@dmckee where? :: looks around ::

@dmckee Is there supposed to be a link under "here"?

@0celo7 lol, I never understood what people might need more than one or two pipes for

Er ... no, I was going to make it the next line ('cause there was some talk of one-boxing arXiv links...)

arxiv.org/abs/quant-ph/0202158

Nice box.

@ACuriousMind I can see 24 pipes from my current position.

Yeah, yeah...I didn't know if it had actually been implemented or not.

(They're on stands in the basement man cave.)

@0celo7 That's a lot of pipes.

@ACuriousMind I have an eccentric father.

Interferometry is awesome

Is he a pipe collector?

@StanShunpike Perhaps.

@0celo7 Not the worst thing to have :)

@DanielSank So really, the coin analogy is a good one, particularly when teaching newbies, because we actually do have a qubit that behaves like that --- spin. But in principle, if one could extract the information, the coin would be a qubit as well. It just isn't available for obvious reasons.

@ACuriousMind @bolbteppa I'm still not convinced that $dz\wedge d\bar z$ is not valid. There is nothing stopping me from writing $$dz\wedge d\bar z:=dz\otimes d\bar z-d\bar z\otimes d z$$

@0celo7 First, I have said thrice that I am not sure of what I said and you should find someone else ;) Second, the first summand there lives in $\bar \Lambda \otimes \Lambda$ and the second in $\Lambda \otimes \bar \Lambda$, as I understand it, so you can't write it in my understanding.

@ACuriousMind Hmm.

Fource is not enough. I still think you know it.

@StanShunpike I'm hesitant to agree because I'm not exactly sure what you mean here.

@ACuriousMind I will...make a post on Math.SE.

Unless you think we can get a string theorist in here.

@0celo7 I'm not stopping you

@0celo7 I don't think so

@ACuriousMind I was hoping for sympathy, not you stopping me.

Well, I just meant that I thought the coin example was a joke. An analogy only. And it is. But I thought that just meant coins werent quantum mechanical. So I just think its an interesting example because, it works both as an analogy but also as a demonstration for how complex our macroscopic quantum info world is.

@0celo7 Sympathy? What on earth is that?

@DanielSank it was more me just appreciating a new perspective not asking a question.

@ACuriousMind It's that thing that normies have when others do stuff like post on Math.SE.

@ACuriousMind "How do we wedge the complex differentials $\mathrm{d}z^i$ and $\mathrm{d}\bar z^{\bar j}$?"

Has anyone else found math chat inconsistently helpful?

^ I want to know this too.

dangit

@0celo7 I mean this.

@0celo7 I think you've got your question title there

@0celo7 I am 100% sure the link I posted explains this, I am too busy to get into the details right now but I will come back to you on this

Just ask on SE actually

@bolbteppa Currently writing the post.

I'd love some upvotes, oh dear h Bar.

@innisfree I don't understand what you wanted from my answer and what you want with Einstein's non-locality. There are many non-localities, the Newtonian mechanics is also non-local. But you can place a better answer referring to Einstein, to all the types of locality. And if the correlations are perfectly understood, maybe you'll tell by which forces act the particles measured at huge distances from one another. I am always glad to learn. An explanation like this is worth a Nobel prize.

LOL

All I saw was the last sentence when I came back to my screen

are u kidding?

@0celo7 sigh... ::logs in to math.SE::

@0celo7 wow mathworld.wolfram.com/ComplexForm.html

@ACuriousMind You've said yourself how bad Math.SE is...

Hi Sofia, I thought your answer was misleading, and misstates ideas about locality in physics and quantum mechanics. Einsteinain locality isn't an example of locality; it's a definition of locality.

@0celo7 Qiaochu's comment is weird - anticommutativity would not be a sensible requirement for the same reason we can't define the wedge by the tensor algebra ideal, imo

@0celo7 +1 homie

To be honest, I find your manner quite combative and defensive. I don't like your answer to a question on this site, and that's it. No big deal.

@0celo7 maybe you should add my example to your question and see how someone else works out the (1,1) part

I am so glad we don't have these problems...

@bolbteppa You still have the TEX code?

::logs out of math.SE::

@ACuriousMind Huh?

@0celo7 Oh, lol, wrong link :D

There

If you find my post from yesterday and click permalink and open in a new window you'll get it, sorry I have to get back to my work, but I think mathworld.wolfram.com/ComplexForm.html has an example of the C x T(m) question you asked us yesterday as well as a cool <dz/\dz*> notation

@bolbteppa Ugliest Lambda/Wedge I've ever seen right there :D

;)

@innisfree can you explain the mechanism of non-locality? You say that I don't understand. I am always glad to learn from experts.

@sofia, there is no non-locality in nature. there is no mechanism of non-locality,

^this.

@innisfree Black holes?

@innisfree ::looks intensely:: Have you solved the firewall paradox?

