The h Bar

5:37 AM

Last night dream, a micro blackhole is formed in a shopping centre and slowly grows as surrounding get pulled in. The whole dream repeated once, and both times, I fell into the black hole, get speghettified and died as everything turns black after feeling my whole body is being stretched in less than a second

Other than certain death where the black hole grew from the southern end of the city to the northern end, it appears that throughout the city, micro black holes keep popping into existence, merged and then grew. However, by firing some kind of negative energy beam to the position before the hole grew bigger than a baseball, it can be shrunk and removed. There's also one park scene where I felt the ground goes up in a ramp, and then there's a massive drop and then go up in a ramp again

probably due to gravitational waves sent out as the huge black hole merged with other small ones (often became building sized) nearby

This scene however, makes me wonder about a theoretical question: Is it possible to surf on the peaks of gravitational waves radiating out from a binary and hence get a boost in acceleration away from the binary?

dsm

6:14 AM

Merzbacher is a hundred times better than Sakurai -- wish I had been crossreading this the whole time.

at least the later chapters.

Anonymous

9:54 AM

I need some LA halp! Any idea why the OP's question paper asks them to find the square root of $\operatorname{CNOT}_{12}$ by writing it as $H_2Z_{12}H_2$ first? $H_2$ and $Z_{12}$ don't seem to commute and are thus not simultaneously diagonalizable.

Anonymous

So even if they manage to find a spectral decomposition $Q\Lambda Q^{-1}$ for $H_2$, and get to $Q\Lambda Q^{-1}Z_{12}Q\Lambda Q^{-1}$, unless $Q^{-1}Z_{12}Q$ is a diagonal matrix, they won't be able to use the method of "taking square roots of the diagonal elements" to reach the final answer i.e. $\sqrt{H_2Z_{12}H_2}$.

@Blue Is $H_2$ by any chance self-inverse or self-adjoint or unitary?

Anonymous

@ACuriousMind Yep, it's a quantum gate i.e. $I\otimes H$ ($H$ is the Hadamard gate)!

Hmpf, not obvious enough for me on a Sunday morning ;)

Anonymous

10:10 AM

@ACuriousMind Well, so it's basically unitary... :)

Anonymous

Does that mean something?

No, I mean that the answer to your question is not obvious to me :P

But I wouldn't discount it being a mistake in the exercise or OP misunderstanding the exercise

Anonymous

Same here. I have absolutely no idea why the OP is expected to go about finding the square root of CNOT in such a roundabout fashion. I'd probably just directly diagonalize the CNOT. :/

It's just a 4-by-4 matrix, right?

Shouldn't need a trick to diagonalize that

Anonymous

Yep, $\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&0&1\\0&0&1&0\end{bmatrix}$

10:16 AM

ahhhh

The hint is probably meant to get you to realize it's block-diagonal so you just need to diagonalize the two 2-by-2 diagonal blocks

Is the $Z_{12}$ a tensor product, too?

Anonymous

$Z_{12}=\begin{bmatrix}1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&-1\end{bmatrix}$ (it's a controlled Z gate)

Anonymous

@ACuriousMind Ah, I forgot how diagonalization works for the blocks and how that translates back to the diagonalization of the whole matrix...gotta revise...

Anonymous

I don't think we can write $Z_{12}$ as a tensor product directly, but we can use the outer product representation. It goes like:

Not, it's indeed not a tensor product (the lower block would have to be a multiple of the upper block for that)

Anonymous

$|0\rangle\langle0|\otimes I + |1\rangle\langle 1|\otimes Z$

Anonymous

10:27 AM

@ACuriousMind Oh, right

Anonymous

10:42 AM

Anyway, so the point is that $\operatorname{CNOT}$ can be simply block diagonalized, if I can diagonalize the individual $2\times 2$ blocks, and that's probably the simplest way to proceed. It's be like this: $\begin{pmatrix}P&0\\0&Q\end{pmatrix}\begin{pmatrix}\Lambda&0\\0&\Lambda'\end{pmatrix}\begin{pmatrix}P^{-1}&0\\0&Q^{-1}\end{pmatrix}$ where $P$ is the diagonalizing matrix for the upper left block and $Q$ is the diagonalizing matrix for the lower right block.

Q: Would photons "riding" a gravitational wave appear different to an observer?

