The h Bar

@Slereah It depends on what you want to do, there's no "correct" one. Sometimes you want to count immersed submanifolds among them, sometimes you don't

0celo7

There’s five different kinds of submanifolds

Slereah

But he doesn't seem to define them

Only specific types of submanifolds

0celo7

1:05 PM

Yes, because there’s many types

Slereah

yeah but is there some commonality to them all

ACuriousMind

Oh, are you under the mistaken assumption that all these "specific types" are a subset of some general definition that would be useful?

They aren't :P

Slereah

Am I mistaken

Dang it

shakes fist at the heavens

Because I can't say "a subset of $M$ with manifold structure", since that's not true of the figure 8 either

ACuriousMind

Again, what you want to define as a submanifolds really just depends on what "nice" properties you need for what you intend to do with them

Slereah

Is "some mapping $f : N \to M$" not true for some types submanifolds?

It is fairly generic

ACuriousMind

1:08 PM

@Slereah Sure, but without making some demands of that mapping that's gonna be completely useless

Slereah

yeah

But I'd like to have something to go on in the intro

Before going into more specific submanifolds

ACuriousMind

Why

Slereah

Also more importantly

Do we care about immersions in GR?

bolbteppa

What are the 5 types of definitions

Slereah

Are they useful

There's immersions, submersions and embeddings

@ACuriousMind I dunno

ACuriousMind

1:09 PM

Why would you introduce a notion of submanifold that's literally useless? If you're just in an intro, you likely can get away with just saying the embedding/immersion/whatever has "nice properties".

Slereah

If I have a chapter for three different things

I guess I want some commonality explained?

Even if it's just a throwaway line

bolbteppa

The gist is, an injective map between two manifolds will initially only have one topology, and so you have immersed one manifold into the other, however if the topology then matches up with that of the image manifold you have an embedded submanifold

ACuriousMind

And as soon as you want to do something technical you have to pick one of the proper definitions, anyway

bolbteppa

Where are there 5 variations in this picture

Slereah

But not all such maps are injectives

As mentionned

0celo7

1:10 PM

@bolbteppa Embedded, immersed, “standard”, regular, proper (something like that)

Slereah

@0celo7 which one is submersion

0celo7

That one too

ACuriousMind

@Slereah I don't think there's any "commonality" to explain, beyond there always being a map.

0celo7

Check R.W. Sharpe’s book

Slereah

@0celo7 are there any useful non-injective immersions in GR

bolbteppa

1:11 PM

They are small differences compared to the picture above

Slereah

I can't really think of a context where that would pop up

Except maybe like

Path of a particle mapped to the spacelike hypersurface

0celo7

For the proof of the positive mass theorem for locally conformally flat manifolds they’re very important

Slereah

Guess I'd better do 'em then

Also which one is the submanifold

Is it $N$ or $f(N)$

0celo7

@Slereah only if you want to read my thesis

@Slereah the image

But you can abuuuuuse that language

bolbteppa

A great example of why manifold theory is ridiculously pedantic if taken seriously

Slereah

1:14 PM

Like calling the vector components the vector???

bolbteppa

Even if $N$ is a subset, you have to call it's image under the identity map $I(N)$ the submanifold

(This is why physicists often advance math)

Slereah

Is it because they don't care about math

Speaking of which

bolbteppa

Adapting to the language of set theory just sometimes leads to convoluted ways to say obvious things, makes sense, but still

Slereah

If wavefunctions on $I$ admit wavefunctions that aren't $0$ at the boundaries

But not as part of the spectrum of $H$

Are they part of the spectrum of $e^{itH}$

and if so what's the link between the basis of the two

0celo7

confused Jackie Chan

ACuriousMind

1:19 PM

@Slereah what

Slereah

The eigenbasis of $H$ is like $\{ \sin(kx) \}$

ACuriousMind

Please use terminology properly - the spectrum of an operator is a subset of the complex plane, not a bunch of wavefunctions.

Slereah

Which should generate the whole domain of $H$

ACuriousMind

Has this anything to do with submanifolds or have you abruptly switched topics without giving any context? :P

Slereah

But on the other hand, wavefunctions with $\psi(0) \neq 0$ should be valid

@ACuriousMind It doesn't

0celo7

1:21 PM

@ACuriousMind German detected.

Slereah

But they're not part of $H$'s domain

0celo7

Hat das irgendetwas to tun...

