The h Bar - 2017-03-06 (page 3 of 4)

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6:06 PM

@Slereah Kato Perturbation Theory for Linear Operators is highly recommended

He's explaining physicist QM math better than any physicist

Hey Kato

i remember that guy

He did the Lie Trotter Kato formula

is that the $\lim_n (e^{X/n}e^{Y/n})^n=e^{X+Y}$ one?

yeah

it's used to prove the correspondance between operator formalism and path integrals

it probably doesn't work for unbounded operators :P

It works if you're a physicist

Is the propagator well defined?

$e^{iHt}$

it's pretty unbounded for the most part

6:09 PM

If $H$ is legitimately self-adjoint, yes.

You can define it using the so-called "functional calculus"

Making a roast

gunna be good

I thought you were poor

I'm aight

poor people can make a roast

also it was a special offer

6:11 PM

once in a while

10€/kg sort of offer

how much is a kg in real units?

Pretty fine for meat

how many big macs?

$0.482 kg = 1 lbs$

6:12 PM

eight big macs per slurpee

I'm a terrible American, I've never had a big mac

I do love the dollar burgers

Are those your famed "sliders"

400cal/$ is pretty sweet

I'm surprised you're American, you always seem to be on in my time zone

I'm sure you'd get a better ratio eating butter

6:13 PM

@Slereah you can get a double cheeseburger for $1 here

@BenNiehoff It's 1:13 here

I'm sitting in my office

Pretty sweet for the morbidly obese

doing "work"

19:13 here

having dinner :)

@Slereah yeah but you need sum protein boyy

CET?

6:14 PM

central eastern time

No way

@Slereah Maybe learn some topology from Jost before reading Kato. He uses compactness a bunch

I have no idea what the name of my timezone is

@BenNiehoff Are you Israeli?

I'm American living in Belgium

6:16 PM

Close enough.

pretty close from here

I could be in your house if I wanted

Watching

why would anyone think I'm Israeli? never heard that one before

@BenNiehoff Name

I guess because "ben"

classic component of jewish names

it's also a common first name

6:18 PM

I've never met anyone with that name who wasn't full-on orthodox jew

that's weird

I've never met a Jew named Benjamin

they're always Noah or something

Schmiel

6:30 PM

@Slereah Oops. I just realized you were saying that to Ben, not me. Sorry if I came across as rude.

by the way, dollar hamburgers are not sliders. Sliders are even cheaper :D

sliders are smaller, like 1/4 the size of a normal hamburger

it is what americans feed their babies

gotta get them started early

them hamburgers ain't gonna eat themselves!

of course, Belgian food is wonderfully fattening, too :P

they put fries on everything...or maybe it's everything on fries

the national dish is beef stew on fries

Also la mitraillette

Are you in french speaking belgium or dutch speaking belgium

or the weird german speaking belgium

Dutch speaking

6:39 PM

I don't know what a mitraillette is in dutch

Let's see

or, nederlandstalige, I think

you could describe it?

Een mitraillette (uitgesproken als 'mitrajet', letterlijk een kleine "mitrailleur" of "machinegeweer") is een Belgisch snackbargerecht dat ontstaan is in Brussel. Het bestaat uit een half stokbrood met daarop vlees van de snackbar, friet en één of meerdere sauzen. Meestal wordt er ook nog salade en soms kaas aan toe gevoegd. Een mitraillette komt hoofdzakelijk voor in Franstalig België, maar ook in Vlaanderen en Noord-Frankrijk. In Frankrijk wordt het gerecht ook wel een américain genoemd. Met name in Brussel wordt het veel verkocht in de kebabzaken en is het gerecht een combinatie van de b...

I know a few French words for food

so it's cheese on fries on sausage on a baguette

quite a belgian dish

yes, it seems to be having an identity crisis ;)

6:44 PM

Why is this chat so good at making me hungry? :P

I have yet to actually get anything to eat at a frietkot

I should probably try, but I'm never hungry at the times I'm walking past them

Hi, everybody.

@DanielSank hello

Yo, @ACuriousMind, I'm staring to understand how to think about path integrals.

@0celo7 hello

You basically define the path integral to give the answer you want in some "known case" and then define everything else in a reasonable way using linear combinations.

@ACuriousMind Nice try, fellow human ;)

6:47 PM

i.e., your functionals can all be written as linear combinations of simple functionals on which you know what the path integral should be.

...and since integrations is linear, you know the answer.

