to have a truly unique frame, sure

but the actually important thing in relativity is to have a reference velocity, and this is what the ether would have provided

the orientation of the axes or the origin don't really matter that much

note that you've already been thinking like this - in your example with the person and the car, you talked about "the car's reference frame" when in fact there are infinitely many different reference frames in which the car is stationary and non-rotating - take one frame in which the car is, and just rotate its axes or put the origin elsewhere, and it's still a frame in which the car is stationary and non-rotating

it's just that this sort of difference where the frame just differs by the position of the origin by a constant vector or a constant rotation of the axes is not all that relevant/interesting

I see

What do you mean with " reference velocity"

But I get your point regarding the origin

basically whether one can say, in some definitive sense, that they're at rest

if the ether had existed, one could have just said "i'm at rest with respect to the ether" and that'd have been as good a reference as any

the person at rest?

yes?

ok

just to make sure I understood it correctly

@Semiclassical @ACuriousMind For a 2-component spinor $\psi = (\psi^1, \psi^2)$, if I write $\sigma(\left <\psi \right |) = \left |\psi \right >\left <\psi \right | - 1/2 \left < \psi \middle | \psi \right >$ in matrices, I get:

$$\frac12 \begin{pmatrix}(\psi^1)^2 - (\psi^2)^2 & 2{\psi^1}^\dagger\psi^2 \\ 2\psi^1{\psi^2}^\dagger & (\psi^2)^2 - (\psi^1)^2\end{pmatrix}$$

Which is traceless skew-symmetric. So I can write it as a linear combo of $i, j, k$, or its representations in $M_2(\Bbb C)$

I think I get $\sigma(\psi) = (\psi^\dagger i \psi) i + (\psi^\dagger j \psi) j + (\psi^\dagger k \psi) k$

This looks an awful lot like the current density for the Dirac equation, right?

For a 4-component spinor, the current density is the 4-component spinor $\psi^\dagger \gamma^\mu \psi$

Unlike the usual 4-vector

does $(\psi^2)^2$ mean absolute value squared here?

Yeah, thanks

Sorry about that

i'm a bit distracted right now, but for writing it out it's helpful to let $\psi^2=(x+i y)\psi^1$

Good idea

a cute observation. suppose your spinor is normalized, i.e., $\langle \psi|\psi\rangle=|\psi^1|^2+|\psi^2|^2=1$. then $P=|\psi\rangle\langle \psi|$ is projection onto $|\psi\rangle$ and $\sigma=P-1/2=1/2(2P-1)=1/2[P-(1-P)]$

that makes it explicit that $\sigma$ is reflection around the $|\psi\rangle$-ray up to a factor of $1/2$

Hi All...

Hello @JohnRennie Sir...

hallo @123

@DanielUnderwood did you use dedicated software for that?

What the hell is up with nlab lately

They seem to have server troubles

Yesss

Just what I was looking for

that's the link between the stationary condition and the splitting bundle being principal

Well, it always is, but the connection may not be

@Slereah oh yes, that's why I don't understand most of it

I seem to get stuck in "how is that defined again?" link chasing, must be a server issue

@fqq It's an $\infty$ stack

It's always a stack

@NiharKarve yeah I've been using obsidian.md for a year or so and am a big fan. I used roam before that, but it was expensive and had a lot of problems. Notion also seems pretty popular, but didn't fit my use case

there's a ton on youtube about each of those, though the videos can get a bit fanatical at times

@ACuriousMind do you recall when we discussed about the phase space, and you told me that $\rho(\vec x,t)$ which is the pdf in phase space, can be interpret as the macrostate, and can be thought of cluster of points, each representing a microstate of the system?

@imbAF sure

If we can leave aside, whether the system we observe is in equilibrium or not, and whether we are considering a MCE/CE/GCE

if you evaluate $\rho$ in different points in phase space at different times (just to generalize)

I assume, since mathematically is a pdf, it's value should variate for different inputs

is that an acceptable assessment ?

