The h Bar

Slereah

20:00

Hm, let's see

⊢ ((Fun ◡𝐹 ∧ (𝐹 ↾ 𝐴):𝐴–onto→𝐶 ∧ (𝐹 ↾ 𝐵):𝐵–onto→𝐷) → (𝐹 ↾ (𝐴 ∖ 𝐵)):(𝐴 ∖ 𝐵)–1-1-onto→(𝐶 ∖ 𝐷)) means...

Fun ◡𝐹 > the converse of $F$ is also a function

(𝐹 ↾ 𝐴):𝐴–onto→𝐶 > The restriction of $F$ to the subset $A$ is a function from $A$ to $C$

Anonymous

@Slereah What type of notation is that :P

Slereah

(𝐹 ↾ 𝐵):𝐵–onto→𝐷 > The restriction of $F$ to the subset $B$ is a function from $B$ to $D$

Anonymous

@ACuriousMind That's it indeed!

Slereah

Then the restriction of the function to $A \setminus B$ is a function from $A \setminus B$ to $C \setminus D$

It's a mix of logic/set theory notation with machine-readable notation

Anonymous

@ACuriousMind So I'm a bit confused now. If continuity preserves connectedness why do they say that:

Anonymous

20:04

Deleting a line in $\mathbb{R}^2$ doesn't necessarily mean deleting a line in $\mathbb{R}^3$. — Qiyu Wen Jul 3 '16 at 6:51

ACuriousMind

@Blue They mean that a line in $\mathbb{R}^2$ can get mapped to an arbitrarily convoluted curve in $\mathbb{R}^3$ instead of a straight line.

Nothing to do with connectedness

Anonymous

Oooh, I see

Anonymous

Basically like we parameterize a curve in $\Bbb R^3$ using real values from $\Bbb R$

enumaris

seems like I got time to do some more reading...

still freaking warm in here tho

Anonymous

That curve would be homeomorphic to the real line

Slereah

20:07

though homeomorphisms also preserves compactness

Hence it cannot be a circle either!

Only some line

Anonymous

However, I can't imagine that curve being so weird that it divides $\Bbb R^3$ into two different connected components

Slereah

It cannot, no

Well, it can

But not if it's via a homeomorphism

enumaris

there are space filling curves...dunno how you map to them tho

Slereah

Usually by sequences

but those are not homeomorphic

23

Q: Is it true that a space-filling curve cannot be injective everywhere?

Here, it is said that a space-filling curve cannot be injective because "that will make the curve a homeomorphism from the unit interval onto the unit square", since every continuous bijection from a compact space to a Hausdorff space is a homeomorphism. That does mean there can be no injective ...

enumaris

I made a pretty tableau plot

seems kinda useless tho

Anonymous

20:12

Reading those answers I'm still not clear whether $\Bbb R^2$ and $\Bbb R^3$ are homemorphic

Anonymous

Are they?

Slereah

they are not

enumaris

would be weird if they were, can you imagine

Slereah

Generally, an $n$-dimensional manifold is never homeomorphic to an $m$ dimensional manifold unless $n = m$

(Or both are empty)

Anonymous

@Slereah And what's the exact reason?

enumaris

20:13

in what situation can you have n!=m but they are both empty?

I guess n is undefined for empty set so that's why the caveat?

Slereah

Well for $\mathbb{R}^2$ and $\mathbb{R}^3$ you can use algebraic topology

enumaris

hmmmm

Slereah

Otherwise there's a fancy theorem

bolbteppa

Not obvious how that reproduces the Poisson bracket as anything but a formal thing

Slereah

@enumaris The empty set is a manifold of dimension $n$ for any $n$

enumaris

20:14

hmmm

Anonymous

@Slereah Lol, I know nothing of algebraic topology. Is there anything from first principles which I can read through?

Slereah

Every point of the empty set is contained in some neighbourhood homeomorphic to a subset of $R^n$ (true because it has no points at all)

The atlas of open sets is $\{\}$

Semiclassical

@bolbteppa which book is that from?

