The h Bar

Silly Goose

04:47

@Slereah hm okay

well i will perhaps check out Moretti's suggestion "Varadarajan’s book 'The geometry of quantum theory'" to see what happens to such linear operators

Ryder Rude

10:50

hi

i have some very interesting results connecting Lie theory, physics and Number theory (lemme finish)

theorem : for a quantum system to have periodic time evolution, the energy eigenvalues must be commensurable

proof: for periodicity, we have $e^{-iHt}=1$ at some $t$. for the energy value $E_n$, we have $e^{-iE_nt}=1$ at $t=\frac{2\pi a}{E_1}, a\in Z$. therefore, for all $e^{-iE_nt}$ to simultaneously be $1$ at some $t$, we require $\frac{2\pi}{E_n}a=\frac{2\pi}{E_m}b$, or $\frac{E_m}{E_n}=\frac{a}{b}$ which is a rational number

this means all $E_m$ and $E_n$ are commensurable

this idea is relevant in the quantum harmonic oscillator and the angular momentum generator

the converse of the theorem need not hold in infintie dimensional spaces. e.g. consider a Hamiltonian with eigenvalues : 0.1,0.01,0.001.....

Silly Goose

this paper seems interesting: "Convergence of Dynamics on Inductive Systems of Banach Spaces" link.springer.com/article/10.1007/s00023-024-01413-6

Ryder Rude

now we connect this to Lie theory. for a Lie group to be compact wrt a generator, we require $e^{-iAa}=1$. this means the compactness of a Lie group is related to the commensurability of the eigenvalues of a representation of the generator

Silly Goose

"Many features of physical theories become really clear only in some limit- ing situation. Quite often, the limit is not just the limit of some parameters in a fixed framework, but the structure of the theory changes in the limit."

"The structural limits are essentially obtained via inductive-limit construc- tions for a sequence of Banach spaces that describe either states or observables of the approximating systems."

Ryder Rude

11:07

to check for commensurability, we can divide the generator by an unknown constant $a$ and assume that the resulting eigenvalues are rational

then we form the characteristic polynomial of $\frac{A}{a}$ and look for rational solutions using p-adic numbers

so this connects compactness of a Lie group to p-adic numbers

i am finished

let me know if u think this is correct

Ryder Rude

11:31

and ur thoughts on these results

naturallyInconsistent

13:56

@RyderRude The two results that you are so eagerly putting forth, the former applies equally to classical systems and is inside Goldstein, and the latter is trivially obvious once you consider the former. They are necessarily correct. Not sure if it is necessary to require $\mathrm e^{-\mathrm iAa}=1$ though.

Ryder Rude

14:15

@naturallyInconsistent how can it apply to classical mech when classical mech always has a continuum of energy values? the theorem requires commensurable eigenvalues.

@naturallyInconsistent $e^{-iAa}=1$ implies that the subgroup generated by $A$ is compact... and then i relate this to commensurability of eigenvalues, which is related to rational solutions of characteristic polynomial, which relates to number theory

naturallyInconsistent

@RyderRude Goldstein was considering celestial mechanics. For periodic orbits to be closed the periods of x-y must be commensurate. Same with for $\vartheta$ and $\varphi$ orbits.

Ryder Rude

@naturallyInconsistent it makes sense i guess but it's unrelated to energy

naturallyInconsistent

That's just because it is in the classical context. It is perfectly analogous.

Ryder Rude

yes... there seems to be a general relation between commensurability and periodic phenomena

naturallyInconsistent

There is no extra insight that you are bringing in, once a person understands what Goldstein is saying, what you are saying is obvious.

Ryder Rude

14:19

it is still connecting Lie theory to number theory

it brings the structure of rational numbers to continuum mechanics

naturallyInconsistent

It is literally a trivial observation and noted in the literature. What "deep" are you talking about?

I suppose you are entitled to have a little celebration whenever you learn about a connection between topics.