Well you can work out path integration by hand for a few cases, too

more or less rigorously

this works great until you have an interacting theory :P

@DanielSank What do you mean by "what the path integral should be"? Are you talking about expectation values?

@ACuriousMind In this particular case I'm thinking about probabilities for a diffusion process starting at $(t=0, x=0)$ to wind up at $(t=t_f, x=x_f)$.

Also, I don't think the linearity is a special "feature" of path integration at all - you want a way to assign expectation values, after all, and since those are linear, the tool you use to assign them should better be linear, too.

6:49 PM

So, transition amplitudes, if you want.

@ACuriousMind Uh, yeah, ok whatever.

What is the conclusion? Dan has to think more?

Hm? No conclusion. Just me not entirely grasping what insight he wanted to convey, apparently.

@ACuriousMind I am conveying that one can construct path integration algebraically.

@DanielSank so like defining differentiation algebraically?

1

Q: Spherical symmetric solutions for gravity+dirac

Concerning only $g_{\mu\nu}$ and fermion $\psi$, I can write Lagrangian as \begin{align} \sqrt{-g} (R + \bar{\psi} \gamma^\mu D_\mu \psi + m \bar{\psi}\psi) \end{align} Now i am curious about static-spherical solution of this Lagrangian. (I.e, (?) extended version of Schwarzschild solution) ...

general-relativity fermions

6:57 PM

@0celo7 Yes.

You know what the derivative of x is, and the product rule, and you can take derivatives of polynomials

That guy writes papers for 40 years but he doesn't know how to write $\hbar$ in latex

That works fine until you have something that's not a polynomial :)

Suppose I declare the following $$\int_{\mathcal{C}(0,0;x,t)} dW = \frac{1}{\sqrt{4 \pi D t}} \exp \left( - \frac{x^2}{4 D t} \right) \, .$$

@0celo7 Still works fine as long as functions are analytic, but yeah

6:59 PM

$\mathcal{C}(x_0, t_0; x_1, t_1)$ means "all paths starting at $(x_0, t_0)$ and ending at $(x_1, t_1)$.

@DanielSank I doubt that; or rather, you're probably using the word "path integration" for something it is not. The path integral, by definition, is some sort of integration over paths. What you can do is get out of doing the actual integration by using algebraic relations between different evaluations of such an integral, just like I can get out of doing an actual integration by using the fundamental theorem of calculus. That does not mean one can meaningfully define the integration by that.

$dW$ means "weight each path the right way". It's a measure and I don't want to get into the technicalities because it's really complicated.

@BenNiehoff doubtful, because you still have to take limits somewhere

@ACuriousMind Oh but I disagree.

You can define normal integration entirely by algebraic relations.

@0celo7 Ah, yes, that's true, but the point is that you can take limits of things that have well-defined limits ;)

All those math people

7:01 PM

I mean hell if you want to be a smart ass you can use Stone-Weierstrass and write any function as a uniform limit of polynomials

Ah, actually I seem to remember that you can integrate analytic functions term by term, but differentiating them may cause the series to become non-convergent

But you still have to use limits

Obviously the path integral is just $$\mathscr D x(t) \propto \prod dx_i$$

@BenNiehoff that's true

the real physicist way

7:01 PM

@0celo7 Fair enough.

@0celo7 The point is that I don't have to use limits to define the integral itself.

I only need them to figure out what happens when I put in nontrivial functionals.

@DanielSank if you define a random algebraic thing it's not clear how it behaves wrt. limits

"Through the symbol $A_1 \times A_2 \times \ldots \times A_n$ the product of sets $A_j \subset \mathcal{J}, j=\overline{1,n}$ is defined". What does the overline mean in this case (set theory)?

@0celo7 Sure it is, because again, I know the value of the integral for trivial functionals!

Like if you define integration algebraically (without a theory of measure) then you get the Riemann integral which is pretty terrible.

So, when I have a nontrivial functional, I wind up with a limit of things that I understand how to handle in limits.

7:03 PM

Riemann integral is fine

@0celo7 The Reimann integral is totally fine.

@DanielSank again, Riemann integral

Although it doesn't work in path integrals, ironically

Can't define the action with a Riemann integral in the path integral

Or a Lebesgue integral, then again

@DanielSank try doing any kind of serious analysis with the Riemann integral

The Reimann integral fails only on functions that I really, really do not care to integrate. I guess Lebesgue is also better for proving stuff about how integrals behave under limits too...

7:04 PM

You won't be able to take limits

@0celo7 Yeah, i know.

