06:00 - 17:0017:00 - 20:00

6:17 AM
I thought the conditions of Stone Von-Neumann theorem to hold is equivalent to checking whether $[x,p]$ has the same domain as $x$ and $p$...Is this wrong?
For particle in a box with momentum restricted to act on square integrable differentiable functions which vanish at the boundaries of the box, the commutator bracket has the same domain as that of the operator...then why does the SVN theorem fail?
I can see this in other ways ofcourse---exponentiated weyl relations hold only for unbounded operators...here $x$ is bounded. SV theorem holds for self adjoint operators only, but here $p$ is merely symmetric.

2 hours later…
7:48 AM
@Sanjana The SvN theorem is strictly speaking about the Weyl relations, not the CCR. If your representation of the CCR on some space does not exponentiate to the Weyl relations, the SvN theorem does not apply.

Okay...Is there a case in QM where momentum is bounded? From the CCR one of the operators must be unbounded...I have seen cases (e.g. particle in a ring) where the position is bounded so momentum must be unbounded...but never a case where momentum is bounded.
In harmonic oscillator the phase space trajectories are bounded but still harmonic oscillator momentum operator is unbounded...
@ACuriousMind I am thinking of crystal systems but i am not sure.

I don't know of any example off the top of my head

8:05 AM
@ACuriousMind Okay, I have another doubt then---In Hall's book on quantum theory for mathematicians here it is told that $p$ is defined on space of smooth compactly supported functions on the line then it is self adjoint...I can't see why it is even symmetric---how will the boundary term get cancelled? Does compactly supported imply in some way $\psi(a)=\psi(b)$ for some $a,b \in \mathbb{R}$

@Sanjana of course - compact support means there is an $[a,b]$ outside of which the function is 0

Yes...but on the boundary i.e. at $x=a$ and $x=b$ nothing guarantees $\psi(a)=\psi(b)$ right?

why do you think you need to take the boundary?
just look at $\psi(a-\epsilon) = \psi(b+\epsilon) = 0$

Is this because the integral limits are $\pm \infty$ and not $a,b$?

I don't exactly understand the question but I think the answer is yes :P
what you really need for the boundary term in partial integration to vanish is that the function "falls off fast enough" towards infinity, and compactly supported functions fall off the fastest possible: They're just identically zero after some point.

8:13 AM
@ACuriousMind Thanks, now it is clear...I was making the mistake of integrating from $a$ to $b$...
So, if the space was itself $L^2([a,b])$ then it would not be symmetric right?
If I am not putting extra conditions like $\psi(a)=\psi(b)$ or $\psi(a)=\psi(b)=0$ then the momentum operator is not symmetric, right?

@Sanjana see e.g. physics.stackexchange.com/a/671931/50583, the self-adjoint version needs boundary conditions $\psi(a) = \mathrm{e}^{\mathrm{i}\theta}\psi(b)$.

@ACuriousMind So there are many possible self adjoint extensions! Hmm...

The $\theta$ shifts the zero of the momentum spectrum, so we usually choose $\theta = 0$ - physically it makes little sense for the momentum spectrum to be asymmetric about 0

@ACuriousMind Hmmm...Can you tell what is the easiest way to check boundedness of an operator? It seems pretty hard to check directly from the definition (I have to construct some bound which might not be obvious for an arbitrary operator). I am looking for maybe evaluate something and be able to infer from that quantity.

@Sanjana just look at the spectrum
the norm $\lvert \lvert A\psi\rvert\rvert$ is the eigenvalue of $\psi$ when it's an eigenvector, if the spectrum is unbounded, so is the operator
conversely, if the spectrum is bounded, you can do the same argument as for the finite-dimensional case since then the maximum of all eigenvalues exists

8:29 AM
Yeah I know that if the spectrum is unbounded then the operator must be so...
But unbounded operators may also have bounded spectra right?

@Sanjana no, for self-adjoint operators this can't happen
and I just gave the argument in the "conversely" part above
of course you can have non-self-adjoint operators that are unbounded with bounded spectrum but in physics we rarely care about boundedness of arbitrary operators - all our observables etc. are self-adjoint

Oh...wow...that makes life really easier, and is the proof 2-line or something/intuitive?
I would go and have a look...Thank you so much.

