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20:02
@Blue Their similarity is that almost everybody has to take the GRE (for grad school, in the US) just like almost everybody takes the JEE (for IIT, in India).
@JohnRennie I think mom shouldn't be capitalized here neither.
Quantum question which I should know and it's bothering me.
Anonymous
@lılostafa Well, I suppose SAT would be a better comparison in that case (since high school passouts appear for SAT). But then again, SAT is waaaaayyyy easier. The problem students face in getting admits to the top schools in the US is that they have to show their extra-curriculars,academic achievements like olympiad medals, social service etc. Only SAT is never enough.
most people have never heard of olympiads
@0ßelö7 True, but (unfortunately) people here (like India) mostly think of olympiad medals as a very great scientific achievement! I hate that.
The Robertson uncertainty relation states that, if $\hat{A}$ and $\hat{B}$ are Hermitian operators, then the standard deviations of these operators satisfy the bound $\sigma_A\sigma_B \geq \frac12 |\langle [\hat{A},\hat{B}]\rangle|$.
Anonymous
20:08
@0ßelö7 Same here. Very few people actually take part in the olympiads.
Anonymous
JEE is more hyped
can we please not talk about JEE
If you take $\hat{A}=\hat{x},$ $\hat{B}=\hat{p}$ then this gives back the usual statement $\sigma_x \sigma_p \geq \frac{\hbar}{2}$.
@Semiclassical yes
what is $\langle A,B\rangle$ though?
@Blue Do general public in India have more respect for a gold medalist in the international math olympiad or the JEE 1st rank?
Anonymous
20:10
@lılostafa Well, it atleast shows you're hard working and passionate towards that subject if nothing else. I've never heard anyone calling olympiads a "scientific achievement"
should that not be a commutator
yeah, fixed
Anonymous
@lılostafa The latter, obviously.
Anonymous
But then most JEE top rankers are olympiad medalists too.
but, uh, I just figured out where the source of my (as yet unstated) confusion was
so never mind :/
20:11
hmm, ok
actually, I might as well say where I was going.
just to make clear why I was wrong :P
@Blue Really? I got respect for JEE then :P
As a particular special case, take $\hat{A}=\hat{H}$ to be the Hamiltonian of some system and $|E\rangle$ to be an energy eigenstate.
are you trying to get the time uncertainty principle?
Nah
this is much simpler and much dumber :P
then $\langle H\rangle =\langle E |H|E\rangle=E$ and $\langle H^2\rangle =E^2$, so $\sigma_{E}=E^2-E^2=0$.
20:14
@Blue Well, I mean I hate it when I see people years after they got these medals or rankings still use them when introducing themselves. things like this.
which makes sense: you've got a determinate value of $E$, so there's no possibility for variance.
Anonymous
@lılostafa That's strange. Those guys are usually quite humble (from personal experience).
Anonymous
I've never seen anyone boasting about their olympiad medal, lol. :P
@lılostafa As a way of remembering a certain time in ones life, that kind of thing isn't so strange. But using it as evidence of ones talent is a bit strange.
@Semiclassical ok
20:17
the above uncertainty relation would then require that $0\geq \frac12 |\langle [H,A]\rangle|$, and since the RHS is nonnegative real it must be zero.
stop
now, is that so strange? ...well, no
you can have $\sigma_A=\infty$
sure, but you hardly need to.
or you accept the fact that your uncertainty principle needs to be more rigorously stated :P
20:18
The above is perfectly rigorous. There's a proof on Wikipedia.
The principle is not the issue.
@Semiclassical For finite dimensional Hilbert spaces.
For interesting spaces, it needs to be done more carefully.
You're making it too complicated. The resolution is just this:
$\langle [H,A]\rangle = \langle E|HA-AH|E\rangle$.
@Semiclassical I'm actually not, you can "violate" the uncertainty principle because of this. There is an easy counterexample.
@Blue What I said isn't true for everybody, and it usually happens by others (who think too much of these olympiads and exams) rather than themselves. Also, it depends on what you call boasting.
@Semiclassical Yes
20:20
ffs. I'm saying why, even if you grant that $\sigma_A$ is finite and that nothing weird is going on, having $\sigma_H=0$ is not a problem.
