Hi. I am getting a little confused about some basic things. When we say that our theory has a certain translational symmetry (say, of translations by some "a") then we only demand our wavefunction to be the same after translations of "a" up to a phase. But when we impose periodic boundary conditions on our theory, say of a period "L", then we demand our wavefunction to be exactly the same (not just up to a phase) after shifts of L. Why do we do this?

In particular, why do we make this distinction?

I can see the validity of this distinction based on the requirement of a single-valued wave-function. But how does an ant living on a circle know if it is living in a space which has a translational symmetry of 2*piR or it has periodic boundary conditions with a period of 2*piR?

Just getting back into running & the Fitbit didn't even track it 😫

Why can't I post a image:/ here

@DvijMankad if a wavefunction overlaps itself then it will interfere with itself and this places constraints on the phase. That's why with typical periodic conditions like a particle on a circle we require that the phase be the same at the boundary.

@KyleKanos I hate it when that happens.

I've been robbed of wrist-mounted-robot-overload points!

Robbed, I tell you!

@JohnRennie Just to reply to what you said earlier about the pendulum - I realize it's kind of an elementary thing to go through, but I'm still curious as to how (numerically or analytically, but most likely the former with an elliptic integral) we could approximate a pendulum wave (coupled and uncoupled). PM actually gave some pretty nice tips for this, especially the useful algorithms involved.

I feel like I'm going to need to do a bit more work to add the damping coefficient into account though.

@kylecampbell in the small angle approximation? Or taking into account the anharmonicity?

@JohnRennie We don't need small angle approximation if we're using the elliptic integral, it apparently has less than 1% error even at an initial angle of 163 degrees. But mostly taking into account the anharmonicity.

Really, I just want to figure out how long it will take for a pendulum to stop (in a realistic, lab-like setting).

I did a bit of finite difference approx. methods for a nonlinear pendulum at some point, but yeah I'm just curious.

I have no idea how to do the damped pendulum with elliptic functions

@JohnRennie I don't either.

Do you have any other suggestions?

I really don't care the method, if you give me something I can approximate that's just as useful as an analytic solution.

fun timesssssss

prolly gonna be a restless night...fun timesssss... somebody shoot me -.-

why? do you have finals?

finals are long behind me bruh

I'm too old fo dat

what are you so busy with tonight?

not busy lol

restless is diff

crap internet is slow

crap internet company overcharged me and hasn't fixed my bill

Getting my wisdom teeth pulled on the weekend

no good things this week I think

just shoot me lol

@enumaris ouch :-(

also got rejected today so there's that I guess

fun times

How odd. physics.stackexchange.com/questions/473272/… Maybe it was accidentally created by a bot. Or a prankster with access to their account. Or maybe they're rage-quiting after the massive downvotes on their recent question. I hope they aren't cracking up.

@PM2Ring what sort of language?

lol

@kylecampbell Here are some Dancing Pendulums I did several years ago, hosted on a friend's blog. The are just fake pendulums, executing pure SHM, but they make nice patterns. :) You can do this with real pendulums, if you keep the angle smallish, although the damping obscures the alignment effects a bit. Sadly, the forums mentioned on that blog page no longer exist.

Very nice

@Akash.B Sorry, what do you mean?

@PM2Ring what the headline of the question say?

it is full of letters without a clear meaning

@kylecampbell Thanks! I haven't written in Javascript for a few years, I mostly do Python these days. But I'm glad that script still runs, it's even ok on my phone.

Yeah I'm keeping that. It's nice to watch.

@Akash.B It just looks random to me. And the body of the question looks like a crude Markov chain, as used by spambots. In fact, I assumed it was a spam precursor, until I noticed the user has several hundred rep points.

@PM2Ring oh

@kylecampbell Oh, good. :) It can be rather hypnotic. I actually raytraced the raw ball image, using POV-Ray.

@dmckee Yesterday, on SO, I del-voted a recent HNQ. It still needs a few more votes, but once it goes, several people will lose quite a few points. They ought to know better than to answer a basic dupe... but I won't be surprised if there are a few revenge downvotes. I also found a dupe target for another HNQ, but someone else took responsibility for hammering it (& thus kicking it off the HNQ list), and it doesn't need deleting.