16 mins ago, by ACuriousMind
conversely, if the spectrum is bounded, you can do the same argument as for the finite-dimensional case since then the maximum of all eigenvalues exists

@ACuriousMind hey . I wanted to hear your opinion on something. if you got the following image - ibb.co/YNY3J61 such as you say: y = y'+a, that must surely mean that we're using passive transformation

conceptually it's literally the same argument as in the finite-dimensional case
I do not recall how ugly it is to make it rigorous :P

8:44 AM
because the same point is just observed from different frames, which means passive transformation. active transformation would be y' = y+a. do you find anything wrong in here ? and to do homogeneity check, we must use active transformation method. that's why we're doing y+a switch instead of y in Lagrangian :P @ACuriousMind

@GiorgiLagidze I don't know what the image is supposed to tell me about the "activeness" of the transformation. Maybe the image depicts someone actually actively lifting the stick figure up, maybe it depicts someone passively changing their coordinate system.

@ACuriousMind Hmm...great...But for the case of particle in a box, doesn't unbounded momentum contradict the fact that Fourier Plancharel transform (being unitary) take $x$ to $p$...how can one be bounded and the other not?

@RyderRude i think i have everything figured out :P and better not to go to rabbit hole. :p
we can leave it as it is

ok :)

@Sanjana what do you mean by it being "unitary"? It's a transform between function $L^2([a,b])$ and Fourier series $L^2(\mathbb{Z})$. These two spaces are not the same but usually we define unitarity only for automorphisms of a single space.
all you can say here is the Fourier inversion theorem for Pontryagin duality, i.e. the forth-and-back Fourier transforms are adjoints of each other

8:58 AM
@ACuriousMind I got this idea from a comment by Valter Moretti which I think I may have misread.

@Sanjana unless I'm missing something, that answer is about the operators on $L^2(\mathbb{R})$, so I don't know what it has to do with the situation you were discussing

@ACuriousMind So isn't what was said there not extendible to the case I was discussing? What's the issue with Fourier transform on a closed interval or the circle?

@Sanjana hello. the fourier series of a bounded x-dependent function contains discrete but unbounded frequencies
to have a bounded momentum, u need an $x$ space which is unbounded but discrete @Sanjana
but i dont know if there exist quantum theories on such spaces
or you can also have a bounded but discrete x space. the discreteness puts a limit on the fourier frequencies @Sanjana
there do exist QFTs on such discrete spaces, called lattice QFTs
in such QFTs, the momenta of the particles have an upper limit @Sanjana

9:14 AM
Oh yess...this is nice...are there other possibilities? Much like the crystal case I was thinking of where you have discrete translational symmetry...

Is it impossible to have bounded momentum without making the $x$ space go discrete?

@Sanjana yes, many possible extensions corresponding to different physics - I think it's pretty cool to be suddenly drawn back to reality while worrying about the difference between symmetric and self adjoint :D

@RyderRude I asked this question a few minutes ago on the main site...You might want to add an answer there, also.

great :D
@Sanjana ohh but the boundedness is only for the single particle states of the Fock space
but the Fock space states can have an arbitrary number of particles
so this is technically not an answer

I also already said it with the $L^2([a,b])$ and $L^2(\mathbb{Z})$
@Sanjana Yes, you have a direct correspondence between boundedness of one operator and discreteness of the other; this also follows from the Pontryagin stuff I already linked :P

i think my QFT example technically does not fit because of the Fock space. I will try to add an answer using single particle discrete space QM
if we get rid of the multi-particle states, then the momentum and energy become bounded
@ACuriousMind hi. this need not be seen as a transformation between different hilbert spaces. one can think of these two as different bases of the same hilbert space

9:32 AM
@ACuriousMind We've already been at this but maybe not so directly: how much of a hindrance do you think it is not having a supersymmetry course, a QG course, a ST course etc. to learn about each of these?
@Amit I'm inept when it comes to philosophy, so I'd better shut up :P

@ACuriousMind @Mr.Feynman is there a path integral analogue for the finite dimensional Quantum Theory?
i couldnt find anything on PhySE or google

@RyderRude Wait what do you mean? $$\int Dq e^{iS[q]}$$ isn't enough?

