The h Bar

0celo7

15:00

I don't remember him saying anything about the convergence, but I'll reread.

Ben Niehoff

the point is the "Taylor" series has constant coefficients which grow factorially

so it obviously can't converge

but for a while, it looks like it's going to

0celo7

Aha

DanielSank

Hi, everybody.

y u no answer my path integral question?

Emilio Pisanty

@DanielSank gimme teh path integralz?

=P

0celo7

because no one has the link

DanielSank

15:05

4

Q: What is the link between path integral source terms and the propagator?

Consider a diffusion process defined by $$\frac{\partial \phi}{\partial t} = D \frac{\partial ^2 \phi}{\partial x^2}$$ where $\phi$ is the probability density of a particle's position and $D$ is the diffusion coefficient (or whatever it's called). Denote the transition probability to go from $(0,...

@0celo7 False.

0celo7

it's false now

DanielSank

Hey, @dmckee @ACuriousMind @DavidZ @JohnRennie or anyone who knows about chat:

0celo7

but you can't overturn causality

DanielSank

I just clicked "hide messages" on a user in one of my rooms and all their messages disappeared. Is there a way to undo that?

Secret

@DanielSank do you happen to be familar with github stuff, cause I am trying to download a library archive from a specific github ruby code and I have no idea how to do it

DanielSank

15:08

@Secret Yes, I know github reasonably well. I know almost nothing about Ruby though.

What OS are you using?

0celo7

@Secret my sister used to work there

Secret

The background is that there's a program that requires tcllib1.18tar.gz to install but none of the tcllib1.18 download sources (the tcl library hub, the sourceforge mirror etc.) has the correct sha256 checksum. I figure if I can somehow extract that download from that tcltk.rb code in github it will be my final chance to get the file

I am using a mac, version Sierra

0celo7

@DanielSank So what's the issue with computing the inverse of $A$? Is it not the direct sum of inverses of blocks?

Secret

@DanielSank This is the Link somehow I need to get that particular sha256 tcllib1.18 there

from this:

resource "tcllib" do
url "https://github.com/tcltk/tcllib/archive/tcllib_1_18.tar.gz"
sha256 "6a87881f545afb69c1130f60984b5d35cc22f1593b0835b982871c188fde3de8"
end

0celo7

apparently not

hmm

Block matrix

In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. Any matrix may be interpreted as a block matrix in one or more ways, with each interpretation defined by how its rows and columns are partitioned. This notion can be made more precise for an n ...

Secret

15:11

btw the direct downlad by using the url above in the code also give me a tcllib1.18 with a wrong sha256 checksum

0celo7

@DanielSank I'm missing something. Why can't you use that to invert it?

Secret

There's a guy managed to did it somehow described here

yas375 commented on Oct 3, 2016
Ended up by copying the content of that formula into a gist, fixing SHA there and installing from the gist. It did work!

0celo7

direct sum of a bunch of those guys

Secret

but I don't know how to download from a gist. I also have gist and homebrew installed on my mac

(basically, most of this sounds like alien language to me thus I don't really know what to do to get that file with the correct sha256)

DanielSank

@0celo7 Where in the question am I asking for the inverse?

@Secret You're asking a lot of stuff all at once. Can you narrow the question?

0celo7

15:15

@DanielSank You said you don't know $A^{-1}$. There seems to be some comments trying to invert $A$. I don't understand why it's not trivial.

DanielSank

@0celo7 I want to know how $A^{-1}$ is related to the Green's function for the problem.

If you can answer that by solving for the inverse explicitly, I'll be very happy and probably give you a large bounty.

Secret

@DanielSank Ok the short version is that I need to get tcllib-1.18.tar.gz with the folllowing checksum 6a87881f545afb69c1130f60984b5d35cc22f1593b0835b982871c188fde3de8 and I don't know how to get that from the ruby code posted in this github link via mac terminal commands:

github.com/Homebrew/homebrew-dupes/blob/…

and all other sources of tcllib-1.18.tar.gz (including the official site) all have the wrong checksum

DanielSank

I don't understand why you mean by get a file from ruby code.

0celo7

@DanielSank because a greens function is an inverse linear operator on a Banach space of functions, and you're approximating that linear operator with a matrix

So the approximate inverse (approximate green's function) is the inverse of the matrix

Secret

If I interpreted the code correctly, it seems somehow it grabs the tcllib-1.18.tar.gz from a directory, and then somehow put this checksum on it as shown

0celo7

15:20

@DanielSank is that what you're looking for or am I still off base?