Slereah

@Blue Basically this stems from the fact that loops are contractible in $\mathbb{R}^3 \setminus \{ 0 \}$

But not in $\mathbb{R}^2 \setminus \{ 0 \}$

and this is a property which is preserved by homeomorphism

96

Q: Elementary proof that $\mathbb{R}^n$ is not homeomorphic to $\mathbb{R}^m$

It is very elementary to show that $\mathbb{R}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>1$: subtract a point and use the fact that connectedness is a homeomorphism invariant. Along similar lines, you can show that $\mathbb{R^2}$ isn't homeomorphic to $\mathbb{R}^m$ for $m>2$ by subtracting a...

general-topology

Anonymous

Okay, thanks. I'll read through that! :)

Slereah

20:19

Also the same is true for $\mathbb{R}$ and $\mathbb{R}^2$, by considering removing a point and the number of connected components

I... I think the generic proof might be removing a point and the homotopy group?

bolbteppa

@Semiclassical books.google.com/…

Semiclassical

kk

apparently Dirac has a similar discussion?

bolbteppa

ll

The thing is it seems they are saying you can get it from the fact $\psi \sim e^{iS/\hbar}$ somehow

Or else they are saying it's purely formal which I doubt

If you can actually literally derive poisson brackets as the first order approximation to commutators that is nuts

(In this case with $\psi \sim e^{iS/\hbar}$ at least)

vzn

Classical Mechanics and Poisson Structures / Esposito

Slereah

Like I think you have $\pi_n(\mathbb{R}^m) = 0$ if $n \neq m$

or something like that

Semiclassical

20:23

well, one property of that is that $\hat{p}_x\psi = -\partial_x S\psi$

bolbteppa

The Hamiltonian has no explicit form yet which is interesting

Semiclassical

in which case you've got something very similar to ye-olde Hamilton-Jacobi equation, since there $\vec{p}=-\nabla S$

Anonymous

@Slereah BTW can there be any such situation where removing a subspace from a topological manifold corresponds to removing a subspace in another topological manifold (under homeomorphism), such that the number of connected components in the second topological manifold is variable depending on the particular homeomorphism?

ACuriousMind

@bolbteppa It is well-known that the Poisson bracket is the first-order approximation of the Moyal bracket that corresponds to the commutator in the phase space picture.

So this is really not an outlandish claim.

Slereah

@Blue Sure

Oh wait, no

I don't think so

bolbteppa

20:25

That's interesting

ACuriousMind

@Blue No, that is precisely forbidden by "continuous maps preserve connectedness"

Anonymous

@ACuriousMind How exactly?

Anonymous

It isn't obvious to me

Anonymous

Could you suggesting some material for reading this?

Slereah

For topology?

Munkres I guess?

Semiclassical

20:27

the intro to this is pertinent re: Poisson bracket: projecteuclid.org/download/pdf_1/euclid.jdg/1214437787

ACuriousMind

@Blue If you have a homeomorphism $f: X \to Y$, then it restricts to a homeomorphism $f: X - A \to Y - f(A)$

Semiclassical

And, in fact, it gives a pretty good reason for wanting to think in terms of a bracket

ACuriousMind

So $X-A$ and $Y-f(A)$ must have the same number of connected components for any $A\subset X$, since otherwise they cannot be homeomorphic.

Slereah

@ACuriousMind Put some \to's on that shameful display

Semiclassical

Namely, suppose $f,g$ are constants of the motion. That requires that $\partial_t f =\{H,f\}=0$ and $\partial_t g=\{H,g\}=0$

ACuriousMind

20:29

@Slereah lol, no idea what happened there

Emilio Pisanty

@BalarkaSen thx =)

@Slereah @ACuriousMind also, learn to use \setminus, you barbarian

enumaris

who chose $\{\}$ to represent both an anti-commuatator and a poisson bracket?

whoop

Semiclassical

I've seen people do $[A,B]_{+}=AB+BA$

which is a nice solution imo

ACuriousMind

@EmilioPisanty Never

Emilio Pisanty

@Semiclassical Cohen-Tannoudji does that, as I recall

@ACuriousMind blasphemy

Slereah

20:31

there's only so many symbols in math

Semiclassical

The Jacobi identity then implies that $\{H,\{f,g\}\} = \{\{H,f\},g\}+\{f,\{H,g\}\}=0+0=0$

