The h Bar

12:01 AM

Thus we're just analyzing critical points of $f(x_1,\dots,x_n)$ in new variables, $f(x_1,\dots,x_n) = f(h^{-1}(y_1,\dots,y_n)) = (y_1,g_2(y_1,\dots,y_n),\dots,g_p (y_1,\dots,y_n))$ so that $f$ sends hyperplanes $\{y_1 \} \times R^{n-1}$ into hyperplanes $\{y_1 \} \times R^{p-1}$ whose jacobian is $(p-1) \times (p - 1)$ as we expected,

0celo7

@IceLord I can be bribed.

bolbteppa

Now you're supposed to know by induction this proves step 1

0celo7

@IceLord Hardback Lee Smooth Manifolds.

bolbteppa

and so we started from the set $C$ of critical points, noticed critical points can either have all derivatives zero $C_1$ or some non-zero $C \backslash C_1$ so that in $C = (C \backslash C_1) \cup C_1$ we first analyze points in $C \backslash C_1$, yada yada

Assuming we had that part proven, we then work inside $C_1$ and repeat the same proof, decomposing $C_1 = (C_1 \backslash C_2) \cup C_2$ and so the first two steps of Milnor's proof more or less make sense

repeating the same proof implies we use second derivatives, and then doing it again use third derivatives etc... so then step 3 will already involve some Taylor expansion

Don't like this induction thing merging with Fubini at all

That's the problem

0celo7

12:21 AM

@bolbteppa why

did you prove Fubini

Obliv

12:34 AM

@0celo7 whats up d00d

dmckee

I may be officially an old fart.

I just told my lab students that vernier scales are "elegant weapons for a more civilized age".

7

0celo7

@dmckee Took you how long to figure that out?

@dmckee Just? Good god you'd better be on central time

Obliv

@0celo7 hey i just finished my calc class like 30 minutes ago

0celo7

@Obliv Smokin, chillin.

dmckee

@0celo7 We have a lot of wroking students. I teach the night class most semesters. But yeah, central time. And class is breaking up.

Obliv

12:43 AM

it's not that abnormal having class this late.

0celo7

@Obliv you're in Indonesia, right?

Obliv

yes. south-west of eastern indonesia.

that's in europe right?

0celo7

No, South America.

Learn topography.

Obliv

i'd rather learn topology first

0celo7

you probably won't ever

maybe 5% of physicists have a (good) knowledge of topology

Obliv

12:46 AM

i am the 1%

0celo7

@Obliv lowest 1%

user218912

1:09 AM

lol

user218912

@0celo7 for what?

user218912

@Obliv wait what?

DanielSank

Hi, everybody.

user218912

hey

user218912

@DanielSank if you have an n-particle wave function is that the same as many particle quantum theory?

DanielSank

1:21 AM

@dmckee Well, if any of them start whining about going to Tosche Station to pick up power converters, that's when you throw in the towel.

@IceLord That's like asking "Is having a basket of strawberries the same thing as farming?"

user218912

xD

user218912

I mean in a many particle quantum theory (which my qft prof said is inconsistent for several reasons) do you have n-particle wavefunctions? or something else?

user218912

because he didn't show what a multiparticle quantum theory looks like.

0celo7

@IceLord my birthday

DanielSank

@IceLord There's a stupid way to to multiple-particle quantum mechanics, and there's a not stupid way to do multiple-particle quantum mecahnics.

What have you learned so far? Basic quantum?

user218912

1:25 AM

yes

DanielSank

Have you done anything with two particles?

0celo7

what are you, a chemist

user218912

well I remember a chapter in shankar on identical particles did that

user218912

like bosons and fermions with 2 particle kets and stuff

DanielSank

So you did symmetrization and antisymmetrization of wave functions?

user218912

1:26 AM

yes

DanielSank

Oh boy

Ok read this:

86

A: Which is more fundamental, Fields or Particles?

This is a tricky question because it asks about the meaning of words. People use the word "particle" to refer to various, not always well defined, notions in physics. In the end, I think the simplest and more correct single way to categorize the terms is to interpret "particle" as "excitation of...

Then ping me.

user218912

thanks

Sir Cumference

Not to sound stupid, but is there an upper-mass limit for Schwarzschild-solution black holes?

user218912

you learned GR?

Sir Cumference

A tad bit

user218912

1:33 AM

like, real GR?

Sir Cumference

@IceLord As opposed to?

0celo7

the fuck is real GR

as opposed to quaternionic GR?

user218912

like the stuff @0celo7 knows about GR.

Sir Cumference

I thought 0celo didn't know GR as well as he did QM?

0celo7

@SirCumference do you know what a manifold is

user218912

1:34 AM

versus superficial GR

user218912

@SirCumference the opposite actually

0celo7

@SirCumference pretty sure QM is a farce

Sir Cumference

@IceLord What?

0celo7

@IceLord the test is simple...

@SirCumference what is a manifold

user218912

@DanielSank I read your post, makes sense except for the use of numbers in the kets

DanielSank

1:35 AM

@0celo7 Oh don't start this.

user218912

I think we'll learn about second quantization notation tomorrow.

Sir Cumference

@0celo7 Isn't it just a set of points on a surface?

user218912

@DanielSank except he uses bosons as examples not numbers.

