People say there's no classical analogue of spin but if you take the opposite limit from the quantum to the classical i.e. replace operators with functions, surely you get spin too

There are classical analogues to spin

But (for spin 1/2 anyway) they do not matter much because you won't observe them in non-quantum regimes

Due to Pauli's exclusion principle, there are no "big waves" of fermions

Unlike for electromagnetic waves

@DIRAC1930 you're imagining the "classical limit" as far too simple

"replace functions by operators" works somewhat well because you can figure out the space of states as representations of your observables

Just because you can set up 'classical' representation theory involving spin, it doesn't mean it applies to the real world, there's no reason to think spin is classical physically, it goes away as $\hbar \to 0$ as far as anyone knows

"replace operators by functions" doesn't work because how are you gonna know what the phase space is?

@ACuriousMind ancient magic

as for "spin has no classical analogue", [here's](physics.stackexchange.com/a/594504/50583( my take on that

How do we go from a quantum theory to a classical theory?

very carefully :P

@DIRAC1930 a very good question

there's no universal procedure of "classical limit"

We talked about it a bunch certainly here and I don't think we know still

You can have the classical limit of the observables, that's not too hard, but reconstructing the actual theory from it seems difficult

but in a way, that is to be expected - some systems just "vanish" classically, like a point particle with spin

If you want the classical equivalent of an electron it's called the spinning point particle, though

@ACuriousMind When we observe an ideal gas of N particles, and consider the sphere in phase space (I don't know why we take the volume of a sphere), when we consider a spherical shell in this N-dimensional sphere, we find the nr, of microstates with energy between E and E + dE , and as we said the probability density function is constant, and corresponds to a macro state or it represents one.

It's just a little point that has an orientation associated to it

:59797721 that's just a thermodynamic limit

If we increase the thickness of the shell, which corresponds to more microstates, does \rho change?

it may or may not have "classical" properties

A spinning point particle isn't actually a real classical theory though?

@bolbteppa Define "real"

Newtonian mechanics

Not complex

lol

There's certainly no physical object we can observe that obeys that EoM

@bolbteppa You can do the classical version of it no problem

@imbAF yes?

It's not gonna be super useful but it exists

(I'm waiting for the question)

You can set up some model of something that doesn't exist sure

But I'm pretty sure this is purely quantum stuff that hasn't been observed yet

even though the previews microstates belonging to another macrostate, still are included in this new macrostate?

So the only states classically are functions of $x,p$?

Also I guess it probably describes an electron wavepacket okay?

@DIRAC1930 classically a state is just a point $x_0,p_0$ in phase space

If you don't do anything too wild with it an electron at large scales is probably just a little ball that go along carrying a spin vector on its shoulders

All you need are $(x,p)$ classically, QM is so different because it says they don't exist simultaneously, it's like an on-off switch

@ACuriousMind could we be in a private chat for like 5 min? You got time. The multiple texts confuse me

@imbAF I don't really want to commit to that because statistical mechanics is not my strong point and here at least people have the chance to correct me if I say something wrong

Ok

One question. Let's consider the case of the system in equilibrium, which means we have a certain region, (volume) where \rho is constant

forget that

Are you familiar with the N-dimensional sphere for an ideal gas?

sure

first of all, why do we consider the volume, as that of a sphere and not ..idk a rectangle

is there a reason for this?

Forget that as well, since a sphere

with a certain radius

represent an energy value

for the ideal free gas, you have that the Hamiltonian/energy is $\propto \sum_i p_i^2$

yes

and your gas is contained in some box with volume $V$

I know

Ok my question

ah, sorry, I overlooked that you already understood why it's a sphere

First of all, we don't care about whether this ideal gas is in eq. or not

Yeah, we basically

confine ourselves to a constant energy

Now, as I said we don't concern ourselves about the state of this system

eq or not eq.

When you consider the spherical shell, ofc inside this volume, \rho has a value, represents a macrostate and all the points inside are the microstates, whose energy is between E and E + dE

am I correct until now?

I'm afraid we'll end up having to have a philosophical discussion about infinitesimals, but yes, that's how many people present it

no no

don't worry

actually I realized my stupidity xD

But I will tell it to you as well

so you can confirm it

Ofc this dE difference is small enough that the energy of the macrostate is defined and at the same time big enough so you can have many microstate in the energy interval, correct ?

it's infinitesimal :P

yes

but big enough so you can have microstates

I mean...yes, that's how people who write stuff like that want you to think about it

This question is out of curiosity nothing more

I'm not a fan, but sure, it's fine

How would you express it , correctly then?