But that's what we're talking about! Limits!

you can't even integrate a stochastic process with a Lebesgue integral

@DanielSank I think what you mean is something like the Daniell-Lebesgue process where one defines the value of the integral on a class of "simple" functions and then extends it to functions that can be well-approximated by these simple functions. That's not what one would call usually "algebraically" because the hard part is showing that the limits involved in the approximating work out fine.

The thing is though, that if you think a bit, you really don't need Lebesgue to figure out what those limits should be. You do fine by regularizing integrand.

@ACuriousMind Yes but the answers are exactly what you expect, so I don't care about proving the $\epsilon$'s and $\delta$'s.

7:05 PM

I think @DanielSank is rediscovering basic measure theory :P

@0celo7 That could be true.

It's been a while.

@ACuriousMind who is Daniel?

The only real method for integrals is the method of exhaustion, sorry

Archimedes was right all along

Look, basically my point here is that it seems like you can forgo the formal measure theory and instead look for self-consistent algebraic relations. My guess is that doing this implicitly involves swapping order of limits somewhere.

@0celo7 Percy John Daniell

7:07 PM

friends don't let friends swap the order of limits!

As we know from analysis, limit swapping is only defined integrands which behave properly at infinity (and a few other restrictions). We can for "bad" integrands to behave well by adding regularization factors, e.g. $\exp(-|x|)$ in $\mathbb{R}^1$.

So then the question is when it's ok to put in those regularizers. My experience is that it's fine as long as you think about what those regularizers mean in terms of boundary conditions.

Is this making any sense?

@BenNiehoff It's fine if you make sure you know what the swap implicitly means you're doing to your integrand.

@DanielSank There is a sort of duality between the measure-theoretic viewpoint and just viewing the integral as a linear functional on a class of functions, most evident in the case of e.g. Radon measures where you can actually drop all references to measure theory as such

@ACuriousMind Nice.

This is the only Radon measure I know

user image

Measure theory is horrible.

7:09 PM

So, you seem to be thinking about how to define integration in the first place, not specifically about path integration.

@ACuriousMind You could say I'm re-discovering something I once heard about for regular integration but now in the context of path integrals, yes.

@ACuriousMind strange, where did you see this integral?

@DanielSank ...what basis do you have for this claim?

@0celo7 One of the analysis profs here likes to take the "integration is a linear functional" approach to integration and calls the way of defining integration in this way the Daniell-Lebesgue process.

@0celo7 Having studied it in college.

I did not find it interesting at all.

I encountered it in that course of unitary reps of the Poincaré group

7:12 PM

@0celo7 I found my analysis course a bit like lifting weights.

It strengthened me as a mathematician, and in fact greatly developed my ability to think about mathematics, but the actual subject matter itself has not ever proven to be useful to me ever again.

@ACuriousMind ah. I call that the Riesz integral

What I really go from analysis was that it's ultimately not needed. After painstakingly constructing the reals and going through all this measure theory stuff, one finds that on the set of objects I actually care about, we could have started from higher level axioms that make sense and have gotten much the same theory in the end.

Ultimately?

You had a shitty prof, I'm sorry

@0celo7 I'm pretty sure he just doesn't care about the same things you care about :P

@0celo7 Crappy book, actually.

7:15 PM

@ACuriousMind Did you read what he wrote?

The prof was pretty good. His style helped me get over a lot of anxiety about analysis being hard.

@DanielSank which one?

@0celo7 Beals.

Don't read it.

@0celo7 No, I'm just contributing random tidbits to the conversation that don't actually relate to what's being said :P

@ACuriousMind I think you're on the money though.

7:16 PM

@DanielSank never heard of it, not a good sign

It's crap.

Measure theory or analysis?

@0celo7 They are centered for a bound volume. Some space is absorbed by the binding. You should see that the extra space alternates sides (on the left on odd-numbered pages on the right on even numbered pages).

Let me see if I can rephrase: I learned really useful stuff about how to do mathematics in analysis.

However, I have not ever needed to prove the uniform convergence of anything in my entire research life.

I know of people who do, though

or at least, analysis-like things

7:19 PM

@BenNiehoff Sure, and there are people who prove abstract algebra theorems for a living.

...but that doesn't make algebra interesting to me.

No, I mean people who are doing physics with it

(I like algebra, by the way)

@loltospoon A museum (art or otherwise), sculpture park, scenic walk (make sure you know your collective fitness levels and it's better to underestimate), etc.

