@Obliv Heya

I wish I could go to sleep and wake up in 3 weeks.

@3075 Tired?

:v

well, no I'm excited for uni admissions.

which will be in 3 weeks (for me specifically)

wait I misread.

it's from now until 3 weeks from now.

:D

I feel your pain man

I was freaking the hell out last month too

what places did you lads get into

nowhere so far.

admission decisions are made in may.

are you a transfer applicant?

or fresh out of highschool

fresh out of high school xD

@user507974 VA Tech, UCSD, UMN

Pretty shitty results

GA Tech and Waterloo to go

GA is a pretty good school

Right now Im trying to convince a friend to come to my school over SD

I dont see what they see in that school for EE

What's your school?

SD is a good school, sadly I can't study what I want to study there

and I'm convinced I'll suck balls at everything else

@BernardMeurer which is

CSE/CEng

how is that different from ECE?

It isn't every school calls it some different bullshit so to make my life more difficult

Computer Science Engineering, Computer Engineering, Electrical and Computer Engineering, Electronic and Computer Engineering

for the love of god

I've been reading forums for the past 3 hours.

about admissions.

@3075 Don't do that man it'll drive you nutsie

last month I was doing the same thing

almost set my kitchen on fire a fourth time

lol

my average is 85 idk if i'll get in.

Really though, go eat ice cream and get fat

it's the best strategy

It's the same average I have

but then I have NO clue how Waterloo works for internationals

because I had 14 different parallel subjects to attend

@vzn anything seem to pop up in your head from reading that

ANYONE CAN HELP ME??

@MAFIA36790 I have 5 stars now ;-)

@yuggib you like terry tao?

he's an extraordinary mathematician

but with a very nice attitude (at least it seems so)

hey @JohnRennie mind if I bug you with a paper and a question

hello

Symmetry operators are done by unitary operators

@MAFIA36790 so theyre symmetric when theyre symmetric

@user507974 no problem, I'll help if I can

@JohnRennie so basically I had someone who came presented to our department about a type of high precision interferometry. Here's a paper on it: arxiv.org/pdf/1410.8486v3.pdf

@user507974 OK

If I've pieced together an understanding of this right, he essentially injects a signal (bloch oscillations) and later combines this with a PI feedback loop and this gives orders of magnitude improvement in accuracy (~1500 fold) and suppresses diffraction phase shift.

I don't think symmetry transformation operators commute with $H$ generally

Lorentz transform doesn't, for one

I don't know, I am not Feynman

@JohnRennie So I'm basically trying to understand if this is just an application of some kind of common method in processing signals, if theres any kind of starting point for looking into trying to do similar signal injection precision improving methods into other systems

@user507974 I would have to go off and read the paper to anything useful about it, and at the moment I'm afraid I'm tied up with another job. Sorry :-(

np,

have you ever heard of signal injection systems like this before out of curiousity

I guess it's some kind of convolution method when you inject signals right?

@MAFIA36790 It depends what is the interpretation you want to give to it; the idea of symmetry transformations is to have something that modifies the description of the system, but not its "physical content"

in classical theories, this is crystal clear when formulated in terms of the action

and Noether's theorem

in quantum theories, we essentially would like that the corresponding classical conserved charges are still conserved when quantized

i.e. they commute with the Hamiltonian

this is essentially what I said ;-P

(apart that there are plenty of mathematical caveats in the procedure, but let's forget about that)

In relativistic theories, the hamiltonian is (supposedly) one of the generators of the Poincaré group

supposedly because there is no rigorously defined QFT in high dimensions (from $3+1$)

yes, but there is really nothing special there

the point is the following: the Heisenberg equation of motion for observables is

$i\partial_t Q(t)=[Q(t),H]$

with, pedantically, $Q(t)=e^{itH}Q(0) e^{-itH}$

(i.e. the solution)

it is then clear that if $Q(0)$ commutes with $H$, or equivalently with $e^{-itH}$ for any $t\in\mathbb{R}$, then $\partial_t Q(t)=0$, i.e. $Q(t)=Q(0)$ for any $t\in\mathbb{R}$

there is nothing special, it is just the definition of the dynamics of operators

provided that everything is sufficiently regular, and I am omitting tons of domain problems etc.

