The h Bar

0celo7

15:08

@skullpetrol IT WORKS

THAT'S WHY JOST SAYS WLOG

Bass

p.64 of Zee's QFT nutshell: $\langle\vec{k}_3\vec{k}_4|e^{-iHT}|\vec{k}_1\vec{k}_2\rangle = \langle\vec{k}_3\vec{k}_4|e^{i\int d^4x\mathcal{L}(x)}|\vec{k}_1\vec{k}_2\rangle$. Shouldn't there be a path integral $\int Dq(x) e^{i\int d^4x\mathcal{L}(x)}$ on the RHS?

Secret

where is Slereah?

yuggib

@Secret out there living la vida loca? he's unemployed...

Bass

if Zee's equation is true for all $\vec{k}_i$, then $-iHT=i\int d^4x\mathcal{L}(x)$, so the Hamiltonian would be equal to the Lagrangian (up to some constants)

ACuriousMind

@Bass I don't see why a path integral, but that equation is definitely fishy.

yuggib

15:17

@ACuriousMind I think he means a path integral without a scalar product...the left hand side seems to suggest something like the integral kernel of the evolution operator

anyways, it is fishy indeed (whatever that means)

Bass

@ACuriousMind well previously he showed that $\langle q_F|e^{-iHT}|q_I\rangle = \int Dq(t) e^{i\int_0^T dtL(\dot{q},q)}$, which is (as I understand it) the essence of the path integral formalism

$q_F, q_I$ being some initial and final state

if it's fishy anyway I'm just going to ignore it

ACuriousMind

@Bass Yeah, but there you have position states. I'm honestly not sure what the equation Zee wrote there is supposed to be

0celo7

Zee? Lemme get on my laptop, where I can get the book

ACuriousMind

The accompanying text is not very helpful :P

Bass

@ACuriousMind ok never mind. maybe I really shouldn't look too closely while still in that chapter 1

0celo7

15:23

@ACuriousMind You read it?

ACuriousMind

@0celo7 I looked it up in my completely legal copy, yes.

0celo7

I own the book

So my legal copy is actually legal

At least moral ;)

@ACuriousMind Indeed

I never understood that section

QFT sucks

::goes back to Jacobian determinants::

yuggib

@0celo7 You should start studying operators...are not so different from matrices and could become useful if you plan to be on PDEs

0celo7

Doesn't that require analysis

yuggib

I had a PhD course on matrix theory (for numerical analysis), and ended up doing all the proofs using operator theory :-D

@0celo7 not so much as you may think

Slereah

15:29

I am here

yuggib

of course it is needed to do operators in concrete spaces (e.g. $L^p$)

0celo7

My goal right now is to learn geometry that's not physics

Including this really poorly written proof

yuggib

why on the world would someone do that??

0celo7

But I refuse to ask Bajoran, I won't give him the satisfaction

Slereah

@0celo7 Try Euclid

It's a classic

Hilbert wrote a modern day version of Euclid's elements, too

With actually consistent axioms :p

Euclid's axioms aren't consistent

you can prove weird things with them

0celo7

15:35

Well, the Jost proof still makes no sense.

@yuggib Because geometric analysis seems interesting

yuggib

@0celo7 then you should immediately aim high:

web.math.princeton.edu/~seri/homepage/courses/Analysis2011.pdf

0celo7

Why

yuggib

because you will learn something useful

0celo7

Does that even raise GDP

yuggib

of course

it doesn't

0celo7

15:46

Seriously

What does actually increase GDP

yuggib

studying at the university

by means of paying tuition fees

0celo7

That's deficit financing

ACuriousMind

@yuggib Only in countries that have those :P

yuggib

@ACuriousMind every country has a small fee to pay for university enrollment I think

Italy and France have those

Slereah

Yeah

yuggib

15:50

they are very cheap compared to US, but still are non-zero

Slereah

But it's 400€ in France

yuggib

per year?