11:34 AM

1

Gravitational waves travel at the speed of light. A photon travelling in exactly the same direction as a gravitational wave will therefore remain in exactly the same position relative to the wave - at the peak, the trough, or somewhere in between. So my question is: if we're observing light emitt...

general-relativity electromagnetic-radiation photons spacetime gravitational-waves

I wonder what it will be like to ride on the wavefront of a gravitational wave

user351417

11:45 AM

Well, we finally have a nomination from an involved user of Physics SE.

user351417

0

A: 2019 Moderator Election Q&A - Questionnaire

Describe the current challenges specific to the PhysicsSE community. Largely floods of low-quality questions, especially homework questions. How much time do you expect you can commit to moderating the site? (A couple hours a month? Ten hours a week? Ten hours a day? Take a good guess) A...

Still about 32 hours to go, it's not over yet ;)

Captain Bohemian

@Secret what a dream! I never feel it's likely that I would fall into a black hole, and have never made that kind of dream. Nevertheless, I feel I have fallen into a region where my time flows differently from it in the outside world, so that I often feel confusion of time and made related dreams---often I feel the time observed by the outside world just crazily races so that I let unusual quantity of time lost without rationalization.

user351417

Frankly, I wasn't particularly fond of a lot of those answers to the questionnaire, but I do think that Chris'd be a nice mod.

Captain Bohemian

I can't remember clearly what I dreamt today; it's like I am together with some people I may not know in reality on a bridge.

user351417

12:06 PM

1

A: Why we can't write in air?

I write in air all the time, it is quite feasible. I also play air guitar.

user351417

Yikes. And it got 3 upvotes, presumably because of the HNQ effect.

How have 5 people voted on that answer and no one flagged it?

user351417

I did a downvote.

user351417

But I don't know if that qualifies as NAA

user351417

Technically, it answers the question. In a sarcastic, unhelpful way.

12:08 PM

@Chair It's a joke answer.

Also, that question is HNQ?

::sigh::

Thoughts on whether it should stay there?

user351417

Yep. But I don't think that the close vote makes sense.

@Chair These days, mods can remove HNQ status without closing

user351417

Obviously, it's a poor question and I wouldn't like to see it as an HNQ, but I can't see any good solid reason why it's invalid.

user351417

@ACuriousMind Yep, I had a terrible suggestion under rob's meta post about the conditions under which our mods should kick questions = )

user351417

But I don't think any of the conditions described there are really relevant here.

12:10 PM

they aren't, but that could as well indicate that the conditions there are incomplete

user351417

It does indicate that the conditions are incomplete.

I think I don't dislike the question so much because of its topic but because of its vagueness and effortlessness. What does OP mean by "write in air"? Are they imagining moving a pencil and leaving a trace of graphite suspended in air?

user351417

Aah well I somehow automatically thought about pens, not pencils. The answer to why pens don't work in the air should be easy to google: it's as simple as searching for "how do ball-point pens work" I just did that and found a rather nice article which answers the PSE question perfectly.

user351417

But yes, they're presumably thinking of a trail of graphite/ink in the air.

@Chair So, when they ask when we "can't write in air", do they mean why the writing implement doesn't emit the ink/graphite/whatever or why it doesn't stay there? The former doesn't seem to be a question about physics or "writing", but about the specific implement used, the latter should be so obvious that I can't imagine anyone asking it.

user351417

12:18 PM

> So, when they ask when we "can't write in air", do they mean why the writing implement doesn't emit the ink/graphite/whatever or why it doesn't stay there?

user351417

Yes, presumably?

-.-'

user351417

I guess it's eventually a question about how pens work, but it's a pretty strange question to be asking.

user351417

It would be so much easier to just google that.

user351417

And then there are the answers which so deliberately miss the point. I'm thinking about the one which mentions skywriting with planes.

user351417

12:20 PM

(physics.stackexchange.com/a/476556)

@Chair Well, the question is so vague that I'm not even sure they are missing the point!

Maybe OP will be like "Wow, didn't know about that" and accept it!

user351417

Anything can happen with HNQs.

Anonymous

12:45 PM

@Chair Yeah. Hopefully he'll be able to pick up things over time. It's unfortunate that this time almost none of our more experienced users are interested.

user351417

@Blue Oh, I think he's experienced enough. He's done plenty of reviews on the main site and he has a decent amount of meta participation. I imagine that he falls right in that bracket of ~10k rep users who've been around for 1-3 years.

With this example, why is it that the total number of micro states doesn’t take into account that each molecule removes an available state?

user351417

It's just that a lot of the answers are... standard and snappy. There's nothing technically wrong with that.

I.e why is it not N!

user301074

2:05 PM

wtf... physics.stackexchange.com/questions/476544/…

user407574

2:33 PM

@Chair would you please have a look at my question? It is active now "why we can't write in air?" I have edited it.