DanielSank

@0celo7 Ok well the Green's function for this case is $$\sim \exp \left( - \frac{x^2}{4 D t } \right)$$

Can you explain how this is related to that $J_i J_j (A^{-1})_{ij}$ term in the problem?

@Secret I don't know Ruby.

Emilio Pisanty

@Secret what's wrong with cloning the repo?

0celo7

Oh, the greens function and solution operator are not the same. The solution operator, your A^-1, is essentially the distribution associated with the Green's function, i.e the integral weighted by the Green's function

Do you want to see an explicit demonstration of that?

@DanielSank am I wrong to think that your matrix is block diagonal with [[2,-1],[-1,2]] as the blocks?

DanielSank

@0celo7 It's not block diagonal.

It's got stuff on the diagonal and both first off-diagonals.

Otherwise inversion would be trivial.

@0celo7 Yes plz.

0celo7

Then you wrote the question very poorly.

You wrote a block diagonal thing

DanielSank

15:25

@0celo7 Thank you.

Did you read the title?

0celo7

Yeah?

If I'm trying to show this explicitly I need to invert A, so I need to know what A actually is

Secret

@EmilioPisanty that repository only contains the ruby code files, and I am not terribly sure what to do with them to get the library archive file I need after downloading them

Emilio Pisanty

this? $$A_{ij} = \left[
\begin{array}{cccc}
2&-1&&&\\
-1&2&&&\\
&&\ddots&&\\
&&&2&-1\\
&&&-1&2
\end{array}\right]$$

obviously not a block matrix

0celo7

Yeah. That looks like it's suggesting a block diagonal matrix

Emilio Pisanty

@Secret you want to be looking here

github.com/tcltk/tcllib

also here github.com/tcltk/tcllib/releases

DanielSank

15:27

@0celo7 I'll add more elements to make it more clear...

...working...

Emilio Pisanty

@0celo7 no it doesn't

0celo7

@EmilioPisanty you're being obtuse on purpose

Emilio Pisanty

it's quite clearly a tridiagonal matrix with 2s on the main diagonal, -1s on the upper and lower subdiagonal, and zeroes elsewhere

0celo7

If you write two blocks and dots, what the hell else is it supposed to be

Emilio Pisanty

presumably that's what you meant, @DanielSank?

write it as $$A_{ij} = \left[ \begin{array}{cccc} 2&-1&&&\\ -1&2&-1&&\\ &&\ddots&&\\ &&-1&2&-1\\ &&&-1&2 \end{array}\right]$$ if it makes you happier

DanielSank

15:29

@EmilioPisanty Correct.

@EmilioPisanty do you understand what that post is trying to ask?

0celo7

Aha

Much better

Emilio Pisanty

@DanielSank have yet to read the post

DanielSank

Everyone seems to be caught up with the form of $A^{-1}$ whereas I'm trying to ask more generally how this matrix relates to the propagator.

I don't care about the exact form of $A^{-1}$ beyond how that form helps me understand how it's related to the propagator/Green's function. The problem I used to introduce the subject is incidental.

Emilio Pisanty

@0celo7 either way, you've seen how the original notation did its job by communicating what was meant

DanielSank

Is there a way to delete messages from a room I own?

Emilio Pisanty

15:31

@DanielSank Do you particularly care about your boundary conditions?

i.e. would introducing -1s on the top-right and bottom-left elements be bothersome?

DanielSank

@EmilioPisanty Wellllllll, the boundary conditions are what lead to the presence of the $J$ term in this case.

Emilio Pisanty

if you do that then maybe you can solve explicitly for the inverse

DanielSank

@EmilioPisanty Periodic conditions?

Emilio Pisanty

@DanielSank yes

DanielSank

@EmilioPisanty I don't reaaaally care about solving for the inverse.

I keep saying this :-\

I want to know how the $JAJ$ term is related to the propagator, if at all.

0celo7

15:33

@EmilioPisanty no, he wrote a block diagonal matrix then complained he couldn't invert it. I know that's not the question, but it was written in there.

Secret

The file in https://github.com/tcltk/tcllib/releases has checksum=9988b4385403c2aac78743fd3fce2d22e82686a56e6ca25942cb83c7d9e641db but the required checksum is 6a87881f545afb69c1130f60984b5d35cc22f1593b0835b982871c188fde3de8

the library archive is not found in the repository itself github.com/tcltk/tcllib

0celo7

Complained isn't the right word there

But whatever

Physics Meta

0

Q: Should discussions on topics which disturb long time users of the chat room be banned?