Slereah

But since $e^{itH}$ is a bounded operator, its domain should be all of the Hilbert space

Including those wavefunctions

ACuriousMind

@Slereah Yes. What's the problem?

Slereah

What's the eigenbasis of $e^{itH}$ then?

And what's its link to the one of $H$?

0celo7

1:22 PM

He’s asking if the eigenfunctions of that guy (if they exist) are somehow eigenfunctions of H

Slereah

that guy is called the time evolution operator

0celo7

I can’t remember, is that guy even compact?

Slereah

In physics you usually just write $e^{it E_n}$ for the eigenvalues

But if their basis is different, it seems weird

ACuriousMind

@Slereah Whose basis?

Slereah

The basis of $\text{Dom}(H)$ and $\text{Dom(e^{itH}) = H}$

ACuriousMind

1:25 PM

That doesn't make any sense :P

These are different spaces, how could their "basis" ever be the same? One is dense in the other, though

Slereah

Though I guess $Dom(H)$ is dense in $H$ so I dunno

@ACuriousMind Because in physics that's what you do :p

You pretend they're the same

ACuriousMind

What is what you do

Slereah

So I'm wondering what's up with that

ACuriousMind

I get that you're confused about something relating to energy eigenfunctions, but I can't quite make out what exactly the issue is

Slereah

You just decompose the wavefunction in the basis of $H$ for the time evolution operator

0celo7

1:27 PM

@ACuriousMind they could (and should) have the same Hilbert basis

Well, “the”. They should have “a” Hilbert basis that works for both.

Not that it will be an eigenbasis of course

Slereah

@0celo7 How do you write a wavefunction that doesn't vanish at the boundaries from functions that do, though

ACuriousMind

@Slereah I don't know what that's supposed to mean.

Slereah

Like $$e^{itH} \psi = \sum e^{itH} \psi_n = \sum e^{itE_n} \psi_n$$

0celo7

@ACuriousMind Huh. TIL the harmonic oscillator is bounded.

ACuriousMind

@0celo7 Hm

bolbteppa

1:29 PM

Seems like you're saying that $\{ \sin (kx) \}$ is a set of eigenfunctions for $\hat{H}$, so you can expand $\psi$ in terms of them, but then saying that even though the solution to Schrodinger $\psi = e^{-i\hat{H}t}\psi_0$ so that $e^{-i\hat{H}t}$ has to act on the same basis, this $e^{-i\hat{H}t}$ operator can also act on other functions for some reason there is an issue?

Slereah

(This discussion isn't helped by the fact that I'm using $H$ for both the Hamiltonian and Hilbert space)

0celo7

@Slereah Use $\mathscr H$ for the Hilbert space.

Slereah

That's a lot of characters to type

0celo7

I have a single fluid motion for \mathscr

ACuriousMind

@Slereah Ah, yes. There's an implicit assumption there that the $\psi_n$ are located in the Hamiltonian's domain, of course. Is that your problem?

Slereah

1:31 PM

Yes.

ACuriousMind

What is the problem with that, though?

Slereah

Well, shouldn't the basis be wider?

ACuriousMind

Why?

bolbteppa

You are just considering a Lie group vs. a Lie algebra acting on the same vector space

ACuriousMind

What do you mean by "wider"?

Slereah

1:32 PM

I'm pretty sure you can't generate all wavefunctions of the Hilbert space with that basis

ACuriousMind

The domain of definition of the unexponentiated operator should be dense.

Slereah

Yeah that is what I'm thinking as kind of the solution, but only in a vague sense

I have no idea of how it would actually work

ACuriousMind

@Slereah Why? Do consider that the Hilbert space consists of equivalence classes, not functions, so you need to treat physicist statements like $\psi(0) = 0$ very carefully.

Slereah

That could be the problem, yes

0celo7

@ACuriousMind Explain the boundary conditions in the particle in the box please.

That is what you need to explain to him

He's bad at telling you that

bolbteppa

1:33 PM

Think you're asking about whether in exponentiating from a Lie algebra to the Lie group acting on a vector space, this process should make the basis wider i.e. add more functions in the process?

Slereah

Well apparently not!

0celo7

@bolbteppa You're talking complete nonsense

bolbteppa

@0celo7 yes to someone who hasn't studied basic lie theory

0celo7

Oh no, you called me out!

