@JakeRose $\delta a_j = [\delta \mathbf{a} \cdot \mathbf{P},X_j] = [\delta a_i P_i,X_j] = \delta a_i [P_i,X_j] = \delta a_i \delta_{ij} = \delta a_j$

'The Sad Story of Heisenberg's Doctoral Oral Exam'

> When an angry Wien asked how a storage battery works, the candidate was still lost

@NovaliumCompany financial advice is generally worth what you pay for it. This applies to my advice as well :-)

@FakeMod You must start with “Rubycon”: en.m.wikipedia.org/wiki/Rubycon_(album)

Please have a look at the recent meta post of mine. (I need some help)

Can anyone tell me why in the right the starred message by ACM "Stop annoying other users" is stagnant there for almost one week. Why it's not getting away like all other starred messages?

@Knight because it's been pinned (super star) :-)

@Knight it has been pinned.

@JohnRennie hello sir :)

@user8718165 hi :-)

@FakeMod I was trying to render mathjax in chat but it isn't working. What should I do?

@JohanLiebert if you use Chrome there is an addon-on to render MathJax.

@JohnRennie I went here and clicked add to the desktop. But nothing has changed now. What should I do?

That isn't what I use.

Use this

@JohnRennie I searched for it(and some others) on meta. Can you tell me what and how to use?

This seems unnecessarily complicated to me.

@JohnRennie I have added that. Now what to do?

What I suggest is that you uninstall TamperMonkey. Then just add the addon I linked and MathJax will automatically render without you needing to do anything else.

If it's working you should see an extra button (outlined in red)

You may have to refresh the page ...

@JohnRennie how to uninstall that add-on? (I don't know anything about it!)

Click the "three dot" menu then More tools then Extensions. This opens a page that shows all the add-ons you have installed.

@JohanLiebert it should look something like this

@JohanLiebert to remove an add-on click the Remove button.

I haven't yet

@FakeMod that Google link just opens a black page for me ...

@FakeMod yes, listening now.

It's good, but my taste is for the trippier end of the spectrum.

A lot of electronic music from the 60s and 70s was heavily influenced by LSD. It tends to have less of a beat and be more meandering.

I think Sia is a great singer.

I'm not a huge fan of EDM but her collaborations with David Guetta are excellent.

I guess he's good if you're into EDM ...

I heard his singles, but i don't know his music well. I'd guess the singles were the poppier end of his work.

@JohnRennie Dya srsly telling me u want money for advice?

@PM2Ring tenk

How do we know what the cosmic microwave background is? I mean, as far as I know it's just EM radiation in some range of frequency (I suppose not the visible range). Like, where does it exists, where did it came from and how do we capture it?

@NovaliumCompany no, what I'm saying is that my advice is probably worthless.

@JohnRennie What does that mean?

@Knight a room owner can pin a message so it always appears at the top of the star board.

The message stays there until it is unpinned or it disappears after some set time that I think is about two weeks.

I have unpinned the message now since I didn't see any point in leaving it there.

@FakeMod click the star icon a second time. The star icon alternately stars and unstars the post.

@FakeMod You don't, stars are "locked" after a while just like votes.

@JohnRennie Thank you so much for unpinning it.

Thanks sir.

:P

@NovaliumCompany You're doing that thing again of asking about stuff that can be easily Googled. Take a look at the Wikipedia article, and if you need help understanding something there, ask a focused question about it.

Yo

How to become world's best theoretical Physicist ever born in this universe ever!? just like Landau and Lifshitz?

Lev Davidovich Landau:

. i.sstatic.net/OlUch.jpg

Evgeny Mikhailovich Lifshitz:

. i.sstatic.net/Djdmw.jpg

Please don't post giant images inline without good reason, it pushes the rest of the chat off-screen unnecessarily.

And to become a good physicist I'd suggest a little less person cult and a little more studying the actual subject.

@ACuriousMind You are being disrespectful to those great physicists!! :P This won't be tolerated.

@ACuriousMind You are not a true physicist then! XD :)

@FakeMod Congratulations, you are now a physicist!

@skullpatrol He's a good guy, I've seen some of his stuff on AI.

Is $\langle \psi | \hat T | \psi \rangle = \langle \psi | \hat T \psi \rangle$?

For some operator/linear transformer $\hat T$?

@Knight How's your day?

@AbhasKumarSinha Fine bro. You get scolded again?