@Mr.Feynman but this is for an infinite dimensional hilbert space quantum theory

Oh, finite in that sense

yes

9:43 AM
Like e.g. a fixed spin $1/2$ particle in a magnetic field

yes

@RyderRude this might help I guess

thanks
27 mins ago, by Ryder Rude
@ACuriousMind hi. this need not be seen as a transformation between different hilbert spaces. one can think of these two as different bases of the same hilbert space
i think i am incorrect about this as a change of basis cannot be seen as a transform on the abstract Hilbert space

10:00 AM
@ACuriousMind Thanks...this feels like the best thing I have read in a while and I don't understand it :(
I mean...it is said there that the functions are mapped to the Pontryagin dual space via the Fourier transform. So how can I say anything about the behaviour of the operators?
(Even from a beginner's perspective...the position space and momentum space wavefunctions are related by a Fourier transform and not the operators themselves!)
Actually I am interested in knowing what (other than the functions) are getting mapped via FT? Because...then---
What went wrong with the particle on a circle? I have a bounded position variable, I am getting discrete momenta..all good. But following Valter's comment why can't I say analogously that I get bounded momentum from bounded position...
You might say "boundedness" is the property of the operator...It doesn't get mapped...that's why my question is "what gets mapped?"

@Sanjana I have always pointed out that introductions of Fourier transform should be starting from the finite discrete case, and then taking the appropriate limits, so that the relationships become trivial to see, and students will never be asking these questions.

@naturallyInconsistent Can you guide me to a reference? I didn't know a full world of abstract stuff with Fourier transforms also exists!
And...All of this seems to work without a choice of the kernel...as in the wiki page

@Sanjana I don't know how to give it to you; in the end I got fed up enough about it that I wrote my own little pdf about it, from which I teach.
The mathematicians have a whole literature on the subject and I have no conception of the depth with which they cover the subject, but mine is meant to be most student-friendly.
mathematicians, of course, need the depth because they want to prove a lot of important properties of the Fourier transform, given that it is so central to so many branches of analysis, but physicists don't need that, and instead want a simple-to-understand basic bit.

10:18 AM
@Mr.Feynman I don't think supersymmetry is so complicated that it needs its own course :P QG and ST courses are certainly nice but you will have to read papers anyway to actually understand what people in the field are talking about
@Sanjana Here we have a Fourier transform $F : L^2(G) \to L^2(\hat{G})$ and its adjoint $F^\dagger : L^2(\hat{G})\to L^2(G)$ and $FF^\dagger = 1_{L^2(G)}, F^\dagger F = 1_{L^2(\hat{G})}$. What exactly is your question about these operators?

For the case of a particle on a circle $G$ is $S^1$ and $\hat{G}=\mathbb{Z}$
For the case of a free particle both $G$ and $\hat{G}$ are $\mathbb{R}$

Why can we say $p$ is unbounded because $x$ is, in the second case and not in the first case?

@Sanjana but we just told you that the boundedness or unboundedness of p is due to discreteness or not of x; boundedness or unboundedness of x is a totally different thing to talk about
:64499719 not a typo; it is the central confusion you are having

@Sanjana Because in the second case $L^2(G) = L^2(\hat{G})$ and so $F$ is a unitary operator that maps $x$ to $p$, and operators related by a unitary operator have the same spectrum
I already explained why this doesn't work for $G\neq \hat{G}$ here

10:34 AM
Okay, so the reason why $p$ is unbounded is because $x$ is discrete?
Which statement implies this? I see from the wiki page and also from ACM's reply that the F.T. is a map between the Hilbert spaces...and here we are speaking of operators...what exactly is the correspondence?

@Sanjana unbounded p because x is NOT discrete

@naturallyInconsistent oops...sorry, typo.

@Sanjana $x$ is the multiplication operator on $L^2(G)$ and $p$ is the multiplication operator on $L^2(\hat{G})$. so naturally the spectrum of $x$ is $G$ and the spectrum of $p$ is $\hat{G}$.

if you started with discrete & bounded x = finite x, then p is also discrete and bounded. Then you get to take limits as you like, and derive the relationships between boundedness and discreteness.

to have a quantum theory with a bounded P, u just need to take a quantum theory with a bounded X and discrete P, and then re-name the X as P and P as X

10:35 AM
all 4 possibilities will be covered, and there will be no more possibility of confusion

this gives u mathematically a quantum theory with a bounded P. whether actual systems are modelled by this theory is another busienss
the closest thing to a real world theory with this stuff is lattice QFT @Sanjana
but lattice QFT still has the whole Fock space, which makes it unbounded

Okay...But there's also a line in the wiki page you linked "Since the complex-valued continuous functions of compact support on $G$ are
$L^{2}$-dense, there is a unique extension of the Fourier transform from that space to a unitary operator..."