Now this conserved $Q$ is interpreted as the conserved charge related to a symmetry

and if everything works well (it does in many actual cases)

in the limit $\hbar\to 0$, to the operator $Q$ will be associated a function $q$ of the phase space, and to $H$ the energy function $h$; with the property that $i$ times commutators become Poisson brackets

therefore you get $\partial_t q(t)=\{q(t),h\}=0$, i.e. the corresponding classical function $q$ is conserved as well

and should define the charge of a symmetry of the action (if we want the picture to be coherent)

@yuggib is there any $Q$ such that $[Q,H]=0$ but is not the charge of a Noether current?

since we construct the quantum theory based on our classical experience, the picture is often seen in reverse

@AccidentalFourierTransform I don't know

it may be possible

I mean, even provided that $Q$ "converges" to a function $q$ that is conserved in time

that does not automatically mean that $q$ is the charge of a Noether current

(right?)

@yuggib perhaps by covariance if $\partial_t Q=0$ it must be the charge of a current

but no idea

Usually the picture is reversed in this sense: we start from a classical charge $q$

but that is somewhat unsatisfactory, since the quantum theory should exist "per se", and have the classical theory just as an approximation

You guys are probably better suited to answer it, so advertising it here

in fact, maybe I should migrate - would it be a fit here?

to me it seems fit; but I am no mod

yeah, pretty much so

no prob

What's the deal with this no randomness in QM POV? physics.stackexchange.com/questions/253840/…

Is it defensible?

if you want to borrow terms from mathematics, then observables in quantum mechanics are actually random variables of a non-commutative probability theory ;-P

and quantum states are non-commutative probabilities (positive measures with total mass one)

but that's just the way we would formulate things in mathematical (probabilistic) terms

it does not mean that there is or not "randomness" from a physical standpoint (whatever this means)

anyways, never listen to curious one ;-P

@yuggib I agree, but it surely means that declaring authoritatively that there is no randomness in QM, expressing surprise that other people disagree etc is disingenuous

that's his usual style

when he's ignorant on something, he just dismisses the others being 100% wrong and stupid

anyways, in my opinion there is a distinction between random variables and randomness in the context of quantum mechanics

Fair enough. I was indeed told that my understanding was 100% wrong :)

How so?

I agree that observables may be thought of as non-commutative random variables, and that is actually the way probability theorists define them essentially

but in physical system there is something that in probability theory is not so important, and it is dynamics

or in their terms, a one-parameter group of automorphisms of the random variables

that describes how they are modified when time flows

How do dynamics affect your interpretation of randomness?

this dynamical evolution is not random, but deterministic if you want

even if it acts on random variables

Sure

I think that many people misunderstand the statement on randomness as the evolution of observables being random (not you, but maybe curious one)

so it may be worth to stress the point to avoid confusion and discussions

;-)

Fair enough.

so what I would say is that QM describes the deterministic evolution of non-commutative random variables/probabilities of a given system

Sure, I completely agree with that.

@yuggib aka, you deterministically evolve your wave function, which encodes information that is indeterminate

at least until you collapse it

@MAFIA36790 well yea, and i could assert that i dont need to shower for a week but thats not what physical reality necessarily corresponds to

anybody knows where I can find worked out examples of tree level amplitudes for QCD?

I'm having a hard time with them color factors...

I know wiki uses bots to fix spelling, formatting etc. Why doesn't SE?

Hello

'lo

@MAFIA36790 that is some weird choice of words indeed

im not sure what they mean

maybe read translation operator instead :/

Every time I write a private message to someone through forums or emails, why am I so careful and take a large amount of time thinking and perfecting my response?

@yuggib cc @MAFIA36790 you don't have to defer to mods, you know. Moderators aren't the ultimate guardians of what is on- or off-topic for the site.

@rumtscho I guess it's okay. Not a great question for us IMO but I think it's just over the threshold of on-topicness.

What can cause the asker of a question (not migrated here!) to be deleted merely an hour after posting the question?