@Knight when I was scolded? Except ACM, everyone likes me ^_^

@AbhasKumarSinha HAHHAHHAH

@Knight :)

Wanna hear more Raps?

yeah man of course

XOXO

yo yo yo

No, please don't.

@Knight Let's fire this room then!

@AbhasKumarSinha what's the r.h.s. supposed to mean?

@ACuriousMind Why?

@ACuriousMind $\hat T$ operator applied on psi function

@AbhasKumarSinha ...and the difference to the l.h.s. is?

@AbhasKumarSinha Hahahaha. You can't sing now.

@ACuriousMind No, is LHS = RHS? that's what I'm asking, that's a simple one equation question!

@Knight He's playing his physics card!

@AbhasKumarSinha Who is starring all your messages?

@Knight Whoever, doing a big favour to me... I know him :P

@AbhasKumarSinha And I'm saying that I don't understand the question because I don't understand what you think the difference between the two sides is supposed to be

@ACuriousMind See carefully the reposed question then!

@AbhasKumarSinha I must tell you that you have got a golden chance to learn something from ACM, please don't let this opportunity go off, I shall talk to you a little later.

$$ \langle \psi | \hat T | psi \rangle = \langle \psi | \hat T \psi \rangle \tag{*}$$

@Knight Awkey!

@ACuriousMind I've applied $\hat T$ to psi on the RHS

Is $(*)$ correct?

To me these are just two different ways to write "Apply $T$ to $\lvert \psi\rangle$, then apply $\langle \psi\rvert$ to the result".

@ACuriousMind You are totally wrong then!

Writing the $T$ inside the ket is just an unusual stylistic choice.

@ACuriousMind Okay, wait for asecond

@ACuriousMind Do you know Heisenburg picture?

Heisenberg, but yes.

@ACuriousMind If these were according to you, then we would not have got third term inside the bracket =0, there in the picture.

That's only possible when $(*)$ is not true, then!

I don't understand how you arrived at that conclusion.

@ACuriousMind that operator A derived once wrt time....

@ACuriousMind Do you understand question?

@Knight I think ACM again got confused, let's continue our stuff!

@AbhasKumarSinha You can't say that.

@AbhasKumarSinha No. In the Heisenberg picture, the first two terms in that picture don't make sense because there the state is not time-dependent. So this must be in the Schrödinger picture, where you discard that third term if and only if the observable is not explicitly time-dependent.

@Knight Why?

@ACuriousMind I first asked you, if you understand my question or not.

You don't.

:(

@AbhasKumarSinha I believe in him. He is trying to answer you even when you are not willing to accept an answer.

@Knight I too believe him.

@FakeMod There are some people who are genius from birth, you are one of them!

@JohanLiebert @Knight @FakeMod yo yo yo, my Gang's back!!!

@AbhasKumarSinha We are almost nothing in front of ACM.

@Knight We 'individually' but when we combine, we are everything >:)

yo yo yo!

@FakeMod What do you think? Am I crazy? believe me, I've done ton of research for it.

ya

@FakeMod You are doubting my skills! *)

@AbhasKumarSinha I'm not sure what was unclear about me asking you to not start copy-pasting lyrics again.

@AbhasKumarSinha How much Linear Algebra and Hilbert Space you know?

@AbhasKumarSinha Your whole Gang is here. Even NC

@Knight gang yo yo yo!

@AbhasKumarSinha wher?

@ACuriousMind I agree

→ 9 messages moved to Trash

→ 1 message moved to Trash

I will agree with @ACuriousMind as well. I mean, @AbhasKumarSinha you can be weird sometimes

@ACuriousMind @Knight @FakeMod @NovaliumCompany good evening!

@AbhasKumarSinha This time you were ....

@JohanLiebert To you too bro

i demand to be greeted.

@NovaliumCompany I have greeted you.

Bye guys. @AbhasKumarSinha Leader just ping me if you want someone to be beaten, okay?

@JohanLiebert yeh, thanks! :D

@AbhasKumarSinha @NovaliumCompany @FakeMod yesterday it rained but today rather than being cold it was kind of warmer.

I can't wait 11th grade to be over... people say 12th grade is totally easy, teachers don't give a damn and it's completely unnecessary so... 11th grade ends and I'll finally be able to devote my time to what I wanna to.

bye

@NovaliumCompany contrary to it I think it is a bit tough. For me I found that while studying about the topics given in syllabus you can easily get into weird quantum stuff!