As some of you might know about the "argument" which took place between me and another long-time user of the hBar about discussion of "JEE related stuff". As far as those discussions are concerned, they are mostly related to Physics, Math or Chemistry at the high school/ Olympiad level (you can c...

discussion

Emilio Pisanty

@Secret yeah, I get the same checksum

if that makes you feel like you're not going crazy

DanielSank

@0celo7 That comment you just posted in meta is ridiculous.

Emilio Pisanty

15:36

but that is patently the file that was intended, or at least, that is patently the canonical homepage for the file that was indicated, delivered over https

0celo7

The meta post is ridiculous.

DanielSank

The only effect of that post is that someone (me) had to flag it, and now a moderator needs to deal with the flag. You have succeeded in nothing but creating work for other people.

Emilio Pisanty

so... you should seriously consider the possibility that the indicated sha256 is wrong?

DanielSank

If your personal amusement is worth more to you than other people's time, then perhaps I am starting to understand why you keep getting suspended.

2

0celo7

No, you created work by flagging it

DanielSank

15:37

@0celo7 Yeah, ok, you can go on with that in you brain. Hope it works out for you.

Secret

I suspect so, but I have no idea how to modify it cause it seemed to be automatically calculated

There's also a guy in the github posting:
yas375 commented on Oct 3, 2016
Ended up by copying the content of that formula into a gist, fixing SHA there and installing from the gist. It did work!

But that's alien language to me, I don't exactly know how he install from a gist

For more info, I have homebrew installed on my mac, however I am not sure how to do this "install from gist" thing

ACuriousMind

@DanielSank My current problem is that I know the relation for a relativistic field theory, but for this non-relativistic version the (integral kernel of the) inverse of the thing in the exponent is different from the propagator. I suspect this might go away if you turn this essentially "Hamiltonian path integral" into a Lagrangian path integral, but I'd need to work that out for this specific case.

DanielSank

@ACuriousMind o_O

Is that even possible for a diffusion process?

ACuriousMind

I don't recall/know

DanielSank

What does "relativistic" mean?

0celo7

15:44

@ACuriousMind is the relation you're thinking of basically what I said above?

ACuriousMind

@DanielSank That the e.o.m. is also second order in time, and that you consider a slightly different path integral. Then you get $\square^{-1}$ (d'Alembert operator; $\partial_t^2 - \nabla^2$) in that exponent and the integral kernel of that operator is precisely the propagator.

@0celo7 I have not read through chat

DanielSank

@ACuriousMind I guess I'm wondering this: if I have a differential equation and I know the Green's function, isn't there some generic relationship between that Green's function and the stuff that shows up in a path integral representation?

I think you're saying "yes, but only if {conditions}".

ACuriousMind

Yeah, I'm saying "yes, but only if you consider a certain path integral". "Path integral" is not unambiguous, there are several things you can compute by these. You are computing the propagator itself in your case, and in the field theoretic cases I'm thinking of you use it to get a partition function and expectation values of observables

(where the two-point correlation function is the propagator, as a special case, but in your case it doesn't look like you're computing the path integral with the insertion of two $x(t)$, and I'm not entirely sure it's the propagator in this instance, anyway.)

0celo7

How does one even compute a propagator for a general PDE

I'm sure there's a Fourier resolution in "nice" cases

ACuriousMind

There are also probably people with a much better understanding of this because the path integral is more of a formal symbol than an actual method ot compute stuff to me, anyway.

I never deal with cases where you can actually compute the path integral - every time you can, it is because it degenerated into some other, more tractable objects.

DanielSank

15:54

@2017 By the way, deleting your account in protest is not constructive. If you want to change some aspect of the site, you must stay and work constructively for that change. Leaving increases the chances that things stay the same. I did not leave when I was frustrated that the users here dismissed applied physics questions. I stayed and worked for change.

If you leave you accomplish absolutely nothing.

Ben Niehoff

@ACuriousMind Considering that no mathematician has yet been able to give a definition of what a path integral generically is, yes :P

DanielSank

@ACuriousMind Yes, but that's always the case.

ACuriousMind

@BenNiehoff Precisely, I think of it more as a good heuristic to find the objects you actually want to compute (such as invariants in many topological and supersymmetric theories)

0celo7

You can compute path integrals explicitly in certain cases.