(diverging from my question)

What is really of interest for statistical mechanics is the density of states, not counting some number of states inside a volume, and that's what you get from the "states inside the volume" in the limit $\mathrm{d}E\to 0$

so we could skip talking in contradictory terms about "how big" $\mathrm{d}E$ is and just say we're only interested in that density and that limit anyway

density of states is the \rho ?

No

\rho is the probability density

sorry I forgot

like, let's say you compute the "volume" between $E$ and $E+\mathrm{d}E$ to be $V(E)$

yes I am aware

DOS and pdf in phase space are 2 completely different thing

things*

then the density of states is $\Omega(E) = \lim_{\mathrm{d}E\to 0}\frac{V(E)}{\mathrm{d}E}$

yes

and it turns out that this is all you need to do statistical mechanics, so I would just prefer not bothering with "how big" $\mathrm{d}E$ is at all - you just need this limit

My initial question (which I realized it makes no sense in the context for find the nr. of microstates for an dE change in energy) was: What would happen if you would increase dE by a lot

I understand

And it's better the way you express it

if you would increase dE by a lot, you wouldn't be in the same macro state right?

well, strictly speaking, you wouldn't be in the same macrostate if it's non-zero at all :P

that's why I suggest not worrying about it

it's just a construction to obtain the density of states

yes

if you insist on keeping it, you'll discover that it just does stuff like add a constant to the entropy

so what is the meaning of considering a sphericla shell of E + \Delta E

and not dE

just curious

if it makes sense

to observe such a big energy difference

If I had to give my input

physically, or what is being done

well, the intended meaning is probably that you stay "in the same macrostate" as long as $\Delta E \ll h$, where $h$ is your coarse graining

is that we are observing a bunch of macro states, and finding all the micro states of all this macrostates which belong in this vase energy interval

what?

well, in order to "count" microstates at all, we've coarse-grained our phase space into little cubes with side length $h$, right?

where we say all the states inside such a cube are "the same state"

in statistical it's a cube yes

in thermodynamics it's a point

with volume (2 \pi \hbar)^3

...and that's my turn to ask: what?

about what?

thermodynamics is just statistical mechanics

well in our class the professor said

there is a distinction between the two

an example would be the entropy

entropy in classical thermodynamics can be expressed as a function of state variables such as E,V,P

etc

while in statistical mechanics, its equal to the log of the multiplicity of a macrostate

that's a very strange statement

and also, a point in phase space is how we represent a microstate classically

and a cube with the volume (2 \pi \hbar)^3N in quantum mechanics

since you have plank's constant

which is only used in qm right?

(2 \pi \hbar)^6N*

should be right?

also a very strange statement

In order to relate a phase space integral to a number of microstates, you need to consider a microstate as a tiny cube

since then phase space volume divided by the cube volume gives you the number of states inside a volume

Yes

but that's what he said

that the entropy is a state function in thermodynamics doesn't mean it's not also the phase space volume of the macrostate

Didn't confuse me anyway

"it's not also the phase space volume of the macrostate"

another way of saying multiplicity ?

for a particular macrostate?

yes

like, the "multiplicity" is just the phase space volume divided by your cube size

yes

anyway cuz we diverged from my question

considering E +dE you get the nr of microstates for a macrostate

what do you get if you take the spherical shell with half the radius of the sphere?

basically if dE is to big

you get nonsense

hehehe

that's why I said, I was thinking of something stupid :D

thx for the confirmation

Is any 'operator' a dynamical variable?

Any Heisenberg picture operator obeys the Heisenberg equation which is just the time evolution of classical objects transformed to quantum objects by changing the Poission bracket

again, this isn't special to QM

What do you mean?

you also have classically that $\dot{f} = \{f,H\}$

where by $\dot{f}$ I mean the time derivative of $f(q(t),p(t))$ where $q(t),p(t)$ is a solution to Hamilton's equations of motion

Ah so all observables classically are functions of (q,p)

And we just quantize by changing $f \rightarrow \hat{f}$ for every function

i.e. this "Schrödinger" or "Heisenberg" picture choice exists also in classical Hamiltonian mechanics - you can either consider observables $f(q,p)$ time-independent and look for time-dependent states $q(t),p(t)$, or you can directly consider time-dependent observables $f(q(t),p(t))$

But doesn't that imply that, as an example, operators such as the spin $z$ operator $\hat{S}_z$ can be written classically as $S_z(p,q)$ and then transformed to the operator $\hat{S}_z$ when producing the quantum theory

no

there is no "dequantization"