Slereah

At some point you're gonna get overlap

enumaris

unacceptable

I demand unique symbols

Semiclassical

So $f,g$ being constants of the motion $\implies\{f,g\}$ is also a constant of the motion

enumaris

I should petition math to do it

Anonymous

20:32

@ACuriousMind Interesting, I see

Semiclassical

And the fact that this is directly related to the Poisson bracket satisfying the Jacobi identity seems like a pretty good reason to think in terms of said braket

Slereah

@enumaris $🤾‍A,B🤾‍$

enumaris

are those squares or is latex not rendering for me

Anonymous

@Slereah Munkres is boring to read from (albeit it's good) :P Also it's a bit difficult to hunt down exact theorems there

ACuriousMind

@EmilioPisanty I've mistaken the setminus too often for a quotient - somehow my brain doesn't really parse the direction the slashes are pointing in well :P

Semiclassical

20:34

by contrast, just writing down $\displaystyle \partial_t f = \frac{\partial f}{\partial x}\frac{\partial H}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial H}{\partial x}$ really doesn't suggest 'I satisfy a nice antisymmetric identity'

Slereah

If you want I have a big table of topological conserved quantities

Anonymous

However, I should probably pick up that book sometime in the future

Slereah

somewhere

Lemme see

Anonymous

@Slereah Yay!

Emilio Pisanty

@ACuriousMind if that's the case then you might as well confuse minus signs for slashes

Anonymous

20:34

That would be helpful :)

Emilio Pisanty

they're also just lines

Semiclassical

I suppose I should've pinged @DanielSank regarding the above

Anonymous

@EmilioPisanty The angle matters :P

Anonymous

(to an AI's brain)

enumaris

$\setminus$

$\\$

$\$

hmmm

Semiclassical

20:36

I'm not sure what else to say about the Poisson bracket beyond it being a convenient machine for creating new constants of the motion from old ones

enumaris

$\\\$

Semiclassical

though I will note that the commutator bracket of QM also satisfies the Jacobi identity

so that's another parallel right there

Slereah

enumaris

I took a course that developed Hamiltonian mechanics from symplectic manifolds

Slereah

enumaris

20:37

then I promptly forgot that course

Slereah

I don't remember what the symbols mean

Anonymous

Bookmarked them for quick reference: chat.stackexchange.com/rooms/71/conversation/topology-tables

Anonymous

Thanks!

Anonymous

I don't know what most of those words mean yet though

Anonymous

@enumaris What text did they use?

Slereah

20:40

The most important thing to remember is that under homeomorphism, open sets are open and closed sets are closed

and connectedness and compactness are preserved

those are the big 4

enumaris

@Blue can't really remember lol, some math book about geometric mechanics

wasn't super helpful tho

Semiclassical

Marsden maybe?

Anonymous

@enumaris I'm looking for a book which nicely shows the transition from classical mechanics to quantum mechanics including the deformation quantization and moyal bracket and stuff. Couldn't find anything similar so far

Slereah

Moyal brackets aren't really commonly taught

enumaris

don't think I've ever seen those lol

Semiclassical

20:43

Moyal brackets are firmly in the category of stuff I've heard about but never touched myself

Slereah

I did go into them because I had to use them for my master thesis

DID U KNOW

Anonymous

Also all the QM books simply begin with operators like $\hat{H}$ and say that it corresponds to total energy analogously like $H$

Anonymous

It doesn't make sense to me, other than the fact that it is a weak analogy

Slereah

There is a relation between the stochastic integral used for path integral and the ordering used with the Weyl quantization

bolbteppa

Maybe they are saying something like that while on the one hand we have
\begin{align}
\frac{d\hat{\overline{f}}(t) }{dt} &= \frac{d}{dt} \int \psi^*(q,t) \hat{f}(t) \psi(q,t) dq \\
&= \int \psi^*(q,t) \frac{\partial }{\partial t} \hat{f}(t) \psi(q,t) dq + \frac{i}{\hbar} \int \psi^*(q,t) [\hat{H} \hat{f} - \hat{f} \hat{H} ] \psi(q,t) dq
\end{align}
on the other we have that $\hat{\overline{f}}(t)$ should in the quasi-classical limit represent a scalar function $f(t)$ and so from
\begin{align}

Anonymous

20:45

How exactly do classical variables get converted to operators? And what are the extra conditions or assumptions imposed? Is there any mathematically precise account of that?