@BenNiehoff If someone needs to prove uniform convergence of something in order to do physics, I speculate that their brand of physics is rather far away from anything I'd find interesting.

basically, since perturbative series are usually asymptotic, there are people working on how to (Borel) re-sum them using instantons and such

7:20 PM

Activities where the fun doesn't come from people but from where you are or what you are doing.

and it involves caring very much about the details of how/whether something converges

1+2+3+... = -1/12 man.

@BenNiehoff Maybe you can help me understand this...

who cares about any convergence at all?

If perturbative series are asymptotic, why bother with them at all?

I've never understood this point.

7:22 PM

because we haven't reached the turning point yet where they begin to diverge

Is the question to understand how best to get numerical approximations?

@BenNiehoff Ok so the game is to prove how far out you can go and still get useful approximations?

If so, why is that interesting? Is it for precision metrology?

not quite...I think it's more that it's hard to do anything else

@DanielSank Because being asymptotic means the first terms still are a good approximation and no one knows how to get better answers in many cases

AccidentalFourierTransform

> divergent series converge faster than convergent series because they don't have to converge.

@ACuriousMind Right so I'm asking why we're interested in ever-more-accurate perturbation series.

What is the end goal of going past, say, two or three orders?

7:25 PM

because we'd like to find something wrong with the Standard Model

AccidentalFourierTransform

precision tests on the Standard Model

very very important

@DanielSank Detecting whether e.g. the anomalous magnetic moments of fundamental particles are correctly predicted by the Standard Model, for instance.

How about the fact that all of QM is a lie

nah, just the "observer" part ;)

@AccidentalFourierTransform Ah!

7:26 PM

Or, for instance, comparing the perturbative answers to non-perturbative results from e.g. lattice models to see whether the lattice model actually models the same theory

So it is all about getting really accurate numerical predictions. Ok, that makes sense.

@ACuriousMind Yeah, got it.

(It's surprisingly tricky to know what your lattice is really doing)

Are perturbative expansions the state of the art there?

really accurate predictions + measurements make the difference between little bumps in the data being 2-sigma or 5-sigma

@BenNiehoff Yeah, I totally get this. Thanks for explaining it.

7:28 PM

@DanielSank For high energy predictions, yes. For things like low-energy QCD or condensed matter, perturbation theory is often much less useful.

AccidentalFourierTransform

@DanielSank and not only about accuracy. The weak contribution to some problems is very weak, so you need the electromagnetic contribution to a high accuracy just to check the weak contribution properly

that's when you go lattice field theory

AccidentalFourierTransform

i.e., just to check that the weak contribution is there, you need to understand the EM contribution to a very high accuracy

lattice QCD is highly non-trivial and super hard...that's about all my understanding of it :P

I know a guy who works on SUSY lattice computations...apparently that's even harder

AccidentalFourierTransform

fermions dont like lattices

7:31 PM

apparently not

I don't think I understand why

something something fermion doubling

but what about all of these QCD computations I've seen that have quarks in them?

AccidentalFourierTransform

they lied

@BenNiehoff Depends on whether the quarks are static or dynamic

oh, heavy quarks, just sitting there

7:33 PM

Also, I recall that there are several techniques that can control the fermion doubling in certain situations, but nothing that always works

@DanielSank Do you think it is important to know that the Einstein equations can be uniquely solved given some initial data (whatever that means)

Pick your favorite PDE, really

@0celo7 Yes, it certainly is very important.

@0celo7 they don't

AccidentalFourierTransform

$$f''(x)+f(x)=0$$

@Slereah I said (whatever that means) to counteract your smart ass

7:34 PM

right, so my friend has some papers where he exploits some extra symmetry to do it...probably the SO(6) R-symmetry of N=4 SYM or something like that

I don't know the details

I wonder what's a good example of a non-unique development with a matter field that makes sense

@DanielSank You don't think that needs the concept of uniform convergence?

I mean c'mon, uniform convergence is incredibly useful

You need it to prove the spectral theorem

I mean if I pick a matter field that's $T = $something versus $T = $ something $+ \theta(t)$ then obviously it's gonna be non-unique

But that's a pretty fake example

AccidentalFourierTransform

Norton's dome

Norton's dome isn't very GR

7:36 PM

@BenNiehoff This answer indicates it's closely related to anomaly cancellation - since a single fermion species could contribute to an anomaly a lattice theory cannot have, you cannot really put a single fermion onto a lattice.