@JohanLiebert I'm not a physicist nor do I want particularly good grades so... that justifies my opionion. (and other students I know, here in Bulgaria)

@NovaliumCompany then it's OK!

"nearest major galaxy Andromeda is a staggering 2.573 million light-years away" Boy we never getting out of this solar system :( we either have to transfer our brains into computers or create AI to tell us smarter ways to do it lol

did you see my video link about AI?

@NovaliumCompany What has leaving our solar system to do with the distance to another galaxy?

A few man-made objects are already leaving our solar system.

@Loong My mistake, I mean, to leave the solar system and go somewhere else, another inhabitable planet or something

@skullpatrol I'll check it out now but that guy Fridman scares me

The video is too long, isn't there a summary or something

There is a table of contents.

and also that channel has a lot of good content

@skullpatrol I'm sure the video is interesting but the way both of them are speaking... I'll play this when I'm having a hard time falling asleep

lol

cya, pal

@NovaliumCompany Proxima Centauri b might be habitable and is "only" 4.25 lightyears away

What does "Supress Dialogs" mean when given as an option to leave a chatroom?

@ACuriousMind :O

I'm gonna be the first person to step on Centauri b

lemme invent light travel first, see u in a week

I hv a question

hv?

high voltage... duhh..

riiiight

It's a shocking question

deservant of a shocking answer:

I hv no answer

What is a shocking question?

a high voltage question

where is the question?

I want to talk something about Young's Double Slit experiment, anyone interested?

> Don't ask about asking, just ask.

askaway

If a slab of width $w$ is put after the slits and the refractive index of the slab is $\mu$ then what will happen?

@skullpatrol @ACuriousMind Please reply.

@Knight Don't ping random users to answer your questions.

@ACuriousMind You told me to ask. I just did that.

Especially not so briefly after having asked it. I was in fact already typing an answer.

hahahhahhahahha

I'm sorry. But why do you hate being pinged?

yeah, and don't call me "random" :P

@skullpatrol Yes. We are Pals, we are friends, we are everything to each other.

@Knight The sound notification can be quite annoying and if one doesn't react to the ping in a certain amount of time, it goes to the inbox on the main site.

oh I'm sorry. I think this pinging sound can even destroy the computer's sound system. I know you get pinged so many times in a day so it is quite frustrating to get pinged.

But in any case, you'll see the interference pattern "spread out", i.e. the distance between the bright spots will get larger, and increasingly so as you go outward.

How? Some people say that path difference gets larger because wavelength gets changed. Please explain.

Sir ACM are you busy somewhere else?

Light experiences a lateral shift depending on the incidence angle (cf. e.g. physics.stackexchange.com/q/199995/50583) that gets larger with increasing incidence angle.

How does this depend on the width of the slab?

The question (or rather its answers) I linked show exactly that

But it should be fairly intuitive that the thicker the slab, the larger the shift, since the path the light must travel inside the medium gets longer

The angle I have made with yellow is vertically opposite to $\theta_1$ but why he is writing it as $\theta_1 - \theta_2$?

It's not vertically opposite to $\theta_1$, look again

Yes. Aha!

But how it is $\theta_1 - \theta_2$?

The red line (inside the slab) divides the angle that is opposite to $\theta_1$. So the two angles adjacent to it together are $\theta_1$.

Yes. I have got it. So, what the lateral shift would do the double slit experiemnt?

Are you putting the slab behind one of the slits or behind both?

I'm putting the slab between the slits and the screen

Now that I think about it, I think I said it exactly the wrong way around earlier

The gray box that you see is the slab with some refractive index $\mu$ and width $w$

Sir ACM please reply.

Sir are you here?

The central maximum stays where it is, and the pattern "shrinks", i.e. the other maxima get closer together. This is because at spots beside the center you now get a longer path for the more angled incident ray compared to the less angled ray, i.e. the path difference at these spots increases.

Give people time to reply - they won't answer faster just because you asked a third time (also no need to call me 'Sir', I have never been knighted :P)

I request you to please explain this path difference thing a little more elaboratelt.

Dobur den na vsichki. Ako chetete tova znachi mojete da izpolzvate gugal transleitor.

Well, we have established that the path length the slab adds increases with the incidence angle, right?

yes .

At the central spot, this does nothing - the rays from the two slits have exactly the same angle

@NovaliumCompany Да аз съм

But at each other spot, one of the rays has a greater incidence angle than the other, and so its path length is increased more than that of the other by the slab.