@Infinitesimal If you agree with svavil, you should edit in that you mean writing on paper in the air, that's different than just "writing on air"

@Infinitesimal Uhh...if that's what you mean, none of the existing answers actually answer the question. Edits that invalidate existing answers are usually not allowed.

This is what happens when people answer an unclear question.

fixing the question will invalidate upvoted answers, but closing it while rejecting edits that make it well-defined also seems unjustified

user407574

@ACuriousMind Would you help me to answer the question?

user407574

@JMac I have edited it.

@Infinitesimal So wait, do you want to know how to write in the air without any paper, or do you want to write in the air on a piece of paper?

user407574

2:48 PM

@JMac I want to write in air without any paper.

@Infinitesimal Okay, that's confusing, because in the comments you agree with svavil, but he was talking about using paper

user351417

@Zober I don't understand that... you appear to have linked a deleted comment.

user407574

@JMac I edited it now.

user407574

@ACuriousMind please look at the question again

user407574

@JohnRennie please have a look at my question.

3:00 PM

@ACuriousMind Just curious - was this account removed because of age concerns?

user351417

Frankly, that user didn't sound like a 12-year-old.

@Infinitesimal I'm looking at it. Why should the ink stay in the air? What's supposed to be different about waving a pen than just raising a ball into the air and letting go of it? Are you confused about why the ball falls to the ground instead of hanging in the air, too?

@Infinitesimal This type of indiscriminate pinging is not appropriate on this chat room, unless you have had previous interactions with that user about the subject, or very specific reasons to believe that they will be interested.

user351417

More generally, I think plenty of users will see the physics SE font page; whenever you make an edit to your post, it will be bumped to the top there. So frequently pinging users in chat isn't necessary: they'll open it on that page if they're interested in the edits which are being made to make the question clear.

@EmilioPisanty That's between the user and SE. (It's possible for moderators to find out, but given that it is rather difficult without a userscript or manual URL manipulation, I think it's not intended to be public information)

user407574

3:04 PM

@ACuriousMind I agree with @Chair who commented well on the question.

And before that you agreed with svavil who said something completely different.

@ACuriousMind ok, thanks.

user351417

3:19 PM

Hmm, it is sometimes surprising when starts chatting actively just after crossing the 20-rep requirement.

yeah, the account is gone now too :S

user351417

@Chair Eh I'm missing the word "someone".

user351417

Or should that read "you're missing..."?

user351417

@Chair Replying to oneself never fails to feel interesting.

dammit, @knzhou, learn to queue. I was in line to set a bounty to the book-bending question.

user351417

3:30 PM

I'm the one you informed though: you didn't tell knzhou.

user351417

(Or I don't think you did)

user351417

It's unfortunate that one can't put bounties on questions which have been marked as duplicates though. Is there anything one can do in situations where there's a bounty-worthy answer?

@Chair oh, no, indeed I hadn't

@EmilioPisanty Just curious - why did you mention that account just now?

user351417

@ACuriousMind Eeek this is going to spark off a lot of speculation.

3:33 PM

@EmilioPisanty ha, I'll catch up to you in bounties awarded yet!

@Chair still, as someone living in the UK, Kevin should've caught up with detecting invisible queues

@knzhou you want to give away 30% of your rep?

I guess your privileges would still be safe

Oof, now that you put it that way, I dunno.

@knzhou Two high-rep users fighting over who can give away more rep? Now that's a race to the bottom I can get behind!

I'm kind of tempted to hit 50k and just retire.

Or maybe donate it all as bounties.

user351417

@ACuriousMind Mess with me. I'm on -750 for the last month.

3:35 PM

@knzhou I have to say, giving away more than, say, 3k rep, as bounties, takes a surprising amount of dedication

Because if I don't stop at 50k, I'm probably gonna gun for 100k, and there goes half my PhD right there!

I'm currently at 5.5k rep given away.

@ACuriousMind I was reviewing my old flags, which sent me to this answer. I'm trying to remember what the flagged comment said, but I don't care that much either.

@knzhou there's other gamifiable metrics beyond raw total rep

total offered bounty rep being one of them

true. I won't quit until I have the QM gold badge, too.

you have a decent chance at hitting a good second place there

how do you check?

3:39 PM

@EmilioPisanty it was just run-of-the-mill rudeness (like "think before saying something")

@knzhou the scoreboard on total rep offered?

SEDE

lemme find the link

@ACuriousMind ah, yes. thx.

user351417

There were three comments saying that same phrase ("think before...")