No I mean that you start off with an $S_z(p,q)$ and then quantize it

quantization just gives you a prescription for how to associate a quantum theory to a classical theory, it does not claim that all quantum theories have to be produced by quantization

and in fact, any quantum theory with a finite-dimensional Hilbert space (like pure spin!) cannot be a theory with $[q,p] = \mathrm{i}$

i.e. it's not the quantization of a classical phase space theory with $\{q,p\} = 1$

What if you have $\hat{S}_z$ with the position forming a complete set of observables

but spin is not a "function" of $q$ or $p$

since $[S_z,x] = 0$ and $[S_z,p] = 0$

I'm finding it confusing how some of the quantum operators seem to arise out of nowhere

Okay so $\hat{H}$ is dependent on $\hat{x},\hat{p}$ so there exists a classical analogue of the Schrodinger equation

Does every operator that doesn't have a classical analogue just arise because we have to consider projective representations and not the normal representation?

the "non-classicality" of spin doesn't really have anything to do with projective representations

spin would still be "weird" if we only allowed integral values for it

Quantum mechanics is confusing

@ACuriousMind I have one question about the N-dimensional sphere. For an ideal gas of N particles, we have a 6N dimensional phase space, but when you find the formula for the volume of it, you have powers of 3N and not 6N. Is it because we consider an ideal gas, and positions play no role in the energy?

@imbAF yes

this is a new for me

so position is included only if somehow contributes to the energy, which would mean interaction between particles, that depend from their relative position ?

well, if your text does it right, it should include a factor $V^{3N}$ for the position part for the volume your gas is contained in

but yes, you'd only have to figure out some integral there if the positions occured in the energy

I am doing it myself

And I have this (1/2)^(3n) factor that I have to explain

in the sphere, we only consider positive values of the momentum

so for p_1 for example you take into acc only 1/8th of the sphere

each momentum has 3 components

then you have N particles, each has N momentums, and each momentum has 3 components

(1/8)^N

why do so many papers have the model space of a G structure be $\mathbb{R}^n$ but then a bunch of them famously are not

de Sitter geometry doesn't have $\mathbb{R}^n$ as its model space!

tel.archives-ouvertes.fr/tel-00409501/document

thank god I am French

The only thing I remember from my french class

Pase compose

misspelled probably

can we write $L^2(\mathbb{R}) = H_1 \otimes H_2$$ where $H_i$ are both Hilbert spaces of dimension greater than one?

@fqq sure, $L^2(\mathbb{R})\otimes \mathbb{C}^2\cong L^2(\mathbb{R})$. Just pick any orthonormal Hilbert basis $\psi_n$ of $L^2$ and then map $\psi_n \otimes \lvert 0\rangle \mapsto \psi_{2n}$ and $\psi_n\otimes \lvert 1\rangle \mapsto \psi_{2n+1}$

infinite-dimensional spaces are annoying because they just "absorb" finite-dimensional ones :P

"In this book we will elaborate three and a half models for Mobius geometry"

@ACuriousMind right

I was thinking about something like that

more generally, any separable Hilbert space is isomorphic to $L^2(\mathbb{R})$ and so any choice of $H_1$ separable and $H_2$ finite-dimensional works

even $H_2$ separable, come to think of it

of course

you even have $L^2(\mathbb{R}^n)\cong L^2(\mathbb{R})$ as abstract Hilbert spaces

though one should see this as the equivalent of saying that there are bijections of $\mathbb{R}^n$ and $\mathbb{R}$ - you're not trying to preserve the right structure if you're just thinking about Hilbert spaces

@ACuriousMind I am stuck on a certain point in the calculation of the volume. I calculate the volume of a microstate V_{micro}=(\frac {\habr \pi}{\sqrt{2m}})^(3N) \cdot \frac {1}{V^N}

And I know that we are observing a certain energy, which is the radius of the sphere in phase space

y'know, if you put dollar signs around your math I perhaps could actually read it :P

But I can't get the total volume, since I don't know the amount of squares inside of it

Oh i didn't?

lmao

$V_{micro}=(\frac {\habr \pi}{\sqrt{2m}})^(3N) \cdot \frac {1}{V^N}$

much better

Is spin different to say, $SU(3)$ because spin relies on the structure of the Lorentz group? When we start off with a classical theory, the fields are already stated to transform under $SU(3)$ right?