@MAFIA36790 I am very honoured to be included in that list ^_^

@user507974 what area are you working in? its very specialized. afaik its entirely different than LIGO technology which afaik doesnt use atom beams but could be used for similar purposes someday with higher sensitivity/ precision but probably greater expense. its an atom interferometer... they are attempting to overcome intrinsic "oscillation" in the interferometer with a dampening method, but details are not clear to me...

ah, ok reading LIGO it basically uses lasers in very high-purity vacuum and measures very minute frequency shifts... very precise... not sure how atom interferometry would compare in theory/ practice, but its another option/ alternative to explore for gravational wave detecction... ligo.caltech.edu/page/faq

@ACuriousMind Hey man

@MAFIA36790 Heya

@BernardMeurer Hi there

@ACuriousMind Tell me something I don't know

\\\\\\o//

@BernardMeurer In high dimensions, the volume of a d-dimensional ball is almost entirely concentrated at its surface - that is, a thin shell of the same radius does not differ markedly in volume from the full ball anymore.

Whoa that's cool!

Define 'high dimensions', and why does that happen?

@BernardMeurer For example, in ~500 dimensions, a shell of thickness ~1% of the radius has ~99% of the volume. And it happens because the difference between the ball of radius $r$ and the shell with thickness $s$ goes as $(1 - s/r)^d$, and since $s/r < 1$, this becomes very small if $d$ is large.

@ACuriousMind Ah, I see, and who came up with $(1 - s/r)^d$ ?

@ACuriousMind not convinced. I need experimental confirmation

I'll be in my garage checking it up. I'll be back in a few minutes

@BernardMeurer Well, that's not that hard - the volume of a $d$-dimensional ball goes with $r^d$ (it's just $r^d$ times the volume of the ball with radius 1). The volume of a shell of thickness $s$ is thus $\propto r^d - (r-s)^d$, and the ratio of that to the volume of the full ball is $\propto\frac{r^d - (r-s)^d}{r^d} = 1 - (1- s/r)^d$.

What's $\alpha$?

it's $\propto$, not $\alpha$, and $\propto$ means "is proportional to"

Oh I see, bloody hell those look so alike hahaha

But I understand it now, thanks man :)

I'm starting a list, every time ACM makes me understand something I owe him a beer

Prepare your liver @ACuriousMind

Don't worry, my liver is in a state of perpetual training :P

Well, you are German, so that's nothing but your obligation :p

@BernardMeurer Sorry man I'm not gonna be able to work on any of those challenges this week. I have a couple papers due and finals at the end of the week. I'll give it some time next week though for sure :)

@Obliv That's alright, no need to rush it, it's just a fun project :)

Yikes, finals

bless your soul

lol thanks

So ehh... Please stop starring all messages by ACuriousMind @BernardMeurer (+other person?). No offense to anyone, but they're not that interesting to the rest of us.

@Danu Last time I checked people had the right to star what they want to :)

@Danu You jelly? :P

@BernardMeurer You do! I'm not threatening any sanctions.

I'm also allowed to remove stars if I see a good reason to.

@ACuriousMind Always.

I cry myself to sleep over your stardom

Btw, speaking of volumes of balls, have you ever seen that 1-paragraph proof of Liouville's theorem?

@Danu Even so, is the purpose of stars really to 'flag' a message as "This is interesting for others"? I always thought it was just a 'flag' for "Hey I found this funny,interesting,informative,..."

@BernardMeurer No, it's meant to signal that this is worth reading for the others in the chat. That notion often coincides with "I found this funny/interesting/informative".

But sometimes it doesn't.

@BernardMeurer Yeah, it was once meant to be to flag things as interesting/worth preserving for other. Then chat met reality ;)

One way to see that this makes sense is that stars are anonymous: If you intend to show "hey, I found this interesting" it should be clear who's doing the starring. It's not, so all you can conclude is that the starring is supposed to draw the attention of others, and nothing more.

@Danu My discussion earlier with ACM however was informative, for anyone curious about ball volumes at least

@Danu The Hamiltonian one?

@ACuriousMind Eh?

The one that starts like this: Take two points, and consider balls of radius $r$ around them

@Danu There's like six things that are called "Liouville's theorem"

The constant functions one

@ACuriousMind Sorry, I mean the one from complex analysis

@Danu You mean the standard proof that's also on Wiki‌​?

@ACuriousMind No, no Cauchy

Much nicer

I know exactly one proof of complex Liouville, and it's that one.

