The h Bar

Slereah

15:00

which is fairly easy in a vector space

Though I guess it is multiplication by $x$ at every point

Oh wait

Mithrandir24601

@glS not so much yet, in the future, yeah

@0celo7 That's... Huh? O.o

Slereah

Are the ones that aren't $L_2$ the wavefunctions that do not converge to $0$ at infinity?

0celo7

@ACuriousMind his other student was there too, and he calls him by his first name

ACuriousMind

???

Anonymous

@0celo7 He doesn't like you ;P

ACuriousMind

15:02

I...can only come to the conclusion your advisor doesn't like you? oO

Slereah

If they don't converge at infinity they might grow too big on the edges to be $L_2$, I guess?

0celo7

@Blue the opposite

Slereah

@0celo7 am I correct in my assumptions

Mithrandir24601

@0celo7 what would your advisor do if you just started calling them by their first name?

0celo7

I'm staying at his house during a conference over the summer and also at his parents' in Rio

glS

15:02

@Mithrandir24601 when you meet him, say Stefano someone from Belfast says hello, he'll know who I am!

Anonymous

@0celo7 He loves you so much that he wants a personal pet name from you?

0celo7

@Blue ...fuck

ACuriousMind

Or maybe he's so impressed by you that he feels first name basis would be disrespectful of you :P

0celo7

what is wrong with you

Anonymous

lol

0celo7

15:03

@ACuriousMind Should i insist he calls me Mr. Unger?

@Slereah what?

Mithrandir24601

@glS will do! Although that might be January

ACuriousMind

@0celo7 He doesn't?

0celo7

@ACuriousMind No

ACuriousMind

That is, he calls you by your first name but you're not allowed to use his?

0celo7

Yes

ACuriousMind

15:03

that's weird

0celo7

well, yes

Slereah

@0celo7 That $\hat x \psi(x)$ may not be $L_2$ if $\lim_{x \to \infty} \psi(x) \neq 0$

Anonymous

@0celo7 Sounds like the typical Indian professor

ACuriousMind

Such a dynamic is completely unheard of in Germany outside of school - whatever two people call each other is usually reciprocal

0celo7

sure, but I'm not sure that's guaranteed

@Slereah

@Slereah also take a function that's $O(x^{-2})$ (but not better) at infinity

that's not in the domain of x

ACuriousMind

15:05

And I would usually take someone else's usage of my first name as an implicit invitation to use theirs in return

0celo7

@ACuriousMind Well Morwen asked for me!

Slereah

Are $O(x^{-1})$ functions not $L_2$?

ACuriousMind

Yeah, I got nothing for this situation. It's very odd

0celo7

@Slereah oh shit $L^2$

take the square root of what I said

Mithrandir24601

@ACuriousMind it does seem to be the system in China - even the postdocs call the profs 'professor' when using English and I could swear I heard 'sifu' more than once

Slereah

15:06

Yeah I s'ppose

0celo7

@Slereah $O(x^{-1})$ integrates to $O(\log x)$

Slereah

Which makes sense what with the whole unbounded operators aren't defined on the whole Hilbert space

Secret

So the philosophical question goes like this:

Slereah

Right

0celo7

@Mithrandir24601 one of us calls profs over the age of 50 "professor" unless they're women and she likes them

very complicated system

Slereah

15:08

M'lady

Anonymous

It's way simpler here. Just call everyone "Sir".

0celo7

@Slereah exactly, but I think you have to wait till chapter 9 to see that proof

@ACuriousMind See I don't want to go nuclear but insisting he call me Mr. Unger would be hilarious

Slereah

@0celo7 is there a relation between $x\psi$ not in $L^2$ and $p \psi$ not in L^2$

Or are they not in $L^2$ independantly

ACuriousMind

@0celo7 It would be equal parts justified and petty. I like it.

