The h Bar

Anonymous

6:01 PM

I can't understand Schroedinger Equation......how is $$(\frac{\hbar}{i}\frac{\partial}{\partial x})^2 = \frac{\hbar}{i}\frac{\partial^2}{\partial x^2})$$ ?

Anonymous

hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

Anonymous

0celóñe7

@Blue where does it say that

Anonymous

Sorry, I meant how is $(\frac{\hbar}{i}\frac{\partial}{\partial x})^2 = \frac{\hbar^2}{i^2}\frac{\partial^2}{\partial x^2}$ ?

0celóñe7

by definition of square???

Anonymous

6:05 PM

@0celóñe7 Derivative's square doesn't give second derivative

0celóñe7

proof?

Anonymous

$(dy/dx)^2 \neq d^2y/dx^2$

Anonymous

Isn't it?

0celóñe7

@Blue you cheater

that's not what you initially wrote and not what you wrote above

Anonymous

@0celóñe7 Sorry. My browser is having some issues. I meant $$(\frac{\partial y}{\partial x})^2 \neq \frac{\partial^2 y}{\partial x^2}$$

0celóñe7

6:09 PM

correct

but again, no one is claiming that

we are claiming $$(\partial_x)^2f=\partial_x^2 f$$

which is, pretty much, the definition

Anonymous

@0celóñe7 What does $\partial_x$ mean?

Kyle Kanos

$\partial_x=\partial/\partial x$

0celóñe7

@Blue I think you can guess :)

Anonymous

@0celóñe7 I couldn't understand this statement. You mean $(\frac{\partial}{\partial x})^2 f=\frac{\partial^2 f }{\partial x^2}$ ?

0celóñe7

Yes

What is hard? Applying a derivative twice is the second derivative.

How else are you defining the second derivative?

Anonymous

6:15 PM

But that's not what they are doing in the hyperphysics webpage

Anonymous

They are writing $p^2$

0celóñe7

Yes

They are applying the $p$ operator twice

Kyle Kanos

$p^2=p\cdot p$

0celóñe7

@KyleKanos What is that $\cdot$ supposed to be?

Mithrandir24601

@Blue yeah, finding the answer to 'what is $p$?' will explain this

Kyle Kanos

6:17 PM

@0celóñe7 Times...what else does it mean?

Anonymous

I thought $p$ is plain and simple "momentum"

0celóñe7

@KyleKanos But it's not. It's supposed to be operator composition.

Kyle Kanos

No, it's times

Anonymous

Does applying momentum operator twice differ from squaring the momentum ?

0celóñe7

@Blue When you do first quantization you take classical variables $\omega^k$ and take them to quantum operators $\Omega^k$, where the $k$ now means "compose $k$ times"

Kyle Kanos

6:19 PM

@0celóñe7: I think your want for rigor is probably above the discussion level at the moment

0celóñe7

So $p^2$ in the classical energy becomes $P^2$, where $P=-i\hbar \partial_x$

And the square now means $P\circ P$

So $P^2=P\circ P=(-i\hbar \partial_x)\circ (-i\hbar\partial_x)=-\hbar^2 \partial_x\circ\partial_x=-\hbar^2 \partial_x^2$.

Anonymous

All of a sudden converting $p$ from simple momentum vector to a quantum operator and then "composing" it twice seems a bit hand-wavy to me. That was the reason I was confused.

Anonymous

@0celóñe7 That makes sense

0celóñe7

@Blue These are physical postulates, they are only hand-wavy if they are defeated by experiments :)

@KyleKanos Does the dark theme carry over to non-Windows apps?

It's not making Chrome dark

Anonymous

@0celóñe7 All of a sudden they change "multiplication" of vector magnitudes to "composition" of operators. Phew..=P Is there a more rigorous proof of this? (Now I am beginning to see why mathematicians call physicists hand-wavy)

0celóñe7

6:24 PM

Proof?

My comment about $\Omega$ is basically what quantization is, by definition.

It's how you get quantum mechanics. It cannot be proven.

Anonymous

@0celóñe7 I mean the logic behind converting momentum vector to quantum operator

Anonymous

What is "quantization" ?

Anonymous

I need to read more abt this

0celóñe7

@Blue Taking a classical theory and formulating it in terms of quantum mechanics.