@ocelo7 hmmm

So tempted to use this picture in my explanation of scattering matrices... britneyspears.ac/physics/transmtx/transfermtx.htm

@ocelo7 you raise a good objection.

@bolbteppa lolwut

@innisfree would you be so kind, 1st of all, not to give me grades to my understanding in QM? I don't know how much an expert you are, but I have a few tens of years in the domain. So, 1st of all, would you kindly take away the offense?

@bolbteppa lmao

I'll settle for the cool one on page 4 here arxiv.org/pdf/1205.1318.pdf

@0celo7 Only a valid objection if you believe the semiclassical hocuspocus of QFT in curved spacetime is a good theory ;)

@ACuriousMind The horizon is nonlocal in pure GR.

I don't fully see the link between 1-D and 3-D quantum scattering tbh

@ACuriousMind It's not clear if string theory eliminates this nonlocality or if something else, deeper, is going on.

@0celo7 How? I mean, what does it mean for a surface to be nonlocal?

(My lack of GR is showing, I know^^)

@0celo7 So you were right. AFuriousMind is old.

@ACuriousMind It's nonlocal in the sense that you could be inside of a horizon right now and wouldn't know it, but a dude at Alpha Centauri could be freaking out that a black hole is eating you (4 years in the future ofc).

@sofia I think you have issues with criticism. my judgement about your answer on a Q and A website shouldn't bother you much. what is the problem?

There is also the issue if one sees Hawking radiation being produced at the event horizon as one falls through it.

@0celo7 Semiclassical hocuspocus again ;)

@sofia i'm sorry if my critiscism of your answer has upset you, but really you'll have to deal with it. fwiw it's a single answer, i've written bad answers, made mistakes etc in physics. i'm not building an evaulation of your potential or ability as a physicist from your answer.

@ACuriousMind YFW gravity turns out to be semiclassical.

@0celo7 I know. I'll humbly hang my head in shame then.

Then I'll get mad at nature for being so inconsistent.

@0celo7 twistor59 seems to disagree that black holes have to be non-local.

I don't even know what I just said.

Haha

I think it was an abstrusegoose reference ;)

@ACuriousMind I listened to a Carroll talk in late 2014 and he said we don't really know.

As much as you hate his QM interpretations stuff, he is a string cosmologist.

@0celo7 I hate to ask that question, but do you know whether the local/non-local descriptions predict anything different?

@ACuriousMind I don't know enough string theory to answer that question.

I.e. do we not know because we haven't had a black hole to experiment on or do we not know because it is a meaningless question?

@ACuriousMind Carroll stressed this. We can't test this stuff, because we'd have to make measurements right at the horizon.

@ACuriousMind This is also connected to the issue of information loss in a black hole. IMO, the idea that information simply isn't conserved seems like the best answer.

@0celo7 But that's not in principle impossible, right? It's just that we never see the guy making the experiment ever again, and he can't transmit data to us, right?

There are plenty of other things that seem like conserved quantities at first, but aren't.

@innisfree whatever I want is if you kindly take back the phrase that I don't understand what is non-locality. It is discrediting of me in the eyes of the user. It's big harm, because in the future this user will say Aaaa! Sofia - she doesn't know what she says. Shall I do you the same?

@innisfree shall I post there a comment that you don't understand what you say?

@0celo7 I think I read some SE post saying that there can't be a symmetry associated to information conservation. If you can't get Noether to conserve it, it's not conserved! ;)

@Sofia sorry to jump in here - tell me it's none of my business and I will shut up - but I think the majority of people here will not jump to conclusions based on a single thing that one user says about another. The body of work - insightful questions asked, well constructed answers given - will over time build your reputation. I think that fighting over a single comment will do more harm than good. Like Elsa sang "Let it go". If you can...

@Sofia Please, stop bringing this up every time someone disagrees with you. The hypothetical future user will be able to judge for themselves. It's the Internet, people know better than to go by a single example.

@ACuriousMind Yeah, the problem with nonlocality is that you really have no idea how to get to the horizon. In principle, you just look at planetary orbits to find the mass and thus the radius. But really getting there is hard -- you can't communicate with your base back home because your signals get redshifted to oblivion. Suppose further that you accidentally cross the horizon. You have about $\pi M$ seconds before kaputt.

@alarge - you and I both, eh?

@sofia this is getting petty and childish. do as you please, i won't reciprocate/participate in any destructive/unsociable behaviour. i think your answer is wrong and that's the end of it.

@0celo7 Oh, well. Gotta love modern physics. Is there anything we can test anymore? :D

@ACuriousMind We really need a guy to somehow get to the horizon and tell us if it's hot or cold there.

Hawking radiation is actually colder than the CMB, so we have no chance in hell of measuring it from where we are.

Some physicist will destroy the world just by creating a black hole in their lab to finally measure Hawking radiation.

@ACuriousMind Carroll mentioned that micro black holes could provide some insight.

@Floris You were faster, as you usually are (more than once have I seen an interesting fresh question only to find that you've already answered it).