Veche e vreme vsichki da govorim na tozi ezik - bulgarski. Hehehe mnogo e qko yess

@ACuriousMind Yes I agree. It's due to the isosceles triangle.

So the path difference between the two rays incident at that spot has increased compared to the situation without the slab

How can we analyse it mathematically?

@NovaliumCompany Sorry, please only English in here

@NovaliumCompany Какво е общата религия в България?

@Knight Well, you draw the rays I'm talking about and compute by how much their path difference has increased

Okay i'm drawing..

I can't find any place where that's worked out easily right now and I'm not very interested in doing it myself :P

@Knight Christianity

@NovaliumCompany oh

Where's the bend in the rays from the refraction?!

Warren Buffett

I don't know what has happened, how to refract them, they are just starting from the slab

my house looks like a wave pattern.

They will refract after they leave the slab, I think.

I'm Spongebob's neighbour.

AcuriousMind Is this one right?

@ACuriousMind Are you around?

That looks more useful, yes

Now, what we should do?

What do you want to do?

I mean, I don't know what quantity exactly you're looking for

That looks like a nice combination of squares and lines that I'm supposed to understand.

I want to derive this $path~difference ~= d\sin\theta + \mu(1-w)$

I want to how the thing $\mu (1-w)$ was born.

Wasn't $w$ the width and $\mu$ the refractive index?

Yes.

Then whatever that formula describes, it does not describe the situation you've explained to me here, since here, at the central spot ($\theta = 0$), we get a path difference of $\mu(1-w)$.

That's exactly what it says.

I too don't get it.

But that is obviously wrong here. The situation is completely symmetric about the central axis, there cannot be a path difference there. So either whatever source you got that from is wrong, or it is looking at a very different situation from this one.

Okay. So, how can derive something from this situation. I want to derive the equation for path difference in the situation that i have just described you.

The path difference is the difference in length between the two blue lines you've drawn. You now need to do some geometry to compute their lengths.

Please just order and I will draw each and very thing.

Sorry, I'm not really interested in doing this exercise myself

Okay, so just tell me how should I begin?

Hi. Any idea what's the correct way to normalize the one-particle distribution function ( in this example for a system under brownian dynamics)?

@ConstantineBlack Do you have any more context here? Or do you just want to normalize so that the integral over all space comes to 1?

Is it possible with 5G that we can literally get rid of our computers and phones and just have screens that can connect to the internet and our files, games, photos... everything that was before on the computer will be on the cloud and it will be all just fast communication?

Like, I pay 1$ a month for a company to literally "have my computer" on the cloud and my PC will be just a monitor with a mouse and keyboard and everything that I do will be just fast communication?

That way people won't have to setup computers in their homes blq blq, just pay a company 1$ a month and you have your own cloud computer which you can access by a simple send a receive device and a monitor (mouse and keyboard).

What do you guys think?

That way you literally won't have to pay thousands of dollars for a computer. You just pay initial setup fee and from there a monthly fee and you have your cloud computer.

It will be 10 times faster than now, and even now is quite fast.

@NovaliumCompany I do not wish access to all my data to be dependent on a) me paying a monthly fee and b) the company that hosts it not vanishing.

@ACuriousMind Then you simply buy a monitor, keyboard, mouse and a hard drive (or some storage device) and your worries are gone.

@NovaliumCompany What's faster is the total data transferred, but not the latency as such

It would be quite annoying to potentially have lag on every application

@ACuriousMind You won't? I mean, the latency is 10 faster.

And it would increase the lag for online gaming and such (once lag for the connection of the cloud computer to the server, then for the connection of the cloud computer to you)

"Latency is the time that elapses since we give an order on our device until the action occurs. In 5G the latency will be ten times less than in 4G, being able to perform remote actions in real time."

@ACuriousMind Wouldn't it be fast enough to not feel the lag? I mean... Google Stadia.

@NovaliumCompany Some people report latencies of 100 - 200 ms using Stadia, which is certainly enough to feel it

@NovaliumCompany ...but what do I plug the hard drive into?

@ACuriousMind The send/receive device.

But then I would still have to transfer my data to some company over the internet in order to access it

@ACuriousMind Not a big deal. Google Stadia have taken the risk.

@ACuriousMind So?

@NovaliumCompany I...do not want to have to send all my data over the internet when I want to do something with it?

@ACuriousMind The company will request only the data it needs.

@NovaliumCompany My point is that I cannot even display the data on my drive without first sending it back and forth if I don't have a computer anymore.

