@ACuriousMind Physicists love to talk about "classical limit" of quantum mechanics as "$\hbar \to 0$". This is a lovely idea, but what does it mean? For example, can you explain to me how $1/i\hbar \cdot [-, -]$ goes to the Poisson bracket under $\hbar \to 0$ (this is going in the opposite direction of quantization).

Hi All......

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@BalarkaSen I think we had that discussion before here

and our verdict was that there didn't seem to be an actual proper reverse quantization

At least not one where you get back classical mechanics

You may get back an operational-level classical mechanics but not the full phase space shenanigans

at least not without some fairly hard waving of the hand

but in the general case, Slereah and you are of course correct - the classical limit is poorly understood, and $\hbar \to 0$ is merely a heuristic, not a well-defined limiting process

which is weird, considering that our universe is quantum in nature and we discovered classical mechanics first!

So obviously there is some process to do it

it's not as if we don't know how to take a meaningful limit for specific systems

it just seems it's not a generic procedure

you have to know a lot about the quantum system to decide which "part" of it is classically relevant

Also the range of observables depend on $\hbar$, so it gets a little tricky

in the limit $\hbar \to 0$, some discrete observable may become continuous

I don't even really know what the rigorous math process for "$\mathbb{N}$ in some limit is $\mathbb{R}$"

@Slereah you just make the 1 really small :)

@ACuriousMind I assume there's some topology stuff to define

ncatlab.org/nlab/show/…

@BalarkaSen From $\frac{d}{dt} \hat{f}(t) = \frac{\partial \hat{f}}{\partial t} + \frac{i}{\hbar} [\hat{H}, \hat{f}]$ we see that the Taylor expansion of $[\hat{H}, \hat{f}]$ in $\hbar$ about $\hbar = 0$ must be $0$ at order zero, the PB at order $\hbar$ so that in the limit we get classical mechanics which is defined as the $\hbar \to 0$ limit, and then whatever at higher orders

Uh just something that hadn't occurred to me, the correlation functions are Green's functions, the 2-point function is the Feynman propagator which is the Green's function of the appropriate derivative operator for the field theory, $(\partial^2+m^2)$, $(\gamma_\mu\partial^\mu-m^2)$, etc. But which derivative operators are the higher order correlation functions Green's functions of?

@Charlie they aren't Green's functions for any operators, it's a misnomer :P

D:

Is it still wrong for the 2-point function? I've definitely seen $(\partial^2+m^2)\Delta_F(x-y)=\delta(x-y)$ or something like that derived before

the 2-point function is a Green's function

the other n-point functions aren't

ah ok

I see, that's quite an egregious misnomer in that case :P

@Charlie that question was asked a couple months ago, here

I see, so there is a sense in which they somewhat resemble Green's functions, but aren't strictly Green's functions

yeah it's a bit of a stretch

"Green's function" is a bit vague also because QFT uses a lot of different functions as the solution of the same equation

Depending on which pole you integrate around

My book says ∆H=∆U+P∆V+V∆P+∆P∆V is this correct?

Roughly so, yes?