"variate"?

do you mean vary?

vary

sorry

I suspect that there's a different question you really want to ask instead of this one :P

ofc xD

but sure, it's a function, its values can be different for different inputs

I am just preparing the play field

ok

if that is the case

since $\rho$ vary, and now we consider it's physical interpretation, that of a macro state

does it mean that different $\rho$, different values of $\rho$, represent different macrostates?

what do you mean by "different values of $\rho$"

$\rho$ is a function that represents a macrostate. A different choice of function represents then, of course, a different choice of macrostate

not a different choice of function

rather different set of inputs

I don't know what that means

in the simple case, mathematical case that is

f(x)=x+3 can have different values, for different inputs

but the function is the same

mathematically, $\rho$ should behave the same

or I am wrong?

yes, but the macrostate here is $f(x) = x+3$, not its value at any individual $x$

There is no particular meaning to the value of $\rho$ at a point

at a region yes

in fact, since it is a probability density, its value at a point is completely meaningless, what matters are its integrals over volumes

yes

and I recall I asked you, that as long as the system is not in eq. you can integrate wherever in the phase space

I don't know what equilibrium has to do with anything

until eq. is reached which means that rho is constant at a certain region, and zero elsewhere

"equlibrium" means just that $\rho$ has a certain time-invariant form

@imbAF careful, that's only true in the MCE!

yes

only in MCE

this is what i was going to ask you

ok

I am asked this question

and I want to answer it myself. but I am not sure

therefore I need your assistance

what is the qm eq. of phase space pdf

that would be the density matrix

right?

if "eq." = "equivalent", then yes

ok

the what si the eq of $\int \frac{1}{N!}\Pi_{i=1}^N \frac{d\vec p_i d\vec q_i}{(2\pi \hbar)^d}$

nothing

that would be $\Sigma_r$ where r indicates an eigenstate

you have no phase space in QM, so there's no phase space integral

there is no something eq. to it?

well, if you look at a concrete expression, I could tell you how to compute it in QM

but there's nothing to say about the generic integral

the closest analog would probably just be the trace

yes

which would imply that we use the eigenstates of the system

hm?

what would imply that?

what does it mean to "use" eigenstates of the system?

that's how we have taken the trace

the trace is basis-invariant

yes you can change the basis

it doesn't imply "usage" of any particular base

one sec

$\langle \hat O \rangle= Tr(\rho \hat O)= \Sigma_n \langle n| \rho \hat O | n \rangle $

this how we defined the average of an operator, when we use the density matrix

and $|n\rangle$ are the eigenstates of the system

ofc you can rename them, the trace is invariant

sure

what's the actual question?

you asked me "what does it mean to "use" eigenstates of the system?"

oh, but you don't have to compute the trace that way

this is the only way, I am aware

is there another?

the $\lvert n\rangle$ can be any basis, they don't have to be eigenstates and they don't even have to be orthonormal

that's what I mean when I say the trace is "basis-invariant"

there's nothing special about it that you'd have to use eigenstates

I didn't know that

in all my calculations, we used to consider the eigenstates of the system

simply because they are orthonormal

anyway in my solution it is said that the eq. of the phase space in QM is the hilbert space

which If I am not mistaken, is the space of the eigenstates of a system?

it's just a basic linear algebra fact - it follows directly from "cyclicity" $\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$ and basis changes acting as $A\mapsto PAP^{-1}$ on matrices.

@imbAF what do you mean by "space of eigenstates"?

also, what are "eigenstates of a system"? Do you mean eigenstates of the Hamiltonian of the system?

yes that is a nice question

because every operator in a system can have it's own eigenstates right?

I don't know what an "operator in a system" is, either :P

well hamilton operator

and a random other operator

Each of them, has it's own eigenstates

not every operator has eigenstates, but sure

at least more then 1

more the the hamilton operator having

I have a feeling you're going about this from the wrong starting point

I can't articulate properly

but I know how things operate in this case

if operators commute, then we can use the same basis of kets

We describe a physical system in quantum mechanics by a) a Hilbert space that tells us what the possible states for the system are, b) a Hamiltonian operator on that space that tells us how states evolve in time according to the Schrödinger equation, c) a bunch of other operators that represent physically interesting observables of the system in question

Nothing about this changes when we do statistical mechanics.