@knzhou Psh, QM gold is easy. Go for the QFT one

@ACuriousMind see, talk like this is why I'll never be able to quit!

@knzhou Users by total bounty given

you're in fourth place as of the current cut

3:41 PM

@Rishi Photographic memory, wayback machine, or giant folder full of SE screenshots?

wow, that's remarkable. the 3rd place guy I've never heard of, and gave just about all his rep away...

@knzhou that's not unheard of

user351417

@ACuriousMind My head takes screenshots of punctuation. I have no idea why.

there's a similar case on either maths.se or MO, but with a much higher total rep

@knzhou Mostly on his own questions, though - not that there is anything wrong with that.

3:44 PM

remarkable, John Rennie is in the top 10, but his total rep is so high that the rep given as bounties rounds to "0%".

user351417

meta.stackexchange.com/users/165773/… gives away a lot of bounties.

user351417

Looks like it'd be about 90% of the total earned.

user351417

But when you've earned 43k, 90% is quite a bit.

@Rishi this is Meta we're talking about, though

user351417

Yeah, I guess it isn't a real site, so earning rep is quite different.

3:46 PM

@Rishi try the same metrics for that user on his top non-meta sites

user351417

@EmilioPisanty Yep, it's much smaller for his top site (programmers)

@knzhou I ought to give away more bounties really. The trouble is that 300k is getting tantalisingly close :-)

@JohnRennie c'mon, John, bounty away

you know you should.

@EmilioPisanty didn't I give you a big bounty once? Maybe we could just cycle rep between us :-)

A: Why won't you spend your reputation (on bounties)?

@JohnRennie indeed you did, and no, we shouldn't.

I've argued about this before

7

I'm the anomaly that rob pointed out in the comments, so I'll weigh in. As far as I'm concerned, any rep above 25k is fairly gratuitous, given that there are no more privileges to be earned (despite some excellent proposals for the 30k tier), and it might as well be spent constructively by award...

but as a recap:

- To incentivize the right types of questions, i.e. the types of in-depth, well-thought-out, highly non-trivial (but still within the realm of the answerable) questions that, because of a narrower subject matter or some other factor, are less likely to attract upvotes or answers.
- To reward the highly-detailed, high-quality answers that we all wish got a ton of votes but (again through e.g. a narrow subject matter) don't.
- To give users with a record of high-quality community moderation and content a faster access to the moderation privileges that will make them most effective at helping …

user351417

3:52 PM

@EmilioPisanty The paper-folding question got 200 votes (though the first ~180 were probably rep-cap-restrained). And we're still fighting to put bounties on that.

in Discussion on answer by tfb: Why do we bend a book to keep it straight?, Apr 18 at 16:41, by Emilio Pisanty

I need to join this chorus. This is an outstanding piece of science communication. Congratulations!

is why

The trouble is identifying candidates for the bounties. So much of my time is being taken up by the Problem solving chat that I don't have a lot of time for the main site any more. I don't have the time to review the days answers looking for particukarly good ones.

user351417

@EmilioPisanty Wait, if you gave my Bell telescope answer a bounty for reason 3, I'll actually be irked. I thought it was a cool thing to write about.

@Rishi ah, I just understood what happened. I think you've made a good choice.

user351417

@EmilioPisanty Yes, I'm sure all of us see comments saying "Congratulations!" getting ~10 upvotes every day -.-

3:56 PM

@Rishi It's never reason 3 on its own. It guides the emphasis, but if the answer isn't worth a bounty then I don't set it.

user351417

@EmilioPisanty Wait, what choice?

user351417

@EmilioPisanty Oh thenx = )

@Rishi name change.

user351417

@EmilioPisanty Oh yes. I've gotten bored of Chair. It was a meme in 11th grade, but I'm done with that. Rishi is more convenient; I get the point when people direct comments at me.

@ACuriousMind what've you got against my group leader?

he is the cure to every problem.

3:59 PM

Can anyone recommend interesting science blogs?

user351417

I got somewhat hyper and irrational, so I decided to name my flash drive "Chair". And then I named my PC Chair just because it was weird. I think I created my SE account under my real name, and I switched to Chair after a week or so.

Part of my Sunday afternoon relaxation is reading my favourite science blogs, but a lot of them have gone quiet or got uninteresting for various reasons.

user351417

@JohnRennie I like Sean Carroll's, but it's abandoned right now.

user351417

So I just read old things, some of which I barely understand.

@Rishi yes, that has happened to a lot of my favourite blogs.