Unitary? From $G$ to $\hat{G}$?

however, if ur dynamics dont allow for particle creation and annihilation, then u can restrict this theory to a subset of the hilbert space with bounded H and P @Sanjana

@naturallyInconsistent Yes..that clarified things a bit.

10:41 AM
@RyderRude You won't believe me...I was also actually thinking of this. @ACuriousMind What do you think? Would this work? (mathematically I mean)

@Sanjana ofc it works mathematically lol. whether u interpret P as position or momentum in ur experiments is later business

@Sanjana why would you be interested in something so silly when you already know of a situation where it is handled better?

but an actual example of a theory with discrete space where the bounded P is interpreted as momentum is lattice QFT restricted to a subset of the Fock space @Sanjana

@naturallyInconsistent cz I am silly :p btw...handled better where---lattice theories?

@Sanjana sorry I went down the wrong path

10:45 AM
@ACuriousMind The real mess seems to be that stuff beyond the Standard Model are so dispersive. Like, just for gravity one has many types of QG (strings, LQG etc.) but then I find out there is also supergravity which is related to supersymmetry D:
I wish I had a more detailed map like this :P

@Sanjana no, crystalline physics. unbounded x but with periodicity, so p is bounded by 1BZ

@Sanjana "unitary" here just means that $F^\dagger$ and $F$ are inverses of each other, and what's true is that operators $O_1$ and $O_2$ that are related by $O_1 = FO_2F^\dagger$ have the same spectrum
you seem to be trying to argue that this should somehow imply that $x$ and $p$ have the same spectrum, but it just doesn't (it is not true that $x = F p F^\dagger$ - the Fourier transform of $p$ is $\mathrm{i}\partial_x$) - try to figure out why you think that

@Mr.Feynman you cannot be hoping that research can be mapped out before we know what they are actually going to end up becoming...

@Sanjana calling one of these operators "p" and the other "x" is completely arbitrary; it's only on physical grounds that we decide which is which, mathematically they are completely interchangable
@Mr.Feynman so what? the real world is messy
no one knows where the journey really goes beyond the SM so there's paths in all directions that intersect at random points
that's just how research is

@ACuriousMind I was just confused by Valter's comment "Since $x$ and $p$ are transformed into each other through a unitary map (Fourier-Plancherel transform) none of them can be bounded."...This is somewhat loosely phrased, right?

10:50 AM
@naturallyInconsistent I mean, I can for existing research :P

@Sanjana What Valter means is that the Fourier transform shows that they can both be represented as the multiplication operator on $L^2(\mathbb{R})$
so their spectra have to be the same

Well, at least in 100 years things will look more organized, will they? :P
If the human race is still a thing

@Sanjana one way we can get rid of the "re-naming example" is by having an independent definition of momentum. we can define momentum as the generator of the translation symmetry of the Hamiltonian. then the re-naming would no longer work, as only $P$ (and not $X$) commutes with $H=P^2/2m$

@Mr.Feynman The frontier will just be somewhere else

with this independent definition of $P$, lattice QFT is the example that works @Sanjana

10:53 AM
I don't think it's reasonable to believe that physics - or any science - will be "finished"

@Mr.Feynman that is really just not possible. For example, string theory was originally meant for studying QCD stuff, and then they realised it has to include QG. Life is crazy

@ACuriousMind No I didn't mean that. Of course I meant that it will shift somewhere else and this will be settled :P

I mean, you can hope for it to be settled :P
I personally currently see no indication either way that we'll have settled quantum gravity by then

And I can also hope that there is still civilization :P

im just saying that whatever operator we interpret as momentum need not be decided in experiments. it can also be decided by the symmetry of the Hamiltonian. only one of X and P commutes with the Hamiltonian
so the way to make this question more well defined is : are there theories satisfying : 1. bounded P, 2. [X,P]=i, 3. P is a symmetry of H
the real world theories satisfying this are non relativistic lattice QFTs
with the dynmaics restricted so that particles cannot be created. this is so we can restrict the Fock space to have a bounded P
the QFT must also be non relativistic, as there's no $X$ operator in relativistic QFTs, which means the $[X,P]=i$ requirement becomes meaningless

