@0celo7 Yes.

I knew about the residual theorem, that if your contour encircles some singularities of a function, then the value is just the values of the singularities added together (or smoething like that)

@DanielSank Well, if you find what you're searching for please show me! I have a feeling we're talking slightly past each other here, but I don't think what you're asking for has currently any chance to exist.

@ACuriousMind That's awful if true.

While that is correct, I'm not so sure Stokes theorem lifting to complex manifolds is so trivial. But maybe I'm wrong.

Dammit, this might mean I wind up writing something, and I don't have time for more of that.

@0celo7 It works find in the plane.

@ACuriousMind probably does not know off the top of his head

@DanielSank I mean the proof.

@ACuriousMind Well, it is good to know. From now I will only do this, if I think I can make an enough important change to significantly shift the review decisions.

The proof of residual theorem strongly reminds of green's theorem because in one complex analysis lecture notes they started drawing circles around the singularities and that reminds of circulations

@0celo7 You're taking your mathematician's expectations to his physicist's explanations. I'm not getting into that fight :P

@ACuriousMind You mean engineer's expectations.

speaking of poles in complex analysis, that reminds me of one question...

@0celo7 No, dude, you can literally translate complex contour integrals into a pair of two integrals of 2D real vector fields.

@0celo7 For 1 dimensions? Yes. For higher dimensions, no idea.

Stoke's theorem goes through verbatim.

@DanielSank That's because $\Bbb C$ is nice.

Here, look:

^ that is the proof for 1 dimension, what @DanielSank said.

@DanielSank Again, you might be misunderstanding me: I'm saying the generic QFT path integral is not understood well enough to write what you're asking for. It's not that no one has bothered to write that down, it's that we mostly just stumble our way through specific applications of a general case where we have no idea what the general case actually does (beyond being an integral).

I'm thinking about an arbitrary complex manifold.

i.e. is it true that $$\int_{\partial M}\omega=\int_M\partial \omega+\bar\partial\omega$$

@ACuriousMind Yes but you can usually do something like lattice-ify your space, work things out, wave your hands at the limit as the lattice spacing goes to zero, and get perfectly happy answers. If there are divergences it's rather clear why they're there, etc.

@DanielSank Oh, you haven't seen what the lattice QFT people have to deal with, have you? If I remember correctly, no one even knows whether the continuum limit of 4D lattice $\phi^4$ theory really is the continuum $\phi^4$ theory. In some dimensions it is known to not be the case - the continuum limit just becomes the free field.

@ACuriousMind All I know is that in many body physics course we discretized stuff and computed useful quantities.

It didn't seem that weird.

@0celo7 Er. $d\omega = \partial \omega + \bar{\partial} \omega$.

Are you sure these problems you're talking about don't come from lack of proper regularization?

You know, I really wish that answer of mine about contour integration were accepted.

That follows from how $\partial$ and $\bar{\partial}$ are defined, not?

@BalarkaSen The complex differential is not simply $\partial_i \omega\wedge\mathrm dx^i$ though.

@DanielSank The point is that the "proper" regularization on the lattice tends to produce the free theory in the continuum limit

It bugs me when good question/answer don't get an accepted answer.

Maybe if you work everything out

But $\mathrm d\omega$ on a complex manifold is slightly different than $\mathrm d\omega$ on a real one

I...think.

@ACuriousMind Even if you add in some kind of loss channel?

Maybe I'm wrong

I duck out as I don't know this stuff.

@BalarkaSen I duck out too

I mean, when we bump poles off the real line, all we're doing is adding damping.

@DanielSank I don't know what that means

Maybe I'm wrong too.

@BalarkaSen Maybe we're both wrong

I'll check Kobayashi

Maybe I agree.

@ACuriousMind I mean that isolated oscillators/fields/whatever often behave poorly, but that behave well if they're coupled to something else which can wash out singularities.

@DanielSank Which answer?

@BalarkaSen The one I posted above. Scroll up.

Ah, OK, I didn't note that was posted by you.

I think it's a pretty good answer. Take a +1.

@DanielSank ahahaha, well, then you're in the realm of thermal field theory or even non-equilibrium theory, which requires different techniques still. Most non-equilibrium simulations are fine in this sense, although they might only be because we have no chance in hell to compare them to an analytical result, anyway...

@ACuriousMind You laugh...