Kyle Kanos

@0celóñe7 No idea

Anonymous

6:27 PM

@0celóñe7 Hmm. Can you suggest me a book/website which covers this topic of quantization? Feels like I have some gaps in my knowledge

Kyle Kanos

Don't have Win 10

0celóñe7

@Blue I would vote to close that as opinion based. There's many ways people justify the postulates of quantum mechanics in admittedly post hoc fashions.

@Blue Get the book by R. Shankar.

You will need some serious linear algebra first.

Anonymous

@0celóñe7 I have it. I've (in)formally started the MIT OCW lectures (at home) and am reading the recommended parts from Shankar. But my electronics prof has started with Schrodinger's equation without covering these basics first. Hope to catch up soon.

0celóñe7

@Blue Without going into ridiculous math, the best answer is that the postulates of quantum mechanics work "as expected" for plane waves. But things like the Born rule (rule for calculating probabilities) have no a priori theoretical justification.

@Blue Why does an electronics professor need the Schroedinger equation?

Anonymous

@0celóñe7 She was teaching free electron theory of metals

Anonymous

6:33 PM

@0celóñe7 Hehe. Hope to learn the ridiculous maths someday

0celóñe7

@Blue It seems pretty backwards to me to talk about the Schroedinger equation in a freshman course.

Anonymous

I find it strange that people in my class nod their head for whatever the teacher says. I'm sure most of them didn't understand the conversion of classical parameters to quantum operators and accepted it without justification =P

Anonymous

@0celóñe7 Agreed.

Anonymous

They should first teach the basics of QM properly

0celóñe7

To really understand quantum mechanics (for physicists) you need a lot of linear algebra.

Avantgarde

6:36 PM

They won't. Electronics courses won't cover that

so it's good you're reading the physics yourself

Mithrandir24601

@0celóñe7 I just looked at some of the additional problems from my first year... shudders in horror

Anonymous

@Avantgarde I'm a physicist, man =P

Mithrandir24601

@Blue What are the basics of QM? Historical? Experimental results? simple but inaccurate QM? Maths? QFT? Classical Waves? (Classical) Lagrangians etc. leading to quantisation?

Anonymous

@Mithrandir24601 QM which is taught in 1st or 2nd year in Physics courses in most good universities

Kyle Kanos

Sheesh...Why would anyone not bring a chemist's combine ability to fight XDeath?

Mithrandir24601

6:44 PM

@Blue Just because a Uni is 'good'/has the best method of teaching doesn't necessarily mean that said method makes any sense whatsoever to those starting out

Anonymous

@Mithrandir24601 Anyhow, that's at-least better than teaching Schrodinger's equation without explaining what quantum operators are and how they behave

Mithrandir24601

e.g. do you want to teach them QM or do you want to teach them the fundamental principles that lead to QM? Or... Do you want to teach them the problems that people in the past had with QM and directly show the students by example why that's not the way that it's done anymore and so that they understand what good physics is?

Anonymous

@Mithrandir24601 I'd prefer teaching everything in chronological order starting from Classical Mechanics

Mithrandir24601

@Blue OK, yes, they should explain what a quantum operator is, at least to some extent

Anonymous

Anyway, I'm not complaining =P This is after all engineering :)

Mithrandir24601

6:50 PM

@Blue This would have both positives and negatives as well - if you teach everyone classical mechanics, then it might just be more confusing when they get to QM

Although, yeah, the best method of teaching doesn't seem to be an easily solved problem :P

Anonymous

What is Energy "eigenvalue" now?

Anonymous

I don't understand eigenvalue :/

Avantgarde

Do you know what the eigenvalue problem is, in math?

Anonymous

@Avantgarde Nope. I just know about eigenvalue of matrices...characteristic equations and stuff

Avantgarde

6:54 PM

Yeah, that's about it. What's the problem here? $E$ is the eigenvalue of the operator $\hat{H}$

Anonymous

What does that mean? ;-;

Anonymous

I'm a noob

Anonymous

How would you explain it to a layman?

Avantgarde

$\hat{H}$ is an operator, which in some basis (say position basis) is a matrix. $\Psi$ is a vector (ray, actually, but forget that for the time being) in Hilbert space. In general, when an operator acts on a vector, it results in a different vector, which is unrelated to the original vector. In some cases, however, the resultant vector is just the old vector times a constant multiple. Such special cases are what observables are about in QM. That constant multiple is called the eigenvalue

And the constant multiples are the experimentally measurable values in a lab

Makes sense?

Anonymous

@Avantgarde What does "basis" mean?

Anonymous

6:59 PM

googles

0celóñe7

You don't know enough linear algebra.