Yes I know all of this

We just start considering density matrices instead of only single elements of the Hilbert space

@imbAF sorry, but you're not talking like you do :P

Yes

I know

But anyway this is what the answer says, that the qm eq. of the phase space is the hilbert space

this is the answer given

sure, elements of the classical phase space represent the state of the system classically, and elements of the quantum Hilbert space represent the state of the system in quantum mechanics.

And are the states in the Hilbert space, eigenstates of the system, or it's not always the case

For example

The hilber space of the hamilton operator, contain it's eigenstates

it's= hamilton operator

@imbAF If you have two eigenstates with eigenvalues $\lambda_1$ and $\lambda_2$, then what about $\lvert \lambda_1\rangle + \lvert \lambda_2\rangle$?

that's very clearly not an eigenstate unless $\lambda_1=\lambda_2$, but it's clearly part of the Hilbert space

because it is a state, that is a superposition of the eigenstates

?

Hm?

no, just because a Hilbert space is a vector space!

really, a lot of QM is just slightly fancier linear algebra

Ok but help me understand this

and it's a fact that for any operator that's not a multiple of the identity, not all vectors in a vector space are eigenvectors, because you can always add two eigenvectors with different eigenvalues to get a vector that can't be an eigenvector

multiple of the identity ?

a multiple of the identity operator

Ofc I understand

if you have the hilbert space of the eigenstates of the hamilton operator, here you have also superposition of this orthonormal states

but they are not eigenstates of the hamiltonian

even though they belong in the same hilbert space

If I understood you correctly

really, it's not "the hilbert space of the eigenstates of the hamilton operator"

you have the Hilbert space first

then you write down the operator

and then you can go look for the few vectors in this Hilbert space that are eigenvectors

ahaa

but doesn't orthonormality play a role here?

the spectral theorem guarantees that you'll find an orthonormal basis of eigenvectors for a bounded self-adjoint operator

if you have two different operators, which do not commute , can you still say that :"you can look for the few vectors in this hilbert space that are eigenvectors"?

well, you can look at the eigenvectors for each operator

doesn't commutation play in role in this?

if you're lucky, you can find some vectors that are eigenvectors of both, but you'll never find a basis of such shared eigenvectors if the operators don't commute

ofc

then what happens

wdym "what happens"

nothing happens

you just have two operators that don't commute

and how many hilbert spaces

they're all operators on the same space!

there is just one vector space here

and it has a bunch of different operators (=matrices)

if you have problems imagining what's going on here, really just think of some 3d vector space and rotations on it

Maybe I am failing to explaining what I want

Now I do understand

the different rotations are just different operations on this same 3d space

Ok, I can understand that

they have different axes they leave invariant - those are their eigenvectors

but the rotation around the z-axis and the rotation around the x-axis both act on this same 3d space

the eigenvectors of each are orthonormal among them right?

that's where the analogy breaks down because each rotation only has 1 eigenvector :P

because it's a real vector space the spectral theorem doesn't work

freaking hell

but I wanted to talk about something you can visualize and you can't imagine even a 2d complex vector space :P

Before we continue to what I want

I want, once more to try and "explain" you, in caveman terms, how i view all this

you don't have to wait for confirmation to continue - I'll shout "No, I can't take it anymore!" if I want to quit ;)

We have system K and operator A and B. Operator A has it's own eigenstates (Like the hamilton operator with $|n\rangle$. These eigenstates belong to a hilbert space and we can express them via ket vectors, who are orthonormal and therefore can be used to express every other state that the system can be in, physically as a superposition of the eigenstates, and mathematically as a linear combination of the ket vectors multiplied with some constant.