Slereah

It's thousands of dollars in the US

yes

0celo7

Tens of thousands

yuggib

in italy is more than 1000€

still cheap compared to US

ACuriousMind

15:51

@yuggib Hm, well, I would not call the ~150€/semester I pay here for access to the university's facilities and a cheap universal ticket for public transport a "tuition fee".

yuggib

@ACuriousMind Well ok, anyways it is a fee that you have to pay

call it a tax, or a (very partial) recovery of university's costs

Secret

@Slereah
gah, he refused to talk because he thought I have relayed all the info to you and that you disagree with all of his points

I should not have been the relay person, that way I won't miss this important opportunity to extract infomation from that potential conversation

However, he said he wants your facebook to talk later

yuggib

there are still people very convinced that the wavefunction has to be continuous...

how disappointing

in general, it does not even make sense to evaluate a wavefunction at a point

Bass

@0celo7 here's how to raise GDP: goo.gl/hckzgj

Secret

@Slereah
Is this your facebook?
https://www.facebook.com/samuel.lereah?fref=ts

ACuriousMind

15:58

@yuggib What?! Next you'll tell me there are no position eigenstates or that I can't apply the momentum operator to everything... ;)

yuggib

:-D

damn oversimplifying books that end up saying just false things

0celo7

What

How does quantization of angular momentum work

yuggib

The statement, “in general, wavefunctions are technically meaningless when evaluated at a point”, is not correct, IMHO; please provide a proper reference material. Many local physical quantities are obtained from the wave-function at a give position — as I have mentioned above. Notice also that not all valid wave-functions are $L^2(\mathbb{R}^d)$; for example, the eigen-states of the momentum operator (which are $\propto e^{\mathrm{i} k x}$). Consult the reference I have given above: Ballentine, section 1.4. @yuggib — Khwārazmi 40 mins ago

0celo7

Does the wave function have to be single valued?

yuggib

@0celo7 technically, it is not a function

ACuriousMind

16:01

@yuggib The reluctance of many physicists to actually admit those simplifications never cases to astound me. I recently suggested to someone that if they want to pretend everything is linear algebra, they should teach finite-dimensional systems, not particles. They indignantly refused because apparently doing QM on $\mathbb{C}^n$ is not "intuitive" to students.

Slereah

@Secret I guess

I don't use it though

yuggib

@ACuriousMind Ahahahah

Secret

In that case it is not very usefull, but I'll just tell him anyway
After all, it is HIS problem

Slereah

Spinor wavefunctions are multivalues

yuggib

I am always puzzled as well

Slereah

16:03

Also who is that guy @Secret

And why doesn't he just come here

And why is he mad

Should I get a gun

Secret

facebook.com/yatsen.sun.7?fref=ufi

yuggib

@ACuriousMind Anyways, QM on $\mathbb{C}^n$ would be so nice...it is a pity that then with second quantization you still end up on $L^2(\mathbb{R}^n)$

Secret

Well, if we think of a logical peerspective, he does not even have a stack exchange account

But the main reason, according to him, is that he think you shot down all his points

but there's a recent bit of the discussion we have that have not been relayed to you

but he said I have already done that, and we had a small argument

Slereah

Did I shoot down anything

I don't remember saying much

ACuriousMind

@yuggib Well, some people doing quantum mechanics never get to second quantization!

And they don't need it, either

And with quantum computing and the like, finite-dimensional quantum systems are actually not that abstruse a topic to deal with

Slereah

16:11

quantum computing is pretty low key

Basically all eigenstates are finite

(basis)

yuggib

@ACuriousMind Yes, it's true

but matrices are so boring

ACuriousMind

@yuggib Well, look at the projective spaces of states as complex manifolds/varieties and do geometric things with them if you're bored by the Hilbert spaces ;)

There are interesting relations between entangled states of qubits and special fibrations of spheres, for instance

Secret

I am tired of being the relay perosn, keep on getting infromation wrong, his reason is shown here, apologies for the previosu misinformation

yuggib

@ACuriousMind I am bored at matrices, not Hilbert spaces. Anyways, I am well enough with doing topology on non-separable spaces

Secret

nvm, that (removed) does not make sense

0celo7

16:22

@ACuriousMind is there any result that hinges on QM not being linear algebra

ACuriousMind

@0celo7 $[x,p] = \mathrm{i}\hbar$ is impossible in linear algebra, for one.