ACuriousMind

@0celo7 The domain of definition on which the momentum operator is self-adjoint is the one with apropriate boundary conditions.

0celo7

1:35 PM

@ACuriousMind Don't tell me that

(also Neumann conditions get you that)

Why did the caveman physicists originally give that problem Dirichlet boundary conditions?

key word being physicists

ACuriousMind

Is @Slereah's problem he doesn't believe that domain of definition is dense?

0celo7

I hope not

ACuriousMind

Then I don't know what the problem is.

0celo7

@Slereah $C^\infty_0(\Omega)$ is always dense in $L^p(\Omega)$, $\Omega\subset\Bbb R^n$ open.

Slereah

I do believe that it is dense

It's more that, from a naive perspective

if $\sin(kx)$ is the basis

I don't see how you could get something without those boundary conditions by summing them

ACuriousMind

1:37 PM

@Slereah Are we in a box or on the real line?

Slereah

In a box

$[0,L]$

0celo7

@ACuriousMind how does one justify the boundary conditions phyically?

ACuriousMind

@0celo7 "Continuity of the wavefunction" ::cringes::

Slereah

the eternal struggle between math and physics

0celo7

@ACuriousMind I was going to say you're not allowed to say that.

ACuriousMind

1:38 PM

@Slereah You're not summing functions, you're taking the limit of a series in $L^2$-space

Slereah

How would I get, for example, the wavefunction $\psi(x) = 1/L$ from the basis, for instance?

Is there such a series

ACuriousMind

@Slereah Sure. Its Fourier series, obviously.

Slereah

or $\psi(x) = 1$, if we just pick $L = 1$

Ah yes, makes sense

Basically you just get the slope at 0 steeper until the limit is just $1$ at $0$?

ACuriousMind

Note that the Fourier series of a constant is a constant, though.

Slereah

true

But the same applies for any other function that doesn't vanish at the boundaries, I guess

btw is $\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$ in a Hilbert space?

It would help for a proof if it did

Otherwise I'll have to expand inside

ACuriousMind

1:46 PM

@Slereah The product literally behaves like the dot product of ordinary vectors. Is that true for ordinary vectors?

Slereah

Trying to do the exercize with $$\frac{d}{dt} \langle \psi, \phi \rangle = \langle \frac{d}{dt}\psi, \phi \rangle + \langle \psi, \frac{d}{dt}\phi \rangle$$

@ACuriousMind I guess not!

Guess I'll have to do the Taylor expansion of $\psi(x + h)$

0celo7

@Slereah I've done that one!

use the definition of derivative

Slereah

That's what I did yes

ACuriousMind

I think you just mimick the proof of the standard product rule for that one :P

0celo7

@Slereah wtf

@ACuriousMind yeah and also use the fact that $\langle f,-\rangle$ is continuous

Slereah

1:49 PM

I get $$\lim_{h \to 0} \frac 1 h (\langle \phi(t+h), \psi(t) \rangle + \langle \phi(t), \psi(t+h) \rangle - 2 \langle \phi(t), \psi(t) \rangle)$$

Hence the need to expand

ACuriousMind

You don't need to expand anything, but I have no idea how you arrived at that expression.

Slereah

From $ \langle \frac{d}{dt}\psi, \phi \rangle + \langle \psi, \frac{d}{dt}\phi \rangle$

Put in the definition of derivatives

Then by linearity

Or do I need to start from $\frac{d}{dt} \langle \psi, \phi \rangle$ instead

ACuriousMind

Oh, you're starting at the r.h.s.

Slereah

Yes

But I guess that using the same proof as the product rule sounds like a good idea yeah

0celo7

you're gonna add and subtract things if memory serves

Slereah

1:56 PM

Yeah the old $ + a - a$ trick

CooperCape

Hate that so much

(only cause I'm too basic to figure out that on ma own...)

Physics Meta

0

Q: Using @ with any username

I was wondering if attaching @ with any username sends them a message in their mailbox or does it have no significance

discussion

Slereah

Most analysis is basically the same tricks over and over

Just gotta remember to do them

Second exercize is $AB - BA = cI$ in finite dimension

Which ironically I saw a proof for the specific $=0$ part a few days ago

But no idea why $= cI$ first

Oh wait, maybe it's the same proof?