The same applies to the B operator. And if they do not commute, then each operator has it's own basis ket, which abstractly/physically is a hilbert space containing the eigenstates and every possible state that is a superposition of them.

whether operator A has it's eigenstates or the system, idk which is the correct way to say it

I would just remove the phrase "eigenstates of the system" from your vocabulary

most people will guess that you mean the eigenstates of the Hamiltonian of the system, but it's unnecessarily confusing

correct

Because most of the times, that is the set of basis ket we consider

other then this

?

..looking for music suggestion of the general downbeat ambient genre, a la Rubycon by Tangering Dream youtube.com/watch?v=BVkRl0sXjjY...

I was trying to find the classical partition function for the 1d harmonic oscillator, and while I got the same result in the solutions, I did something for which I have no explanation as to why I should. When I integrated I took $-$ and $+$ infinity as my boundaries.

While in phase space that makes sense,

physically how do I justify this?

The harmonic oscillator cannot have infinite displacement or momentum

@imbAF why not?

if the body moves in one direction in infinity

don't think "infinite", think "arbitarily high"

is it really oscillating ?

all real numbers are valid positions/momenta for the oscillator

so your phase space is $\mathbb{R}^2$

there's no "maximum" displacement or momentum

in phase space yes, you can have every value of the 2d phase space

I was trying to justify this physically, which is not possible, infinite momentum

ok

the phase space doesn't include $\infty$

when you write $\int_{-\infty}^\infty\mathrm{d}x$, you're not saying $x$ can take the "value" $\infty$

you're saying it can take every real number

and that's certainly what's physically possible here

ok

@ACuriousMind One question. In the case of particles in box, the condition for classical interpretation of the situation is when $\frac V N >> \lambda$. For the classic harmonic oscillator I found for the entropy $S= kln(\frac{kT}{\hbar \omega}) + k$ and it is said that this makes physically sense if $kt >> \hbar \omega$

Why it makes sense only in this case?

and why is this the classical limes for the oscillator

Ofc if $kt<< \hbar \omega$ we have a negative value, but that doesn't mean that the whole entropy is negative, which is not supposed to happen

didn't you answer your own question there?

partly, because it depends from the negative value

if it is bigger then k obviously it's not allowed

but you can have kln(\frac{kT}{\hbar \omega})= -1/2k

which would give you a + 1/2 k for the entropy,k which is a positive value

the $\hbar \omega$ essentially is the "size" of one coarse-grained microstate here (remember, entropy is "supposed to be" the logarithm of the number of states)

yes i rremember

its related to the multiplicity of the macrostate

if I remember correctly

and if your $kT$ is not much larger than that, then this approximative coarse-graining fails - e.g. you'd be saying there's "half a state" for $kT = \frac{1}{2}\hbar\omega$, which is absurd

so we need positive integer values then, even when it's positive

not something of the sort 1/2 etc

well, not exactly

"(remember, entropy is "supposed to be" the logarithm of the number of states"

this is a continuous function, so you almost never get integers

it's all an approximation

yes

but when you have a million states, it doesn't matter if your approximation tells you you have a million and a half states

yes so it's close to a number

that's still "about a million states"

and you can approximate accordingly

I got it

when it tells you you have half a state, it's a bad approximation

of what?

in which scenario you would get such a result?

of reality/the underlying quantum statistics

the underlying quantum statistics?

you said yourself this is the classical limit of some quantum treatment of the system!

for low $kT$, it's just not a system that behaves very classically

so there the classical limit becomes a bad approximation for what actually happens

gotcha

One more thing, but this is a mathematical one that has been bugging me for quite sometime

When we consider the quantum 1d harmonic oscillation, with the Hamilton operator etc. When we want to find the quantum canonical partition function, we use the following formula (which was never explained how do we get it, but it's not important rn)

$Z=\Sigma_n \langle n| e^{-\beta H}|n\rnagle$

this can be written as

$Z=\Sigma_{n=0}e^{-\beta E_n}$

can you tell me how does the change happen?