Ryder Rude

i think, after further investigation, this can bring deep results by relating group theory/matrix theory/operator theory to number theory

i was wondering if this particular connection has already been made

Goldstein's observation seems to be relatively limited in scope, as they merely relate periodic trajectories to commensurability

to have classical trajectories, u have to solve differential equations. maybe Goldstein's idea can relate differential equations to number theory

but this idea relates operator spectra and group theory to number theory, which has more scope

imbAF

19:11

Hi! I have a question about the phase space, and the wigner function.
In the classical case, if a system is represented with a point in phase space, then what do we represent an ensemble with? collection of points, which translates to a volume in the N-dimensional phase space ?

ACuriousMind

Feb 10, 2023 at 8:31, by ACuriousMind

@imbAF You're making this again more complicated than it needs to be: The ensemble is represented by a probability density $\rho(q,p;t)$ on phase space. Equilibrium is $\partial_t \rho(q,p;t) = 0$ (and $\partial_t\rho = \{H,\rho\}$ due to Liouville's theorem)

we've literally talked about this over a year ago :P

imbAF

Yes I do remember the lengthy discussion but I have forgotten some stuff

And I was hoping you would reference that :P

ACM I will ask 2 questions, but they will sound stupid because of the way I will construct them

So sorry in advance

I make a distinction between probability distribution and pdf. With an exampe. The normal distribution f(x) is a function bell-shaped, and for every x value we have a corresponding f(x) value. In terms of probability , when we consider a continuous variable , the probability for a specific x is zero. The pdf, mathematically speaking is a corresponding value of f(x) for a given x. And the totality of all f(x) for a range of the x, is the probability distribution.

I am describing this, for a reason

But first I want to hear your opinion on what I just said

It's not what I am aiming to understand, but I believe it's necessary later

ACuriousMind

I don't really understand what distinction you're trying to make here

imbAF

for example

for the normal distribution

ACuriousMind

"pdf" means "probability distribution function", it's just a function that defines a probability distribution

imbAF

19:19

the pdf is the function itself f(x). While the probability distribution are the corresponding values for all the x values of the considered interval

ACuriousMind

the only context in which I would consider the two not synonymous would be in a rigorous math context where there may be distributions (e.g. the $\delta$-distribution) that are not functions

imbAF

What I am trying to get is that, in order for you to draw the bell shaped curve

you need, mathematically speaking, to calculate the f(x) value for every x in a considered interval of values

would you agree with that

And also would it be correct to say that the ensemble is represented by a probability distribution?

ACuriousMind

@imbAF That's not how people use these terms. If you want to draw a distinction here, here's how the terms are formally used: A probability distribution $P$ on a space $X$ is a measure on $X$ obeying certain condition, i.e. it's a function on the $\sigma$-algebra of measurable sets on $X$

A probability distribution function is a function $f$ on $X$ that induces such a measure $P_f$ by $P_f(S) = \int_S f$, i.e. assigning to each measurable set the integral of the function over that set

imbAF

Ok

ACuriousMind

importantly the abstract distribution does not have values "for every x", or rather, in the usual case, those values are just all 0

imbAF

19:25

yes

in the continuous case

ACuriousMind

I'm not quite sure what you're trying to get at, but you should definitely not use the "distribution vs. function" terminology for that

imbAF

Ok

is this accurate

And also would it be correct to say that the ensemble is represented by a probability distribution?

ACuriousMind

An ensemble is more or less defined by a particular probability distribution, yes

imbAF

Ok. Now the issue I am having is with the wigner function

So I need to be as accurate as I possibly can, so that you can see where my problem is

ACuriousMind

the Wigner function is not a probability distribution (function)

imbAF

19:27

I know

I will get at what I am trying to say

First of all the physical setup I am considering, for simplicity, is an electromagnetic mode in some setup i.e waveguide, resonator or w/e else

In my lecture the following is said:
The WF is a quasi-probability distribution function used in quantum mechanics, to represent the quantum state in phase space.
In wikipedia:
WF is the analogues of the Liouville density.

