The h Bar

DanielSank

3:00 PM

Ok, that's a very helpful thing to know.

Thanks.

I will now retreat to a cafe with my notebook and a pen, and will likely return later for a more rigorous go-over.

I'm also interested to understand this without the discrete approximation step.

Sir Cumference

Player-Baby Come Back

►

DanielSank

I assume there are "operator-y" ways to think about all of this stuff.

2017

@SirCumference I have an astronomy question. Do you have a few minutes to spare?

Sir Cumference

@2017 Sure :)

DanielSank

Heh, actually, I think this extra source term is going to wind up being a normalization factor that I don't care about...

We'll see.

ACuriousMind

3:02 PM

@DanielSank I don't know what you're doing, of course, but usually, the term without the source ends up being the normalization factor

2017

@SirCumference Yeah, so I was learning the applications of Doppler's shift recently. Most books seem to mention that light from cosmic objects is redshifted/blueshifted. But what is the reference for the light? w.r.t what is the redshift/blueshift measured?

Sir Cumference

@2017 Not sure if I understand your question. Do you want to know how to calculate redshift?

2017

Suppose the received wavelength is $\lambda_R$

Then what is the $\lambda_o$ ? Such that $\lambda_o-\lambda_R=\text{shift}$

Sir Cumference

OK, so the formula is

$$1+z = \frac{f_{\mathrm{emit}}}{f_{\mathrm{obsv}}}$$

Or $$1+z = \frac{\lambda_{\mathrm{obsv}}}{\lambda_{\mathrm{emit}}}$$

2017

@SirCumference Yup, how do you know the $f_{emit}$ ?

HDE 226868

3:06 PM

@2017 Many observations are made of emission/absorption lines. We know where in a spectrum certain ones should show up, and if the observed spectrum can be matched with a spectrum at $z=0$, we can tell how much the wavelength has been shifted by. We then use @SirCumference's formula.

Sir Cumference

Yup

2017

@HDE226868 "we know where in a spectrum certain ones should show up"....but for that you need to know the composition of the cosmic object beforehand, isn't it?

HDE 226868

@2017 Sure, but you can expect certain lines to show up in most cases, depending on what the object is.

(This is for spectroscopic redshift, not photometric redshift.)

2017

@HDE226868 "expect certain lines to show up in most cases"...could you give some examples? I'm not sure I get you...

Do you mean for planets or stars you expect some spectral lines for sure?

HDE 226868

@2017 The Balmer series is commonly observed.

@2017 Yes, you'd expect something like that to show up in most stars.

Stars obviously have quite a lot of hydrogen, and there should be plenty of that in the atmosphere. This leads to Balmer lines.

2017

3:13 PM

@HDE226868 We now know that stars have a lot of hydrogen. But, before this technique of doppler shift was discovered, did we know it? Or it was just an educated guesswork initially ?

I suppose the wavelength of each different type of material is shifted by nearly same amount....so perhaps some extrapolation would do the job

HDE 226868

@2017 I don't know that off the top of my head. I believe spectroscopy of the Sun (and thus theories of its composition) slightly predates the proposal of the Doppler shift, but I'm not 100% sure.

There were plenty of developments in the 19th century, though.

2017

@HDE226868 Ah, that's seems an interesting topic :-) Thanks for helping...I need to read a bit more about this! Thank you!

HDE 226868

@2017 No problem.

2017

@SirCumference Thank you too :-)

Sir Cumference

@2017 Oh, np

Slereah

3:36 PM

0

Q: Stings in the loop space of timelike curves

In Smith's paper on homotopy groups for Lorentz manifolds, he builds the loop space of all timelike loops in the following fashion : Consider all piecewise continuous timelike curves which start and end at point $x$. This include timelike curves with $q$ changes in time orientation (the tangen...

plz respond

I kind of suspect that when he says "this isn't enough to have a group structure", he was refering to the absence of the constant path

But that doesn't really explain why stings are there

DanielSank

@ACuriousMind can you expand on the idea that $J A^{-1} J$ has something to do with propogation?

In general, we have some linear equation $T|\psi\rangle = |J\rangle$ and we want to solve for $|\psi\rangle$.

To me, a "Green's function" is the solution to $T |G \rangle = |e \rangle$ where $|e \rangle$ is some basis vector.