I said "yes"
I'm listening
But $H|E\rangle = E|E\rangle$---it's an eigenstate---and since $\hat{H}$ is Hermitian we also have $\langle E |H=\langle E|E$.
Ok, I see what you're doing.
Sure, that's fine.
Anonymous
@lılostafa Well, other people do hype about it. Sure. Nothing can be done about that.
And from that it follows that the expectation of the commutator vanishes.
So there's nothing actually strange about this.
Anonymous
20:22
Other people hype anything they find extraordinary.
@Semiclassical now read this physics.stackexchange.com/questions/233266/… and explain it to your students
because that is legitimately strange
ugh, that one
is annoying
@Semiclassical it's the basis for many careers :P
I know I've thought about it before, but I can't remember the full of it
I think my usual manner of attack was to consider it as a limiting case of a confining potential.
and argue that we shouldn't be surprised if things go weird when we take the limit
@Semiclassical That's a physical way of looking at it, I guess
20:25
this is pretty handwavey, to be sure
Amusingly, the Wiki page I mentioned earlier actually does address this 'counterexample'
I'm pretty pragmatic about it and simply don't expect $[x,p]$ to make sense everywhere on the Hilbert space
@Semiclassical I own the book they reference
Relevant bit: "Although this result appears to violate the Robertson uncertainty principle, the paradox is resolved when we note that $\psi$ is not in the domain of the operator $\hat{A}\hat{B}$, since multiplication by $ \theta$ disrupts the periodic boundary conditions imposed on $\hat {B}$. Thus, the derivation of the Robertson relation, which requires $\hat {A}\hat {B}\psi$ and $\hat {B}\hat {A} \psi$ to be defined, does not apply. "
yes, I'm aware
that's what I meant by being more careful with the statement
20:28
Right.
to be fair, it's correct on a dense subspace of the Hilbert space
so it's almost correct :P
I do think it's interesting to consider the case where the system is an annulus rather than a straight ring
it should be generic boundary condition behavior
@Semiclassical I'm gonna do something dangerous and pull a physicist :o
On the physical side, making it an annulus doesn't change the fact that the angle is restricted to $[0,2\pi)$ and therefore must have finite variance in $\theta$.
So maybe it doesn't help.
If $H$ is a nonnegative self-adjoint operator I expect that for $\xi\in (0,\infty)$, $$||(H+\xi)^{-1}||\le 1/\xi$$
because $H+\xi$ is larger than $\xi$
seems legit
that might actually be a valid proof lol
@Semiclassical see, it's things like this that I expect someone to have done experimentally
has anyone actually measured HUP violations due to boundary conditions?
20:34
dunno, haven't read up on it. I know there's a lot that's been done, though.
8
Q: Does mathematical sloppiness in quantum mechanics ever produce incorrect predictions?

0ßelö7Does mathematical sloppiness in standard quantum mechanics ever produce predictions that don't pan out? I'm not talking about things like the WKB approximation, but instead subtle functional analytic issues, such as assuming every Hamiltonian is self-adjoint, has an eigenbasis of bound states, do...

I think the way people would interpret the ring example is not so much "HUP is violated" as "HUP is being misused"
I mean, when we usually think of an observable, it's a real number and as such there's no upper bound on its variance. With the ring case, the observable is only defined mod 2pi.
@Blue Kalpit Veerwa, who secured the 1st rank in SC category of JEE (Advanced) exam, topped the JEE Main exam and created history by scoring 360 out of 360 in the exam ^
Holy shit I'm terrified just by his look :O
20:38
So the claim might be that one can't equate uncertainty with variance in that case.
Anonymous
Dude, 0celo will kill you for talking about JEE now.
@Semiclassical yeah
@Semiclassical hmm
Anonymous
@lılostafa Heh?
@lılostafa ...is there an "a" missing in the slogan behind him? Because "Educating for better tomorrow" seems ungrammatical
I think there's probably a better statement than this, though
Educating for better tomorrows?
20:40
@Semiclassical Hm...sounds weird to me, but it's possible, yes
you don't actually see the end of the word, so it could be plural.
google time, methinks
nope, google images has it as just "tomorrow" as well.