I've delvoted lots of stuff, but I most admit I feel a little nervous at being party to wiping out around 170 votes.

I just used this in an Astronomy.SE answer. From the USGS, here's my favourite image to show there's not a lot of water on the Earth:

The larger blue ball is the salt water, the smaller one is the fresh water. You may need to zoom in to see the fresh water...

@JohnRennie This OP just edited, but I guess it's still pretty confused. And confusing. physics.stackexchange.com/questions/472308/…

@PM2Ring I'll look again, but it still seems like an incomprehensible mess to me.

@JohnRennie Questions like that scare me. If I read it too closely, I might end up with less knowledge than when I started. ;) I was just giving you a heads-up because at least the OP tried to improve their closed question (& probably doesn't realise that close-voters don't get alerted to edits).

I would appreciate it if someone could explain-to-OP/mediate/step-in/vote-to-reopen/vote-to-close here.

@Qmechanic since they are keen on the work fuck I would just let them fuck off

@dmckee I think the issue is that I used an app on the thing that want made by Fitbit & it didn't use the GPS or trigger a workout (which is weird since my heart rate was through the roof)

@Qmechanic I'm on that

well, to some extent

Can someone protect this question? physics.stackexchange.com/questions/473061/…

We could, but it's only gotten one answer from a low-rep user so far and it doesn't seem especially likely to get more. Personally I'd defer protecting it until we see at least one more problematic answer which protection would have prevented.

Idk, maybe I'm just irked because the second highest voted answer doesn't do a single thing to answer the question, and the other answers seem to rehash what tfb explained in much more vague terms

Yeah, but protection doesn't help with that. All but one answer were posted by high-rep users.

@JMac protection only prevents users with <10 rep from answering. There could be something said about protecting any/all HNQ, but this one doesn't seem to have drawn that much negative responses.

Speaking of protection, Dvij Mankad recently proposed an extra level of protection for pending dupes. It hasn't got much feedback, so far.

@PM2Ring interesting case. That definitely looks like it received a massive pile of abusive or spam flags after it was closed. @Qmechanic, is that the right conclusion?

@EmilioPisanty Spam / abusive flags wouldn't surprise me. I did actually spam flag it, but retracted 30 seconds later, when I noticed the user's rep, and recalled a slightly odd interaction with them a few days ago. Plus I figured that a flag was pointless, since a mod was obviously already aware of the post, but I was a bit puzzled that it wasn't deleted.

And I guess it's not Markov text, just a text dump of the front page.

huh

i called travelocity customer support to get helped on something, and i'm still waiting

and, of course, there's hold music. so far, nothing too strange

@EmilioPisanty : Yeah, I don't recall seeing any spam flags when I closed it.

...but then "never gonna give you up" came on

the travelocity customer support literally rick-rolled me

@Semiclassical I wonder if that's an intentional song that plays just to mess with people

I gotta wonder.

The falling body is falling into the black hole because the speed of light is reducing. It falls faster and faster, until at some point it's falling at the local speed of light. Then it erupts into a gamma ray burst.

Sigh. And we can expect the same soon.

GRBs are due to something falling into a black hole!?

@JohnRennie ow

oh no

noooo

@Semiclassical :-)

@JohnRennie I was just reading that too. But I've given up on responding to him. It's like mud-wrestling a pig. You can't win, you get covered in dirt, and the pig enjoys it.

Well, if the trend of his low quality answers continues, would his suspension resume? And for longer?

I wouldn't have thought it's low-quality answers which would get someone banned, at least not in chat

@KyleKanos that's up to the moderators ...

@Semiclassical I imagine it's basically something along the lines of it's assumed they're just trolls and/or spammer

yeah, there's a fine line between "consistently wrong" and "spams low-quality answers"

Guys like the "I've derived the fine structure constant from numerology first principles" guy that I linked yesterday aren't really a problem. It doesn't take much expertise to recognise that he's pushing a crackpot theory, and the occasional comment is enough to make that clear to newbies without that expertise.

But JD is pretty subtle, and is an expert at smoothly transitioning from mainstream material to his pet theory, so he manages to accrue upvotes.