11:01 AM
Lattice QFT is not a "real world theory". There is no system actually modeled by lattice QFT, it's purpose is to obtain continuum QFT in the continuum limit/make numerical computations for continuum QFT.

i am generalising "real world theory" to mean "theory with real world applications"
it's not supposed to mean a fundamental theory, as no fundamental theory has discretized space
one may also ask "why do i need to explicitly restrict particle creation if the QFT is already non relativistic"
this is because quasi particles are created in Non relativistic QFTs in cond matter physics

11:19 AM
@ACuriousMind Okay...What about energy quantization? Is the Hamiltonian spectrum discrete because it's dual is compact?

@Sanjana No, the Hamiltonian has nothing to do with this
The Hamiltonian is some function of $x$ and $p$, there is in general no relation between such a function of multiple operators and the spectrum of the individual operators
of course when it is free as $H=p^2$ then you can get its spectrum directly from the spectrum of $p$, otherwise not

there is a little bit of relation, I think. for the energy values where the particle's trajectory is classically bounded, the spectrum is discrete

11:34 AM
in fact, trying to apply this analysis to the Hamiltonian is how we traditionally ("Pauli's theorem") conclude there is no time operator (which "should" be the conjugate operator to H), since the Hamiltonian should physically always be bounded below but the time operator should be continuous and unbounded in both directions

@ACuriousMind I thought that this was for the energy operator and not the Hamiltonian operator.

@Sanjana ...what's the difference?
do you really want to throw gauge theories into this mix where the Hamiltonian might not be "energy" :P

Oh no...no constraints...
Well if I define $E$ as $i\partial_t$ and $H$ is what we write on the other side of the schroedinger equation $[E,t]$ and $[H,t]$ are not same for starters...

Which out of $H$ and $E$ is the conjugate to $t$ ?

@Sanjana $\partial_t$ is not an operator on the Hilbert space, it's an operator on trajectories

wh-whaaat?

11:42 AM
the idea is that the time derivative is undefined on the vectors on the hilbert space

But in TDSE we write $i\partial_t |\psi\rangle=H |\psi\rangle$ right?
It is acting on the states of the Hilbert space?

@Sanjana no, it is acting on a time dependent state
operators act on just states
id/dx can do that, for e.g.
@ACuriousMind there are non-gauge theory cases too. see this post : physics.stackexchange.com/a/614821

12:08 PM
@Mr.Feynman I got myowself confused on the pictures nonsense. In QM, the path integral is computing the propagator $\left<x,t\vert y,0\right>$. In this expression, both bra and ket are in Heisenberg picture, and this is part of the confusion: In the Schrödinger picture whereby everybody is much more comfortable with, the states $\left|\psi(t)\right>$ evolve with time and thus carry the time variable, whereas operators $\hat x$ do not.
The important part is to realise that the time-independence of the operators means that $\hat x$ has eigenfunctions that are ostensibly also time-independent. i.e. $$\forall t\in\mathbb R\qquad\hat x\qquad\implies\qquad\exists\left<x\right|\forall t\in\mathbb R\ |\quad\left<x\right|\hat x=x\left<x\right|$$

@naturallyInconsistent you don't have to think of that bra and ket as states (which do not evolve in Heisenberg picture) but as eigenvectors of the position operator. Think of the spectral decomposition $$\hat{x}(t)=\int x\lvert x, t\rangle\langle x ,t\lvert$$
The eigenvalue is fixed, it's the same you have in Schrödinger picture, while the eigenkets have that time label as the operator

and the horrible old notation $\psi(x,t)$ is really $\left<x\vert\psi(t)\right>$ in Schrödinger picture. The resolution of identities go as $\int\left|x\right>\mathrm dx\left<x\right|=\hat{\mathbb I}=\int\left|p\right>\frac{\mathrm dp}{2\pi\hslash}\left<p\right|$

Truth is you don't have to think of that $t$ as time dependence, because it is not. Nothing is evolving here. The label $t$ is there to tell you that since the operator is evolving, it is different at each slice of time, thus is has different eigenkets at each time

@Mr.Feynman I'm typing that part up, give meow time. Also, this is somewhat an abuse of notation that should be highlighted, because the eigenfunctions of the position and momentum operators live in Rigged Hilbert space, are thus invalid as quantum states.