0celo7

@ACuriousMind besides that

ACuriousMind

What do you mean besides that!? That means you cannot have operators with the same properties as $x$ and $p$ in linear algebra at all!

0celo7

@ACuriousMind damn, I might need your help on this proof

Secret

Also apologies for my strong language on top (the caps), I am not having a very stable mental state today because of the exictement x worry of tomorrows party

and that so many things happening all at once

0celo7

16:23

I'm soooo close to understanding it

but I want to stop giving you the satisfaction of explaining things to me

@ACuriousMind what

So?

It's linear algebra with strange commutation relations

I don't see the issue.

yuggib

every irrep of the Heisenberg group (over a finite dimensional vector space) is unitarily equivalent to the schrodinger representation

here you go with something that is hardly provable with linear algebra

0celo7

but why does that matter

ACuriousMind

@0celo7 Because if not all representations are isomorphic, how do you know which one to pick?

0celo7

@ACuriousMind what?

just pick the one with $[x,p]=\mathrm{i}\hbar$

I'm not trolling here

ACuriousMind

No, they all have that.

0celo7

16:27

I just think you guys are overthinking this

@ACuriousMind pick $x=x$ and then it's obvious $p=-\mathrm{i}\hbar \mathrm{d}$ does the trick

ACuriousMind

Yes, that's one possible representation. How do you know it is the physically correct one?

0celo7

@ACuriousMind did....

you referenced a post before it was posted

holy shit

ACuriousMind

@0celo7 No, perfectly correct order on my screen.

0celo7

you're a time traveller

yuggib

correct order on mine too

0celo7

16:29

Physics Meta

1

Q: Modification by Community to question in Recently active tab

I was reading the questions on RECENTLY ACTIVE tab which I don't often do (I usually look up the newest questions tab) and found this. But I had viewed the question just a few minutes before that and after this so-called modification by the Community, I re-viewed the question and found it exac...

discussion

0celo7

@ACuriousMind does it make verifiable experimental predictions

answer: yes

ACuriousMind

Okay, if you only want the theory to be post-hoc justified, alright. Still, there's a lot commonly used that doesn't make sense in linear algebra. Like: WTF is $\int \lvert x \rangle\langle x \rvert \mathrm{d}x = 1$?

0celo7

It's the identity operator, duh

ACuriousMind

The LHS is the issue.

0celo7

16:34

It's an integral

Maybe you should take some calculus

ACuriousMind

I can't tell if you're being obstinate or if you genuinely think this isn't an issue.

0celo7

I'm being obstinate

But I also don't think it's an issue.

AFAIK no particular measurement issues hinge on what exactly $\int |x\rangle\langle x|\mathrm{d}x$ means

if you can find one, great

but physics is about measurements, nothing more, nothing less

yuggib

what do you mean by measurement issues?

ACuriousMind

@0celo7 No, the hackjobs you can do all give the right answer. The problem is that there is a lot of formal "linear algebra manipulations" that simply give nonsense answers, like $[x,p] = \mathrm{i}\hbar$ giving $0=1$.