0celo7

@ACuriousMind (I have the same situation with a German btw)

Slereah

15:09

Can't rly think of a link between bad functions at infinity and discontinuous functions

Mithrandir24601

@0celo7 you can say that again...

0celo7

@Blue how very Indian :P

Sid

@0celo7 Yep. No one gets any issues with it

ACuriousMind

@Blue What do you call women?

Emilio Pisanty

@ACuriousMind It does also include a rocket landing

ACuriousMind

15:10

Ma'am?

0celo7

@Sid In America we use our names a lot, which is an issue.

Russian spies used to get tripped up by this

Secret

We knew that the wavefunction in continuous variables (such as position and momentum wavefunctions) resides in a hilbert space where the fourier basis is an orthonormal basis. Thus given any integrable and continous $\psi (x)$ it can be expressed as an infinite sum of plane waves. i.e.

$$\psi (x) = \frac{1}{2\pi}\sum_{n=1}^{\infty} a_n\sin (nx) + b_n\cos (nx)$$

Anonymous

@ACuriousMind If you (we) don't know their gender (online), just call them Sir unless they clarify. And once you know they're female then call them Maám. Weird but simple

Secret

Now consider the following figure:

Anonymous

It's sort of sexist though

Slereah

15:12

Not all $L^2$ functions can be expressed by plane waves!

0celo7

@Blue I agree with this. Assume male until proven otherwise. Yeah sexist but it works.

Slereah

I think they can differ at a set of measure $0$

ACuriousMind

@Secret Stop right there. If both position and momentum are continuous, you cannot have a discrete Fourier series.

0celo7

Until genders in STEM normalize you'll be right more often than not.

Slereah

unless you're on a compact space

0celo7

15:13

If/when they do normalize, rotate!

Slereah

I did a statistics class at uni once

Emilio Pisanty

@Slereah ... or you're OK with a non-square-integrable $\psi$

Anonymous

@0celo7 lol

ACuriousMind

@Slereah The point is that if position is compact, then momentum is discrete, and vice versa.

Slereah

it was where all the ladies were hiding for some reason

@EmilioPisanty not a very good wavefunction

Emilio Pisanty

15:13

@Slereah indeed it wouldn't be

0celo7

@Slereah There were a lot of women in my algebra classes. Not passing judgement.

::looks at @ACuriousMind::

ACuriousMind

::looks at @0celo7::

0celo7

@ACuriousMind Alicia

Emilio Pisanty

$$
\begin{array}{ccc}
\begin{pmatrix}
\text{Fourier transform}\\
t\color{blue}{\text{ unbounded}}\text{ and }\color{blue}{\text{continuous}}\\
\omega\color{blue}{\text{ unbounded}}\text{ and }\color{blue}{\text{continuous}}
\end{pmatrix}
& &
\begin{pmatrix}
\text{Fourier series}\\
t\color{green}{\text{ bounded}}\text{ and }\color{blue}{\text{continuous}}\\
\omega\color{blue}{\text{ unbounded}}\text{ and }\color{green}{\text{discrete}}
\end{pmatrix}
\\ & & \\
\begin{pmatrix}
\text{Time series}\\

as they say

0celo7

@EmilioPisanty did you have that prepared

Emilio Pisanty

15:15

@0celo7 nope, typed it just now

0celo7

@Slereah what is the question now?

Emilio Pisanty

four keystrokes, ctrl c ctrl v

from

6

A: Why is $S(\omega)=S(\omega +2\pi)$ true for a frequency spectrum?

The relationship $S(\omega)=S(\omega+2\pi/T)$, which you can simplify to $S(\omega)=S(\omega+2\pi)$ if you have some reference timescale $T$ to compare to, is false in the general case. It should be intuitively clear that the arbitrary function has an arbitrary Fourier transform, which means that...