@ACuriousMind what structure am I loosing here? I think I only care about the Hilbert space

@fqq yeah, then it's okay

I was thinking about e.g. equivalences of representations of some operator algebra

V is the volume of the box/potential well

@ACuriousMind I don't think that's a problem at least for now

thanks

@imbAF I'm not sure what you mean by "I can't get the total volume" - the volume is the volume between your sphere and the infinitesimally thicker sphere at $E+\Delta E$ we were talking about earlier

But doesn't spin naturally arise if you consider classical Grassmann valued spinors that transform under $SU(2)$? Is the word 'classical' used wrongly in this example?

@ACuriousMind Or you could use the formula for the N-dimensional sphere and for the radius you substitute the value of energy expressed in the quantum nr.

No @ACuriousMind see here lies the problem

imagine a quarter of a sphere in the cartesian system

that's what I am asked to find

@DIRAC1930 If you're considering spinors, I think spin has already "arised"

@DIRAC1930 you could start with a classical theory that transforms only in some group covered by SU(3)

but no one would ever do that because you have to pass to SU(3) anyway when considering the quantum theory and we don't really have "classical" field theories that work like that

i.e. you can write down such classical theories but they aren't actually relevant for modeling important real-world classical systems

@imbAF if your problem is that some boxes might be cut off by the spherical shape instead of lying inside or outside of the volume, just ignore that :P

The problem is the following

Is there any reference that explains all of this simply?

take your volume, divide by the volume of your microstate, that's the number of microstates

yes

but I need the volume

Let me explain

imagine 1/8 of a sphere in the cartesian system

I need that volume

@DIRAC1930 all of what - the questions you've been asking lately have been all over the place from spinors to classical Hamiltonian mechanics to quantum field theory!

@imbAF sooo...compute the volume of a sphere, divide by 8?

which I did

Look

I will take the time and actually write the equations

cuz there is no other way for me to explain wth is this book doing

it makes 0 sense

now that sounds promising :D

@ACuriousMind Well they've all been to do with spin essentially which is what I'm trying to properly understand at the moment

@fqq I should probably point out that the "canonical" separable Hilbert space mathematicians would use is the space of square-summable sequences $\ell^2(\mathbb{C})$ because it's easy to map every separable space to it - pick any Hilbert basis $\psi_n$ and write $\psi = \sum_i c_i\psi_i$, the map is mapping $\psi$ to the sequence $c_i$

@ACuriousMind

the appearance of the $h$ suggests that's already the expression for the multiplicity, i.e. the volume divided by the volume of a phase space cell

apart from that, the relevant structure of the expressions is the same - the gamma function is equal to your factorial, and the dependence on the $E$ is also the same, so it's not too bad I'd say

then why it is asked to find the volume

and not the nr. of microstates

and the $V^N$ is because that's the confinement of your gas to the volume $V$ in position space

the gamma is (3n/2) \cdot (3n/2)!

yes

@imbAF no, you have $\Gamma(n) = (n-1)!$, so they're really the same

you're thinking of $\Gamma(n+1) = (n+1)\Gamma(n)$, I think

no

gamma ( x+1)=xgamma(x)

ah, yes, sorry

still, $\Gamma(\frac{3N}{2} + 1) = \left(\frac{3N}{2}\right)!$

and why not

(3n/2) \cdot (3n/2)!

uh, that's just how the gamma function works - it has $\Gamma(n) = (n-1)!$

I'm not sure where you get the extra factor from

aaa

so you chose that

as your expression

you could as well choose what I initially wrote

but you take the whole

3n/2 +1 as one

therefore minus 1

I'm not sure where there's a choice here, but yes, that's what I mean

I mean

\Gamma (x+1)=x\Gamma(x)

but if x+1=u

\Gamma (u)=(u-1)!

=(x)!

yes but on your l.h.s. you have $\Gamma(u-1)$

so you get $\Gamma(x+1) = x\Gamma(x) = x(x-1)! = x!$, it's all consistent

aha yes

Show that the phase space volume Ω (E, V, N) available for the gas is given by:

is what the exercise is asking

and the solution should be

$\Omega (E)=\frac 1 {N!}\frac 1{(2\pi \hbar)^{3N}} \frac {V^N (2\pi mE)^{3N/2}} {\Gamma(\frac {3N}2 +1)}$

But that is impossible to be the volume

that must be, as you said and as I also thought, the nr. of microstates in this volume

it might just be using "phase space volume" to mean "volume divided by volume of a phase space cell"

really just depends on how you defined $\Omega$

I don't think you and the book disagree about anything except what that word/symbol means

Maybe

but do you know how in the exercise, in the following sub point the nr. of microstates in the sphericla shell is expressed

$N= \frac {\partial \Omega(N,V,E)}{\partial E} \Delta E$

THE first expression gives you the surface

and multiplied with \Delta E

the volume of the shell

sure

(it's getting at what I was talking about earlier - what matters is the density of states $\partial_E \Omega$, not the size of $\Delta E$, but let's not get distracted again)