Anonymous

@Slereah Hehe, but sounds like good stuff to learn :)

Slereah

No, there's no mathematically precise method to convert classical theories to quantum theories

Because it's always only defined up to ordering

You can get the classical theory from the quantum one, but generally the converse isn't true

There's hints on how to do it, generally

Anonymous

But can it be shown that classical variables can be obtained as a classical limit of quantum operators?

Slereah

sure

It's a classic thing

Although of course it works best by considering fairly localized states

Anonymous

@Slereah How?

Anonymous

20:48

I couldn't find any well written material on it so far ;_;

Slereah

Generally you just do $\hbar \to 0$

Semiclassical

Part of the subtlety, it should be said, is that the limit $\hbar\to 0$ is typically a matter of singular perturbation theory

Hence you have to be rather careful.

enumaris

nobody got time fo dat

just do it

Anonymous

@Slereah And that comes from the path integral stuff?

Slereah

@Blue You can use this method for any quantum formalism

It works with canonical QM, path integral, deformation

Semiclassical

20:49

For a comparison: The energy levels of a harmonic oscillator are given by $E_n=\hbar \omega (n+1/2)$, and that at least is a nice function of $\hbar$

Slereah

with path integral it's a stationary phase argument

Stationary phase approximation

In mathematics, the stationary phase approximation is a basic principle of asymptotic analysis, applying to oscillatory integrals I ( k ) = ∫ g ( x ) e i k f ( x ) d x {\displaystyle I(k)=\int g(x)e^{ikf(x)}\,dx} taken over n-dimensional space ℝn where i is the imaginary unit. Here f and g are real-valued smooth functions. The role of g...

Semiclassical

on the other hand, the tunneling amplitude across a barrier is typically on the order of $e^{-\Delta E/\hbar}$

and that's rather badly behaved as $\hbar\to 0$

in particular, it doesn't have a meaningful Taylor series about $\hbar=0$

Anonymous

@Slereah Interesting. So is the whole reasoning behind $\hbar \to 0$ explained properly in any standard book? Sakurai, etc?

Anonymous

And how that converts operators to variables

Slereah

Basically using the stationary phase approximation, you can show that the classical path's probability is $\approx 1$ for $\hbar \approx 0$

Semiclassical

20:51

in the same way that $e^{-1/x}$ in general is problematic about $x=0$

Slereah

For path integrals it's in Feynman Hibbs

Anonymous

@Slereah And for canonical QM and deformation?

Anonymous

amazon.com/Quantum-Mechanics-Integrals-Richard-Feynman/dp/…

Anonymous

This ?

Slereah

Yes, although there's a modern version of the book

Semiclassical

20:52

This is all in the realm of semiclassical methods, hence why I know of it

Slereah

For canonical QM it's in most books I think

For deformation QM it's like

Semiclassical

Shankar has a good bit of detail on it iirc

Slereah

Trivial

Since $a \star b = ab + O(\hbar)$

Semiclassical

mathematically, the real subtlety is that the kinds of differential equations you do in QM tend to have irregular singular points at $x=\infty$

which makes things rather tricky

Anonymous

Okay, I got to read about it

Anonymous

20:54

thanks

Emilio Pisanty

0

A: Planned maintenance scheduled for July 14, 2018 at 13:00 UTC (9AM US/Eastern)

Make sure you double-check that you've brought a charge-only USB cable You know, in case you feel like doing this again: (More background.) Though heck, I imagine you do all of this remotely?