0celo7

@EmilioPisanty so you had it prepared

Secret

Correction:
$$\psi (x) = \int_{-\infty}^{\infty} \tilde{\psi} (p) e^{-2\pi i p x} dp$$

0celo7

> putting $2\pi$ in the exponential

Emilio Pisanty

15:16

@0celo7 yeah, you could say that =P

0celo7

you're gonna get your butt whooped

Emilio Pisanty

@0celo7 putting $dp$ in the exponential is kinda worse, innit

0celo7

@EmilioPisanty that was a typo, he fixed it

ACuriousMind

@EmilioPisanty I'm afraid my excitement for that is limited, too

Emilio Pisanty

but yes, $2\pi$ in the exponential won't make you many friends

@ACuriousMind each their own and all that, then =P

Sid

15:17

duh, Python Data Structures are easy and boring

0celo7

@EmilioPisanty I define $dx$ to be Lebesgue measure times $(2\pi)^{-n/2}$ :)

my Fourier transforms have no $\pi$s in them

Emilio Pisanty

@0celo7 your convolutions do

Anonymous

@Sid Start solving CLRS

Emilio Pisanty

and/or the convolution theorem

@ACuriousMind you seen any SpaceX landing? granted, after the first they're not that different. but if you haven't seen one you oughter.

0celo7

don't think so @EmilioPisanty

Emilio Pisanty

15:18

heck, you don't even need to be convinced the rocket went anywhere fancy to appreciate the landing

ACuriousMind

@EmilioPisanty Yeah, I did (not live either, though)

Emilio Pisanty

@ACuriousMind fair enough

Secret

Before I started elaborating using the picture, for clarify, let me first state his question so if I happen to fail at my explanation, at least we know what area we are dealing with:

Emilio Pisanty

@0celo7 don't think what?

@Secret "his" question?

0celo7

@EmilioPisanty I think the fact that there's only one FT on one side of the convolution theorem cancels out the normalization in my definition of the convolution

Secret

15:20

This is a question by a philospher I met today in the complexity criticality and computational symposium

Hari Prasad

Hey guys. After a very long time

0celo7

if you do that normalization I think it's equivalent to putting the thing in the exponentian

Emilio Pisanty

@0celo7 hmmmm. That might work? I'm not sure, though.

Secret

I found it intriguing enough thus I decided to ask it so

0celo7

@EmilioPisanty when I teach functional analysis in a few years I'll get back to you

Emilio Pisanty

15:21

@0celo7 you do that

0celo7

in any case some people (Swedes) define a dbar x like $\hbar$ that's normalized by $(2\pi)^{-n/2}$ and it works out well

then they have two measures and never need to worry about it

Emilio Pisanty

@0celo7 is there a dbar in tex?

0celo7

@EmilioPisanty tikz

Emilio Pisanty

$\dj$?

nope

0celo7

Emilio Pisanty

15:23

c'mon, detexify, you're better than that

Sid

@Blue I am now trying to go for C++ after finishing Python Data Structures.

0celo7

@EmilioPisanty see the $d\eta$

it's the natural measure for Fourier transforms

Anonymous

@Sid I don't understand. Why would you do data structures for C++ and Python separately?

Emilio Pisanty

$\text{\dj}$?

nope

Anonymous

Start solving the problems instead in both languages

Sid

15:25

@Blue Nah, not completely confident in C++. So, will try to brush that up now...

Slereah

How does one pronounce BRST quantization

Balarka Sen

@0celo7 Approximately

Slereah

I want to say "breast"

0celo7

burst

Emilio Pisanty

@0celo7 reputedly all of those issues go away if you do like Secret and put the $2\pi$ in the exponent

0celo7

15:26

@EmilioPisanty sometimes you have to have integrity instead of going for the easy solution

Emilio Pisanty

my thoughts exactly

0celo7

@BalarkaSen how many 50-minute lectures would it take to get through the first 30 pages of little Milnor

@EmilioPisanty Also introducing another standard is stupid at this point

Balarka Sen

Intro to diff top? hm

Secret

His question: I cannot understand how can true randomness (meaning the process is not pseudorandom) will ensure localisation will be possible. For example, the fourier transform of a continuum of frequency of plane waves gives a sinc function,

but if quantum mechanics is truly random, that means in theory the phase difference from one frequency to the next does not necessary all line up at the origin to produce the peak, and even if peaks were still being produced, the particle will not be localised in a certain region of space. How does true randomness ensure the classical limit behaviour where particles have well defined positions be arised. Does the law of large number holds for sample spaces that are infinite?