Take the trace

$Tr(AB - BA) = Tr(AB) - Tr(BA) = Tr(AB) - Tr(AB) = 0$

I guess that doesn't help for the $cI$ part

Can you express all operators as $A = \sum c_{ij} |\psi_i \rangle \langle \psi_j|$ in finite dimensions

ACuriousMind

@Slereah Yes.

That's quite literally what a matrix is, the $c_{ij}$ are the matrix coefficients in the $\psi_i$ basis

Slereah

What is this """matrix""" you speak of

Is the commutator of all matrices $I$?

ACuriousMind

2:06 PM

@Slereah No, why would it be?

Slereah

Because that's the exercize

Oh wait

I misread the exercize

That's a hypothesis, not the result

So I guess the trace argument works

$Tr(AB - BA) = 0$, $Tr(cI) = cN$, hence $c = 0$

Now to show that $(cA)^* = \bar c A^*$

$\langle \phi, (cA)\psi \rangle = c \langle \phi, A\psi \rangle = c \langle A^*\phi, \psi \rangle = \langle \bar c A^*\phi, \psi \rangle$

There we go

0celo7

@Slereah are you becoming a functional analyst?

Slereah

God forbid

just trying to go through Hall unscathed

It's weird that he asks to prove that $[A,B]$ is self-adjoint if $A$ and $B$ are but not that $(AB)^* = B^* A^*$

bolbteppa

2:28 PM

Symplectic geometry & classical mechanics, Lecture 1

►

Guy goes through Arnold's mechanics book apparently

Slereah

a bold choice for a class

"We're going to show you why a ball falls to the ground using symplectic geometry"

DanielC

For A,B s.a., then [A,B], if properly defined, is skew-selfadjoint.

Slereah

Well, $1/i\hbar [A,B]$

Not the commutator proper

DanielC

@Slereah Then you should quote the problem fully, not approximately.

Slereah

You're not my mom

ACuriousMind

2:39 PM

@DanielC We've been telling him that for quite some time ;P

DanielC

A piece of advice is fortunately not mandatory, so you have the option to do as you please, of course.

Slereah

@ACuriousMind who has the time to know what they're doing

I just stumble down physics like some very steep stairs

DanielC

That is not really physics, but mathematics.

Balarka Sen

3:00 PM

@Slereah You're right. I am

Slereah

D:

Balarka Sen

You should quote the theorem fully, not approximately

A correctly stated theorem a day keeps the cranks away

0celo7

@BalarkaSen that’s what I keep telling these physicists but they insist on being wrong

Slereah

If we had to wait until theorems were correctly stated we'd never do physics :p

Balarka Sen

The world would be a better place

Slereah

3:06 PM

well no atom bomb I suppose

Balarka Sen

That too

But above all, physics is an abomination of science

4

Slereah

Good thing you don't read psychology

0celo7

@BalarkaSen you are learning well

Balarka Sen

a-bombination

0celo7

Hadron

Slereah

3:11 PM

@0celo7 so is non-relativistic QM real

Or is it fake like QFT

0celo7

It’s very real

Completely rigorous for the most part

The only thing that’s fake is geometric quantization

Slereah

What about deformation quantization

0celo7

Good lord how is r/furry on the front page

Slereah

Or path integral quantization

Also what about measurement

0celo7

@Slereah deformation?

Slereah

3:13 PM

Measurement sounds p. fake

@0celo7 the weyl stuff

0celo7

@Slereah I don’t touch that

Slereah

With the Weyl star operator

0celo7

@Slereah well, it’s inconsistent

Slereah

what's wrong with weyl

0celo7

All of the quantization techniques fail somehow

It’s a Theorem in Hall

Slereah

3:14 PM

o

I guess I'll see it later

0celo7

They fail for cubic terms

Slereah

a lot of the measurement stuff sounds p. handwavy

0celo7

Maybe not path integrals.

Well definitely fails though

Slereah

Cubic terms doesn't sound very bad

0celo7

*Weyl

Slereah

3:15 PM

Not a lot of cubic potentials

0celo7

If you like this stuff you need to check out Reed and Simon later

Slereah

probably, but it's like

800 pages

Also I think "like" this stuff might be a bit of an overstatement

I like sunny days and ice cream

0celo7

These questions are imporant for us PDEers too

How to represent differential operators is a big question

Balarka Sen

Hey, Vsauce Michael here. What is random geometry?

0celo7

No clue

@BalarkaSen it’s actually “Hey Vsauce, Michael here”

He is talking TO Vsauce

Slereah

3:19 PM

It's a random shape

Is it a square???