4:00 PM

(joking aside, there's plenty of hilarious moments if the PI's initials match one of the group's research topics)

I guess people just run out of things to talk about.

user351417

@JohnRennie He has an article about Physics SE; it has comments by David Z : P

user351417

preposterousuniverse.com/blog/2011/01/13/physics-stack-exchange

2011! Gosh it's been a long journey.

What a long strange trip it's been.

@JohnRennie I frequent Shtetl-Optimized (by Scott Aaronson) and In The Pipeline, by Derek Lowe

the density of hilarious posts on the latter is relatively low, but it's still worth it

and if you haven't gone through the full Things I Won't Work With column, that's your Sunday evening right there.

4:05 PM

I've read Shtetl-Optimized occasionally but he tends to talk about quantum computing rather than physics so I haven't become a devoted fan. I'll have a look at In the Pipline. Thanks :-)

@EmilioPisanty is that still going? I've read a lot of excellent articles on it but for some reason I thought it had gone quiet.

it's pretty quiet nowadays

“Sand Won’t Save You This Time” was the classic post :-)

I think I'd just read the book on rocket fuels when I read the “Sand Won’t Save You This Time” article and that made it even more enjoyable.

user351417

@EmilioPisanty Was Chris the person (or one of the people) you were thinking of when you were talking about users with ~10k rep who'd make good mods? I thought JMac would be a good mod, but IIRC he mentioned that he wasn't interested.

user351417

@JohnRennie Chem SE noticed our string of Ignition!-related posts.

@Rishi :-)

4:10 PM

@Rishi which Chris?

user351417

@EmilioPisanty Um... the one with a nomination up?

ah, there's a nomination? good.

user351417

Yep, I'll take you're response as a 'no'.

user351417

But he looks like a pretty good candidate.

@Rishi I had a broad population in mind, more than a set of specific users.

user351417

4:14 PM

user351417

@Rishi My friend (who's a biochem major) just sent me this; I don't know what to make of it.

The bartender should be asking Why is your face Terrell rotated, though admittedly that kind of spoils the joke :-)

user351417

@EmilioPisanty Aah. You probably didn't see that conversation with ACM and I speculating about whether or not you were deliberately avoiding around a name to avoid pressurizing them.

user351417

@JohnRennie There's a more fundamental problem: if you're moving at 99% the speed of light with respect to the bartender, you probably won't be walking.

@Rishi well, depends

4:19 PM

I'm currently suffering post-huge lunch sleepiness. I think I'll clock off and settle into my armchair to read this week's New Scientist.

gamelab.mit.edu/games/a-slower-speed-of-light

2

well worth a look if you haven't seen it

@CaptainBohemian Actually, that's not the first time I fell into a black hole in dreams. Why I kept on dreaming about falling into black holes I have no idea

Anonymous

@ACuriousMind Heh, we were overthinking it. I think I've got it now (check the update).

user351417

@EmilioPisanty Oooh that's really cool! Thanks = )

Anonymous

The $Z_{12}$ is a really special case. It's a controlled phase gate gate, so the square root can be easily determined.

4:25 PM

@Blue Well, you didn't tell me $H_2$ was self-inverse! That's much more than "basically unitary" :P

Anonymous

Yeah, I sorta missed it that time. The Hadamard is an interesting gate i.e. $H.H=\Bbb I$ from which it follows that $H_2.H_2=\Bbb I$ where $H_2=I\otimes H$. :P

Anonymous

Anyway, so it was much ado about nothing. I was trying too think too hard about various matrix decomposition methods and this one skipped me.

Anonymous

At least I can sleep peacefully now. :P

user351417

I'm surprised by the fact that a reasonably large proportion of people in my grade want to do astrophysics research.

user351417

I always thought that HEP was the glamorous one which everyone would kill for.

Anonymous

4:38 PM

@Rishi Eh, how do you even define glamourous? :P

user351417

@Blue Flashy?

user351417

Capital-intensive?

Anonymous

::shrugs::

Anonymous

I'd choose astrophysics (especially if it's theoretical cosmology) over HEP anyday.

Anonymous

@Rishi Ah, that's a different measure.

Anonymous

4:42 PM

With CERN and particle accelerators stuff, it is capital-intensive indeed. But more fun? Dunno.

user351417

@Blue In the context, I think it's related.

Anonymous

@Rishi BTW heh, I thought you're a new user. If you're changing to your real name, why not use your full name? :D

user351417

@Blue Hehehe. I considered my full name, but then I decided I don't want it to be too searchable. I haven't told many people about my SE account (and i left it out of college apps), and I think I'll switch to my full name once I start contributing more actively, and I've gotten some nice answers written down.

user351417

However, I don't want a pseudonym because it's just confusing, and a real name makes one look like a more normal human.