Slereah

but what is

the red function

DanielSank

Slereah

"Smith's understanding of the relationship between corporeal and physical objects extend to his interpretation of biology, where he has become an opponent of Darwinian evolution, as the fundamental element in a species would be its form, not its causal history, which evolutionists favor."

"Smith has also taken a stance towards a relativistic rehabilitation of geocentrism."

Wew lad

I'm starting to have doubts about that guy

I hope he's not crazy about math either

ACuriousMind

@DanielSank It's classical propagation, not quantum: If $A$ is a linear differential operator, then one can often solve the equation $AG = \delta$ for $G$, where $\delta$ is the usual delta distribution. Then, the equation $A\phi = J$ is solved by $\phi = \int G J$ (that's a convolution, I don't want to botch the variables here).

So the operator $\int G$ is inverse to $A$.

Slereah

3:51 PM

"According to Smith, such an interpretation of quantum mechanics allows for the usage of the hylomorphic concepts of act and potential to properly understand Quantum superpositions."

DanielSank

@ACuriousMind (I use $*$ for convolution)

@ACuriousMind Ok so $J A^{-1}J$ is like $\langle J | \int GJ \rangle$.

Slereah

I could send him an email, he's still alive

DanielSank

...as we said earlier.

ACuriousMind

@DanielSank Yes, exactly

Slereah

If he knows how to use a computer

(He's 87)

Oh wait

I know what's going on

His math papers are from the 60's

His crazy papers are from the 90's

ACuriousMind

3:54 PM

And you can get surprising mileage out of the idea that now taking functional derivatives w.r.t. $J$ of that gives you back combinations of $G$, depending on how many you take.

Slereah

He got the physicist disease

DanielSank

Interesting. I wrote $|\psi \rangle = \sum_i |G_i \rangle J_i$, but you wrote $\phi(x) = (G * J)(x)$.

ACuriousMind

That's behind the "generating functional" that forms a large part of the path integral approach to QFT.

DanielSank

These must be the same thing.

@ACuriousMind That sounds familiar.

$$\langle x | \psi \rangle = \sum_i \langle x | G_i \rangle \langle i | J \rangle$$

@ACuriousMind I think your form requires translational invariance, or something.

ACuriousMind

@DanielSank Yes, the Green's function only acts as a simply convolution if $A$ is translationally invariant

Otherwise it's a true function of two variables and you don't convolve, you simply integrate $\int G(x,y) J(y)\mathrm{d}y$.

DanielSank

3:58 PM

@ACuriousMind Yeah ok.

That matches what I have up there in bra-ket notation.

Thanks.

I might have written $G_y(x)$ where you have $G(x,y)$, but whatever.

2017

4:10 PM

@JohnRennie If we pull an elastic rod with a constant force $F$ upwards (gravitational acceleration still stays g), then to calculate the extension in the length of the rod (using the Young's modulus method) do we consider the net effect of $g$ and $F/m$ or do we only take $F/m$ ? My book neglects the $g$...but I think that is wrong!

DanielSank

4:37 PM

@heather we should chat about your multi qubit basis question.

heather

@DanielSank, I saw your comment, but I'm not sure how the Kronecker product relates (I read through the wikipedia article).

DanielSank

Ok let's start from the beginning. What are you trying to do?

Sir Cumference

@EmilioPisanty Thanks for pointing out the copyright though, I would've screwed up.

Emilio Pisanty

@SirCumference No worries, just think it through next time

To be honest, I'm not entirely sure whether the logo infringes that image as it is

Sir Cumference

Yep :)

Emilio Pisanty

4:45 PM

it's probably covered under fair use (?), but that's always a tricky area

In any case, I still think the question is not particularly well formulated (and indeed it's pretty entitled) in its current version, but I've seen that you just won't be persuaded so oh well

Sir Cumference

@EmilioPisanty Well if you have any suggestions, I'm all ears. I'm just hoping my suggestion will be heard.

Emilio Pisanty

@SirCumference I've said it something like five times already

heather

@DanielSank generate a list of the basis vectors that will be eventually used to calculate the state of a qubit, which will be manipulated in the program.