But dude, isolated oscillators are pathological even in the absolute simplest case.

Try analyzing a driven simple harmonic oscillator. It's totally pathological. However, if you add damping everything is fine, and it's almost always the case that you can take the damping to zero at the end of your calculation and get a reasonable result.

@DanielSank I laugh because you kinda remind me of this guy. Yes, the things you say work, but they spawned entirely new subfields instead of fixing the original problem.

@DanielSank what do you mean by pathological for a single SHO?

@ACuriousMind Based on my experience, which is limited, your position that this is all very complicated is caused by having learned the techniques in the context of complicated systems. I guess the techniques themselves are not so hard.

I have this nitpick about proving Cauchy's theorem by applying the Green's theorem though, but I am not sure if a physicist would appreciate it. If you apply Green's theorem, you're assuming the real and imaginary parts are differentiable, hence that $f$ is $C^1$. That's slightly stronger than assuming $f$ is holomorphic, in a way.

I know that sounds presumptuous, but this has happened to me in the past more times than I can remember.

@EmilioPisanty The Green's function has a singularity which goes away if you introduce some damping.

@BalarkaSen Wait really? I thought holomorphic was stronger than C^1.

Yes, holomorphic means analytic, thus the "in a way" :)

@DanielSank fair enough, and I'll look it up, but what does "damping" mean, exactly?

But the proof of that uses Cauchy's theorem, so you're in circles if you want to apply that :(

I don't know of a proof of holomorphic implies continuous differentiable which is Cauchy-free.

@EmilioPisanty Basically this.

@DanielSank oh god

too much math for right now

@EmilioPisanty What?

@EmilioPisanty Dude, it's turbo-simple integrals :|

@DanielSank believe it or not I'm still thesising at this time

=/

This is the reason most complex analysis book introduces the proof differently, by Goursat's lemma (Cauchy-theorem on a triangle) and then constructing an antiderivative explicitly.

Awwww, it's just a few Fourier transforms.

@BalarkaSen I don't understand.

Why geometrically a complex integral of this form $\oint \frac{1}{z}=2\pi i$ nonzero. We knew that a closed contour encircling a singularity cannot be continously deformed into one that is located in a simply connected region, but

What's wrong with working with differentiable functions?

@DanielSank The point I made at the beginning was that in this case the technique doesn't really generalize. The QM path integral is well-defined, usually well-behaved, and well-understood, and you can in principle directly evaluate it to compute stuff. The QFT path integral is horribly ill-defined, usually assumed to behave similarily to the QM path integral, and a purely formal object that you always need to translate into something else to even have a chance at computing it.

what geometric properties of that loop determines the value you get for a singularity. E.g. if I have two loops that are encircling a pole of 1st and 2nd order respectively, then it is easy to find out what they integrate to, but why are these loops different, because they both seemed to be just circling a hole of genus one?

@DanielSank $C^1$ is stronger than differentiable.

@DanielSank Nothing. But most people introduce holomorphic functions by the derivative existing, not assuming that the derivative is also continuous.

@Secret Geometry of the loop doesn't matter. All that matters is what singularities it encloses.

You need $C^1$ to use various integral theorems

@BalarkaSen Meh, ok.

@Secret The genus isn't enough. A second order singularity is basically two first order singularities sitting on top of one another.

@DanielSank Heh, I knew you wouldn't like it :)

I give you all your points about the shifting of poles or the other things where you dislike that the QFT people act as if they invented it, but this is a different thing: We're desperately trying to generalize a well-known technique and are still, in general, failing at making it precise.

So really, a second order singularity isn't genus 1.

@DanielSank How general is the technique? Basically what you're doing there is adding a damping term so the Fourier integral will converge, but if memory serves, you do in some cases have to choose the damping terms carefully (I forget the details, but working out some periodic Fourier transforms, basically). I guess you could use distributions to do things more rigorously?

@DanielSank Ah now I start to get why $\oint \frac{1}{z^n}=2\pi n i$

@alarge The various ways of moving poles have various physical meanings.

@DanielSank No, but seriously: what is that damping physically?

It's not in the actual QM of an isolated system

Pushing them both up (or down depending on your Fourier transform convention) is damping. Pushing one up and one down is Wick rotation.

So you need to be coupling to some external system and then tracing over it.