Anonymous

Ah, fair enuff...

user351417

4:50 PM

And I got bored of Chair....

Anonymous

I love my pseudonym, but I changed it to my real name on QCSE (a couple weeks back) to see how much time the search engines will take to crawl my user profile. I see it has already crawled, but it's still on the second page (while a 2 year old abandoned account with 1 rep ranks higher). ;)

user351417

So you're switching QC back to Blue after a while?

sciencedaily.com/releases/2019/04/190419165531.htm

That's a really cool way to use a peltier element

Anonymous

@Rishi Maybe, dunno. Depends on my mood swings and whims. :P

sciencedaily.com/releases/2019/04/190424153446.htm

${}^{124}Xe$ has half life longer than the universe, observed to decay

user351417

4:59 PM

Oooh tpg put up a moderator nomination!

I have a question that asks to find the lowest frequency of allowed transitions of a material consisting of a cubic crystal with an electron in each, which is impinged by an electric field. We have discussed magnetic field Hamiltonians, but how do I handle electric fields?

Anonymous

@Rishi Yay!

@Rishi The second place in the last election was a rather close affair between rob, tpg and Alfred, so one would assume their chances are still good

user351417

5:14 PM

@ACuriousMind I tried to view those using the open STV software a couple of days back, but I couldn't manage it.

user351417

I agree with your comment under tpg's questionnaire though... there doesn't seem to be much hostility in chat these days.

@Rishi I've never tried to use the software

user351417

Good for you : P It's a pain.

user351417

I hate technology way too much for someone who spends so much time on his computer.

@Rishi "Toxicity" isn't exactly the same as hostility, although hostility is usually always "toxic"

user351417

5:18 PM

Ah, yes. But there don't seem to be many toxic comments either...

I sure hope so!

Anonymous

5:32 PM

@Rishi This chat truly used to be toxic once upon a time. I can totally understand where tpg is coming from. :(

Anonymous

And now although the toxicity has reduced, we've lost out on several members (for whatever reason).

user351417

@Blue I think I first saw chat when I noticed a meta post by a user regarding a long chat suspension.

user351417

I think that was before I really knew what meta was though : P

Random thoughts about entanglement:

I think I am starting to understand why the particles have to once interact together before they can became entangled:

Let $\lvert 1 \rangle$ and $\lvert 2 \rangle$ be two subsystems

Initially, the system can be described by the product state $\lvert 1 \rangle \lvert 2 \rangle$ and evolved under some hamitonian following the Schroedinger equation

@Rishi I lurked for about a month before I started posting on meta. It's really not the most intuitive of systems for new users

5:45 PM

When the two subsystem are brought close enough to interact, the interaction changes the hamitonian needed to describe the evolution. This hamitonian may have off diagonal terms, thus allowing the product state to rotate in one of the many entangled states $\sum_{ab} c_{ab}\lvert ab \rangle$

Mostly, I think, because so few other sites work quite like SE.

Now, when the subsystems are physically separated again, nothing happens to their state, thus they stay correlated, hence the entanglement has no dependence on separation

Only when another interaction occurs causing them to rotate out into some other state will the entanglement be broken. The environment can easily do this by entangled with the subsystem such that the two subsystem are no longer in superposition

Therefore, the reason why entangled states cannot be broken by a local unitary transformation is because only one of the subsystem is rotated in hilbert space, thus it does not go to a product state. It will take a unitary interaction encompassing both subsystems to break the entanglement, which such interaction is necessary nonlocal if the subsystem is spacelike separated

But that brought out an interesting observation: When we measure one member of an entangled state, the state get projected out into one of the eigenstates, and thus the system is no longer entangled. Does that mean the interaction caused by measurement is somehow nonlocal?

it is clear that measurement is nonunitary because the wavefunction get projected out, but is it also nonlocal since the whole system get affected?

@Secret questions about whether or not something is "nonlocal" are always questions about interpretation, not about the formalism itself. In interpretations with instantaneous collapse, this phenomenon is certainly non-local. In interpretations in which the "projection into the eigenstate" is modeled differently, it may be different

I see

Chuck

6:46 PM

Hello, I've got a problem I'm trying to solve and I'm trying to do my due diligence in researching existing approaches to the problem, but I'm not sure what this class of problem is commonly referred to.

I have a load that is on a pulley and I'm trying to do motion control by varying the line tension.