Emilio Pisanty

Make it clear that there are in-practice instances of the logo being under-resolved

in the question itself

Sir Cumference

OK, I'll get to that when I get home

Emilio Pisanty

4:48 PM

"vector logo" isn't a goal in itself

it's a tool to get a crisp logo in all real-world instances

Sir Cumference

Which is a few hours from now

DanielSank

@heather the basis vectors are always $\{ (0,0,1), (0,1,0), \ldots\}$.

Emilio Pisanty

the fact that you can zoom in enough to get it to show its pixels is meaningless if under normal conditions it shows fine

DanielSank

I think your real question is what these tuples mean in terms of the single qubit states.

Emilio Pisanty

Either way, there's no "we" in updating the logo file

The design is managed entirely by the SE design team

and a vector logo file is not a trivial change

heather

4:50 PM

@DanielSank not always...i thought for a single qubit system, it was [0,1] or [1,0]; for a two qubit system, [0,1,0,1], [1,0,1,0], [1,0,0,1], or [0,1,1,0]

dmckee

@BenNiehoff In the corners of nuclear and particle experimentation (and instrumentation) that I've been involved in there is an expectation that every paper can be found on arXiv.

Emilio Pisanty

Not all browsers support svg

Sir Cumference

@EmilioPisanty Duly noted.

dmckee

I've had papers with tens of citation by the time they first appeared in "print".

And they were in PRL at that.

Emilio Pisanty

and that means that they need to implement a fallback png

and that means code

and it means testing to make sure it works fine for all relevant browsers

DanielSank

4:52 PM

I can't chat from my computer...

test

dmckee

Nawh, just a quiz.

heather

i can read it @DanielSank

DanielSank

@heather you have to remember what it means to write vectors as lists of numbers.

heather

@DanielSank each of the components of the vector

Emilio Pisanty

It's probably worth implementing if it actually impacts how the site looks, but if it doesn't then it's not worth the extra work. So put the evidence up front.

DanielSank

4:53 PM

Those lists implicitly mean $$(a, b, c) \equiv a |e_1 \rangle + b|e_2 \rangle + c |e_3 \rangle$$ where $|e_i \rangle$ is a particular choice of basis vectors.

Sir Cumference

Yep :)

Emilio Pisanty

The bitmap logo is probably a holdover from way before high-dpi was a thing

DanielSank

If you're trying to write a basis in its own basis, then of course they'll be $(0,0,1)$, $(0,1,0)$...

Emilio Pisanty

and this site is not usually at the top of the list in terms of getting site-design upgrades

dmckee

^ THat.

Emilio Pisanty

4:54 PM

witness e.g. the network-wide CSS update

6

Q: Is there a known timeline for the deployment of the new CSS?

Some six to eight weeks ago, Stack Exchange announced a programme to re-design the CSS and tweak the site styles throughout the network, including the completely redesigned profile pages. Since then, we've made some big requests to come with the re-design, so we've remained "on deck" for some tim...

support status-completed design

DanielSank

@heather I think you're confused because you 1) don't know what your basis vectors actually are, and 2) Don't know how those basis vectors relate to the single qubit basis.

Emilio Pisanty

including a sizeable update to the profile pages

heather

@DanielSank right

dmckee

We've just big enough to be "important" in some way, but still small enough that it's not a very important kind of important.

heather

@DanielSank oh

but...that isn't what i'm looking for, i'm looking for the basis, just not written in it's own basis, i guess. the basis that i'd use to calculate the state of the qubits.

you're right, i'm very confused =/

DanielSank

4:55 PM

@ACuriousMind If $$ \int dx_1 \ldots dx_n \exp \left( -\frac{1}{2} \sum_{ij} A_{ij}x_i x_j \right) = \sqrt{\frac{(2\pi)^n}{\text{det}A}} \, ,$$ then what is $$ \int dx_1 \ldots dx_n \exp \left( -\frac{a}{2} \sum_{ij} A_{ij}x_i x_j \right) \, ?$$

Emilio Pisanty

@dmckee I dunno

vzn

@Obliv an interesting topic for Computer Science chat? swade is one of the worlds leading authorities on babbage machines/ engines, he basically got the designs to really work over ~1½ century after devised, hes got a good book out on it too, recommend it. there have been some exhibitions at museums in recent times.