DanielSank

In the Wiener path integral, we know the answer because we can solve the diffusion equation and all other cases are reduced to that one known case.

@BenNiehoff The Wiener measure is actually well understood!

0celo7

15:56

@ACuriousMind that's a nice way of thinking about it

ACuriousMind

@DanielSank Only in low dimensions

Ben Niehoff

as far as I can tell, a "path integral" is really just a collection of procedures that physicists do, most of them more or less algebraic

DanielSank

@ACuriousMind Yeah but people also use it to actually compute scattering amplitudes!

0celo7

I guess the Feynman rules are just heuristic anyway because the series tend to explode?

DanielSank

@BenNiehoff Yep.

Ben Niehoff

15:57

@DanielSank Have never heard of this Wiener measure, what is it?

2017

@DanielSank I've already sent out the delete request. If there is an option, I'll rejoin later, after seeing the response.

Anyway, thanks for your suggestion.

ACuriousMind

@0celo7 The Feynman rules don't have much to do with the path integral, they're just the "recipe" for computing the terms in the perturbative expansion, which exists in both the path integral and the canonical formalism

0celo7

@BenNiehoff it's the path measure for Brownian motion

DanielSank

^ that

0celo7

The actual definition is very bad.

Ben Niehoff

15:58

ah, ok, I might actually expect that to be more tractable

ACuriousMind

They're not heuristic, they just compute the perturbation series, which does tend to explode - but not because it is heuristic, but because it is a perturbative expansion which you don't expect to converge anyway

Ben Niehoff

because it doesn't require the "cancellation of highly oscillatory phases" that quantum path integrals do ;)

0celo7

@ACuriousMind I learned the rules from Zee who uses path integrals

ACuriousMind

Yeah, you can derive them from the path integral formalism, but they're not bound to it

0celo7

But I think Weinberg uses magic to derive them

DanielSank

15:59

Well, @ACuriousMind, now I'm trying to make sense of your post about why we use density matrices in subsystems and in statistical ensembles of pure states.

0celo7

Fucking S matrix and cluster decomposition or something :P

ACuriousMind

If you've got the LSZ formula and Wick's theorem, you can figure them out without mentioning path integrals

Ben Niehoff

and many people did!

ACuriousMind

It's more of a stroke of genius there, but it's possible

DanielSank

5

Q: How to connect these two formulations regarding the need for a density matrix in quantum mechanics?

I found these two formulations: The density matrix is: 1) "needed if we consider a system that is part of a larger closed system." 2) "needed for a system to be in a statistical ensemble of different state vectors." What is the link between them bringing one to see their equivalence?

quantum-mechanics quantum-information density-operator

What's $\mathbb{C}(|b_j \rangle)$?

Vector space spanned by $|b_j \rangle$?

ACuriousMind

16:01

@DanielSank A mathy notation for the one-dimensional space spanned by $\lvert b_j\rangle$

DanielSank

Ok and what's circle-with-a-plus-in-it?

0celo7

Direct sum

The Dark Side

@EmilioPisanty Yes, that's true. But also, there is a fair chance that he can return when $t > t_{\rm Relaxation (Anger)}$

0celo7

Cartesian product for vector spaces

Ben Niehoff

\oplus

DanielSank

16:02

@0celo7 Ohhhhhhh

0celo7

In the case of Hilber spaces, the completion of the direct sum.

DanielSank

I always thought that was denoted $\times$.

0celo7

Depends on how cool you want to look. They're equivalent symbols.

DanielSank

i.e. $\mathbf{V} \times \mathbf{W}$.

@0celo7 Thanks, I never knew that.

ACuriousMind

@0celo7 They are not (but they are in this case :P)!

Emilio Pisanty

16:02

@TheDarkSide I read Auger and was completely mystified

Ben Niehoff

I thought it depended on whether you were emphasizing their vector space structure or their topological structure

0celo7

@ACuriousMind what?

@BenNiehoff should be clear from context

In theory people give a crap about that but in reality people just do what they want

Emilio Pisanty

Is $t_\mathrm{Auger}$ the time it takes for a site newcomer to fall into Chris White's role, violently expelling some other newcomer in the process?

DanielSank

@ACuriousMind in your answer, you motivate the partial trace by saying that for a state $$|\chi \rangle = \sum_{ij} c_{ij} |a_j \rangle \otimes |b_j \rangle$$ the right way to not care about the second space is $$p(|a_i \rangle) = \sum_j |c_{ij}|^2 \, .$$

Why?