@ACuriousMind I don't think we have any evidence of semiclassical gravity.

@0celo7 Ah, if they'd "evaporate", we'd at least know that there is Hawking radiation, right? (How would we know there was a small black hole and not just a very energetic particle there, though?)

@0celo7 That's my impression, too.

@ACuriousMind Maybe we'd find some massless spin-2 particles whizzing around in the aftermath.

@0celo7 That'd be nice.

@ACuriousMind As Sofia says, that would be Nobel worthy.

@ACuriousMind I just glanced through BBS, and it seems Hawking radiation can be at least motivated by string theory, but that's about all I understood in that section.

@innisfree discrediting is childish? I only asked that you take back the insult that I don't understand non-locality. It's rude.

@Sofia It is childish to get that upset about someone saying they think your answer is wrong, yes.

@ACuriousMind the comment said that I don't understand non-locality. Please don't distort the issue.

@Sofia Well, you told the user asking the question "I don't want to offend you, but you don't understand non-locality.".

Innisfree indicated that they agreed with the user's comment on your answer, and therefore asked you if it could be you who doesn't understand non-locality

If your initial remark wasn't rude, then neither was innisfree's

Conversely, if your initial remark was rude, then you have no business being upset about receiving rude retorts.

@ACuriousMind I made a god damn fool of myself. I told my physics teacher you need both GR and SR to calculate the time dilation of the ISS, but when I sat down to actually do it I remembered as long as I assume the ISS is on a geodesic in the proper time formula, GR perfectly incorporates SR time dilation due to motion.

Good thing she has no clue about GR.

@0celo7 Are you gonna tell her? ;)

@ACuriousMind Aaaa! I can be offended, that we conveniently omit, but if I ask someone whether it cannot be opposite about non-understanding, then there comes the supreme judge inisfree to punish me. Now I understand.

@ACuriousMind Maybe. I'm gonna work out the whole thing and show her. I have to first find the proper time of a dude at sea level stuck to the ground, then of the ISS in orbit.

I have to solve the geodesic equation.

Oh lord no.

::eye twitches:: Christoffel symbols, we meet again.

@ACuriousMind Is $e^{i\theta}$ a bounded linear operator? If so, how do we show it is bounded?

@Sofia I have no idea what you just said there. No one is "punishing" you.

@StanShunpike What's $\theta$?

@StanShunpike If $\theta$ is hermitian, I say yes cautiously.

Um, does the definition of $\theta$ matter? Let's suppose its a real number. That's the example Wikipedia uses

@ACuriousMind of course you have no idea. Someone says that of course the entanglement is local because the particles knew one another in the past, so, of course the correlations are local, and I don't understand non-locality. Not only is wrong, it's also impertinence. If I'll say that 1+1=2 someone would come and say that it's otherwise and I don't know math.

@StanShunpike So the operator is just multiplication with a constant number? Of course that's bounded, because $c\cdot v$ always has smaller norm than, for example, $2c\cdot v$. so $2c$ is an upper bound.

@ACuriousMind but if you don't know, why do you get into the talk? Just for explaining me that I have to accept to be rudely offended, otherwise it's childish? So much you are against me? That is good to know.

@Sofia You are completely misunderstanding what I said. The "I have no idea what you said there" was meant to your last remark here in chat. I did, so far, not comment on the factual accuracy of either yours or innisfree's position regarding locality.

Oh, lol. Okay. What if $\theta = i\delta t \hat{H}/\hbar}$? would it then depend on the boundedness of $\hat{H}$?

@ACuriousMind if you don't know, why do you rise against me? Why are you against me?

For whatever it is worth, though, innisfree is correct - thanks to the no-communication theorem and others, Einsteinian locality is preserved by quantum theories, and it is an explicit requirement for all quantum field theories.

@ACuriousMind Derping. $2\pi/\omega=$?

$T$?

@StanShunpike No, every unitary operator is bounded, so as long as $H$ is properly self-adjoint, it's bounded

@0celo7 Uh, I think so.

@ACuriousMind Time to figure out why I got 5061s as the length of a day...

@0celo7 That's why my days feel so short lately! You GR people are playing with the fabric of spacetime again!

@ACuriousMind do you have any experience with several complex variables in your study of formal QFT, e.g. Bogolioubov's edge of the wedge theorem, or correlation functions with SVC techniques?

@0celo7: You remember the Majorana stuff? Someone just downvoted my answer there, and I hoped we'd get a more authoritative correction. Then the downvote disappeared again, No comment. I was actually glad about that downvote, anticipating the comment.

@ACuriousMind Yes that was an interesting question, although I fully agreed with your answer.

@bolbteppa No, sorry. I'm not that versed in constructive QFT, though I am working to change that.

Okay cool, it's a good motivation for SVC anyway

@ACuriousMind I have no idea what I did, GR does infact predict the correct length of the day.

SVC?

several complex variables