Anonymous

@0celóñe7 Exactly.

Avantgarde

@Blue Basis is what the word implies. For example, in Cartesian coordinate system, $\hat{x},\hat{y}$ and $\hat{z}$ form the basis for any vector in 3 dimensional space. Basically, a basis is the fundamental building block, using which you can construct any other vector in the vector space in question

And yeah, you need to do some linear algebra

Anonymous

@Avantgarde Okay. Got this much

Anonymous

Now, what does "$\hat{H}$ is an operator, which in some basis (say position basis) is a matrix"....what does position basis mean?

Anonymous

7:04 PM

And how can an operator be a matrix?

0celóñe7

He is telling you physicist lies

Anonymous

@0celóñe7 *-*

Balarka Sen

An operator is like an infinite dimensional matrix

Think about $d/dx : C^\infty(\Bbb R) \to C^\infty(\Bbb R)$

Anonymous

@BalarkaSen Went over my head....

Anonymous

What does infinite dimensional matrix even mean.......

Balarka Sen

7:07 PM

It eats a smooth function $f : \Bbb R \to \Bbb R$, and spits out a smooth function $f'$

@Blue That's why I'm explaining :)

Give me a bit to write it all out

Anonymous

Sure. Go ahead

Balarka Sen

$d/dx$ is a linear map, because $(f + g)' = f' + g'$ and $(cf)' = cf'$ for a real constant $c$.

Well, before that, $C^\infty(\Bbb R)$ is a vector space, because same reason: you can add two functions, and multiply a function by a scalar.

Is this making sense?

It can be kind of weird to think of $C^\infty(\Bbb R)$ as a vector space. Usually a vector space like $\Bbb R^n$ is just the space of vectors in an n-dimensional space (where a vector is imagined like an arrow from origin to a point in the space).

But here the vectors are the functions.

Weird, eh?

Anonymous

"A vector space is a collection of objects called vectors, which may be added together and multiplied by numbers, called scalars."...is this definition fine?

Anonymous

I'm understanding bit by bit

Balarka Sen

Yeah, it is a good definition.

In fact those are the conditions I checked up there to justify that $C^\infty(\Bbb R)$ is a vector space.

Anonymous

7:13 PM

Okay. I know that $\Bbb R^n$ represents n-D space. What does $C^{\infty}(\Bbb R)$ represent ?

0celóñe7

Balarka just had to pick $C^\infty$

Why not $C[a,b]$ and use the position operator or something?

Balarka Sen

@Blue $C^\infty(\Bbb R)$ is the set of ALL infinitely differentiable functions $f : \Bbb R \to \Bbb R$

0celóñe7

Then you get to explain how the Dirac delta is a matrix

Balarka Sen

i am a simple minded man

i just use derivative

Anonymous

@BalarkaSen Ok. Got it till here

Balarka Sen

7:16 PM

I want you to get to "Weird, eh?" :)

0celóñe7

@BalarkaSen Unexpected issue: proving that step functions are dense in $L^p$ when the functions take values in a Banach space

Anonymous

Basically you are considering infinitely derivable functions as vectors in n-D space. Yes, it is weird :)

Balarka Sen

But no! It's not vectors in nD space

0celóñe7

The usual proof fails miserably

Balarka Sen

It's vectors in $C^\infty(\Bbb R)$

(The $\infty$ represents infinitely differentiable, by the way)

Anonymous

7:37 PM

Sorry for being late. Father called. Ok, so these are vectors in $C^\infty(\Bbb R)$. Hmm, but isn't $C^\infty(\Bbb R)$ the "set" of infinitely differentiable functions?

Anonymous

Vectors= Infinitely differentiable functions ?

0celóñe7

A function in $C^\infty(\Bbb R)$ is a vector in that space

Anonymous

@0celóñe7 Why?

0celóñe7

It's a vector space of functions

@Blue By definition

Avantgarde

The functions satisfy all the conditions for it to be a vector

Anonymous

7:40 PM

Does the function have a direction?

Anonymous

:O

0celóñe7

what

Anonymous

Vectors have a direction....

0celóñe7

no, you need to think symbolically

Avantgarde

No, don't think of vectors as something that have magnitude and direction. There are a certain number of 'rules' for any abstract quantity to be a vector

Balarka Sen

7:41 PM

@Blue It's only metaphorically a vector in the classical sense.

0celóñe7

vectors in $\Bbb R^n$ can be thought of as arrows

Anonymous

@Avantgarde What are the rules?

