meant like

Idk

If they're all saying the same thing, is the necessity of infinite dimensions for operators and state vectors a consequence of the fact that in wave mechanics wavefunctions are defined over continuous domains frequently?

Like position or momentum

@Phase Well, you've just shown the necessity for infinite dimensions - the canonical commutation relations cannot hold on a finite-dimensional space

@ACuriousMind Conversely, if something gets projected to zero it was in the ideal in the first place?

Ahaha I guess I'm doing the thing I said I try to avoid, i.e. trying to explain things with words incorrectly rather than just looking at maths. Sorry 'bout that!

Don't apologize :p

@0celouvsky Yes. You can show that all kernels are ideals and all ideals are kernels (of the projection to the quotient ring), so "is a kernel of a ring homomorphism" is in fact an alternative definition of "ideal".

@ACuriousMind Ahhh

@Phase Using words is not bad in itself, but the language around "wave mechanics" is often unhelpfully muddled - I tend to avoid talking about "waves" as such

@ACuriousMind Perfectly analogous to normal subgroups

Yeah, I think I proved that

Thank you though for helping and also for teaching me a bit about Traces

@Danu yup

The analogy is even closer, of course, between normal subgroups of Lie groups and the ideals of their Lie algebras :P

@Phase Now if you want to learn about what Hilbert spaces really are...

Sure!

If you think I can understand it : p

pls don't hurt me Danu

@Phase You've surely taken calculus, right?

Yep

@ACuriousMind top kek

It's been way too long

@Phase Do you remember what it means for a sequence of real numbers to converge?

@Danu It has :D

@ACuriousMind By the way, do you know the following Schur-type result

In what context?

...real numbers

Given $(a_n)\subset\Bbb R$, do you know what $\lim_{n\to\infty}a_n$ means?

Given some $\rho:G\to \operatorname{Aut}(V)$, a representation, any $G$-invariant inner product arises as a direct sum (with tuneable coefficients) of $G$-invariant inner products on the irreducible summands

(take $G$ to be a compact Lie group to have existence etc)

If it's a convergent series? Surely $a_n$ tends to 0

Aut(V)? What's wrong with $\mathfrak{gl}(V)$

if you let n tend to infinity

I was just confused if you meant in the context of something specific idk

Or GL(V)

@0celouvsky That's End, not Aut. That's not the definition of a representation.

@0celouvsky What do you care? It's the same thing

@Phase no, not necessarily

Huh?

it could tend to anything, why do you think it would have to be 0

@0celouvsky he's thinking about a convergent series of real numbers. The summands have to tend to zero.

I thought if a series converges it's a requirement for lim n -> infinity of $a_n$ to equal zero

You did specify real numbers?

I said sequence, didn't I?

The sum of course doesn't

@0celouvsky Phase said series, though :P

ayy

So you need to refresh the memory

oh

sequence

my brain melted