It might be very trivial , and i think it is

but right now my brain is fried

what's your definition of the $\lvert n\rangle$?

so I can copy this formula

a moment

somewhere

unless there is a faster way to see what it shows

eigenstate

of the hamiltonian

energy eigenstate

@imbAF ah, and what does that mean to be an eigenstate?

in what sense

the system

written as a simple formula

is at this state the whole time? idk really

what I am supposed to say

or a state that doesn't change over time

what I wanted you to write down was $H\lvert n\rangle = E_n\lvert n\rangle$.

yes ofc

I know this

that's the very definition of what an eigenstate is, and it's the only thing you need to go from your first version to the second

but you have the exponential

so?

I don't know how to operate

if I had the hamiltonian then yes

it would be easy

if $\lvert n\rangle$ is an eigenstate of $H$ with eigenvalue $E_n$, then it is an eigenstate of $f(H)$ with eigenvalue $f(E_n)$ for any function $f$

in particular for $f$ being an exponential

:O

literally first time I hear this

in fact, that's actually how one usually defines what $f(H)$ means

or maybe I had forgotten

(many physicists will do it via Taylor series of $f$, but that's really just asking for trouble if you start thinking about it too hard :P)

taylor huh?

I mean, what did you think $\mathrm{e}^H$ meant?

$H$ is some operator on an infinite-dimensional Hilbert space, you can't just plug it into a function without defining what that's supposed to mean

Man, the more I study physics, the more insecure I become about my choice. I have began to think that only gifted people should study this. Others are simply wasting their time trying to hard and achieving nothing

I thought of absolutely nothing

one way to define this is by saying "multiplying $H$ with itself $n$ times is defined, so we'll define this as what happens when you plug $H$ into the Taylor series of the exponential", the other is to say "it's the operator whose eigenstates are the same as those of $H$ but each eigenvalue is $\mathrm{e}^{E_n}$"

the latter way is nicer, because mathematically you'd actually have to wonder about what convergence of a series of operators means in the first case, etc.

one way to define this , define which? e^H ?

yes

Ok let me read it and try to understand it

I understand the 2nd part

@imbAF to be honest, I have the impression you're "running before you can walk" - you're doing quantum statistical mechanics but you seem to be lacking proper training in both classical and quantum mechanics without the added complication of statistics. This makes this much more difficult than it needs to be

Bachelor program for you

And you need a good math session for all of this

when we took the hilbert space we said something about a bunch of functions that are square intregrate-able

and a bunch of other nonsense

And because work + university is hard, I had to leave some exams, and one of those is classic electrodynamics, which has nothing to do with current quantum physics that we do, so I am forced to spend time on both fronts, for two upcoming exams

one in classical electrodynamics and one in quantum statisticalmechanics part 1, cuz I tookl part 2 without giving one :D

I wasn't trying to blame you - I don't know your life, and I know there are many places where this sort of curriculum that sends you running before you can walk is just the way it's done

Yes

I'm just saying, this is doing physics in "hard mode", and so it's no surprise you're finding it difficult

that's how it is

But it's not that I am lacking because I have all the nice explanation and I fail to understand. It's the contrary, we are given bits of information, and then for the rest : "Search on x y z book"

to be fair, that's just how studying works - you are expected to find the resources best suited for you if the lecture is insufficient

That is time consuming and many times useless, because one concept or a topic that is explained in the lecture, which is hard to understand in the first place, is treated or explained differently in different books, and now you are forced to adjust the explanation in the lecture to that of what you read, while at the same time having minimum understanding of what's going on

hope you understand what i mean

I understand, but this won't ever change again - the more advanced your topics get, the less there will be the "one textbook version" and the more there will be different accounts of the same topic that might seem to say wildly different things at first glance

yes

I also know that's not useful advice for having to pass an exam - but it's also okay to not understand something completely and move on

come back to it later with a fresh mind

that's why I am tormenting you :D

I do that rarely, and hope that that what I don't understand it's not in the exam

yolo

@ACuriousMind I have done a small "project" of mine, trying to explain the canonical / grand canonical ensemble, schematically with ofc mathematic explanation attached to it. When I finish, could you have a look at it?

@imbAF I don't want to commit my future time right now, but you can ask me again when it's ready

ofc

That's what I mean

once I Finish