0celóñe7

that's not true/reasonable in general

@Blue We've been over them above

Balarka Sen

Think of vectors as things which can be added togather by a vector sum rule, and multiplied by scalars

Like you quoted

In that case, functions can be added: given $f, g$ real valued functions, $(f + g)(x) = f(x) + g(x)$

and $(c \cdot f)(x) = cf(x)$

Avantgarde

@Blue en.wikipedia.org/wiki/Vector_space Scroll down a little bit till you see 'Axiom'. Any abstract quantity that satisfies those is a vector, by definition

Anonymous

7:43 PM

@BalarkaSen You're using circular logic there. You can't say vectors as things which follow "vector" sum rules

Balarka Sen

@Blue Not if I don't mention vectors in defining the sum rule.

Anonymous

@Avantgarde Checking

Balarka Sen

The sum rule just means there is a well defined notion of addition of two vectors.

In the case of vectors, that's vector sum

In the case of functions, that's pointwise addition

That's all there is to it

Anonymous

Got that part! :D

Anonymous

7:46 PM

@BalarkaSen Understood :)

Anonymous

So you were explaining why an operator is an infinite dimensional matrix

Balarka Sen

Right, so now that we know $C^\infty(\Bbb R)$ is a vector space

It remains to observe it is an infinite dimensional vector space

Which is intuitively true: dude, there are way too many infinitely differentiable functions $f : \Bbb R \to \Bbb R$

Avantgarde

hahahaha

0celóñe7

The proof is Trivial

Avantgarde

"dude, there are way too many infinitely differentiable functions" :D

0celóñe7

7:50 PM

Just find a bump function and move it around

Danu

@BalarkaSen The standard way is by using bump functions

Right.

Balarka Sen

I know the proof :p

Danu

then say it :P

Balarka Sen

shit pedagogical idea

Danu

It is also intuitively true that $S^2$ is simply connected, yet you spent some time berating my assertion to this effect ;-)

Balarka Sen

7:51 PM

did i? i don't recall anymore

did you know the proof of S2 being simply connected? was i teaching you the proof?

Danu

You did; we even had a separate chat room for it and I spent about half an hour arguing with you

I was teaching it to someone

0celóñe7

I wonder what the most ridiculous proof of that is

Danu

and you interjected

No matter

Balarka Sen

maybe because you said you were proving it and the proof was trivial

lol

Danu

No, I did not. I said that the essence of the proof was not in the technically difficult part.

Balarka Sen

7:53 PM

huh I wonder why I interjected

Danu

me too ;)

Balarka Sen

Oh because I wanted to show off the smartass space filling curve maybe

and perturb it tiny bit

Good reason. I will interject next time you do that again.

0celóñe7

Perturb it to get a smooth curve, use Sard to poke a hole, stereographic projection, and bob's your uncle

Danu

Yeah, and I was just pointing out that it that is not the essence of the proof.

Balarka Sen

Yes it is lolkek

Danu

7:55 PM

@0celóñe7 Perturbing to smooth? I feel like that's more effort than perturbing it "by hand" to miss a point.

Balarka Sen

It is a smartass proof and the essence of cellular approximation

@Danu 0celo is proving it by smooth approximation. We proved it by cellular/simplicial approximation.

Both techniques are useful.

0celóñe7

@Danu It is indeed a lot of work, but I knew Whitney's smooth homotopy theorem before I knew the usual topological proof

Kyle Kanos

Proofs are lame

Anonymous

@BalarkaSen Okay. And how is an infinite dimensional vector space an infinite dimensional matrix?

Balarka Sen

In any case, @Danu, @0celo: Can I not prove that $C(\Bbb R)$ is infinite dimensional by inspecting $x^n$'s?

Polynomials form an infinite dimensional subspace.

Your bump function proof is too complicated. GET REKT

I mean, it's still useful to prove C(M) is infinite dimensional for a general smooth manifold M

@Blue Well, the operator is the infinite dimensional matrix. This is because $d/dx : C^\infty(\Bbb R) \to C^\infty(\Bbb R)$ is a linear map of infinite dimensional vector spaces.

You know that linear map of finite dimensional honest to god vector spaces are matrices, right?

Any linear map $T : \Bbb R^n \to \Bbb R^n$ gives rise to the matrix $A$ with columns $T(e_i)$

where $e_i$ are the standard basis vectors.