Emilio Pisanty

(cf. Auger effect for background)

Ben Niehoff

16:04

well, I rarely describe flat space as $\mathbb{R} \oplus \dotsc \oplus \mathbb{R}$

DanielSank

The entire post hinges on that assertion.

ACuriousMind

@0celo7 $\times$ denotes a product, while "direct sum" is a name for a coproduct that is also a product, or sometimes just a coproduct. In the case of vector spaces, the two notions coincide for finitely many terms, but the infinite direct sum of vector spaces is "smaller" (has an injection that's not a bijection) into the infinite Cartesian product

0celo7

@ACuriousMind sure, but we're talking about two vector spaces here.

ACuriousMind

That's why I said "but they are in this case"

0celo7

Ok. Just making sure

Secret

16:05

Emilo: Lol, my chemistry school just talked about that effect in the context of Gadolinium complexes as binary caner therapy agents in one of the PhD seminars today

Ben Niehoff

maybe I should try sneaking $n \times \mathbb{R}$ into a paper and see what the editors think

0celo7

@BenNiehoff but that isn't wrong

DanielSank

Oh hey, since there are mods here: what is kick-mute in chat?

ACuriousMind

@DanielSank Well, $\lvert c_{ij}\rvert^2$ is the probability that the system is in $a_i\otimes b_j$

Bernardo Meurer

Windows is proprietary malware

There is no system but GNU

DanielSank

16:06

@ACuriousMind Yes.

Bernardo Meurer

and Linux is one of it's kernels

DanielSank

That's a good point.

The Dark Side

@EmilioPisanty :) (Also, thanks for the link but I know a bit of Physics too)

ACuriousMind

So, if I don't care which $b_j$ the system is in, and only care about it being in $a_j$, basic probability says I have to sum up all the distinct probabilities for the way in which it can be in $a_i$. That's what $\sum_j \lvert c_{ij}\rvert^2$ is.

DanielSank

We're summing all the probabilities where the first system is in $|a_i \rangle$.

Yeah I guess it makes sense that you do an incoherent sum.

Fair enough.

Ben Niehoff

16:07

@BernardoMeurer Certainly you can improve that with some sort of recursion

Emilio Pisanty

@TheDarkSide yeah, well, I was a good bit into my PhD when I learned about it

0celo7

@BenNiehoff is string theory plagued by the same kind of unrigorous mysticism that QFT is or is it better?

ACuriousMind

@DanielSank It's...what it says on the tin. It kicks the user from the room and prohibits them from returning for a period of 1,5,30 mins for the 1st, 2nd, 3rd kick.

The Dark Side

@EmilioPisanty :)

Emilio Pisanty

I wonder if e.g. DanielSank/ACM/Ben could belt out a definition from the top of their heads

Bernardo Meurer

16:08

@BenNiehoff Use GPLv3 and avoid Tivoization of your software! Protect the users' freedom!

DanielSank

@ACuriousMind How's that different from suspension?

ACuriousMind

@DanielSank It's only for the room they were kicked from

Secret

@EmilioPisanty I think I figured out a way to fix it: After homebrew get that tcl-tk.rb file, I opened textedit and then replace the (should be correct) checksum with the wrong checksum (because we know the file it downloads is correct, thus we copied that checksum and paste it in where the sha256 line in the tcl-tk.rb file is. Now it managed to get through. Thanks!

DanielSank

@EmilioPisanty Of what?

Emilio Pisanty

@DanielSank the Auger effect

DanielSank

16:09

@ACuriousMind Suspension is network-wide?

Ben Niehoff

@0celo7 It depends which aspects of string theory...obviously QFT still turns up, so it's still there

ACuriousMind

A suspension is for all chat rooms, and can only be imposed by moderators.

Bernardo Meurer

@DanielSank Yep

ACuriousMind

A kick can be carried out by every room owner.

DanielSank

@EmilioPisanty The thing with phonons?

@ACuriousMind Got it.

Ben Niehoff

16:09

and as for string field theory, well, we're not even quite sure what it is yet

DanielSank

AFAICT I cannot, even as an owner, delete messages. Is that true?

ACuriousMind

@EmilioPisanty Definition of what?

Emilio Pisanty

@DanielSank @TheDarkSide I rest my case

0celo7

@BenNiehoff statements like that amuse me

ACuriousMind

@DanielSank You can move them to rooms called "Trash", which are there only for that purpose

Bernardo Meurer

16:10

I love it when people invite me to trash

DanielSank

@EmilioPisanty It's an accident that I knew that.