Semiclassical

I don't know why i'm bringing that up, though, since I really don't want to think about it :P

Anonymous

BTW one more confusion:

Emilio Pisanty

↑ those were interesting times

Anonymous

15

Q: Classical limit of quantum mechanics

I have heard that one can recover classical mechanics from quantum mechanics in the limit the $\hbar$ goes to zero. How can this be done? (Ideally, I would love to see something like: as $\hbar$ goes to zero, the position wavefunction reduces to a delta function and that the Schrodinger equatio...

quantum-mechanics classical-mechanics

Anonymous

20:55

ACM linked to this QA in one of his comments

Anonymous

So it's not true that classical mechanics can be retrieved from QM in the limit of $\hbar \to 0$?

Slereah

Not in the general case, no

As said, you need to have well-behaved systems and localized particles

I think it might work okay with coherent states?

Anonymous

I see. But at the least it is true that classical variables are retrieved from quantum operators in that limit? O:)

Slereah

Yes

Anonymous

Interesting

Anonymous

20:57

I had this confusion for a while

Slereah

You have $$m \frac{d^2}{dt^2} \langle x \rangle = \langle F \rangle$$

Semiclassical

i mean, it really comes down to QM and CM being basically different at the level of formalism

so what you usually look for is not so much QM reducing to CM but rather correspondences between the two

Slereah

And $$\langle \frac{d}{dt}x\rangle = \langle p \rangle$$

Semiclassical

e.g. at the level of expectation values, as slearah is pointing out

Slereah

that kind of stuff

Anonymous

20:59

Ehrenfest theorem

The Ehrenfest theorem, named after Paul Ehrenfest, an Austrian theoretical physicist at Leiden University, relates the time derivative of the expectation values of the position and momentum operators x and p to the expectation value of the force F = − V ′ ( x ) {\displaystyle F=-V'(x)} on a massive particle moving in a scalar potential, Although, at first glance, it might appear that the Ehrenfest theorem is saying that the quantum mechanical expectation values obey Newton’s...

bolbteppa

Classical limit

The classical limit or correspondence limit is the ability of a physical theory to approximate or "recover" classical mechanics when considered over special values of its parameters. The classical limit is used with physical theories that predict non-classical behavior. == Quantum theory == A heuristic postulate called the correspondence principle was introduced to quantum theory by Niels Bohr: in effect it states that some kind of continuity argument should apply to the classical limit of quantum systems as the value of Planck's constant normalized by the action of these systems becomes very small...

Anonymous

I guess that's similar to what this says^

Slereah

Although if your states are weird enough, that's not necessarily true

Semiclassical

yeah

bolbteppa

So the Moyal stuff is trying to formalize 'the correspondence principle' stuff I've been using without using these words

Semiclassical

21:00

and there are features in QM which, while you can approximate them using classical terms, are basically not classical in form

for instance, you can estimate the tunneling amplitude as $e^{-\Delta E/\hbar}$

and that tunneling barrier $\Delta E$ is a perfectly well-defined classical quantity

but classically, you'd never have a reason to care about it: if it's large enough that the particle won't get over it classically, then however high it might go beyond that doesn't matter

so classically there'd be no reason to think about $e^{-\Delta E/\hbar}$ when $E<\Delta E$

Emilio Pisanty

Is Physics going to take six-to-eight-weeks to update long after everyone else gets to hop on the new train, as happened on the last go-round? (asking for a friend.) (the friend is me.) — E.P. 11 secs ago

↑ show that comment some love

Anonymous

Just out of curiosity, does QFT also include notions of operators, wavefunctions, Hamiltonians, etc ?

Slereah

Yes.

QFT uses the same framework as QM

Anonymous

Nice, so what's the basic difference? :)

Semiclassical

ehhhh. i have to push against that a bit, if only because QFT tends to be more in line with the Lagrangian formalism than the Hamiltonian formalism

Slereah

21:07

There's no operator for position and momentum

@Semiclassical Both exist!

You can also say that there's infinitely many operators for position and momentum

Well, similar to them

Semiclassical

one should probably say something about Heisenberg vs. Schrodinger picture as well

Anonymous

Cool! And what are the so-called "fields" in Q "field" theory?