Ok

I am doing it with another method tho

I am finding the nr of states for \Omega(E)

Then i do the difference \Omega(E) - \Omega( E-\Delta E)

which is \omega(E) (1 - \frac {\Omega(E- \Delta E}{\Omega(E)}

now do $(\Omega(E) - \Omega(E+\Delta E)) \cdot 1 = (\Omega(E) - \Omega(E+\Delta E)) \frac{\Delta E}{\Delta E} = \frac{\Omega(E) - \Omega(E+\Delta E)}{\Delta E}\cdot \Delta E$, where for $\Delta E\to 0$, the first fraction is just the derivative

(there's our random manipulation dropping from the sky again! ;) )

In theoretical physics when people have these complicated classical Lagrangians and then quantize the fields, how do they know if their classical theory is realistic in the first place, i.e. deep down that it's just a classical point particle with everything else being an internal degree of freedom? Is that the only real-world theory that makes sense?

yeah

something along those lines

@DIRAC1930 in QFT we don't really care about the classical theory

But shouldn't everything be a classical point particle with internal degrees of freedom?

like, we discovered the process of quantization because it worked for QED starting from the classical EM Lagrangian, and then people just kept guessing Lagrangians until they found QCD etc.

But QCD is just a theory of a classical point particle with internal degrees of freedom right?

no, it's a quantum theory

classical point particles don't exist

classical mechanics is a sham

(which makes it ironic that the only way we know how to produce a quantum field theory is to quantize a classical field theory, but that's just the way it is)

But how is having random Grassmann valued spinor fields a 'classical' theory?

the classical version of non-Abelian gauge theories is completely worthless

@DIRAC1930 I fear you might be confusing the words "classical" and "realistic"

Ah yes most likely

no one is saying weird Graßmann fields describe usefully any real-world classical situation

no one is saying the classical theory of SU(3) gauge fields models anything

a "classical theory" is just a mathematical construct

in this case one, that, when quantized, yields QCD

So how do we check if it's the correct theory? Do we just check if all the expectation values agree with experiement?

yes, that's how we do it for every theory!

Lol

QCD explained the particle zoo

so it won over all the competing attempts

@DIRAC1930 I am, perhaps uncharacteristically, 100% serious :P

I honestly don't believe in anything other than the neutron, proton, photon, and electron anymore lol

if you think you'll find some irrefutable derivation of the Standard Model from first principles anywhere buried in the QFT literature, I have bad news for you

that's what all the string theory types are (not-so-)secretly hoping to find, but as far as we can see the Standard Model is just a random QFT out of an infinity of possible QFTs that happens to model our world

Does quantizing a classical Grassmann valued field that transforms under $SU(2)$ produce the same theory as quantizing a classical non-Grassmann valued field that transforms under $SO(3)$?

anything that transforms under SO(3) also transforms under SU(2), so the question is ill-defined

What about if I say 'that transforms under the fundamental representation of'? Does that change anything?

well, then you have two fields in two different representations and of course get two different theories

But spin arrises out of both of them right?

unclear

what does "spin arises" mean?

I'm not sure entirely

if you're starting with a classical spinor field, you've already "put spin in"!

So they are the same thing?

and that's exactly how it works in QFT

we know we want spinors because we observe spin-1/2 particles

and so we start with spin-1/2 fields and quantize them

But when starting in non-rel physics, we don't start with spinors

Or maybe we do

I think you're looking for some "reason" that's just not there

it's an experimental fact that there are spin-1/2 particles

so we need to put a spin-1/2 into our theories

that's it

you can have QFTs without spinors, they just won't describe our world

it's not necessary that "spin arises", we're just putting it in

@ACuriousMind what is a spinor and how it's different from the spin?

Okay, so Pauli just put it in right? As a sort of postulate and it agreed with experiment right?

sure (I don't know whether it was actually Pauli or someone else)

Pauli actually deserves more credit. Was he the first one to suggest the possibility of any internal degree of freedom?

@imbAF that's a looong story - better try reading stuff on relativistic QM and come back when you have specific questins :P

hehehe

We did the spinor, but the guy who was explaining it to us, just mentioned the name

I think the confusion I'm making is that I keep thinking we are measuring things classically in experiment or real life but of course, we're just one massive quantum system

as if he was talking about the weather

and proceed with the lecture

meanwhile the def. of Spinor, the understanding of it, physical interpretation its extremely difficult, and our professor didn't even try to xD