It could be like a newspaper article title, where articles are omitted
"Dog eats cat which eats bird"
Anonymous
@Jasper lol
I dunno, "educating for better tomorrow" sounds...a bit uneducated to me :P
could also be a slightly faulty translation
20:44
@Blue no, I just blocked
him
@Jasper ...what sort of newspapers are you reading?
@ACuriousMind implying anyone reads the newspaper anymore
@ACuriousMind Zoo news from Mars, not Earth.
@BalarkaSen wearing sandals and chinos right now and it feels bad
@lılostafa I think he has nice hair, that's all.
20:47
How do you do this
@ACuriousMind Happens a lot
Bought this today ^
@0ßelö7 hmm, check out equation (1) of this paper and the surrounding sentences: researchgate.net/profile/Miles_Padgett/publication/…
main upshot being that they don't use the usual HUP in trying to handle the case of a photon on a ring
@yuggib how does one say the Sz in Sz.-Nagy?
@Semiclassical >researchgate
@Semiclassical it ain't loading
says it need to validate my browser
@Semiclassical huh, it was an autodownload that was being blocked
looking now
625 PUBLICATIONS 19,542 CITATIONS
how do you get 625 publications??
@Semiclassical If you know @EmilioPisanty, he would likely have insight here
21:01
(0_0)
That's a bit much
what's that face for
publications or Emilio?
That's a lot of publications
@ACuriousMind I am trying to unthaw some pasta sauce my sister in law made for me but it's not happening quickly enough. Should I just throw the brick into a saucepan at low heat?
21:03
@0ßelö7 yes, that should work
Throw it in at high velocity :P
@Semiclassical with enough kinetic energy it will thaw itself
wait, is unthaw a word?
or is the word thaw
@0ßelö7 "unthaw" would be the same as "freezing", right?
But in a language where flammable and inflammable mean the same thing, I'm not taking any chances :P
21:04
No sure, is "un" an isomorphism?
un·thaw
ˌənˈTHô/Submit
verb
1.
NORTH AMERICAN
melt or thaw.
lol
North American English is weird
The "usage" section here for unthaw is funny: en.oxforddictionaries.com/definition/unthaw
I lost the paper in a sea of tabs
@0ßelö7 The best way is to hold it in water for a few minutes
@lılostafa it's a decent amount
a few minutes won't do it
@Semiclassical Ah, I see.
The usual example is a bit of a red herring because it's angular momentum masquerading as linear momentum
I guess $\theta$ is a proper observable?
Seems off because it's defined mod 2pi.
It does seem a bit goofy
21:14
@0ßelö7 The conjugate variable to angular momentum has some issues, I think
I think Emilio knows more about that than I, though
That's why I pinged him
One easy way to get to a non-goofy version of this is to insist one only worry about the uncertainty in Cartesian coordinates
In that case the observable don't have these mod 2pi issues and everything acts as expected
I'm not sure how satisfying a resolution that is, but it does cordon off the issue
22:02
@0ßelö7 @Semiclassical what's the action?
@EmilioPisanty we're wondering about HUP on the circle
22:26
@0ßelö7 what about it?
I mean, beyond the fact that there's no linear momentum on a ring
22:38
@EmilioPisanty for starters, what is the conjugate variable to angular momentum on a circle?
22:58
@0ßelö7 morally? Angle
Technically? Nothing
What, precisely, do you mean by 'conjugate variable'?
(keeping in mind the restrictions imposed by the Stone-von Neumann theorem)
23:31
@EmilioPisanty Do you mean the von Neumann relations?
23:48
er
Weyl relations
Today in Bored Adventures: How many levels of VMs can I do?
Currently at 4
My coffee has some stuff in it, on the top, and I can see the convection currents
pretty damn neat
extra caramel jizz?
Does someone want to answer me the next question?
residue from pasta sauce that I got in it, most likely
23:58
0
Q: Numerical Tools for Bohmian Mechanics

MadPhysicistI am trying to do some simulations using the de Broglie - Bohm formalism and am wondering if there are computational tools that already exist in this area. I generally use Python, but will consider anything, even if I have to put some numerical analysis theory into code.

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