This has nothing to do with Rigged Hilbert space, in the sense that it applies for any operator
I mean, it is irrelevant for this matter, you could take a nicer observable and nothing would change

12:17 PM
@Mr.Feynman yes I do

@Mr.Feynman It turns out to be important in path integration. In QFT, when we write down the path integral, the resolution of identities is easier written in Schrödinger picture when you have the factorisation of the time evolution operator U into many tiny pieces, that you later Taylor expand and then recombine into exponentials. It must be possible to do it in Heisenberg picture, but it appears differently.

The resolution of identities used in the path integral (let's consider the QM path integral to avoid complication) is the Heisenberg one, as you consider one resolution at each time slice
In the Schrödinger picture only one resolution of identity exists because $\hat{x}$ does not evolve

@Mr.Feynman this is simply not true, as long as you are doing time slicing

Again, you should read point II of this answer by qmechanic. There is no such thing as time slicing in Schrödinger picture (for operators)
Although each time slice is "identical" to a copy of Schrödinger picture, the latter is by definition only the one at $t=0$, where it coincides with Heisenberg picture

12:25 PM
@Mr.Feynman I do not see what QMechanic is saying is disagreeing with what I am trying to tell you.

Mhhh you told me you were confused about pictures, so at this point I'm unsure: are you asking or explaining to me?

@Mr.Feynman hi. the usual derivation of the path integral indeed doesnt have anything to do with the Heisenberg picture
Sakurai may have given a weird derivation

@RyderRude Sure, we discussed that a long time ago
Amplitudes don't have a picture

Discussing Zee

12:27 PM
i wasnt in that discussion :P

"We" in the chat and me asking something like that :P

@Mr.Feynman Im explaining

Because Zee said we were working in Heisenberg picture and that was untrue
@naturallyInconsistent I think I understand now what you mean. You were saying to dig the time evolution out of the Heisenberg ket
Sure, that happens in the derivation and then one manipulates either expanding or doing some other trick

@Mr.Feynman I'm telling you that the expression whereby we write $\left<x,t\vert y,0\right>$, both stuff in this expression is in Heisenberg picture, but the way we typically deal with this is that $\langle x\vert\hat U(t)\vert y\rangle$ i.e. sending it to Schrödinger picture, and as long as you are breaking U into many tiny pieces, that is time-slicing. i.e. you are doing time slicing in Schrödinger picture if you do this.

Yes, that's what I said in the last message. I never disagreed regarding this, it's true

12:32 PM
@Mr.Feynman you were saying that we do not do time-slicing in Schrödinger picture...

@naturallyInconsistent yes. this is the derivation I meant

@naturallyInconsistent I was answering the first part. As to the final sentence, I disagreed that you do time slicing in the completeness relations. The amplitude is indeed picture independent as you correctly point out

Anyway, the point I am making is that, while later on we would pull the $\mathrm dx\mathrm dp/2\pi\hslash$ out of the action exponential and write them as $DxDp/2\pi\hslash$, the fact is that they were originally those small $\mathrm dx\mathrm dp$ initially, and this fact remains the same in QFT; i.e. you get $\mathrm d\phi\mathrm d\pi/2\pi\hslash$ all over the place if you do time slicing path integration in QFT
@Mr.Feynman I was not asking any question so what were you answering again?

"I was answering the first part"="When I said I never disagreed, I meant about the first part of your message; I'll now replying to the final sentence"
> i.e. you are doing time slicing in Schrödinger picture if you do this.
And then I did
@naturallyInconsistent by the way, yes. This was the problem and I understood that discussing with @RyderRude I hadn't realized we were doing time slicing also in QFT (and I thought about spacetime slicing)

Yes, my nice QFT prof really did hammer it in that quantum theory seems to privilege time really much more than space.

12:39 PM
@Mr.Feynman i dont think @naturallyInconsistent is talking about the completeness relation functional integral in the message u replied to
he is talking about the unitary evolution path integral
but anyway, time slicing is involved in both cases

@naturallyInconsistent in fact one only needs to look at ETCR to understand it

Issues surrounding the pictures involved and the interpretation of the path integration integrals and so forth, I had to wrestle with him to make sure he put in a particularly follow-able presentation. I'm perfectly fine with sloppier arguments elsewhere, but this was a confusion point that was bothering meow
And I had been sidestepping all the time slicing nonsense by doing the integrals over the Fourier coefficients trick.
There are also treatments that handle the time slicing without Taylor expanding, i.e. making sure to stay in complex exponentials all the way.