Okay, here it is:

Consider a bounded, harmonic function on $\Bbb R^2$, and take two points in the plane. Consider two balls, centered at the two points, with equal radii. Since the function is harmonic, its value at the points is the average of the values over the respective balls. For large enough radius, the two balls overlap on an arbitrarily large fraction of their volume. By boundedness, the average value over the two balls is therefore arbitrarily close, hence the value is the same at the two points.

Hence it is constant.

(this also works on $\Bbb R^n$, for any bounded, harmonic function)

I got half of that I think

> By boundedness, the average value over the two balls is therefore arbitrarily close, hence the value is the same at the two points.

You lost me here

The point is that boundedness shows that the average value doesn't get thrown off by the small part where they don't overlap

@ACuriousMind You don't like it? :(

@Danu I don't like it any more or less than the one using Cauchy.

@ACuriousMind Pfft :(

It's widely regarded as very nice by mathematicians. Note how it doesn't use any formulae, and is very easy to memorize.

What's a bounded harmonic function?

@BernardMeurer A function that is bounded and satisfies Laplace's equation

Bounded as in has boundaries?

Bounded as in $f(\Bbb R^n)\leq M$ for some finite $M$

@Danu Well, it doesn't involve any formulae because you chose not to write formulae for "average of the values over balls" or "arbitrarily large fraction", etc.

Ah I see

@ACuriousMind What's your proof?

@ACuriousMind ...but that's not possible with the Cauchy proof.

There is no nice, intuitive description for that one

I got 0% of that one, so I prefer Danu's :p

@Danu I don't think of "the average of a harmonic function over a ball is its value at its midpoint" as very intuitive (though it is a very nice statement)

@ACuriousMind Once you accept it for a sphere... But at least it can be written down precisely in a short sentence.

No such thing is possible for Cauchy's formula, AFAIK.

$f^{(n)}(z) = \frac{n!}{2\pi\mathrm{i}}\int \frac{f(y)}{(y-z)^{n+1}}\mathrm{d}y$ is not that hard to memorize :P

Factorials are so cool

...but there is no sentence!

Lol @BernardMeurer

@Danu And why would a sentence be superior?

Because it's much easier to memorize

@ACuriousMind I have to say I stand with Danu here, it's far easier to keep in your head so you can use it later on

But...you don't have to use it. It's a proof, the only reason you might need to have it fully memorized is to impress other people :P

@ACuriousMind And impressing other people may lead to procreating, so I say it's useful

checkmate

The point is that this proof is so clear, intuitive and nice that one doesn't need to try to memorize it at all. It just pops into mind as soon as you know it

falcepalm

...it sure did for me, earlier

@BernardMeurer keep it focused on what matters

Keeping*

@DanielSank You always come in when I'm saying something weird

@BernardMeurer And what does that suggest to you?

Let's use basic statistical inference :-)

@Danu Someone here has to keep the focus on what it's all about :p

@DanielSank Fite me

I'll propose that the distribution of times when I come here is roughly bimodal. I wonder why/if you say weird things at those times.

@BernardMeurer k

More serious bonus @ACuriousMind this proof holds for other dimensions too.

@Danu Did you know that the Pythagorean theorem works for k-dimensional objects in n-dimensional space?

It's the coolest thing in math!

Eh?

Yeah!

@DanielSank I say weird things on a daily basis

@Danu For harmonic functions, not holomorphic ones.

@ACuriousMind Of course---but the $n=2$ case is just a special case now

@ACuriousMind Where did holomorphic come from? I though this was all for hamonics

Consider k n-dimensional vectors (n>k). The k-volume of the parallelopiped swept out by those vectors is given by the square root of the sum of the squares of the k-volumes of the projections of that parallelopiped into all k-planes!

(and it's not so obviously complex-analysis at work)

@DanielSank I love the word parep;lrejapoerjpaowefj;l

^correct pronunciation

@Danu hahahahahahahahaha

^ Dutch pronounciation.

I can't say that word

@BernardMeurer It's only meant to be read and written, never spoken.

@DanielSank :3

@DanielSank VOLDEMORT

It's like if I was a German trying to say Squirrel

@BernardMeurer Liouville's theorem is the statement that every bounded holomorphic function is constant - Danu's proof uses the equivalence of holomorphic functions in one complex variable to harmonic functions in two real variables.