I've found endless pulley problems, but any mention of line tension invariably brings up homework-style questions about how pulleys act as force multipliers.

My ropes are not parallel; they are actually quite far apart angularly speaking, close to 45 degrees. This is more like a zip line than a traditionally pulley scheme, but zip line problems don't return any help.

bolbteppa

I am kind of shocked at how much effort has to be put into even establishing the method of partial waves

Partial wave analysis

Partial wave analysis, in the context of quantum mechanics, refers to a technique for solving scattering problems by decomposing each wave into its constituent angular momentum components and solving using boundary conditions. == Preliminary scattering theory == The following description follows the canonical way of introducing elementary scattering theory. A steady beam of particles scatters off a spherically symmetric potential V ( r ) {\displaystyle V(r)} , which is short ranged so that for large distances ...

Anonymous

@Chuck I'm not quite sure about the setup you've in mind but it sounds like the Capstan equation might be relevant.

Anonymous

You might want to frame as a proper question with diagrams and ask it on the main site; we don't have many mechanics experts in chat. :)

Chuck

@Blue Thanks for the help. I'm not sure where the question fits best; here or Math or Engineering.

Anonymous

Certainly not Math. I'd go with Physics.

Chuck

7:01 PM

I'm also not necessarily looking for an answer to the problem, just what this kind of problem is called. I don't mind doing research or reading up for myself.

7:37 PM

In a Quantum assignment we're asked to find "the lowest frequency of allowed transitions between ground state and excited states." Does this mean $$\omega_0 = \frac{E_2-E_1}{\hbar}$$?

No current flows through R2. Find the current through the 4 ohm resistor.
imgur.com/6kgVnWW
I'm getting two different answers by two different methods.

Can anyone please help?

You have an open circuit in the centre. Then, you have a -15 V source to the two resistors to ground. So just do voltage division @SDFG

What answers are you getting?

The entire top half of the circuit is irrelevant

Yes. I thought about that.

Do you have a minute? I'll try to explain my approach

7:47 PM

15/6 amps?

That's one of the answers I got, but this isn't the answer.

The answer is given as 5 amps

And I get that if

I don't see how that could be

By observing the battery on the top right, I see that the junction there must be at -5 V, because only then does the voltage at the right of the battery equal 0 (earth)

Now, since there's no current through R2, the potential at the lower junction should also be -5 volts

My own way to do this is to start at the left ground/earth and trace the following path to the right ground: left earth, +15V battery, 4ohm resistor, resistor R2, +5V battery, right earth

Which makes the potential diff. across the 4 ohm resistor as 20 volts, thus giving current as 5 amps.

7:51 PM

Yeah, that’s my reasoning as well

With your way, I got 2.5 amps.

I feel like this circuit is not consistent

Look at KCL at the node between the 4 and 2 ohm resistors

@Argon Exactly. I don't understand how KCL can go wrong.

Probably the thing to do is work out all potentials and currents to see if things are internally consistent

If the voltage at that node is -5 V, which it must be to ensure no current through R2, then $$\frac{-5-15}{4}+\frac{-5-0}{2}$$ should be zero.

The circuit cannot exist as such, unless $R2=\infty$

7:54 PM

Absent a piece of paper to work this out on, I can’t say more

This gives the 15/6 A answer anyhow

Hmm.

The fact is, KCL was my first approach. This was an MCQ in my practice test, and when I matched my answers, there was no option saying 2.5 amps. That's why I had to resort to the other way. I too think they set this problem wrong. I mean, KCL shouldn't report wrong, right?

KCL is never wrong, and KVL is never wrong. If they do not agree, the circuit is flawed

Yep.

Yeah, I see the goofiness as well now

Anonymous

8:00 PM

It's an unphysical situation — grounding at two points — you'll have to account for a ground resistance too to make it consistent (cf. Grounding two points in open circuit).

Nice

ive got a QM question

say you have 2 particles described by a single wavefunction. is it possible that the 2 particles are totally independent? I guess yes of course

Sure. The wavefunction would be separable

by independent i mean non interacting

ok

great

@Blue Great. Thanks!

8:02 PM

in solid state physics, the free electron model... is a bit strange to me. the electrons are assumed not to interact at all, but yet they satisfy pauli's exclusion principle

8:23 PM

It’s a model as you say. It’s an approximation to reality. @thermomagneticcondensedboson

Okay random thought: What time is it right now at your place?

@Semiclassical

do you have an explanation for this that doesn’t delve into Hilbert spaces?

@SDFG 9:24 in Cambridge

Anonymous

@SDFG 1:55 am

@Blue Hey! Same! Where do you live? India here.