@Mr.Feynman i just want to say that the resolutions of identity we insert are the Schrodinger picture resolutions
all of them

@RyderRude Mr. Feynman was just pointing out that it is possible to do it with Heisenberg picture resolutions of identity, just that they would look very different. I was making the same point you were making

at least in the derivation where u write it as $\langle x'| U(t)|x \rangle$
@naturallyInconsistent yes. it is possible to do it that way. i agree. i am also talking about the usual derivation like u

12:48 PM
Anyway, I think we are kinda on the same page now; I was just trying to point out some seriously dangerous pain points that I found. I mean, I'm trying to teach this thing and write it as a textbook and there were so many silly confusing things involved that it is way too easy to be dropping into one of the many pitfalls
People who present it seem to think that this is so obvious; they dont seem to realise that the arguments that they themselves received and treated as gospel is actually likely to contain subtle mistakes. We are plagued by bad notation encouraging a lot of confusion and nobody seems to want to give a clean presentation that students might hope to follow.

1:21 PM
@ACuriousMind Cool stuff

2:03 PM
@ACuriousMind How do Hamiltonian eigenfunctions really form a basis of the whole Hilbert space or just a subspace? The Hamiltonian acts on only a dense subset of the whole Hilbert space...how would their eigenfunctions form a basis of the full space?

@Sanjana because the notion of a countable basis in an infinite-dimensional Hilbert space also involves notions of denseness
The claim is that for an orthonormal basis $\psi_i$, you can write any vector as $\psi = \sum_i c_i \psi_i$ with $c_i$ square-summable - the r.h.s. here is an infinite series that can converge to vectors that do not lie in the ordinary linear span of any finite subset of basic vectors
this means precisely that the ordinary linear span of the basis vectors - i.e. their finite linear combinations - form a dense subspace of the full space

obviously every basis vector lies in a finite linear span of basis vectors, so there is no contradiction between the Hamiltonian being only defined on this dense subspace and yet the basis being a proper basis of the full space in the sense of infinite linear combinations

can someone help me with finding out the relation between acceleration of the wedge and the block when the syste is released from rest. Surfaces are smooth pulley is massless string is ideal.

it's again the "maximum problem": When you do $H\psi = \sum_i c_i h_i \psi_i$, the r.h.s. potentially diverges when $H$ is unbound, i.e. this is not well defined when the product sequence $c_i h_i$ is not square-summable, but when the span on the r.h.s. is finite, then there is no convergence issue
so the space where $H$ is not defined is naturally a subspace of the vectors that cannot be expressed as finite linear combinations of its eigenvectors

2:11 PM
i dont understand the definition of energy being used in this post : physics.stackexchange.com/a/614821 . shouldnt energy be defined as the conserved charge of time symmetry?
but this lagrangian has time symmetry, but they still say energy isnt conserved

@RyderRude is it not obvious to you that a damped oscillator will not have conserved energy?

@ACuriousMind yes. but how to get that formula from a general definition of energy?

you won't
the damped oscillator breaks the rules

wtf

it's an accident in the first place that it admits a Lagrangian/Hamiltonian formulation, most dissipative systems don't

2:19 PM
oh. so this "conserved charge of time symmetry" is the definition of energy mainly for non-dissipative systems
if we wrote the hamiltonian for an irl damped oscillator, but also included the energy sink to have a closed system, the perhaps this "conserved charge" definition would give the correct expression of the energy
i also found this post : physics.stackexchange.com/a/11911/156987 . This one isnt dissipative. If the Hamiltonian is time dependent, why dont we just say "energy isnt conserved as there is no time symmetry"?
but instead the conclusion drawn in the post is "Hamiltonian =/= energy"
@naturallyInconsistent do you know whats going on with this post? why isnt energy here defined as "conserved charge of time symmetry"?