0celo7

@DanielSank are you trying to kickmute someone

Ben Niehoff

well, there is one SFT that has been written down (by Witten of course), but it's for bosonic open strings (or maybe it was closed ones? I forget)

Bernardo Meurer

"John Rennie invited you to join "Trash" "

Love it

DanielSank

@0celo7 I'm trying to move rubbish out of my weekly featured post room.

Still haven't figured out how...

ACuriousMind

16:11

@DanielSank If you click on the room menu, you should get a dropdown menu in which you can select "move messages"

DanielSank

You know @ACuriousMind your argument about summing over $j$ is sensible but I still don't see how to recover it from the axioms I know about quantum.

ACuriousMind

Do that, select all messages you want to delete, then choose one of the "Trash" rooms as the destination

The Dark Side

@EmilioPisanty Haha. Nah, @DanielSank is still occupied with his main point right now, kicking someone :)

DanielSank

@ACuriousMind Ahhhhh, thanks.

@TheDarkSide wut?

The Auger effect has something to do with valence band electrons/holes recombining through interaction with the phonons. It's important because it is the reason for efficiency loss at high powers in LED's. I lived with a theorist who explained this theoretically using DFT.

Manos Kioupakis.

The Dark Side

@DanielSank Nope sir. Just a freak parallel discussion. When you've kickmuted whoever is one your agenda, maybe you'll read the transcript again.

DanielSank

16:15

@TheDarkSide I am not kicking someone. I just needed to trash some messages.

The Dark Side

Thank God.

ACuriousMind

@DanielSank You're having trouble because the notion of mixed states uses classical probabilities. The quantum way to see the partial trace is the correct operation is what I do in my footnote: You combine two systems by $\rho_A\otimes \rho_B$ and then you observe that the inverse operation is the partial trace, i.e. $\mathrm{Tr}_B\circ (-\otimes \rho_B) = \mathrm{id}_A$. In fact, this is one of the properties defining the partial trace in an abstract algebraic way.

So you have no other choice - you want an operation that reverses "combine these systems", you get the partial trace, period.

The Dark Side

This doesn't answer the question. — DanielSank 22 mins ago

@DanielSank - Didn;t want to clog the comment section there.

But I think it answers it implicitly.

DanielSank

@ACuriousMind I got stuck on the part where it says that the partial trace is needed because it reproduces (1).

I was trying to understand why (1) has to be true.

Now you're giving a more complex argument, which I will try to understand.

Unfortunately, I forgot to bring paper to the cafe, so I might have to do this later :-(

ACuriousMind

@DanielSank Ah, yes, that's because I was focused on the classical probability notion of mixed states there. Now that you explicitly asked how to do this staying within the quantum formalism, I think the argument I just gave is the way to go.

The Dark Side

16:25

Basically, if I read it right, what @dmckee is trying to convey through his grown up approach is - yes, no one owns it, we won't force you, still it would be nice if you don't do it.

Or perhaps that's what I understood.

Secret

@ACuriousMind Just want to double confirm something here. Is what is happening in this formula

$$\gamma (\vec{x}_1,\vec{x}_2;\vec{x}_1',\vec{x}_2')=N_e(N_e-1)\int \Psi^*(\vec{x}_1',\vec{x}_2',\dots ,\vec{x}_{N_e}) \Psi(\vec{x},\vec{x}_2,\dots ,\vec{x}_{N_e})d\vec{x}_3\cdots d\vec{x}_{N_e}$$ we are basically taking the partial trace of $N_e-2$ of the $N_e$ electrons in the $N_e$ electron density matrix?

ACuriousMind

That's not a density matrix, it's just a probability density, but yes, it's the same reasoning - if you've got a joint probability density $f(x,y)$ and you don't care about $y$, you get the density for $x$ by $\int f(x,y)\mathrm{d}y$. Nothing mysterious there.

Secret

I see

Bernardo Meurer

16:48

@JohnRennie Howdy

John Rennie

@BernardoMeurer Afternoon

Bernardo Meurer

IDLES - STENDHAL SYNDROME

►

@JohnRennie See if you like this, new British band

John Rennie

@BernardoMeurer I struggle with angry bands. I guess I'm an aging hippy at heart.

Bernardo Meurer

@JohnRennie Ha, that's interesting

I quite like angry bands

John Rennie

My brother (18 months younger than me) likes bands that make your soul bleed. His latest obsession is the Violent Femmes. You and he should meet up a gig some time :-)

Bernardo Meurer

16:58

He might like Death Grips then

Is he also in Chester?