@EmilioPisanty Sure it is. Adding damping to Schrodinger's equation means that the system has some probability to disappear. If you're working with bound states, then you're giving the system a probability to escape the bound state. You can take that probability to zero later if you want.

But until you actually do the tracing out, and show that there were no ugly singularities in there to begin with, I'm not sold that this fixes the problem.

@EmilioPisanty Yes, adding damping is just a shortcut.

@BalarkaSen Should I continue with Bott & Tu and learn about spectral sequences, or spend some time on Riemannian geometry and geometric analysis?

@EmilioPisanty Well, in reality, your system almost always actually is connected to some external environment anyway.

I vote for Riemannian geometry & geometric analysis. But you should do what makes you happy.

@BalarkaSen Why not spectral sequences?

@DanielSank Yeah, but that's more of an argument for doing the full coupled-system calculation than it is for jumping to effective dynamics.

I'm getting a little burnt out on algebraic topology

@0celo7 That's dirty algebra.

@EmilioPisanty You can look at it either way. Throwing a damping term in there for the sake of regularization is at least physically motivated by the fact that in reality you probably do have some coupling to the environment there.

And the concept of "presheaf cohomology" might not be something I'm ready for.

If you want to learn algebraic topology, look at Hatcher.

@BalarkaSen I own Hatcher

But I don't really feel like that right now

Or learn spectral sequences from Hatcher's notes.

@BalarkaSen I would learn them from Bott & Tu, presumably.

But I guess that requires more topology than you know, so nevermind.

Daniel, I cannot wait that your physical visualisation technique will allow us to visualise general ring algebraic structures, cause that will be really cool (and rigorous enough to make layperson to understand rings in the most general sense)

@0celo7 Yeah, just that Hatcher's writing is more lucid and geometric.

@BalarkaSen Bott & Tu are turning this into an algebra book, I agree.

So it seems.

They define a locally constant presheaf categorically

@Secret What's a general ring algebraic structure?

OK, now I got to sleep. Have a lot of work to do tomorrow.

The arrows are isomorphisms and all groups are isomorphic

Something like that

@Secret Since when is Sakurai an undergrad book

@ACuriousMind @alarge @EmilioPisanty Somewhere I have a nice writeup of how the Caldeira Leggett model works. I should share that in a self Q&A. The model shows how an infinite set of lossless oscillators looks like a lossy element which can add regularization.

@DanielSank en.wikipedia.org/wiki/Ring_(mathematics) sorry I was talking about this, I added the words "algebraic structure" so that we won't be mistaken for other ring like things e.g. superconductor rings (sorry mind mind tend to do such things to avoid any possibility of misconception)

@0celo7 Crap, another mistake! I really have to organise my mind a bti more

And I'm not sure Shankar is undergrad either

It's just as advanced as Sakurai, which is supposedly a graduate book

I own both

Shankar is great for undergrad.

At least it explains wtf is going on, unlike Griffiths.

@DanielSank I'm using Sakurai for my course but I've read Shankar already

@0celo7 Sakurai has the best chapter on angular momentum of the books I've read.

@DanielSank Great.

Sadly neither mention quantum computing at all

The rest of it is meh, although it does have a homework problem that I found extremely important.

oh?

In the chapter on perturbation theory, there's a homework problem asking you to do perturbation theory in the case where the unperturbed system has degeneracy, and where the perturbation does not lift the degeneracy to first order.

This is a critical problem to solve.

You have to think carefully to get it right.

@DanielSank I'll keep it in mind.

I figure $\alpha$ and $\beta$ are all the momenta of the in and out state particles, but what is that commutator looking operator doing in the middle?

@Secret What do you mean "what is it doing there"? It appears when you compute the creation/annihilation operators in terms of the field.

@DanielSank I'm not sure why exactly I should be interested in that model, but I usually enjoy reading your posts once written, so go ahead and write it ;)

@ACuriousMind It's just an example of a very "complicated" topic, originally explained via path integration, being easy to understand in a relatively simple system.

@ACuriousMind in the article, f and g are the flat waves of the field, thus there should be a way to interpret what $\overleftrightarrow{\partial}_0$ is doing to the flat waves?

I also noticed that $\overleftrightarrow{\partial}_0$ look very similar to a term in the probability current of QM, i.e. $\psi^*\nabla \psi - \nabla \psi \psi^*$

so is it some kind of current of the field?

@ACuriousMind what is QOGS good for

@0celo7 what?