Slereah

@Semiclassical Schrodinger is best

@Blue The quantum version of classical fields

Anonymous

Like electric field, magnetic field, etc?

Slereah

yes

Semiclassical

21:11

also something something second quantization

Slereah

Well second quantization isn't strictly necessary

But it is a nice feature

Instead of fields associating values to points in space, you associate quantum operators to points in space

So that the expectation value will associate a value to a point in space

Anonymous

Sounds interesting. Maybe I'll get about learning it someday. Thanks :D

enumaris

something something darkseid....

Slereah

something something complete

Semiclassical

one subtlety implied by this: in QM, you've got one canonical coordinate per particle, and thus one corresponding observable

enumaris

21:14

have you heard the story of darth plageus the wise?

Slereah

It's not a story a jedi would tell you

Semiclassical

for qft, there's an operator corresponding to each possible position where you could measure the field

enumaris

:D

Semiclassical

and since there's infinitely many positions where you could measure the field, you necessarily have infinitely many operators

Slereah

You can find out by doing a Fourier transform of a field configuration, in which case the "positions" are the values of the coefficients

Semiclassical

21:16

the basic point is that a classical field, unlike a particle, has infinitely many degrees of freedom, and you have to quantize all of them in the right way

Slereah

This leads to much misery

Semiclassical

Yep

Slereah

Because a lot of QM theorems relie on the existence of finitely many degrees of freedom

So we mostly pretend that the theorems still work and as a result all of QFT is wrong

Semiclassical

lol

there is a certain extent to which QFT feels provisional to me

enumaris

let the mathematicians figure it out

physicists just do

Semiclassical

21:18

in the sense that we don't know the right way to ground it in sound foundations

enumaris

unnecessary, that's the job of someone else

Semiclassical

I'm not sure how many people actually work on QFT foundations, in math or otherwise

Slereah

Well there's a few

Glimm, Jaffe, Streater, Haag

Those berks

prolly others

Semiclassical

yeah, but compared with the number of people who use QFT

Slereah

I know Wald worked on it

Nguyen

Semiclassical

21:21

the people who work on foundations are decidedly in the minority

Slereah

probably

There is less money in it

Semiclassical

true enough

Slereah

also it sounds less interesting to most people

They want quantum weirdness not analyticity

enumaris

I miscalculated how long this network would take to train

Slereah

Why train a neural network

When you can hire some guy to learn

enumaris

21:22

and now if I don't want to waste 60 hours of training time I gotta wait until tomorrow for it to finish

Slereah

just have a guy look at your training set

He will be your neural network

enumaris

sounds legit

Slereah

Train him by shocking him every time he picks wrong

enumaris

14 epochs left

Semiclassical

as weird as QFT sounds from the above, though, the fact of the matter is that we know how to do perturbative QFT quite well at this point

enumaris

21:23

I'm guessing it has overfit significantly

but I'm not sure until it evaluates on the test set

@Semiclassical do we though

Semiclassical

hence why precision QED is a thing and has stood up so remarkably well

enumaris

where's that thor gif

imagine I linked to the thor gif

Semiclassical

non-perturbative QFT, by contrast...there be dragons

Slereah

Non-perturbative QFT works alright in $1+1$ dimensions

So if you're a line

You're golden

Semiclassical

lol

Slereah

21:25

It works especially well in $0$ dimensions

Since you can get closed form solutions

bolbteppa

I completely forget how I justified the whole 'replace poisson brackets with commutators' thing in the past, but jesus this is actually the 'derivation' of that whole thing

Slereah

Turns out that in $0$ dimensions QFTs all converge nicely!