John Rennie

No, he's a biology teacher in London

Bernardo Meurer

Ha, guess science runs in the blood :)

John Rennie

Well my mother was a bacteriologist ...

0celo7

@EmilioPisanty Here's the neat proof that "bounded sequences have convergent subsequences" can only hold in finite dimensional normed vector spaces. It's weirdly constructive and pretty amazing!

Bernardo Meurer

@JohnRennie Dayum, is everyone in science in your family?

0celo7

17:01

@EmilioPisanty Notation: open ball is $B(x,r)$, closed ball is $D(x,r)$. We want to show that if $(x_n)\subset X$ is a sequence contained in some $D(0,r)$, then $(x_n)$ has a convergent subsequences. Since normed vector spaces are first countable, this is equivalent to showing that $D(0,r)$ is compact in $X$. Since $D(0,r)\approx D(0,1)$, we show that $D(0,1)$ being compact implies $\dim X<\infty$.

John Rennie

@BernardoMeurer No, my dad was just a businessman

And my niece wants to be a journalist

Bernardo Meurer

She could be a science journalist

0celo7

Assume $D(0,1)$ is compact. Consider the covering of $D(0,1)$ by translated open balls $B(0,1/2)$. By compactness, we have $D(0,1)\subset S +B(0,1/2)$, where $S$ is a finite set. Let $L=\vee S$ be the linear span of $S$. Then $D(0,1)\subset L+B(0,1/2)$. Note that $L$ is finite-dimensional, hence closed in $X$.

Emilio Pisanty

@0celo7 this in an arbitrary normed vector space?

Bernardo Meurer

Can you imagine John, your niece working at Popular Science? What a pride eh?

0celo7

17:03

@EmilioPisanty Yes. Completeness is not assumed.

Emilio Pisanty

@0celo7 you lost me at "first countable"

0celo7

Now we can iterate this relation. We have $B(0,1)\subset L+B(0,1/2)$. By scaling, we obtain $B(0,1/2)\subset L+B(0,1/4)$. By induction, we have $B(0,1/2^n)\subset L+B(0,1/2^{n+1})$. This implies $D(0,1)\subset L+B(1/2^n)$ for any $n$.

@EmilioPisanty Do you want the definition or an explanation of what it does?

John Rennie

@BernardoMeurer Sadly, while she's good at science she has no great love for it.

Bernardo Meurer

@JohnRennie That probably makes her a perfect fit for PopSci :P

Emilio Pisanty

@0celo7 not particularly fussed. More of an "if you need to invoke something like that then it's no longer all that simple afaic"

Bernardo Meurer

17:06

I have an unusual amount of friends in journalism

Mostly sports though

John Rennie

@BernardoMeurer being a good science journalist is a lot harder than you think. If you just want to sensationalise I'd guess that's pretty easy because you just make evertything sound like magic.

0celo7

Now that last inclusion means that an $x\in D(0,1)$ is at most $1/2^n$ away from $L$. Since $n$ can be arbitrarily large, we have that $x\in$ closure of $L$. But since $L$ is closed, $x\in L$. Thus $D(0,1)\subset L$, and $L$ is all of $X$.

John Rennie

But if you want to inform as well as entertain, then that's hard.

Bernardo Meurer

@JohnRennie I know, I know, that;s my exact criticism to PopSci

0celo7

@EmilioPisanty First countability is a condition that means convergence of subsequences is equivalent to compactness in the covering sense.

All normed vector spaces, manifold, basically everything are first countable.

Bernardo Meurer

17:08

with reference to SMBC's "The Talk"

Emilio Pisanty

@0celo7 so first countability is this magic property that lets you prove some radically different theorem and pretend that you were talking about the original property?

=P

John Rennie

@BernardoMeurer A lot of my self answered questions are basically an attempt at science journalism and I find it really hard to do

0celo7

@EmilioPisanty Yeah. But it's very easy to prove in most cases :P

Bernardo Meurer

@JohnRennie Science journalism is hard because science nowadays is, mostly, hard. And you need to find a way to explain these complex ideas in a way that is simplified but not too awfully deformed. that's tremendously difficult

John Rennie

@BernardoMeurer Agreed. That's what I found really hard when writing the Q/As

0celo7

17:10

@EmilioPisanty First countable means that any point has a countable neighborhood basis. A neighborhood basis is a set of open sets such that any open set containing the point also contains one of those neighborhoods.