0celo7

@yuggib will you produce different numbers for the outcome of an experiment using your fancy $C^*$ algebras, etc. than some engineer using a linear algebra analogy

Slereah

16:37

It's an operator such that $\int \langle a \vert x \rangle \langle x \vert b \rangle = \langle a \vert b \rangle$ :p

yuggib

I think measurements are quite related to spectral properties of observables

and spectral properties of observables has a lot to do with the spectral theorem

0celo7

I'm not arguing that a rigorous understanding should not be achieved or developed

ACuriousMind

Without the proper formalism, you don't know whether a formal manipulation is allowed or not - you have to guess, and every time you get nonsense you just have to start over without any idea what went wrong.

0celo7

But it's completely unnecessary to bog down intro QM with these details

@JohnDuffield What does that mean?

Why is my name italicized

yuggib

@0celo7 Let's cut it short; justify with linear algebra the existence of observables without a complete set of eigenvectors

or that have an unbounded spectrum

0celo7

16:40

I'm not sure what that question means

ACuriousMind

There are plenty of questions on this site where people derive apparent contradictions from applying naive methods to the operators and states, and it is terrible style to not teach people that there are subtleties, because that leads to people like the friendly fellow berating yuggib that wavefunctions should be continuous.

0celo7

@JohnDuffield I know what you were referencing

But why is my name italicized

@ACuriousMind So QM should only be accessible to people who have taken courses on analysis, functional analysis, measure theory, godknowswhatelse?

ACuriousMind

@0celo7 From my point of view, it is only accessible to those already, and the others are just using tools they do not understand.

0celo7

So there should be no courses in QM unless analysis, etc. are prerequisite?

ACuriousMind

@0celo7 As I said, I am in favour of teaching QM in finite-dimensional systems first.

yuggib

16:47

@0celo7 Sadly, the non-commutative "world" is rather different from the commutative one.

0celo7

Will resume in a bit. Gotta walk to dining hall.

Secret

Here's an interesting question for you guys: How to show that a machine that is claimed to be able to send a signal back to itself 5 mins ago and trigger a stopwatch to count is actually a time machine. What will be the predicted observable consequences?

How will this be modelled mathematically, Will attempt to model the state of the stopwatch in terms of state vectors too risky because there might be relativistic effects have not accounted for?

for simplicity, we can assume the signal travel back via a wormhole with the required time difference (is this the correct term?)

0celo7

@JohnDuffield why

John Duffield

@Secret : I'm sorry, but the predicted consequence is paradox and contradiction and reductio ad absurdum. See articles like this which explain why time travel is science fiction. I know that a lot of people of people have fallen in love with the idea of time travel and don't want to let go, but IMHO you should, because it's woo, and there's some real interesting physics you should get into instead.

0celo7

Man, Coke Zero is like taste bud cocaine

Slereah

16:57

I once was in the Czech republic for holidays

They have an energy drink just called "Cocaine"

That's some marketing gimmick

John Duffield

@0celo7 : because people listen to you, because you're the numero uno popular humorous chatroom guy.

0celo7

@JohnDuffield aww that's so sweet

I think @ACuriousMind has the most credibility here

get him to agree with you!

ACuriousMind

(removed)

Slereah

I wanted to read some HE but then I played Fallout all day long

oops

John Duffield

Coca cola once contained an estimated nine milligrams of cocaine per glass. In 1903, it was removed.

0celo7

17:01

@ACuriousMind NEs have to take a course on quantum mechanics to understand neutron scattering, etc. This QM class is of course one that you said should not exist. So should NEs have to take a lot of math just to "understand" QM?

To be honest you'd kill the program by doing that.

John Duffield

@0celo7 : you're pulling my leg, surely?

0celo7

The average engineer does not give a flying hoot about functional analysis.

Which is a graduate level class, anyway.

What are the undergrads who need some QM supposed to do?

You want to make nuclear engineering a discipline that requires a masters degree in mathematical physics?

ACuriousMind

What do they need it for?

@0celo7 Never said that, you're extrapolating.