Anonymous

8:03 PM

Examples of vector spaces

This page lists some examples of vector spaces. See vector space for the definitions of terms used on this page. See also: dimension, basis. Notation. We will let F denote an arbitrary field such as the real numbers R or the complex numbers C. See also: table of mathematical symbols. == Trivial or zero vector space == The simplest example of a vector space is the trivial one: {0}, which contains only the zero vector (see axiom 3 of vector spaces). Both vector addition and scalar multiplication are trivial. A basis for this vector space is the empty set, so that {0} is the 0-dimensional vector...

Anonymous

Yeah, got to know it now

Danu

@BalarkaSen The linear independence is (even) more transparent when using bump functions. On $\Bbb R$, they are not hard to explicitly write down.

Balarka Sen

Linear independence of $x^n$'s is a one line proof.

Anonymous

Okay. So an operator is an infinite dimensional matrix. Then "In general, when an operator acts on a vector, it results in a different vector, which is unrelated to the original vector."

Balarka Sen

Sorry m8

Anonymous

8:05 PM

What does acts on mean here. How does a matrix act on a vector?

Balarka Sen

By matrix multiplication!

Anonymous

Okay, so matrix multiplication of an infinite dimensional matrix with a vector? =P

0celóñe7

@BalarkaSen Polynomials work, sure

But I am training @Blue to think like a geometer

You're a damn algebraist

Balarka Sen

heheh

0celóñe7

@BalarkaSen But you're right, @Blue probably doesn't know any $C_c^\infty$ functions.

Kyle Kanos

8:08 PM

$C^\infty$ are the best functions

Balarka Sen

@Blue A vector in $C^\infty(\Bbb R)$ is a smooth function. A matrix, that's an operator - like $d/dx$ - eats a smooth function, and spits out another smooth function.

Which it does: $d/dx(f) = f'$

Just like $Av = w$ for a matrix $A$ and vectors $v, w$

0celóñe7

@BalarkaSen ewww

Balarka Sen

lol

notation rekt

Kyle Kanos

That is some ugly notation

Balarka Sen

@0celóñe7 Besides, you and @Danu's proof fails to show the space of analytic functions $C^\omega(\Bbb R)$ is infinite dimensional.

Anonymous

8:09 PM

@BalarkaSen Oh, oh! Got till here....now going back to read what @Avantgarde wrote

Balarka Sen

So you two are rekt anyway.

0celóñe7

@BalarkaSen Hmm, ok.

Balarka Sen

I wonder if you can somehow fix you proof by using decaying functions instead of bump functions

0celóñe7

Probably. Gaussians?

Balarka Sen

Yeah

Anonymous

8:11 PM

@Avantgarde So we measure the energy eigenvalue in the lab? But how?

Kyle Kanos

I work with a function of the form $\exp\left(-\int h\left(\tau\right)\,{\rm d}\tau\right)$ at work all the time. It's very nice to have it so simple.

0celóñe7

I shouldn't trash $C^\omega$, the Cauchy-Kovaleska theorem is pretty nice.

Danu

@BalarkaSen Since when do I care about real analytic things? :P

Balarka Sen

@Danu You are a complex geometer

Complex functions are analytic

Danu

Wow hey @KyleKanos

Balarka Sen

8:12 PM

Don't rek yourself more

Please dont

Danu

hahaha

Kyle Kanos

@Danu Hi Danu

Danu

much more than real analytic

0celóñe7

@BalarkaSen Is $C^\omega(M)$ infinite dimensional for a real analytic manifold?

Balarka Sen

But hey, this is interesting. I don't know how to prove $C^\omega(M)$ infinite dimensional

0celóñe7

8:13 PM

It has to be.

Balarka Sen

@0celo Yeah

I also want to prove the space of complex analytic functions on a domain of $\Bbb C$ is infinite dimensional

0celóñe7

Polynomials fail there?

Balarka Sen

Uh, they don't. Maybe I meant an analytic subset of $\Bbb C^n$?

Like, noncompact complex manifolds.

@Danu Is the space of holomorphic functions on a noncompact complex manifold always infinite dimensional?

0celóñe7

What happens if you restrict polynomials to such subsets?

Danu

@BalarkaSen No idea---I only know things about compact manifolds

Balarka Sen

8:15 PM

In which case the question is dumb m8

Danu

Stein manifold

In the theory of several complex variables and complex manifolds in mathematics, a Stein manifold is a complex submanifold of the vector space of n complex dimensions. They were introduced by and named after Karl Stein (1951). A Stein space is similar to a Stein manifold but is allowed to have singularities. Stein spaces are the analogues of affine varieties or affine schemes in algebraic geometry. == Definition == A complex manifold X {\displaystyle X} of complex dimension n {\displaystyle n} is called a Stein...