For metric spaces, just take balls with radii $1/n$

Secret

Why will $D(0,r)\approx D(0,1)$ i.e. why will a closed ball of arbitrary radii be the same as a closed ball with radius of 1?

0celo7

@Secret the map $x\mapsto \frac{1}{r}x$ is a homeomorphism

Bernardo Meurer

@JohnRennie It's a big part of being a good teacher/professor too

Secret

ah ok

Emilio Pisanty

@0celo7 like I said, you lost me at first countable

more accurately, you lose me at first countable

it's not that I haven't seen the term or that I can't wikipedia it

0celo7

17:11

@EmilioPisanty Ok, then we're just proving Heine-Borel implies finite dimensions.

Emilio Pisanty

it's that my attention span spontaneously shrinks to zero when it comes up

0celo7

:(

Emilio Pisanty

well, not zero, but pretty darn small

That said, yes, your construction is pretty neat.

@0celo7 this includes arbitrary metric spaces?

0celo7

@EmilioPisanty Yep. As I said, $B_d(x,1/n)$ forms a neighborhood basis for each $x$.

Emilio Pisanty

nevermind, it seems so

@0celo7 you lost me at neighbourhood basis

in the above sense

0celo7

17:15

If you give an open set $U$ containing $x$, I can find a $B_d(x,1/n)$ inside of $U$.

Emilio Pisanty

@0celo7 gotcha

now, can we talk about Auger electrons or something physicsy?

=P

0celo7

@EmilioPisanty Btw, first countability implies that the topology is determined entirely by its convergent sequences.

@EmilioPisanty what's an Augur electron?

Emilio Pisanty

Auger effect

The Auger effect is a physical phenomenon in which the filling of an inner-shell vacancy of an atom is accompanied by the emission of an electron from the same atom. When a core electron is removed, leaving a vacancy, an electron from a higher energy level may fall into the vacancy, resulting in a release of energy. Although most often this energy is released in the form of an emitted photon, the energy can also be transferred to another electron, which is ejected from the atom; this second ejected electron is called an Auger electron. The effect was first discovered by Lise Meitner in 1922; Pierre...

ACuriousMind

@0celo7 An electron whose motion follows a prophecy, clearly ;)

Also dubbed the Nostradamon.

0celo7

@ACuriousMind The Augur in Skyrim was pretty damn disappointing

I don't even remember what he did

John Rennie

17:19

@BernardoMeurer: have you ever heard of a band called The Soft Boys?

The Soft Boys - A Can Of Bees (full album)

►

Emilio Pisanty

@DanielSank So there's a solid-state version? Huh.

I mostly meant the atomic process

But presumably you mean this?

> The most prevalent mechanism is a three-particle interaction, two electrons and one hole, or two holes and one electron, in which an electron-hole pair recombines releasing its energy nonradiatively by promoting the remaining particle into a higher energy state.

I guess that's close enough that it earns the name

but really, if you're not ionizing and detecting the electron, you're cheating

Secret

Pictures are extremely misleading in illustrating compact sets:

0celo7

@Secret good, we need more topology talk in here

They are! It's strange that "finite size" sets don't have to be compact

What's even stranger is that there are bounded metric spaces that are not compact

@Secret do you have a specific question?

Secret

Balarka is the topology expert, I sorta get inducted in topology via the maths chat and munkres. Currently trying to be more comfortable in determining convergence of sequences

Noncompact spaces are easy to think about in some topologies, such as (0,1) in the standard topology, you pick a collection so that they overlap pairwise and then it is easy to see that any subcollection will left the set uncovered

No I don't have a question currently, your proof flows fine

0celo7

@Secret You should think about how to prove that a finite-dimensional subspace of a normed vector space is closed. The proof is nontrivial.

I doubt you can find it in Munkres. It's a functional analysis thing

@Secret Actually, think about: if $X$ is finite dimensional, then $D(0,1)$ is compact.

Slereah

17:33

hey hey

Bernardo Meurer

@JohnRennie Have not, I'll give them a listen

2017

17:48

@0celo7 I have removed you from the ignore list. I expect that you won't restart the arguments. I won't, from my side.

rob

17:59

Friendly request: let's not use the chat to discuss what's on-topic for the chat. For JEE stuff there's a current meta post open, and there's also a meta chat room which we've started to use for these meta-discussions.

I'll move some messages to the meta chat room here in a minute.

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