I'm not sure "subtle" is the right word for it, but he writes in a way that presents the appearance of credibility

Well, he doesn't make it blatantly obvious that he's on a mission to push his pet theory to anyone who'll listen. But anyway, I better shut up about this. ACM said we shouldn't discuss people who aren't here to defend themselves.

So to totally change the topic, I tried to get into Bohmian mechanics when I was younger, but I only read his Wholeness and the Implicate Order, which is more of a philosophy book than a physics book. One of my grandads was a Bohm, so I might be distantly related. :)

that was my first exposure to Bohm as well

I had some friends at grad school that studied Bohmian mechanics

at the time I didn't have enough physics under my belt to grapple with the physics part of it, so I stuck to the philosophy part of it

which was interesting enough

As their doctoral research, I mean

nice

later on I did get interested in pilot wave theory itself, though I tend to avoid Bohm's own route to it

I'm more fond of how Bell proceeded

I've always had a soft spot for Cramer's Transactional Interpretation. It's a shame that he hasn't been able to develop it into a viable interpretation, and has been accused of distorting facts to fit his theory.

@JohnRennie sigh

I guess the only thing to do is to downvote incorrect content, and to consistently raise flags when there is a pattern of low-quality contributions stating as much

@Semiclassical I found Bohm really easy to read, almost like reading something written by a distant uncle. Or maybe that was just wishful thinking. ;)

@Semiclassical which number did you dial? The one here or the one directly on travelocity.com?

the former, I guess?

@PM2Ring there are certain parts of it which are pretty easy to get into, yeah

the only section I remember having to wrestle with was the part on the rheomode

@Semiclassical did it work?

insofar as the line connected, yes

but the line mentioned that you ocould contact them via the messenger on their facebook page

so i tried that and got a response

People still use Facebook?

apparently, yes

i had to guess what my password was, lol

guess harder

arrived early to work...but don't feel like workin...

@Semiclassical no, the second site I linked to

@EmilioPisanty ...nicely done

Hi to all. Any idea why in monte carlo simulation binning(averaging between results in a markov chain sampling) will destroy correlations(memory effects) between the results and give the correct error estimates as calculated from variation and standard deviation? Thanks.

I read that: When the the central limit theorem applies, the binned data will becomepractically Gaussian, as soon asNBINbecomes large enough. This allows toapply Gaussian error analysis methods even when the original are not Gaussian

@Semiclassical ::grin::

on a different note, I ran into one of your answers while looking up stuff on partial traces

which is a topic I know less about than I should

I mean

there's only so much to agonize about there

yeah

it hadn't really clicked before tho

not sure why

main annoying right now is that the wiki page on partial trace doesn't cite any sources : en.wikipedia.org/wiki/Partial_trace

there's at least one formula on there which is interesting but I'd like to see more detail

namely, the bit about partial trace and invariant integration

the closest analogue of that which I know about is the resolution of identity for (generalized) coherent states

and it seems like there should be a connection between the two, but I'm not expert enough to see it

@Semiclassical huh

does it come up in Nielsen & Chuang?

that'd be the natural place

good question. i don't have a copy of that yet, but I know someone who does

(sorta flying by the seat of my pants)

@Semiclassical and you're too prudish to use libgen?

mostly I just forget that libgen exists

annoying gif

ooooooooook

lol

I guess

i usually rely on google to find stuff

@Semiclassical huh

I'd never seen that

it makes sense

yeah

I think

lol. that's basically my reaction too

it has nice overtones of Schur's lemma to it

What I'm wondering is if there's a way to derive that expression from the resolution of identity you see when doing (generalized) coherent states, i.e. $I=\int |g\rangle \langle g|\,d\mu(g)$

@Semiclassical possibly, yes.

That one also has a lot of Schur's lemma-type smell

Exactly

Any stellar physicist here?