Emilio Pisanty

being the very last site to get the CSS update was...

not demoralizing, but you do have to wonder

ACuriousMind

@DanielSank Do $x'_i = \sqrt{a} x_i$.

DanielSank

I would think I'm just multiplying $A$ by $a$, but when I look up the result elsewhere we get a new factor of $\sqrt{a^n}$.

@ACuriousMind Right but why can't I think of it as a scaling of $A$? Scale factors factor out of determinants...

@heather, when I write a matrix, what do the elements of the matrix mean?

Emilio Pisanty

4:59 PM

Either way, @SirCumference put your evidence up front, and keep in mind that whenever the design is involved you are asking other people to do work for you.

DanielSank

Oh wait...

If I scale a matrix, the determinant scales by that scale factor raised to the dimension of the matrix, doesn't it?

Emilio Pisanty

@DanielSank yes

ACuriousMind

@DanielSank Indeed it does

DanielSank

Yes, of course, because determinants are volumes of the result of acting the matrix on the unit hypercube.

Ok we're done here. Thanks @ACuriousMind and @EmilioPisanty.

heather

@DanielSank well, the basis vectors of the system you're transforming too.

DanielSank

5:00 PM

@heather you keep getting confused by this :-) We've had this discussion before.

Emilio Pisanty

@DanielSank or, more simply, the matrix scales by $a$ whenever you multiply each row individually by $a$.

DanielSank

^ yes

heather

@DanielSank we have? let me check the transcript then, i don't want to waste your time

DanielSank

It was a long time ago and the connection is probably not obvious to you.

Forget quantum computing.

We need to go over some linear algebra stuff.

Ben Niehoff

@dmckee Meanwhile, I'm not sure if any of my papers have tens of citations at all! :P

Emilio Pisanty

5:03 PM

BTW, now that there's some form of quorum:

given that this exists

Ben Niehoff

in the sense of plural tens

Emilio Pisanty

DanielSank

@ACuriousMind can you unlock this?

Emilio Pisanty

source

dmckee

@BenNiehoff It's all about subfield. That's why bibliometrics are a terrible way to evaluate scientists.

Emilio Pisanty

5:03 PM

Does our site logo infringe the Wikipedia copyright?

DanielSank

or @dmckee

@heather go here

dmckee

I'm a very run-of-the-mill researcher and I'm an author on two topcite-500 papers.

ACuriousMind

@DanielSank Done

@EmilioPisanty Is a Feynman diagram even copyrightable?

Emilio Pisanty

@ACuriousMind Ideas are not copyrightable

specific instances of graphics are

the site logo is a direct copy, just monochromated

user228700

@JohnR: Did u get around to watching the video?

heather

5:05 PM

@DanielSank okay

Emilio Pisanty

and take out the arrow and the t

Ben Niehoff

My most-cited (and coincidentally, oldest) paper has 24 citations...that is the only one with plural tens :P

ACuriousMind

@EmilioPisanty But only if they pass some threshhold of originality. I have my doubts this one does.

Emilio Pisanty

@BenNiehoff I guess I feel better now =P

Ben Niehoff

@EmilioPisanty What do you feel better about?

John Rennie

5:06 PM

@Kaumudi.H Hi :-) yes I did have a look, but I think it was American cooking rather than British. I'm pretty certain we never had Cambell's cheese soup in the UK.

dmckee

I'm quite sure that Feynman diagram had been drawn thousands of times before the version on wikipedia was first constructed.

Emilio Pisanty

@BenNiehoff 34 citations on my most-cited paper

user228700

@JohnRennie Hi :-) Yep, it was primarily American.

Ben Niehoff

And if not, it's easy to just draw another one that looks pretty similar...

Emilio Pisanty

That's the point, it's not a looks-similar case

It's clearly a direct copy

dmckee

5:07 PM

I would consider Feynman diagrams more like words: you assemble them from a finite set of predetermined components.

Ben Niehoff

sure, I'm saying if it bothers you, you could correct the problem easily

user228700

Are ur servers doing OK? (:-P)

Ben Niehoff

Oh, and most of my citations are from collaborators and their friends :P

Emilio Pisanty

@BenNiehoff I can't. Changing the design assets is an SE job.