Well, not all of them

Still needs to be bounded from below

Semiclassical

yes, it is rather easier to do QFT when you only have $\infty^0=1$ values to measure ;P

bolbteppa

I don't think you're able to like 'derive' it analytically by replacing $\psi$ by $e^{iS/\hbar}$ because you'd have to 'derive' going from an operator to a scalar :(

and in fact they derived the Schrodinger equation from this 'correspondence principle' by stating from $\psi \sim e^{iS/\hbar}$ that $\partial_t \psi = \frac{i}{\hbar} (\partial_t S) \psi$ and then using $\partial_t S = - H$ and then saying the quantum Hamiltonian operator is the operator which reduces to the scalar $\partial_t S = - H$ in the quasi-classical limit

Slereah

@Semiclassical In $0$ dimension the path integral is just a regular integral

Very convenient

Semiclassical

21:30

yup

bolbteppa

So you're just supposed to literally take the operator equation $\frac{d \hat{f}}{d t} = \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar}(\hat{H} \hat{f} - \hat{f} \hat{H})$ and then say this should becomes the scalar equation $\frac{d f}{d t} = \frac{\partial f}{\partial t} + [H,f]_{q,p}$ in the limit as $\hbar \to 0$ and so we simply must have that $\frac{i}{\hbar}[\hat{H},\hat{f}] \to [H,f]_{q,p}$,

then generalizing to any (Hermitian) operators because any operator can be thought of as a Hamiltonian for some system

I don't think that's that nice :(

enumaris

conversations here often feel disconnected

Slereah

@bolbteppa well it needs to be hermitian

enumaris

like people are more talking to themselves

bolbteppa

@Slereah right yeah

Semiclassical

21:34

on that note, $\hat{H}\hat{f}-\hat{f}\hat{H}$ is not Hermitian

you need to toss $i$ in there to make it so

bolbteppa

That's a real shame that you can't like directly 'derive' it

Slereah

if you could derive QM from Newtonian mechanics it wouldn't have been a revolution

vzn

@Semiclassical there does not really seem to be a proof that no classical system could have a QM-like formalism & dont think this search has really been exhausted. supposed proofs tend to be related to bell-type results but am skeptical due to misc factors cited in past in here.

enumaris

make it so

time to read about canonical mech

bolbteppa

It's nuts how you have to basically turn an operator into a function and then back again

Semiclassical

21:37

whether or not one can present QM in a 'classical mechanics' form (or vice versa) was not really my point. all I was getting at is that the standard formalisms of QM vs. CM involve different mathematical objects

CM is ultimately a story about phase space

vzn

@Semiclassical agreed but think there is a hidden/ buried correspondence and that seems to be at the center of DSs latest inquiry.

Semiclassical

QM is ultimately a story about Hilbert space

bolbteppa

Yeah

Anonymous

There was some PSE answer by Zachos which said QM can be done in phase space and CM can be done in Hilbert space, but I don't know enough to comment

Anonymous

QM in phase space is more standard I think

Semiclassical

21:40

yeah, wigner stuff

vzn

sometimes the math/ formalism can (inadvertently) conceal rather than reveal...

bolbteppa

Here we have vicious examples of how flawed it is to try assume QM is just some math formalism

vzn

what "vicious examples"?

bolbteppa

We have been discussing turning an operator into a scalar function and then back into an operator in order to derive the Schrodinger equation from the correspondence principle, and trying to use this idea to derive replacing poisson brackets with commutators, this makes math people cry for good reason

Semiclassical

well, tbf, I think if you wanted to try to get QM from CM you'd need to do more than we've said here

for instance, an obvious starting point would be to do a classical Hamiltonian plus a stochastic term

vzn

21:46

have long suspected some SHO-like-interacting classical scenario could lead to "QM-like" formalism... DS seems to be recently contemplating something similar...

Semiclassical

but eh. stochastic formulations of QM don't appeal to me much

enumaris

stochastic

bolbteppa

Even if one is able to derive some structure, i.e. deformations, that really derive the structure of QM from CM, it's artificial to turn on the deformation, and absolutely could be wrong

Isn't there the whole Gell-Mann wave function of the universe stuff

vzn

@bolbteppa think you might be overthinking this. a physical arrangement/ scenario/ setup has a set of eqns to describe it (ie its dynamics). what is the classical system arrangement that results in QM-like formalism? it seems to be jumping the gun to assume a priori it is, if it exists, "artificial".

enumaris

I don't think the physical arrangement really dictates what math formalism you use...

one is certainly free to use a classical formalism even for microscopic particles...the answers just happen to be wrong is all...the more microscopic the domain the more wrong the answer is :D