0celo7

I have enough problems with signs in PDE, don't try to kill me with $2\pi$s

Emilio Pisanty

15:27

@0celo7 I think it's standard in signal processing or some such?

it's definitely around

@Secret That's pretty close to word salad.

0celo7

@BalarkaSen I think the topology 2 prof will let me teach the first 5 chapters if I can do it in a reasonable amount of time. Brouwer fixed point theorem should be doable in 3 lectures

Emilio Pisanty

> if quantum mechanics is truly random, that means in theory the phase difference from one frequency to the next does not necessary all line up at the origin to produce the peak

what?

for one, that's hardly grammatical English. But no obvious grammatical fixes take that into something that makes much sense

Secret

he then argue that quantum mechanics shoudl be pseudorandom. At that time when he discussed that question with me, I don't have answers for him, and I said it is interesting enough thus I will ask the community about it.

Now, my elaboration on what his question is about:

0celo7

@EmilioPisanty none of us are signals people here, so we shouldn't be talking about a 2pi in the exponential

Emilio Pisanty

@Secret the question doesn't make any sense, but to the extent that it does it is based on a misunderstanding of how QM works

if you're going to just build on that without fixing the problems... gulp

Secret

15:31

I also have the same conclusion, but I need to be able to convince him that

(this is the guy btw: sydney.edu.au/sydney_ideas/lectures/2017/…)

He said I managed to understood his question fully, and which is why my elaboration will make it clear on where the misconception is hiding

Emilio Pisanty

@Secret well, produce a version of the argument you want dissected that is grammatically cogent, and we can help with that

Slereah

@Secret that's a sign that he's a crazy man

2

Secret

Ok let me try clean up the question a bit:

Slereah

"Proposition 3.9 For all sufficiently nice functions φ and ψ"

Emilio Pisanty

@Secret good man. and/or woman.

Slereah

15:34

@0celo7 how is niceness defined

What is the set of nice functions

$\mathfrak{Nice}(\mathbb R^n) \subset L^2(\mathbb R^n)$

0celo7

defined in the first line of the proof

Balarka Sen

@0celo7 In the first three lectures you mean?

Emilio Pisanty

@Slereah You put $\varphi$ and $\psi$ in a room together for an hour and you observe whether they get into a fight or not.

Balarka Sen

That sounds about right

0celo7

@BalarkaSen Yeah I want to get through Brouwer fixed point at least

Balarka Sen

15:35

I think 3 is fine

Emilio Pisanty

@Slereah otherwise, that just sounds like the standard hypothesis on any theorem in physics-side QM

0celo7

@BalarkaSen I think 3 can be skipped

4 and 5 probably another 3?

Emilio Pisanty

it's the credit viewpoint on rigour. Get the results first, worry about the hypothesis later. ;-)

Balarka Sen

I usually skip 3

0celo7

@BalarkaSen usually?

you've taught this book before?

Balarka Sen

15:36

lol I mean I usually recommend people to skip 3

im being lazy about my statements

Slereah

Wait

0celo7

I think the proof is beautiful but my advisor is teaching analysis 2 and he'll do Sard there

Slereah

Does it mean $X$ and $P$ aren't symmetric if it's insufficiently nice

0celo7

(using my notes from when I read this book)

Slereah

Or is everyone nice on the domain

Balarka Sen

15:37

Gotcha

Slereah

Yeah I guess it is

Secret

Suppose quantum mechanics is truly random. This means for each frequency component in the fourier transform, there is no restriction (modulo a period of the plane wave in that frequency component) on where it can be placed in space. Our usual results of demonstrating the localisation of the wavefunction of a particle at the classical limit in undergraduate quantum textbooks is by showing that the fourier transform of all frequency modes will give a resultant wave sharply peaked at the origin.