Ben Niehoff

I see. Well, one could write them a threatening letter

dmckee

5:10 PM

@BenNiehoff My sole lead-author paper has six citations. And half of them are from the collaboration I did that work for.

Emilio Pisanty

@BenNiehoff you'd have to be the copyright holder for that to be even mildly threatening

but being a good internet citizen is worthwhile on its own

Ben Niehoff

well, it's possible that Feynman diagrams aren't copyrightable anyway

dmckee

And two of the others are in a context that comes down to "We were careful not to get into the kind of pickle where we'd have to do the work that McKee did".

Ben Niehoff

is there a legal SE where one could ask? :P

dmckee

::sigh::

Slereah

5:10 PM

there is

heather

actually, yeah, there's a legal SE

@BenNiehoff Law

goddammit

google scholar

dmckee

5:11 PM

The last one is a very nice comment in a major review, so that's something.

Emilio Pisanty

stop bungling up my name

Ben Niehoff

what did they do?

Emilio Pisanty

@BenNiehoff That EP Alatorre is Emilio Pisanty Alatorre

and they've gone and put my paternal surname down as a middle initial

Ben Niehoff

ah, I see

bolbteppa

Ignoring the Christoffels (which can't be ignored), this is the shortest I can get Schwarzschild, which is not that short :( :

Ben Niehoff

5:13 PM

are you Spanish?

ACuriousMind

@EmilioPisanty Is Alatorre a second surname or more akin to a middle name? (I have no idea how your name works, as you might see)

bolbteppa

That still seems shorter than what Wald does

Emilio Pisanty

@BenNiehoff I'm Mexican

Ben Niehoff

ah, ok

Emilio Pisanty

Pisanty is my paternal surname, Alatorre is my maternal surname

I normally publish as Emilio Pisanty because maternal surnames confuse English speakers to no end

bolbteppa

5:16 PM

@0celo7 Write up a Straumann/Wald solution shorter than mine above

Emilio Pisanty

but the thesis needed to be in the name I was registered under and that needed to match my passport

and, of course, it's gone on to confuse the hell out of google scholar

now to figure out how to bring it into the fold

Ben Niehoff

one of my grad-school friends was Mexican...he was named José Raúl, but only introduced himself as José

we found out three years later that he in fact goes by Raúl back home :P

but he thought it would be confusing that someone went by their middle name, so he just used the other one with us for three years instead :P

Emilio Pisanty

yeah, that can happen

it's not as bad as needing to invent entirely new names, like many Asian people feel they need to do

Ben Niehoff

I think that happens less often now

Emilio Pisanty

but yeah, there's some pretty intense pressure to change how you interact with your name because English-speaking culture simply can't be bothered to acknowledge that there are other naming conventions

@BenNiehoff I knew multiple people who did that in grad school a couple years back

Ben Niehoff

5:20 PM

sure, not saying it doesn't happen

but I know loads of Asian people who just use their actual names

but backwards of course

Emilio Pisanty

Falsehoods Programmers Believe About Names is a good eye opener

Ben Niehoff

haha...even I have run out of space for my name on a form where little boxes were provided for one letter each

and I don't even have a long name

I dunno wtf they were thinking

ACuriousMind

@EmilioPisanty That's common enough that I have a half-Korean friend who experiences a kind of inversion of that: His name is Otto and it's quite common for people to ask him "No, what's your real name?" since they don't believe an Asian would have such a stereotypically Northern German name.

Emilio Pisanty

Falsehoods programmers believe about time is also really good

one of those was behind some large-scale outage if I remember correctly

ACuriousMind

@EmilioPisanty There's a better version of that somewhere that actually gives examples for each point

Unfortunately I seem to have misplaced the link to it

Ben Niehoff

5:27 PM

I've had Indian people be impressed that I could actually pronounce their names...I was mostly surprised that I guess nobody had bothered trying before

heather

@BenNiehoff i have some friends who are indian, and rarely does anyone pronounce their names right. i had to get them to explain to me how you're actually supposed to pronounce their names (especially their last names).

now I do it correctly, thank goodness, but i also wince along with them when a teacher butchers it especially well.