I don't know if you have any issues with these two definitions

ACuriousMind

no, it's fine

(although I don't think the name "Liouville density" for the phase space density is particularly common)

imbAF

My problem is the following. In the classical case, a state of a system is a point. And we also have the probability density which is an ensemble.

In quantum mechanics, the role of the pdf is played by the WF

and from my poor understanding, at the same time the WF distribution in phase space, is the representative of the state in it

Does that make sense?

Do you understand what I am trying to get at, when comparing both pictures ?

ACuriousMind

I don't know what you mean by the WF distribution being "the repesentative of the state in it", but I think I see your underlying confusion

imbAF

in classic case state = point in phase space

pdf = ensemble

in qm state=????
quasi pdf = wf

ACuriousMind

if the classical theory has as the "real" state points in phase space and then these probability densities that are probabilities over them, and the quantum theory produces also such quasi-probability densities, then what's the quantum equivalent to the point in phase space?

the answer is: there is none, this is precisely the fundamental difference between classical mechanics and quantum mechanics

imbAF

19:35

So, you can't represent the state with something in phase space?

ACuriousMind

in quantum mechanics, position and momentum cannot be simultaneously definite, so there is no quantum equivalent to the single point in phase space

imbAF

Yes

ACuriousMind

@imbAF you are representing the quantum state by something in phase space - by the Wigner function!

note that even a pure QM state results in a Wigner function that isn't a $\delta$, i.e. a pure QM state does not correspond to a point, but also already to a distribution

imbAF

But the wigner function can to a certain extend play the role of pdf

do you see the non consistency ?

ACuriousMind

it's not an inconsistency

it's just the fundamental difference between classical mechanics and quantum mechanics

imbAF

19:37

How not? the distribution in Classic mechanics is the result of considering an ensemble, while here not

Or does the uncertainty plays a role?

ACuriousMind

@imbAF that's not an inconsistency, that's just the way it is

imbAF

Ok, so how would one represent an ensemble in the qm case?

ACuriousMind

@imbAF with a density matrix

imbAF

in the phase space

ACuriousMind

with the Wigner function associated with that density matrix

quantum mechanics is fundamentally probabilistic: even the pure states result in probabilities for measurements, not definite knowledge, and so the difference between pure and mixed states is not the same as the difference between points and distributions in classical phase space

imbAF

19:39

But what if the state of the system, the state itself is a mixed state , i.e thermal state

and additionally you consider an ensemble of the system

ACuriousMind

I don't know what you mean "additionally"

when you declare the system to be in a mixed state, that's your ensemble

imbAF

Oh...

ACuriousMind

e.g. the quantum canonical ensemble declares the system to be in the state $\mathrm{e}^{-\beta H}$

anologous to the classical probability density of the canonical ensemble being $\mathrm{e}^{-\beta E(x,p)}$

as we initially discussed, the ensemble is the probability distribution, so in the quantum case, the ensemble is the mixed state

imbAF

I see

I mean I am a bit rusty, compared to before but it's good that, cuz it makes me think more

So in the qm case a state and ensemble both are represented by the WG, in difference to classic where a state is a point and an ensemble a pdf

is that correct?

ACuriousMind

yes

(assuming WG means Wigner function ;) )

imbAF

19:50

yes

Ok so my 2nd question is regarding the integration of the corresponding pdfs and their meanings. What I mean is

if you were to integrate the pdf, in the classical case, over some region in phase space, the result you'd get, could be interpreted as the probability of the system being in one of the states in the integrated volume

or I remember it wrong?