Slereah

21:52

Classical mechanics isn't contextual

bolbteppa

@vzn my point is basically that if you take Cophenhagen qm as your basis, it is derived explicitly on the correspondence principle which assumes classical mechanics as a limiting case on which to build the QM theory, based on this we are completely in the dark on how to start QM from first principles. Now, people try to set up QM from first principles because this is ugly, but why should anybody believe that?

vzn

@bolbteppa am aware of all this history (& dont outright reject it), but think the point of pilot wave hydrodynamics is that QM like formalism emerges from novel classical setups that were inconceivable to the QM pioneers/ founders. think this can be pushed substantially further.

gateprep

The weight fraction of methanol in an aqueous solution is 0.64. What is the mole fraction of methanol in the solution? Given, mol.wt. of methanol is 32 g/mol and that of water is 18 g/mol.

Can u help i know its from chem

bolbteppa

I think it's really cool that people have been pretty strict on avoiding interpretations for the past 80 years or whatever, it all comes back to this problem, it's justifiable even if not nice

vzn

@bolbteppa copenhagen is an interpretation. or rather its like an anti-interpretation.

bolbteppa

21:55

I should remember that mole conversion :(

enumaris

1/2 right

o.O

imagine you have 100 grams of this stuff, 64 grams is methanol, 36 grams is water <--- correct weight fraction. Then 64 grams of methanol is 2 moles of methanol while 36 grams of water is 2 moles of water <---- mole fraction 1/2

solving by inspection is best solve

Semiclassical

@bolbteppa they were pretty strict on it for the first 40 years at any rate

Slereah

Nah, making up quantum interpretation has been a thing since the beginning

enumaris

fun beans

Slereah

it's been a little cottage industry since like

bolbteppa

21:59

Yeah, I wonder how much of that referred to qft as well as qm

Slereah

early 20th century

bolbteppa

Hmm, the correspondence principle is formulated by specifying wave functions $\psi$ reduce to the form $e^{iS/\hbar}$, they never specified how operators should reduce, maybe this is the operator analogue, god, no wonder you don't plug $e^{iS/\hbar}$ into it

bolbteppa

22:13

@vzn Coincidentally a paper literally from 4 days ago was put up arxiv.org/pdf/quant-ph/0405069.pdf emphasizing the problem with Landau starting from Copenhagen and trying to abstract to general principles

(Oh wait maybe it was just updated 4 days ago)

Emilio Pisanty

@JohnRennie what just happened brexitside?

bolbteppa

3

A: Classical Limit of Commutator

Dirac's argument is on pp.85-86, and it goes like this: the classical Poisson bracket obeys the rules $$\{A,B\} = -\{B,A\}$$ $$\{aA + bB ,C\} = a\{ A,C\} + b\{B,C\} $$ $$\{AB,C\} = A{B,C} + \{B,C\}A $$ $$\{\{A,B\},C\} = \{B,\{A,C\}\} - \{A,\{B,C\}\}$$ Where I rewrote the Jacobi identity in the...

He strikes again

Balarka Sen

the maimed wraith

get maime'd boi

Emilio Pisanty

who, Boris Johnson?

also @JohnRennie what the hell is "checkers"?

this keeps mentioning it and I've not the faintest what it means

bolbteppa

22:29

'What happened at the Chequers Brexit awayday?

May and her ministerial troops headed to the country to thrash out a unified negotiating stance'

Basically planned slogans to pretend a self-laceration to an economy is perfectly fine

Emilio Pisanty

> thrash out a unified negotiating stance

I take it that didn't work?

bolbteppa

Basically, and the past few days a bunch of them quit because of what everyone assumes is a softening

vzn

@bolbteppa thx for sharing, trying to figure out the "core axioms" of QM is a prj that dates all the way to mid 20th century/ von Neumann & is still continuing/ ongoing...

plonk

22:52

Hi! I got a question about physics.stackexchange.com/review/suggested-edits/224178: I tried to fix a mathematical mistake there. Since I don't have enough reputation to comment, I tried to edit it instead. I'm not quite sure I understand the review feedback though.

Although, should that question better be on meta?

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