@ACuriousMind You can send back and forth only the data you need to send back and forth.

Yes, but I still need to send it! Can you not understand that there might be data I do not want to send to anyone else at all while displaying or editing it?

@ACuriousMind Why?

@NovaliumCompany Potential reasons include: 1. It's data that could have negative effects if the wrong people saw it (maybe I fear other people are stealing my ideas, maybe I have a secret double life, whatever...) 2. It's data I'm not legally allowed to possess (hello, people with drives full off illegally copied movies and music!)

The data will be encoded.

plus, anyone can hack into your computer right now. Data breach is not uncommon.

(Or maybe I just plain don't want anyone else to know that I have it. Nothing to hide, nothing to fear is a silly idea)

@NovaliumCompany The endpoint (i.e. the cloud computer at a location I do not have control over) still needs to decode it. How am I to know who has access to the decoded data?

No one else will know. The chances of stumbling upon a hacker so good enough to break a strong encoding system are slim and even right now it's possible for someone to hack your computer.

@NovaliumCompany I'm more afraid of the company I'm paying than any hacker attack in this scenario.

@ACuriousMind The company will be trustworthy... guarded building or something. There are ways to build trust. Google and Microsoft can literally send all your data to their severs right now without you knowing, and you trust them.

Microsoft could have implanted a secret sender in every Windows 10.

@NovaliumCompany The point is that if they did that, I (or people more paranoid than I am) could catch them doing that (the traffic is going through my network, after all). There was already plenty of distress over what data Microsoft telemetry collected without telling the user, and people found out! When I send my data to some cloud, no one can control what happens to the data there except the company.

You trust the company, just like you trust Microsoft and Google right now. (cuz let's be honest, those 2 companies alone can potentially have your darkest secrets)

(if you have dark secrets donno)

@NovaliumCompany Yes, sure - but I "trust" them because I also trust that they would be found out if they violated that trust.

There's no way for me or anyone else to check what happens in the cloud at a location we don't control, so there's no potential oversight there anymore, hence no trust from my part

You can still find out. I mean, what is the company going to do with the data you don't want to send?

When the company do something wrong with the data, then you'll know. Like publishing your naked photos online.

@NovaliumCompany They might just be using it in ways that don't require publishing it (e.g. personalizing ads or general profiling). And in the case of the illegal data, me finding it out because I'm being fined for possessing it is a tad too late.

@ACuriousMind Google, Microsoft and more could be using your data for their benefit without you knowing right now.

@NovaliumCompany Again, in that case I'm extremely certain the paranoid part of the hacker community would long have noticed if they uploaded random data from the hard drives of their users without consent.

A Curious Mind can you please give me just a hint about how to go with that?

@ACuriousMind Didn't understand that one sorry.

@NovaliumCompany You can't just smuggle data out of a network someone else controls unnoticed if that someone's looking for it.

So I'm reasonably confident that the files I have only put on my hard drive and nowhere else (i.e. never consciously sent them anywhere via the web) have, in fact, not left that hard drive.

@ACuriousMind Microsoft could have implanted a sender.

But you trust them.

I've explained at least twice now why I feel that this trust is justified in the case where the user physically controls the computer in a way it is not when they don't.

You just repeating that argument doesn't make me any more likely to accept it.

@ACuriousMind I'm sorry I guess I just don't understand what you are saying exactly.

Okay, let me try again: Yes, I agree that the companies whose software I'm using currently could have made their software upload my entire hard drive to their servers to do as they please with it.

And yes, I agree that I trust them not to do that.

But I don't trust them because I think they're good people or just because I can't see why they would do that

I trust them because there are very paranoid hackers who analyze very carefully what data companies like Microsoft are sending home, and if Microsoft sent entire hard drives home that would be a PR disaster for them. It is technically impossible to send data from a location the user physically controls without the user having ways to notice, so I trust the company doesn't steal that data because the potential of being found out far outweighs any benefits.

But when the data is sent to some cloud, the user has no way to notice what happens to it after it is decrypted because the company controls the entire information flow. So the potential of being found out isn't there anymore (unless you're stupid and just post your stolen data publicly)

Hence, my trust that they won't do that is far weaker.

I just realized that for the data to be in your local hard drive, it must have first been in the company.

ahh... it got even messier.

not to mention the tremendous power consumption of such a cloud solution

Internet data centres already use about 200 TWh per year.

:(

I say it's possible.

And doable.