(The sentence is not finished yet, but does it makes sense to you so far?)

Emilio Pisanty

> Suppose quantum mechanics is truly random.

this makes no sense

0celo7

@Slereah he's only claiming they're symmetric on their respective domains

Emilio Pisanty

it is simply not specific enough to make any sense

Secret

15:39

He define true randomness this way:

Balarka Sen

@EmilioPisanty It means quantum mechanics is the spirit essence of the random universe guided by a probabilistic random variable taking values on a Cantor set

Slereah

@Secret Particles are rarely very spread all over space

The probability of being very far off is very small

Emilio Pisanty

@BalarkaSen ah, gotcha

Slereah

Also particles are indistinguishable, so it's hard to know where particles end up

Secret

A process is truly random if the outcomes are probabilistic and there are no pseudorandom process involved (that is, there exists no seed or determinstic process that can generate a particular probability distribution)

Slereah

15:40

But on the other hand

Balarka Sen

It's really simple, but quite profound

Slereah

It's a TRUE THEOREM that particles can end up arbitrarily far

0celo7

@BalarkaSen actually

Slereah

IN ALL CIRCUMSTANCES

0celo7

the cantor set has measure zero so you can omit it

Slereah

15:41

that is Hegerfeldt's theorem

there's no compact region of space where a particle has 100% chances of being

So a particle can be measured arbitrarily far

Balarka Sen

is that like a

Secret

Which to my understanding (which he checked seemed to match what he had in mind) is that true randomness means there are no hidden variables of any kind, including nonlocal ones

Balarka Sen

statement about the wavefunctions?

i.e., none are compactly supported

Slereah

No

It's not true in non-relativistic QM

You can have a compactly supported wavefunction

Balarka Sen

hm

Slereah

15:43

But it is true if you add relativity

Balarka Sen

ah

Emilio Pisanty

@Secret You haven't actually applied this randomness to anything (yet)

0celo7

these mystical theorems always leave a bad taste in my mouth

very little in the way of definable hypotheses

Emilio Pisanty

You've just stated a theorem about Fourier transforms.

Slereah

The list of hypothesis is fairly simple

Secret

15:44

This will be coming next (I am typing slow this time to ensure we are all clear)

Slereah

It's like

QM

Emilio Pisanty

@Secret sure

Slereah

Some operator of time translation

microlocality

0celo7

honestly QM is not a good hypothesis

what does that even mean

Balarka Sen

lol

Slereah

15:44

that kind of stuff

@0celo7 the usual stuff

Projective Hilbert space

The CCR

0celo7

you mean Weyl relations

Slereah

the actual conditions :

0celo7

is this from the book?

Slereah

fugg

Balarka Sen

reip in rekt

Slereah

15:46

It is!

Although that's Malament's theorem

I didn't include the conditions for Hegerfeldt's yet

Emilio Pisanty

@Slereah There's a Malament theorem?

that's catalan for "badly"

Slereah

There's several Malament theorems!

By David Malament

Emilio Pisanty

also "unwell"

Slereah

maybe his ancestor was a sickly fellow

0celo7

unwell in the head!