DanielSank

Paging Cow-smoothie (@Kaumudi.H)

2017

@BenNiehoff As a fun fact, Indians themselves find Indian names of people (belonging to other states) very difficult to pronounce. For example, I still don't know how to pronounce @Kaumudi.H's name! India has just too many languages and dialects.

Ben Niehoff

I'm not talking about perfect pronunciation...I don't know the exact IPA values of all the sounds. But most people don't even bother to use the right number of syllables and put them in the right order

Emilio Pisanty

ah, yes, this is what it was

arstechnica.co.uk/security/2017/01/…

> Cloudflare hit by leap second “software panic” snafu on New Year’s Day

caused by code that assumed that a datetime function would be monotonous

2017

5:33 PM

@BenNiehoff Yeah, I agree with that. Some of my friends had very bad experience in Singapore when they were forced to select a new name for themselves as the people there couldn't pronounce them properly.

user228700

@2017 Cow-moothie works :-P

2017

@Kaumudi.H just "cow" will be shorter and sweeter :D lol

user228700

Sure :-P

Danu

Hi everybody

user228700

BTW yes, I completely agree. My neighbor calls me "Gomathi" because she still doesn't know how to pronounce my real name.

heather

5:37 PM

@Kaumudi.H, hello =)

bolbteppa

Obviously the EFE for Schwarzschild involve the second derivative of the (logarithm of the) time scale factor with respect to position $\frac{d^2 \nu}{d r^2}$ plus all obvious ways to make the first derivative of this non-linear, e.g. $(\frac{d \nu}{d r})^2$ and $\frac{d \nu}{d r} \frac{d \lambda}{dr}$ as well as the derivative of the (logarithm of the) radial scale factor, per unit radius $\frac{\lambda'}{r}$, up to signs and scale factors :'( :'( :'(

0celo7

Why are you obsessed with the most basic solution

bolbteppa

Because it's the whole content of GR!!!! (modulo tons of stuff)

0celo7

Not the actual derivation though

Ben Niehoff

@bolbteppa where's your metric ansatz?

bolbteppa

5:44 PM

I can copy in the metric and Christoffels if you want

0celo7

@bolbteppa no. Why should I?

bolbteppa

Do you not think it's useful to know the most important calculation in a subject like the back of your hand?

Ben Niehoff

I mean, it's a calculation you can do in a page :P

0celo7

@bolbteppa no

Not even a little

I think you need to reevaluate how you view science

bolbteppa

Engineers don't need to I suppose

(burn)

0celo7

5:54 PM

How is that a burn?

bolbteppa

You can't put Schwarzschild into a bucket of water and just get the answer, need to assume spherical symmetry and get it directly

0celo7

What?

Emilio Pisanty

@bolbteppa that's obviously wrong

you can't assume a cat is spherical

but you can do just fine with a sphere and a cylinder

Danu

^

#getreal

heather

lol, very true @EmilioPisanty

bolbteppa

5:58 PM

5 cylinders, two spheres?

Emilio Pisanty

two spheres?

bolbteppa

head and body

heather

four legs and tail, for the cylinders...hmm, body is probably better as a cylinder

Emilio Pisanty

or this

heather

and you could add in triangular prisms as the ears

0celo7

5:59 PM

The body is not spherical unless the cat is obese

2

Emilio Pisanty

but really, single sphere and six cylinders is plenty

0celo7

@heather the ears make up a very small volume

heather

@0celo7 true =)

@EmilioPisanty that's a very styling cat

Emilio Pisanty

or this

bolbteppa

$\int_{cat}dV = \int_{sphere_{head}}dV_1 + \int_{cylinder_{body}}dV_2 + 4\int_{cylinder_{leg}}dV_3 + \int_{cylinder_{tail}} dV_4$

Emilio Pisanty

6:00 PM

source

14

A: How do astronauts turn in space?

Although this has indeed "worked to bits" on the Physics and other SE sites it's worth looking at, for the sake of Space Exploration, the interesting history behind the analysis of the falling cat. For the fully rigorous description of the cat's righting reflex - perfectly in keeping with conserv...

2017

What is the average density of a cat? :-P

0celo7

@ACuriousMind cat wasn't interested that I was leaving

2017

Is it more than humans?