@SirCumference I like to think that I shine as a physicist

...I'm stealing that one

But just wondering if can verify that I didn't oversimplify anything here

@Semiclassical I suspect Wikipedia's formulation of that one is wrong

at least for starters

it's trivial to prove if $S$ is unitary

That's why I'd like to see a better discussion of it

It seems to be claiming that $$(\mathrm{Tr}_W T) \otimes I_W = \int_{U(W)} (I_V\otimes U^*)T (I_V \otimes U)\,d\mu(U)$$

yes, that's the claim

which seems plausible but

just because it's plausible doesn't mean I know how to validate it

jarrodmcclean.com/integrating-over-the-unitary-group

that gets you a long way

hmmmmm

I've met that guy

at some point

hmm, nice

ages and ages ago

he was a PhD student at Alán Aspuru-Guzik's group at Harvard when I visited

back in the dark ages of 2013

lol

that group does awesome stuff

the day that a quantum computer gets built and then gets used to do something useful for the world, they're strong candidates to be behind the algorithms it'll be using

ah

derrrrrrrrrrrr

suppose $X$ is an operator on $V$ such that $[X,U]=0$ for all $U\in U(V)$

then $X$ shares an eigenbasis with all unitaries

that's pretty much enough to ensure that $X$ has a trivial action, no?

should, yeah

that seems like the basic idea of Schur's lemma

yeah

in the way McClean phrases it, it really is exactly Schur's lemma

$D(X)$ commutes with all $U$s, therefore by Schur's lemma it is a multiple of the identity

I think the main thing that's confusing me, due to my relative ignorance of how group integrals work

@Semiclassical but the proof is much better. Basically, if you assume that $f:V\to W$ is $G$-equivariant under representations $\rho_V$ and $\rho_W$, you can show that both its kernel and its image are invariant under those representations

if the representations are irreducible, then the only way that can happen is if those subspaces are either zero or all of $V$ or $W$

in the wikipedia article, one has $\int_{U(W)} d\mu(U)=\text{dim}(W)$

that gets you the isomorphism.

whereas in coherent state context you see stuff like $I = \int |g\rangle \langle g|\,d\mu(g)$

hmm. I guess the point is that, upon taking the trace of the latter, you get the former

@Semiclassical that does work. Be careful with the normalization, though

yeah

$\int_G \mathrm d\mu(g)$ is essentially arbitrary

ya

I tend to feel anything other than $\int_G \mathrm d\mu(g)=1$ is pretty dodgy

> (An eigenvalue exists for every invertible linear transformation on a vector space whose underlying field is $\mathbb {C}$ , as a simple consequence of the fundamental theorem of algebra.)

what?

I think everything holds up so long as $\text{Tr}[|g\rangle \langle g| ]=1$

though again the normalization is something to be careful about

no, wait, yes, that's obvious

ignoring normalization issues for the moment, one has $\text{Tr}[|g\rangle \langle g|] = \text{Tr}[U^* |\psi\rangle \langle \psi|U]=\text{Tr}[UU^* |\psi\rangle \langle \psi|]=\text{Tr}[|\psi\rangle\langle \psi|]$

but that's just the trace of a pure state and therefore 1.

So modulo the annoying bits, it is all internally consistent

hmmmmmmmmmm

that Wikipedia page looks like it has a truncated argument in the Invariant Definition section

"the maps $E_{{ij}}\otimes F_{{kl}}$ form a basis for $\operatorname {L}(V\otimes W).$", and then?

How much rep does one need to vote in the election

@SirCumference see the Primary or Election tabs in previous elections

physics.stackexchange.com/election/3?tab=primary

say

> In the primary phase, all nominees advance to preliminary community voting. Any community member with 150 reputation may vote in the primary.

> Any community member with 150 reputation may vote in the election. Each voter may select up to three candidates. Please make your selections in order of preference, with the most desirable candidate as first choice.

Awesome, thanks

@EmilioPisanty hmm, so here's a question

Consider a d-dimensional Hilbert space $V$. Suppose I have a state $\rho$ on $V\otimes V$ such that $(U\otimes U)\rho (U^*\otimes U^*)=\rho$.

If I trace out the second Hilbert space, is it obvious that $\rho_1 = \text{Tr}_2 \rho$ must satisfy $U\rho_1 U^*=\rho_1$?

i.e., if the product space is invariant under $U\otimes U$, is the partial trace invariant under $U$?

bah, all of that should be assuming it's true for all $U$

seems like my question comes down to the following properties(?) of the partial trace

I've seen the second one before, but not the first.

Dear all,

physically speaking B is the magnetic field, $\rho$ a electron density and $f$ the scalar field of a gauge.

Sorry, folks, I'm done for the night, I'll have a look tomorrow.