Anonymous

Ah, of course. :P

Anonymous

8:26 PM

Same.

@Blue Where in India?

If you don't mind my asking.

Anonymous

@SDFG Kolkata

Oh nice.

Anonymous

@SDFG You?

Up north. Uttar Pradesh.

Wait....this stuff isn't relevant to physics. Is it allowed here? (I'm relatively new here :P)

Sir Cumference

8:31 PM

@SDFG We sometimes talk about physics. Mostly food :P

@JakeRose I know, this isn't the problem I have. The problem to understand that I have, is that it is claimed that the free electron model consists of non interacting electrons. But yet they do satisfy the Pauli exclusion principle. Doesn't this imply that somewhat each electron is "aware" of the others? I.e. that there is some interaction?

@SirCumference Hahaha

Anonymous

@SDFG Lucky you. We're having a terrible summer this year. It's just April and already the temperature soars to around 40 in the morning. Walking on the streets is a nightmare. ;_;

@Blue So, are you a grad/post-grad/student/etc ?

@Blue "Lucky you"?? Mate, we are not faring better. It's sweltering here.

Sir Cumference

Oh hey we got 4 mod candidates

Anonymous

8:33 PM

@SDFG Heh, I thought Uttar Pradesh is cooler. :P

@SirCumference Yes, I noticed that today. A couple of days back I was wondering if Physics.SE would get even one new mod.

@Blue No way. We are not in the Himalayas xD

@thermomagneticcondensedboson that’s my point. Whoever made the model decided to ignore any other interactions, as it predicted correct experimental results within some limits

Anonymous

@SDFG I guess we at least beat you in the humidity; it's around ~80%, which makes the effective temperature seem higher. ;)

@Blue Yeah, I'll give that to you.

Anonymous

@SDFG Undergrad

Anonymous

8:36 PM

You can read it on my profile.

Oh

Anonymous

Heat index

The heat index (HI) or humiture is an index that combines air temperature and relative humidity, in shaded areas, to posit a human-perceived equivalent temperature, as how hot it would feel if the humidity were some other value in the shade. The result is also known as the "felt air temperature", "apparent temperature", "real feel" or "feels like". For example, when the temperature is 32 °C (90 °F) with 70% relative humidity, the heat index is 41 °C (106 °F). This heat index temperature has an implied (unstated) humidity of 20%. This is the value of relative humidity for which the heat index...

Anonymous

Hah, I just remembered the term apparent temperature after several years. :P

Kolkata is just like Mumbai with a winter thrown in.

@JakeRose I'm sorry, I am not expressing myself clearly enough I think. You are not understanding where I have a problem, i.e. what I do not understand. At leas, that's what I understand.

I know everything you wrote so far and have 0 problem with it.

I agree with it.

my problem resides in the fact that the free electron model is essentially a 1 particle problem. the potential is just constant, i.e. each electron "see" nothing at all

however, quite surprisingly to me, they have to satisfy the pauli's exlcusion principle

8:40 PM

@Blue And I learnt this term for the first time, today, after 17.5 years of existence :P

how is that possible without disturbing the potential?

I see a paradox. The pauli exlcusion principle clearly shows that electrons are aware of each other. However the FEM clearly shows that each electron do not see any other electron. how to resolve the paradox?

is that clearer now?

Anonymous

@SDFG Well, it's not a common term; you must have seen it mentioned in weather apps...usually something like "30C, feels like 37C".

@Blue Yeah.

Anonymous

I'm not sure how exactly that calculation is done though.

Anonymous

There's a table of values...

8:48 PM

Would be interesting info, though.

Like when someone says "Ooof, it feels like 45 degrees today" and you can whip up some calculations and do a fact-check xD

Anonymous

Hmm, it's given in the Formula section on that Wikipedia page.

Let me see.

OKAY. Those are a lot of values.

Anonymous

Yeah, I wonder how they came up with those coefficients and the formula. It's empirical, of course...but that seems way too complicated. :P

Exactly.

Anonymous

> It is the result of a multivariate fit (temperature equal to or greater than 80 °F (27 °C) and relative humidity equal to or greater than 40%) to a model of the human body.

8:53 PM

Ah.

Anonymous

Oh, okay...that makes some sense. So they basically did curve fitting upto second order.

Anonymous

> The Assessment of Sultriness. Part I: A Temperature-Humidity Index Based on Human Physiology and Clothing Science

Anonymous

> The Heat Index "Equation"(or, More Than You Ever Wanted to Know About Heat Index)

Anonymous

Lol