Emilio Pisanty

@0celo7 link doesn't work

2017

lol...yahoo answers answers.yahoo.com/question/index?qid=20081221201848AAaoUCb

0celo7

6:02 PM

@EmilioPisanty hmm. Imgur is doing strange things on my phone

Emilio Pisanty

> The imgur.com page isn’t working

> imgur.com redirected you too many times.

0celo7

It's saying I have to wait 3400 seconds until I upload again.

Never had that happen

heather

i see the cat

it really doesn't care at all, lol

ACuriousMind

It's a cat. Not caring is in the job description.

OBE

@0celo7 sup?

oh because drake is canadian?

@ACuriousMind I like your new pic.

nice blue shadows on your face

0celo7

6:23 PM

It seems people on the chat downvote my imgur posts

Wtf is up with that

DanielSank

How do I make slanted dots in mathjax?

$$\ddots$$

oh

0celo7

What are slanted dots?

bolbteppa

dots ddots vdots

vzn

6:44 PM

The quest to crystallize time — Bizarre forms of matter called time crystals were supposed to be physically impossible. Now they’re not. / nature

reddit.com/r/Physics/comments/5ytn6t/…

DanielSank

@ACuriousMind this is mostly for you.

I'm trying to create an opportunity to carefully explain what we discussed earlier. If you don't have time, I will attempt an answer, but I'd rather not muddy the waters with something less than crystal clear.

WDUK

What are some important maths topics that are common through out physics ? I am looking to get some books to brush up my maths during the summer and want to know what topics in maths to read up on to get a good head start

John Doe

Can someone please check the following QM query please. Given an ensemble of $N$ two-level systems where we define operators $\hat{J}_z := \sum_i \frac{1}{2} \hat{\sigma}_{i}^{z}$ and $\hat{J_-} := \sum_{i} \hat{\sigma}_{i}^{-}$ where $\hat{\sigma}_{i}^{z}$ is the $z$ pauli operator and $\hat{\sigma}^-_{i}$ is the lowering operator on the $i$-th particle i.e. $\hat{\sigma}_{i}^{-}| \uparrow...\uparrow_i...\uparrow \rangle = |\uparrow...\downarrow_i...\uparrow \rangle$.
How does it follow that
$$\hat{J}_{z}(J_-| \uparrow...\uparrow \rangle) = \sum_{i,j}\frac{1}{2}\hat{\sigma}_{i}^{z}|\uparro…

DanielSank

7:02 PM

@ACuriousMind FYI, when I use bra-ket notation, that does not mean I'm thinking quantum.

It just means I'm thinking vectors.

Danu

7:33 PM

So @DanielSank about your question

The identity you give comes from the following (this is the analogous computation for field theory since I happen to have that one written out): If $\mathcal L=-\frac{1}{2}\phi(\square+m^2)\phi+j\phi$ then $$Z[j]=\int \mathcal{D}\phi e^{iS[\phi,j]}=Z[0]\exp \Big(-\frac{i}{2}\int d^4 x d^4y j(y) \Delta(x-y)j(x)\Big)$$ where $\Delta$ is the propagator, i.e. solves $-(\square +m^2)\Delta(x-y)=\delta(x-y)$.

OBE

@WDUK topology and geometry are used a lot in theoretical physics so you could learn about those. but you didn't specify your level. I would say vector calculus is one of the most important areas of math for learning physics.

Danu

So your $A^{-1}$ is $\Delta$ (or its analog in your context)

bolbteppa

Peskin gives a cool discrete derivation of that, also in chapter 2 here desy.de/~jlouis/Vorlesungen/QFTII11/QFTIIscript.pdf

Danu

So I don't really understand the question @DanielSank

ACuriousMind

@DanielSank I'm not sure I can carefully explain it in your discretized case. In the continuous case I could (though I'd have to be a bit careful about what "kind" of Green's function we have here, exactly), but I can't say for sure if the relation between the propagator and the inverse diffusion operator carries flawlessly over.

Discretization has a nasty habit of breaking things when you least expect it :P

Danu

7:42 PM

@ACuriousMind Why can't you just go through the exact same derivation?

For the field theory version, you just do a shift by $-\int d^4y j(y)\Delta(x-y)$---I think I should be able to do the same, no?