Are supposed to be manifolds with "many" holomorphic functions.

@BalarkaSen Your tone is rather annoying, do you realize that?

0celóñe7

@BalarkaSen What is the answer for compact ones?

Balarka Sen

Do any noncompact complex submanifold embed holomorphically in $\Bbb C^n$?

@Danu Yes, intentionally so.

Danu

@0celóñe7 Only the constant ones

Anonymous

Uh, could someone tell me how energy eigenvalues are found in the lab?

Danu

8:17 PM

The point is that they are open maps

Balarka Sen

I am annoying you a little since you interjected in when I was telling Blue that C^infty(R) is infinite dimensional, and with a pedagogically bad proof too

Danu

OK. Bye.

Balarka Sen

Bye, but don't turn into 1st year grad Mike.

0celóñe7

@Danu Can one apply some multivariable version of the maximum modulus principle to charts?

Balarka Sen

He's moved on from being that guy.

@0celóñe7 I think so

0celóñe7

8:21 PM

@BalarkaSen What?

Balarka Sen

You can take bigger and bigger loops, so that maximum lies on the boundary of the loop

And, uh

Ah, well, it's compact

So it achieves maximum somewhere

0celóñe7

Exactly.

Kyle Kanos

physics.stackexchange.com/review/suggested-edits/184733

sigh why?

Balarka Sen

draw a small loop around that in a chart, and it's constant inside

0celóñe7

All the charts touch since it's connected.

Balarka Sen

8:23 PM

Ok, yeah, @0celo

0celóñe7

@BalarkaSen I don't know CA for multiple variables, is the max modulus principle still valid?

Balarka Sen

Yep

0celóñe7

Does one apply the usual one along each coordinate axis or something?

Ah, there is a nice proof in Wells. I thought this was familiar.

@BalarkaSen Do you want to discuss the uncertainty principle?

Balarka Sen

Sure, go ahead

0celóñe7

@BalarkaSen Ok, so what is your argument using Cauchy-Schwarz? Assume $||f||_{L^2}=1$

Sid

8:49 PM

@blue is your last name QHGGN (r-13)?

Anonymous

what?

Sid

Rot 13 that

Kyle Kanos

r-13 is rot-13

Simple letter substitution. Looks like dutta

0celóñe7

@KyleKanos So what are you doing here?

Sid

If so, I think I have successfully stalked you. :P

Kyle Kanos

8:52 PM

Killing time mostly @0celóñe7. Wifey went on a weekend away with half the kids

Anonymous

@KyleKanos I hope you don't have an odd number of kids.

Kyle Kanos

I do

0celóñe7

ahaha I have the proof

Anonymous

@KyleKanos Sounds bad =P

Kyle Kanos

I've got 3 of the kids, the wife with 2. But she's got the PITA kids

Anonymous

8:55 PM

PITA?

Kyle Kanos

pain in the ass

0celóñe7

pain in the *ss

Balarka Sen

@0celóñe7 Sorry, my internet got rekt

Anonymous

ho ho

0celóñe7

@KyleKanos Does one hope that PITA kids fix themselves/get fixed?

Sid

8:56 PM

You have 5 kids?:o

0celóñe7

Does one actively try to fix them?

Balarka Sen

So I want to show $(\int x^2|f(x)|^2)(\int y^2 |\hat{f}(y)|^2) > stuff$?

Kyle Kanos

You have to actively try to fix them

0celóñe7

@BalarkaSen $\ge 1/16\pi^2$ I think

Kyle Kanos

Otherwise they won't change

Balarka Sen

8:57 PM

ok

Let $p(x) = xf(x)$ and $q(y) = y\hat{f}(y)$

0celóñe7

@BalarkaSen And give conditions on $f$ such that every step is well-defined as well

@BalarkaSen q(y)

Balarka Sen

tnx

um so

Anonymous

@0celóñe7 I'm sure PITA kids do grow up to become mathematicians or nuclear engineers too ;)

Balarka Sen

this is $\|p\| \|q\|$, whenever those are well defined

0celóñe7

@Blue I was a PITA kid around age 2-3

I got fixed somehow

Balarka Sen

8:58 PM

in L2 norm

0celóñe7

Why did you remove that?

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