ACuriousMind

that's correct

this interpretation obviously fails for the Wigner function because it has regions where it is negative

imbAF

Yes, but that;s not the issue I am dealing with

if I can say it in a very caveman way

the state in QM, in phase space, has finite dimensions

because the WF represents it in phase space

ACuriousMind

I don't know what you mean by that

imbAF

I mean

ACuriousMind

what's the "dimension" of a state?

imbAF

19:55

I mean

this

that's the state in phase space

ACuriousMind

that's a graph of the Wigner function of a system with 1 spatial dimension yes

imbAF

the graph of wigner function

in what can be considered as the phase space

cuz that's the whole point

we are abstract at this point

As you can see, for picture 1, the distribution is spread over a certain interval in each space (position and momentum)

which wouldn't be the case for the classical scenario, as you'd have a point

so there is no finite dimension of the state, in that case

in the context, as to how it's represented in the phase space

ACuriousMind

I still don't know what you mean by "dimension of the state"

imbAF

forget dimension

I mean

ACuriousMind

then stop saying "dimension" :P

imbAF

20:00

No wait,

I should

How do I explain that the state, represented with a WF, has a spread in phase space, it has a volume,

I mean, think of a state how it's represented in classic case and compare it to here

ACuriousMind

@imbAF well, just saying that is much better than talking about its dimension!

imbAF

dimensions in the sense, that you can see the spread over an interval in x and p

that's what I mean

ACuriousMind

but I still feel we're comparing apples to oranges here: A classical state and a quantum state are, obviously, not the same thing

imbAF

But I haven't reached my point

ACuriousMind

it's not that you have "a state" and you represent it once classically and once quantumly, the two notions of state are simply entirely different

imbAF

20:03

And I am fine with that

@ACuriousMind Having this in mind. I know that integrating the WF for one of the conjugate variables, gives you the probability distribution of the other. Does this marginal probability distribution (if it's correct to say) of one of the conjugate variables, plays the role of the pdf, similar to the classical case? Which would mean, that integrating this marginal distribution over an interval, would give us something. And that something would be? How would you interpret the result of that integratio

ACuriousMind

@imbAF it's just the probability density to measure the variable you didn't integrate - if you just unpack the definitions, you'll find that for a state with position space wavefunction $\psi(x)$, you get $\int W(x,p)\mathrm{d}p = \lvert \psi(x)\rvert^2$

imbAF

Yes

and if you would integrate the marginal distribution

of x, what do you get ?

ACuriousMind

1?

imbAF

Can you say something similar as to what I described above ?

Cuz when I integrated the pdf, the way I described the result was

ACuriousMind

oh, you mean the other way around - of course you just get $\int W(x,p)\mathrm{d}x = \lvert \psi(p)\rvert^2$, i.e. the momentum space probability density

imbAF

20:12

I mean this

ACuriousMind

(where $\psi(p)$ is the Fourier transform of $\psi(x)$)

imbAF

First of all, let's just write $w(x)=\int W(x,p)dp$ as the marginal distribution for x.

In the picture I sent, this marginal distribution would be the shadow case in the position axis

right?

ACuriousMind

yes (but be mindful that it's not literally the "shadow" of the 3d graph, it's just drawn in dark grey)

imbAF

In the first graph

for simplicity

But we are getting to what I want to know

@ACuriousMind I get this part. And let's just write this probability density as w(x)

In the classical picture I said:
if you were to integrate the pdf, in the classical case, over some region in phase space, the result you'd get, could be interpreted as the probability of the system being in one of the states in the integrated volume
Now in the qm picture, where we consider w(x) as pdf, how do you interpret the integration over an interval in the position axis ? You cannot have 1 to 1 explanation with what I said for the classical case, because here we are talking about a single state, while in the classic case, we had an ensemble

Do you understand where I am trying to go with this?

ACuriousMind

no, because just from the fact that $w(x) = \lvert \psi(x)\rvert^2$, you should be able to figure out the interpretation yourself

imbAF

20:18

If I have to

ACuriousMind

there is nothing novel here, this is the very elementary question of what the square of the wavefunction gives you

imbAF

If I have to, then the integration of w(x) would give me the probability of the system having an x value in the range that I integrated it for

or no?