AccidentalFourierTransform

you never said GR

ah, ok

I think Krasnikov wrote a paper on non-unique spacetime developments

I was talking about @0celo7's GR comment

AccidentalFourierTransform

i know

@0celo7 I'm not explaining myself well...

7:37 PM

@0celo7 He didn't say it's not useful in general - just that he never needed it.

The results of analysis are important, but I don't need to do analysis ever.

It's like lifting weights.

@ACuriousMind I understand that

I'm glad I lifted weights. I'm stronger now. But my day job does not require lifting weights, ever.

(Well, sometimes I have to lift heavy stuff at work, but you know what I mean)

But he said "What I really go from analysis was that it's ultimately not needed."

I think that's a pretty dumb comment, no offense

@0celo7 k

7:39 PM

I think you're willfully misunderstanding it to be a dumb comment, no offense :P

If @DanielSank adds a "for me" on the end, sure

I still think you can do a lot of math that presumably depends on analysis without ever doing analysis.

I learned Fourier transforms and series as a thing for which you need Lebesgue integration, precisely because you have to swap two limits to show that the Fourier transform is defined and invertible.

Looking back, I could have convinced myself of that by instead taking a linear algebra approach, noting that $-i D$ is a Hermitian linear operator.

The Fourier transform is a basis change.

Now, I'm glad I learned about Lebesgue integration and I'm glad I learned about uniform convergence, but I can go about my life doing Fourier transforms and all kinds of other stuff, even understanding path integration, without revisiting Lebesgue.

I can even understand pathological cases without revisiting measure theory.

A great example of this is the old thing where an infinite ladder of inductors and capacitors has a real impedance. This makes no sense because each finite ladder has a purely imaginary impedance.

How can the sequence converge to a real number!?

Well, it's because of a limit swap, and I understand what's going on without Lebesgue.

The limit swap is justified only if the LC ladder has some nonzero resistance. This is exactly equivalent to shifting a pole in the real plane by giving it an $i \epsilon$ part and taking that to zero at the end.

So I understand this pathological mathematical situation in terms of integral regulators and boundary conditions and I don't need Lebesgue at all.

::shrugs::

AccidentalFourierTransform

user image

I have no idea if this is making sense to anyone else.

AccidentalFourierTransform

:-P

7:47 PM

Did that case make sense to anyone else?

Maybe I should turn that into a self-answered question.

It's related to the Caldeira Leggett model.

AccidentalFourierTransform

oh yeah do it

and ping me

@DanielSank sorry, studying for a QM exam

reading now

@DanielSank Ok, that's not wrong.

I think the usual physicist proof of that is wrong (it probably assumes the $x$-eigenstates form a basis), but ok

@0celo7 The x-eigenstates do form a basis on the set of functions I care about.

I don't care if my theory doesn't work on crazy-ass functions.

They're not even in the Hilbert space

@0celo7 So what?

7:51 PM

So, basis of what is the question

Forget the hilbert space.

@0celo7 Basis of things with well-defined inner products with things in the Hilbert space ;)

Not even kidding.

@0celo7 ::shrugs:: Physicist proofs exist because physicist care more about agreement with the universe than they do about airtight formalism.

@dmckee I refuse to believe that no physicist has been wrong because he used a physicist proof. You don't hear about the failures

AccidentalFourierTransform

we hear a lot about you

woah there, buddy. Be nice.

Hey, @0celo7 I'm solving a topology problem.

7:53 PM

@DanielSank That's a rigged Hilbert space, I think

But I don't know anything about those

@0celo7 Can we call "rigged Hilbert space" a "Trump space"?

@0celo7 I'm quite sure it's happened, but there is an adage "don't let the perfect be the enemy of the good".

The time lost to the mistakes adds up to less than the time that would have been lost in waiting to make the case airtight in the first place.

@DanielSank Only if you have 5 million illegals in the space

So physics gets ahead better by plunging in with our trousers only half zipped.

@dmckee I go one step farther, actually.

7:55 PM

commando?

Hear me out, please. Coming out of real analysis, I didn't remember the epsilon-deltas of everything. However, I had learned how to think. For example, I learned that one can define e.g. integrals on known cases and then build up the definition on complex cases as limiting sequences of known cases. That was a really useful insight.

That kind of reasoning has been very useful.

I'm not saying you have to remember all of the details

I don't even remember why we're arguing

Hello

'sup?

user image

Topology problem: arrange the parts of this circuit so that the RF traces don't have to overlap.

@0celo7 go.

Show me your skillz.

(I'm joking)

7:59 PM

I'm not a topologist!

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