ACuriousMind

@Danu The relation between the propagator and the inverse continuous diffusion operator relies crucially on the propagator being the Green's function of the diffusion operator. I don't see how it would mean in the discretized case to say that the propagator is the "Green's function", or what its relation to $A$ is.

@Danu What?

This isn't about deriving the value of that integral

It's about how the inverse operator $A^{-1}$ is related to the propagator.

WDUK

@OBE i start uni next september so i imagine the first year is a lot of classical physics which i already know the math for.

beyond that i dunno what will be the next thing to learn

Danu

@ACuriousMind So what I was saying is that to obtain that relation in the case I know about, you can just do that shift on $\phi(x)$ and get the result

I probably didn't understand what's discretized about this variation

ACuriousMind

@Danu Uh...I still think you're explaining how to get the r.h.s of the formula Daniel wrote down in his question after "there's a known formula"

DanielSank

I don't care about discrete versus continuous.

I'm only doing discrete for now because I don't know the continuous case well (yet).

Also, green's functions work fine in discrete land.

Danu

7:51 PM

@ACuriousMind I probably don't understand the question.

I thought the question was sorta, how does $A^{-1}$ relate to $\Delta$

My answer was, they are the same

ACuriousMind

@DanielSank No doubt they do, but I have no clue

DanielSank

If I have $T|\phi \rangle = |J \rangle$, I can solve that with "Green's functions" whether those vectors are continuous or not.

I instead solve $T|G_i^e \rangle = |e_i \rangle$ where $|e_i\rangle$ is a set of basis vectors.

bolbteppa

@DanielSank "However, I am given to believe that there is a link between $A^{-1}$ and the propagator $W$" if your question is - what is the link between the inverse of a differential and the propagator - then the answer is that they are the same thing.

At least for physicists

ACuriousMind

@Danu ...that's exactly what I am not sure about for the discrete case.

DanielSank

Then $$|\phi \rangle = \sum_i \langle e_i | J \rangle |G_i^e \rangle$$

Danu

7:53 PM

@ACuriousMind So why can't I just do a shift by the thing I wrote down to the field, given that propagators still work apparently

bolbteppa

Propagator

In quantum mechanics and quantum field theory, the propagator is a function that specifies the probability amplitude for a particle to travel from one place to another in a given time, or to travel with a certain energy and momentum. In Feynman diagrams, which serve to calculate the rate of collisions in quantum field theory, virtual particles contribute their propagator to the rate of the scattering event described by the respective diagram. These may also be viewed as the inverse of the wave operator appropriate to the particle, and are, therefore, often called (causal) Green's functions (called...

Danu

Doing that shift, all you need is that it's a propagator; the result just falls out immediately after some cancellations

ACuriousMind

@Danu You should probably write down what "shift" you're talking about explicitly because I have no idea what it has to do with anything.

DanielSank

Y U no write answers?

Danu

@ACuriousMind $\phi(x)\mapsto \phi(x)-\int d^4x j(y) \Delta(x-y)$ (I did write it down)

DanielSank

7:54 PM

Also, someone please acknowledge that the equation I wrote above is a discrete version Green's functions.

Danu

where $\Delta$ is defined such that if the Lagrangian is of the form $\phi A\phi$, then $A\Delta=\delta$

ACuriousMind

My issue is that I don't know how $W(x,t;x_i,t_i)$ is supposed to be a "Green's function" for $A$.

DanielSank

$W$ is a transition matrix element/probability/whatever.

ACuriousMind

In the continuous case, it's a Green's function in the sense that $(\partial_t - D\partial_x^2)W(x,t;x_i,t_i) = \delta(x-x_i)\delta(t-t_i)$

Danu

^and that's the (only) crucial thing in my proposed identification

agreed, @ACuriousMind?

ACuriousMind

7:58 PM

@Danu Yes - but how do you get it when you replace $\partial^2_x$ by its discretized version $A$?

Danu

yo I don't know shit

ACuriousMind

To me, that's the crux of the question here. Does that even make sense/hold?

Danu

@Semiclassical Do you happen to know about this? ^

(feel free to ignore if you don't feel like contributing)

ACuriousMind

@DanielSank I think it might well be the case that the relation exists in the continuous but not in the discrete case - in which case the phrasing of your question has led most people astray, see also its current comments

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