ACuriousMind

what does "having" mean?

quiantum mechanical systems don't "have" positions

imbAF

being physically

with un uncertainty

since we cannot pin point position

ACuriousMind

no, you're being too vague

imbAF

20:20

ok

spread of the wavefunction

I am not sure

what to say

ACuriousMind

there is an well-established meaning of what the square of the wavefunction gives you the probability density for

this has nothing to do with statistical mechanics or phase spaces or whatever else, it's what you should have learned and applied over and over again in basic QM

imbAF

Yes

the probability of finding the system at an arbitrary region

ACuriousMind

exactly, finding, i.e. it's a probability for the result of a measurement

imbAF

Yes

ACuriousMind

this is very importantly different from the probability of the system "being" there, since "being" implies that it would also have been there if the measurement had not taken place, and QM does not allow us to make such counterfactual statements easily

imbAF

20:23

I cannot see the distinction

if there's a probability of the system being found at a certain region, isn't that the same as it being there

ACuriousMind

In quantum mechanics, you're not allowed to talk about a system "having" a position or momentum outside of the context of a measurement

imbAF

Why?it sounds deterministic ?

ACuriousMind

what exactly the ontological status of a particle is when no measurement is being performed - if it has a definite position and we don't know it or if it doesn't have one - is a matter of your quantum interpretation

the mere formalism alone does not know a definite quantity like "the position of the particle" outside of the context of a measurement and so you need to avoid language that implies there is (or commit to an interpretation that allows you to speak in that way)

imbAF

I see. Well that's quite the detail

ACuriousMind

@imbAF the technical term for the assumption that definite positions exist when we're not looking is realist, not determinist, and this issue is at the heart of much-discussed things like Bell's theorem, which is why I'm insisting on getting the phrasing right here

imbAF

20:29

Ok I will keep this interaction in mind

I need to be more precise with the way I say things

But

You do see the distinction in the interpretation when integrating the pdf in the classical case and in the qm case for a single state

right ?

ACuriousMind

sure; again, that's just a manifestation of the difference between classical and quantum mechanics

imbAF

Sure

what if the state of the system in qm would be mixed one

ACuriousMind

nothing changes

imbAF

really?

ACuriousMind

really.

imbAF

20:31

Ok, that's quite different than in the classical case

I was thinking something like this:

"if you were to integrate the pdf, in the classical case, over some region in phase space, the result you'd get, could be interpreted as the probability of the system being in one of the states in the integrated volume"

for when the state in qm is a mixed one

But you say no

ACuriousMind

indeed and again - that's because of the fundamental difference that in QM even a pure state isn't a point

imbAF

Yes

ACuriousMind

so talking about the probability to be in "one of the states" in QM doesn't make any sense to begin with

imbAF

This is what I was trying to get at

:P

ACuriousMind

the pure state is already probabilistic - going to the mixed state doesn't change that, you're just adding and additional "source" of probability, but the interpretations of the quantum mechanical probability distributions don't change - they're still probability distributions for the results of measurements

imbAF

20:35

Ok and finally, in the simple case, that we were considering, of a single state. By integrating over w(x) you can make claims about the probable position of finding the system. And can you say anything about the momentum, or nothing at all?

@ACuriousMind RIIIIGHT. This makes sense

But if the system considered is an EM mode, how can I understand position?

the field is spread over space

and I am pretty sure we are not talking about the mode value at some position

but the mode's position

how ?

how to imagine assigning position to a field? the same way you'd do for a particle in some region of space

ACuriousMind

there is no position

imbAF

so amplitude of the mode then?

ACuriousMind

the quadratures of optical phase space are the real and imaginary parts of the complex amplitude, not position and momentum

imbAF

Ok but again what are we getting when we integrate w(x) if not position ?

the quadratures of optical phase space?

ACuriousMind

you're getting a probability amplitude for the other quadrature

it's not a particularly meaningful quantity

imbAF

20:41

wait

I said integration of w(x) and not w(x,p). I acquire w(x) as $w(x)=\int w(x,p)dp$. So I already did this

@ACuriousMind when I got w(x)

ACuriousMind

yes, so this would be the probability density for measuring $x$

imbAF

But you can then, integrate w(x) (the gray colored shadow in the pictures)

ACuriousMind

it's just that $x$ isn't a particularly meaningful quantity :P

imbAF

ahaa

wtf, then what is the point?