Emilio Pisanty

15:48

@Slereah quite possibly, yes

Secret

So, since quantum is truely, random, there is no guarentee that all frequency modes must all line up at the orign to produce the sharp peaked wavefunction as expected to be what happened at the classical limit. Since the space the frequency modes resides in is infinite, it is vastly more likely for the peaks of each plane wave to never line up to produce just one peak,

but that will mean that the particle is actually delocalised in more than one position in space at the classical limit, which is not what we observed for classical particles as they form a continuous trajectory as they move, with no sudden jumps from one observation to the next

Slereah

journals.aps.org/prl/abstract/10.1103/PhysRevLett.54.2395

Balarka Sen

uh oh physics chat session

Slereah

^Hegerfeld's theorem

Balarka Sen

time for me to scurry

Emilio Pisanty

15:49

@Secret This "quantum is truly random" hypothesis has nothing to do with QM as it is normally understood.

David Z

I get to be here for this one, even, although I'm not sure we have anything to discuss

Emilio Pisanty

@Secret All you've done is take a localized state, take its Fourier transform, radically alter it, Fourier transform back, and be surprised that the state has been radically altered.

Slereah

aw man

Sci hub is down

0celo7

@DavidZ these chat sessions used to be involved

Balarka Sen

uck

Emilio Pisanty

15:50

@Slereah try .tw

0celo7

I guess that was before people decided talking about homework was futile?

Slereah

thx

Emilio Pisanty

@Secret There is nothing in QM that tells you that the continuum-representation phases of states will spontaneously randomize.

@Slereah or the telegram bot

0celo7

@BalarkaSen I was talking with the prof last night and he wants to do de Rham but can't be bothered to explain exterior algebra

Secret

@DavidZ I have a barage of phsyics to share in the past 2 weeks which I will do it after resolving the philosopher's quantunm qurestion (which emilo is doing now and it seems to be getting to a conclusion)

0celo7

15:52

and I have to concurr, people not understanding really basic things like tensors/forms makes teaching basic classes a pain in the ass

once you understand them it's second nature

Secret

@EmilioPisanty I see, I think that's might be the answer to his question.

Balarka Sen

@0celo7 exterior algebra should be explained by motivating with the form theory

Secret

and now... preparing physics block

0celo7

@BalarkaSen I don't know how to motivate anything other than 1- and n-forms

Emilio Pisanty

@Secret But perhaps more importantly, the classical limit of QM doesn't mean that all possible quantum states will behave classically (which is the only apparent source of contradiction in your text). It needs to be done carefully, but it essentially says that there are suitable classes of states (not the ones you described) which under certain limits behave like classical systems. The fact that other classes behave differently tells you nothing valuable.

Slereah

15:54

@0celo7 Cross product

0celo7

@Slereah something no mathematician cares about

Balarka Sen

@0celo7 things you integrate. things which are volume elements.

lots of way to think about them

Slereah

Well maybe that's the problem

what about

The curl

0celo7

@BalarkaSen I might just be terrible about motivating math in general

Slereah

Curl is important

And it's a 2-form

0celo7

15:56

because when I think about how to motivate stuff I know very well, I get depressed that it's all useless

Balarka Sen

curl and div are the exterior derivative operators, yes

@0celo7 lol

MatheinBoulomenos

you can explain the cross product via exterior derivative + Hodge *

Slereah

@0celo7 Go to physics

It's full of 2-forms

0celo7

@MatheinBoulomenos so?

Balarka Sen

sanic hodgestar

Slereah

15:57

Sonic collects rings, not stars

That's Mario

0celo7

@Slereah ok, I'll do Yang-Mills to motivate it

I'm sure that will fly

Balarka Sen

@Slereah sanic, not sonic

0celo7

fuck it, Seiberg-Witten

let's go

Balarka Sen

teach me seiberg witten

Slereah

Do math people know the Yang Mill

I dunno

Balarka Sen

15:58

yeah gauge theorists do that

all the time

Yang Mills is like their spirit animal

Slereah

Gauge theorists aren't math people

0celo7

there are math people who do gauge theory

Balarka Sen

they are bro

Taubes is a gauge theorist

he's a math people

0celo7

I'm considering putting some of his stuff in my thesis

Slereah

0celo7

15:59

he worked out expansions for generalized Green's functions on p-form bundles

Slereah

what a nerd

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