0celo7

Calculating cross sections, decay rates

yuggib

@0celo7 Probably they would be ok understanding mock QM in finite dimensional systems

as @ACuriousMind suggested

they have to keep in mind it is almost a fairytale anyways

ACuriousMind

17:04

@0celo7 Can be done in $\mathbb{C}^n$, just introduce a energy cutoff.

yuggib

@ACuriousMind A bit more than that :P

0celo7

Some parts of NE is all about solid state QM, atomic lattices, scattering in lattices

Spectroscopy, MRIs, PET scans

ACuriousMind

Yeah, I am of the strong opinion that all the lattice people could use a good dose of actual maths :P

0celo7

All of that requires QM and NEs take courses on it

In any case, I'm getting a math degree and probably won't know enough to do "proper" QM.

Now I'm sure you could say my school's math program isn't amazing, and I won't argue there.

But the NE program is world-class and no one knows about operator algebras.

ACuriousMind

BTW, you don't have to make it a proof-based course. No need to prove all those horrible theorems you think are so useless. But mention them, and don't pretend you're doing linear algebra when you're not. One never proves the finite-dimensional spectral theorem anywhere in physics, either, and nobody worries about that, so just cite the theorems.

0celo7

17:07

I don't think they're useless. They're useless for most people.

@ACuriousMind The average engineer can't tell you what isomorphic means.

@JohnDuffield No, why would I be?

ACuriousMind

@0celo7 So? No shame in not knowing things. And if you don't have the aspiration to be anywhere near rigorous to begin with, fine, do QM however you like.

The problem is physicists teaching QM as linear algebra and acting as if there were no problems, or as if everyone pointing them out is just an annoying pedant.

John Duffield

@0celo7 : l'll give you an example or two. But not now. It's Friday night, and I've got to go.

0celo7

17:25

@ACuriousMind Please don't get the impression that I don't want to be rigorous with QM. I'm planning to take functional analysis so I can understand the foundations of what I'm going to be doing for a significant part of my life...

ACuriousMind

@0celo7 That wasn't "you" as in "you, personally"

0celo7

@ACuriousMind Ah

ACuriousMind

It was "you" referring to the hypothetical "average engineer" in your statement

0celo7

Yes, yes.

Apparently the post-doc I'm trying to talk to is off at a conference with a prof.

@Fermiparadox Welcome

Never seen you here before

Fermi paradox

hello :)

Well, i rarely drop by.

Been here 3-4 times for a few minutes or so.

0celo7

17:30

Oh no chat star person is about to disappear

Fermi paradox

what do u mean?:P

0celo7

the last thing on the star list

Fermi paradox

There is a vid, with a simulator (its a game actually) showing a big object colliding with earth Which are the unrealistic parts of it that should be changed in the vid?

ACuriousMind

Not the last on my screen

0celo7

it's about to disappear when something else gets starred

Fermi paradox

17:31

Oh, i see.

0celo7

@ACuriousMind did you click on the "see more" option

Fermi paradox

(my linking skills have failed me)

ACuriousMind

...I don't see a "see more" option oO

@Fermiparadox they haven't, multiline messages break markdown - don't press shift+enter if you want links and formatting to work.

Fermi paradox

@ACuriousMind Nice :)

0celo7

@ACuriousMind you don't see a "show 4 more" option?

ACuriousMind

17:35

No, and I never clicked it, but I do indeed see 4 more.

0celo7

strange

maybe it has to do with screen res

i'll goto 1200p

NeuroFuzzy

Have no fear! I'm here!

0celo7

@ACuriousMind it's a screen res issue

when I go to "small text" mode I get all of them

obe

@0celo7 how did it get 10 stars? lol

i feel like you've said similar things before.

Danu

17:52

@0celo7 number uno chatroom pro confirmed

lol

Everybody listens to the almighty cat

Bass

I understand that in QFTs, where fields are operators, one talks about hermitian or non-hermitian fields. But what does it mean for classical fields, like classic solutions of Klein-Gordon? What does it mean for $\varphi$ to be hermitean?

Transcript for