ACuriousMind

the point of what?

imbAF

20:43

of calculating something that has no sense or, no physical interpretation?

ACuriousMind

I don't know, why are you calculating it?

imbAF

cuz it's a pdf, and that's what you do

ACuriousMind

if a quantum optics text shows interest in this function, I'd expect it to also explain what it's gonna use it for

imbAF

All this for something that can't be interpreted in some way

ACuriousMind

@imbAF I have never felt the need to compute a marginal distribution in optical phase space :P

imbAF

20:45

I mean, in the classical case, the integrating of the pdf, has, and correct me if I am wrong, a meaning and interpretation, which I described above and you agreed

here the marginal w(x) distribution

is of no use

ACuriousMind

so what?

shit happens

though I'm really not an expert in quantum optics, there might be a specific use for this function, but at least there isn't a straightforward physical interpretation of it

imbAF

But then, doesn't it mean that

the WF in phase space

the representation of a state

doesn't really, in the end, give us anything of value

I say this because

ACuriousMind

you really should stop saying "phase space" when you mean "optical phase space" :P

imbAF

We started with the WF, and we ended up with these marginal distributions

Sorry

ops

optical phase space

ACuriousMind

the classical phase space of positions and momenta is very different in its meaning from the optical phase space, even if both look the same mathematically

and yes, I'd agree, there's plenty of situations where the optical phase space picture isn't particularly useful

imbAF

20:49

then the WF is not useful in this case, because we started with it in the optical phase space, and we ended up talking about marginal distributions, which physically mean nothing

or perhaps one can make position synonymous with amplitude of the field and momentum with phase

which is what takes place in my lecture

though idk how accurate that is

to do so

ACuriousMind

I mean, that's not entirely true, for instance you can draw pretty pictures about the Wigner function of a coherent state vs. that of a number eigenstate etc.

I know you know this because you were in here asking about those pictures a while back :P

imbAF

Yes

but that's nothing of value

ok, I just have to accept that

@ACuriousMind xD

ACuriousMind

some people find that kind of visualization useful

imbAF

If I consider the electromagnetic field a superposition of modes, in which case the state of the field is a tensor product of the states of all the modes

can I assign a WF to the state of the field

ACuriousMind

I don't understand the question - what do you think the state that you compute a Wigner function for in the optical phase space is?

oh, you mean if you don't restrict to a single wavelength

imbAF

20:53

yes

if you remember I did describe the setup

in the beginning

exactly because I would ask this question in the end

ACuriousMind

no, then there's no useful optical phase space, since you'd have countably many different $x$ and $p$

imbAF

So, the optical phase space is useful for when we talk about single modes

in whatever state the mode might be

And regarding the quadrature ($\hat{X_1}$, $\hat{X_2}$) some of the commentary in my lecture:
1. They represent amplitude and phase of a quantum state (I assume in optical phase space).
2. Correspond to two oscillations of 90 degree phase difference.
3. Describe the amplitude and phase of $E(\vec r,t)$.
4. Play the role of position and momentum

you agree with these statements ?

ACuriousMind

I agree, but especially for point 4 I would caution against reading this too strictly - "play the role of" does not mean "are", it's just about the mathematical equivalence to position and momentum in ordinary phase space

imbAF

Ok, so if I were to have the WF as a function of the quadratures

then can I have a meaningful interpretation of the integration of one of the marginal distibutions?

ACuriousMind

didn't I just say that there is no straightforward interpretation?

imbAF

21:02

Yes

I thought one could

ACuriousMind

so why would my opinion change :P

imbAF

if I were to consider the quadratures

@ACuriousMind Ok

I'll leave it at that

I thank you for your time. I learned quite a lot from this discussion and I am more clear with my thoughts

I was guessing stuff all the time

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