0celo7

@IceLord There's your answer!

user218912

@0celo7 xD

0celo7

1:36 AM

@DanielSank To be fair I would have accepted any number of definitions here.

user218912

@SirCumference so what's the point of asking a question like that when you don't really know GR?

user218912

that's like me asking a random string theory question

Sir Cumference

@IceLord Someone asked me if there were an upper limit to black hole masses on Astronomy chat

So now I wonder if that's a thing in the schwarzschild solution

DanielSank

@IceLord No idea what you mean by that.

0celo7

no

perhaps if you consider stability, etc. but certainly not a part of the basic model.

Sir Cumference

1:38 AM

@0celo7 Okay then how about in the Oppenheimer-Snyder model?

Does that mention anything about black hole upper limits?

user218912

@DanielSank like he writes the 3 particle basis state for bosons as $|k_1, k_2, k_3 \rangle$

DanielSank

@IceLord k

0celo7

@SirCumference is that the collapsing dust cloud

user218912

@DanielSank same thing right?

DanielSank

@IceLord Is your thing 2nd quantization?

user218912

1:40 AM

yes

DanielSank

Did you read the part of my post right before the last section heading?

user218912

yes what about it?

DanielSank

I explicitly mentioned that $|k_1 k_2 k_3 \rangle$ means the same thing as $|k_1\rangle |k_2 \rangle |k_3 \rangle$.

Sir Cumference

@0celo7 Yep

In which the event horizon isn't eternal

0celo7

where do you learn about this crap, it's certainly not in the "usual" books

must be an astronomy nerd thing

Sir Cumference

1:43 AM

@0celo7 Astronomy SE

Dammit...

I spoke too fast...

user218912

@DanielSank oh alright

user218912

mine had commas though so I wasn't sure xD

DanielSank

@IceLord :)

user218912

another question

Sir Cumference

@0celo7 Well then feel free to defer me to the "usual" books

user218912

1:45 AM

@DanielSank why are the position and momentum eigenstates normalized to dirac delta functions?

DanielSank

@IceLord You mean, why are they delta functions?

user218912

yes

0celo7

@SirCumference the usual books do not discuss this, I know nothing about astronomy

@DanielSank Would you happen to know something about axiomatic probability?

DanielSank

@0celo7 I doubt it.

@IceLord Well, what part of that doesn't make sense to you?

user218912

@DanielSank well

user218912

1:47 AM

it just appeared

user218912

didn't say why they are using delta functions

DanielSank

@IceLord Oh, that's not good!

Stupid lazy teachers... stupid lazy books...

user218912

or probably me not reading properly xD

user218912

but anyway can you explain please?

DanielSank

@IceLord Perhaps.

Are you ok with the idea that $-i \hbar (d/dx)$ is the momentum operator?

0celo7

1:49 AM

@IceLord what is the question

user218912

scroll up

user218912

@DanielSank yes

DanielSank

@IceLord Ok what about $[x, p] = i \hbar$?

Actually forget that.

If you know that $-i \hbar (d/dx)$ is how you write the momentum operator in the position basis, then it follows that a momentum eigenstate $|p\rangle$ written in the position basis is

$\langle x | p \rangle = \exp(i k x)$

Right?

user218912

@DanielSank yes I asked @0celo7 about that a few days ago and I thought I knew the answer as to why that is but I realized I don't, if you could explain how that arrises in quantizing classical particle theory I would appreciate it.

DanielSank

I think you're best off taking $[x, p] = i \hbar$ as a postulate of the theory for now, actually.

You can justify it based on different postulates, but then you're just trading some postulates for others.

user218912

1:55 AM

ok

DanielSank

Not sure which part you're ok-ing.

user218912

one sec

DanielSank

gotta go for a bit

0celo7

@IceLord It's better if you look at the translation operator

then postulate that $\hbar P$ is the infinitesimal generator of it

or maybe it's $P/\hbar$

then you can derive $[X,P]=\mathrm i\hbar$

@IceLord where?

user218912

back.

user218912

2:09 AM

ok

user218912

@0celo7 I was asking

user218912

you know how $\langle x | x' \rangle = \delta (x' - x)$?

user218912

why

user218912

maybe I'm just being dumb

0celo7

yeah you are.

user218912

2:12 AM

why

0celo7

cuz every undergrad understands it?

Sir Cumference

2:29 AM

Sigh...does anyone know how the hell I can read this paper?

I see no download button

user218912

@0celo7 yes

user218912

I do too

user218912

but Idk why it works

user218912

oh look

user218912

ACM explains it in one of his answers xD

user218912

2:32 AM

2

A: On the completeness relation in Quantum Mechanics

Let's do a finite example. Suppose we have a vector space $V$ with a subset of vectors $E=\{\mathbf{e}_1,\mathbf{e}_2,\dots\}$. Also assume that $V$ has an inner product, and that this set of vectors is orthonormal. That is $$\langle\mathbf{e}_i,\mathbf{e}_j\rangle = \delta_{ij}$$ If the span of...

DanielSank

2:44 AM

@0celo7 be nice.

Also, be correct.

user228700

4:07 AM

@DanielSank Hello :) Quick question. We've got a box with a partition and we've got 10 indistinguishable molecules/elements. If I were to ask you the number of ways in which it is possible for us to arrange these 10 molecules so that 5 are on one side and 5 are on the other, what would ur answer be?

Transcript for