A triangle???

Balarka Sen

That's the meme. It's memified to imply his first name is Vsauce and last name is Michael

Slereah

A rhombus???

Does Hall talk about quantum probability, btw

Since it's not technically Kolmogorov probability

0celo7

I don’t think he does

Do I even want to know the difference

Slereah

I think there's like one different axiom

4

Q: Probability and Quantum mechanics

I don't quite understand how the probability language of sample spaces, $\sigma-$algebra, random variables, etc, fit into the quantum mechanics' formalism. To wit, we usually say that an observable is a linear operator in a Hilbert space, and afterwards we define the "expected value" of this o...

probability-theory quantum-mechanics

Not 100% sure what's an example where the two differ, though

"Quantum probability theory is a generalization of probability theory in which random variables are not assumed to commute."

how do random variables commute

i do not know

though apparently it's related to the changing of probability theory into algebras thing

Balarka Sen

3:55 PM

@PrathyushPoduval 10/10 death metal album cover on profile pic

Prathyush Poduval

@BalarkaSen Thanks! It's what came to my mind when I took the pic :P

Whats up with your profile pic

you have a wierd hat over there

Balarka Sen

weird hat? it's my finest communist hat

Prathyush Poduval

Seriously though, wtf is that?

Slereah

Top hat is the most capitalist hat, @BalarkaSen

you capitalist swine

Balarka Sen

GULAG

Slereah

4:01 PM

I demand your immediate self critic

descriptions of self-criticism in communist china are quite amusing

0celo7

@ACuriousMind gyazo.com/6cd44cda1fad4d387e62b97bad7d5f4d

Anonymous

@BalarkaSen Are you around?

Prathyush Poduval

@Blue Call him a capitalist, and he'll come in an instant

Balarka Sen

@Blue Yep

Anonymous

@BalarkaSen Could you please come over to this room for a while. I have a few statistics questions

0celo7

4:21 PM

@Slereah did you get ghost recon wildlands?

0celo7

4:39 PM

@Slereah let's read RS Vol 2 Section IX.7

Slereah

@0celo7 I do not know what that is

0celo7

also IX.8

I want to know the Wightman axioms

Slereah

Maybe I'll eat dinner first

Imma hungry

Hm, where did I put RS

I'll just redownload it

0celo7

5:09 PM

@Slereah Theorem IX.28 is interesting

basically just Sobolev but still

Slereah

What's $H_0$, the free Hamiltonian?

0celo7

yeah

Slereah

5:26 PM

they're using $\lambda$ for the Fourier space variable

odd choice

0celo7

5:39 PM

@Slereah but you can see the importance of domain choices

this stuff is important later on too

Slereah

Well I'm reading like

Section 1

So Theorem 28 will have to wait

0celo7

maybe finish Hall first

then go to Reed and Simon

Slereah

Also this week end I'm doing some electronics

gotta give a gift for secret santa and my recipient is an old engineer dude who likes electronics

I'm gonna give him some spare Arduino and decorate with with LEDs and a speaker for Christmas theme

DanielSank

5:52 PM

Hi, everybody.

@knzhou You can see how it's set up in the git repo.

CooperCape

hello...

I'm sure someone's linked this before but...

chat.stackexchange.com/…

DanielSank

Hahahahaha

This is by far my favorite:

Sep 2 '16 at 6:34, by DanielSank

Hey everybody, it's soapbox time:

CooperCape

hah

(I wish I understood what's going on there)

vzn

@DanielSank any reaction to the new silicon qbit news...?

CooperCape

signal processing SE (very specific? or no...)

Ahh classic

your answer is "engineering textbook correct"

DanielSank

5:58 PM

@CooperCape What do you not understand?

CooperCape

@DanielSank Oh no in the answer linked below....

As in

the one linekd under the link you posted back to 2016

(I really didn't specify)

DanielSank

The answer about Fourier transform of periodic signal?

CooperCape

Yeah

DanielSank

@vzn What news?

@CooperCape Ok, well I'm happy to help you understand.

This is one of my favorite topics.

CooperCape

Nah I gotta have dinner

Haven't started anything on fourier things

DanielSank

5:59 PM

Ok, well, if you're interested some time, I'll be happy to help.

Fourier transforms are very easy to understand if you first understand some basic things about linear algebra.

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