2:46 PM
maybe the definition is "the conserved charge of time translations in inertial frames". because general co ordinate systems will not have time symmetry for the same system
to get the energy-momentum in any other co-ordinate system, we transform this conserved charge four- vector with the Jacobian

@RyderRude No please not this again T_T

@Mr.Feynman sorry. i shouldve said u can use either resolutions

The fact that you later write $$\langle x', t'\lvert x, t\rangle=\langle x' \lvert U(t',t)\lvert x\rangle$$ (so express in terms of Schrödinger picture) is true, but the resolution of the identity itself is $$1=\int dx \lvert x, t\rangle\langle x, t\rangle$$, it is Schrödinger only at $t=0$
And even if you write it *in terms of Schrodinger picture, it remains the Heisenberg picture instantaneous completeness relation

maybe if u want to interpret it that way. i just interpret it as a bunch of completeness relations from the Schrodinger picture (i.e. all at $t=0$).

@Mr.Feynman just don't write the resolution of the identity with the silly $\lvert x\rangle$ and you don't have this silly problem :P

3:00 PM
it is very intuitive for me in the Schrodinger picture personally. but yes, another derivation can be written in which we first explicitly introduce the H-picture resolutions

Mathematicians have no problems deriving the path integral without even talking about the $\lvert x\rangle$

You may write
$$1=\int dx \lvert x, t\rangle\langle x, t\lvert$$
as
$$1=\int dx e^{iHt}\lvert x\rangle\langle x\lvert e^{-iHt}$$ of course but it's still Heisenberg picture completeness relation (written in terms of Schroedinger stuff)
@ACuriousMind I'm no mathematician :P
I mean, the discussion about Heisenberg/Schroedinger here is only accidental. I would have avoided it if possible

ok i think discussion is splitting hairs anyway. our thoughts are the same. semantics can be different
lets leave this one

It is indeed, but the language about pictures is fairly established to my knowledge :P

do you know the definition of energy being used here : physics.stackexchange.com/a/11911/156987 @Mr.Feynman
when they conclude Hamiltonian =/= energy @Mr.Feynman

3:11 PM
@RyderRude I would say they're thinking of the mechanical energy defined as $E=T+V$, the sum of the kinetic and potential term in the lagrangian
basically the Lagrangian with swapped sign and expressed in the canonical variables

oh
it was the same thing in the damped oscillator example too
@Mr.Feynman this seems like an unclear way to define energy as opposed to "conserved charge of time symmetry"

@ACuriousMind Is the Fourier transform used in Poyntragin duality a true FT? I mean they don't assume the kernel to be $e^{i \nu \mu}$ or anything...so is that assumed or all the results actually kernel independent?

The problem arises with the explicit time dependence
The mechanical energy function $E=T+V$ and the Hamiltonian are the same (expressing them in the same variables) if the constraints the are not explicitly time-dependent

oh
so this is not a problem with rotating frames but with constrained systems?

(without generalized potentials)
@RyderRude That $\theta\to\theta-\omega t$ adds explicit time dependence

3:17 PM
yes
these examples make the definition of energy very fuzzy

If you will, think it in terms of parametrization and not constraints
@RyderRude If it helps, the Hamiltonian expressed in lagrangian variables is known as "generalized energy"

@Sanjana The $\mathrm{e}^{\mathrm{i}px}$ is, in the abstract language of the Poyntryagin duality, contained in the identification of $G$ with $\hat{G}$. The abstract transform there gives you a $(Ff)(\chi) = \int f\chi^\dagger \mathrm{d}\mu$. The $\chi$ is an element of $\hat{G}$, which are maps $G\to S^1$. So for $G=\mathbb{R}$, we first need to fix an isomorphism between $\hat{\mathbb{R}}$ and $\mathbb{R}$ to get the ordinary Fourier transform.
One such isomorphism is given by associating with any $p\in\mathbb{R}$ the function $x\mapsto \mathrm{e}^{\mathrm{i}px}$ in $\hat{\mathbb{R}}$. Plug this into the abstract transform and you get $(Ff)(p) = \int f(x)\mathrm{e}^{-\mathrm{i}px}\mathrm{d}\mu$.

@RyderRude You may want to read Goldstein as this answer below suggests
That section - which I'm skimming right now - deals with the explicit time dependence of the Hamiltonian and its relation to the total energy (which is what I called mechanical energy)

@Mr.Feynman i know these posts are working with mechanical energy but im unsure how to define mechanical energy in a rigorous way :P
@Mr.Feynman thank you. i will read this :)

3:32 PM
@RyderRude What the lagrangian we use in classical mechanics? $L=T-V$
Mechanical energy is defined as $E=T+V$, what else do you seek?

@Mr.Feynman idk.... i want to know why this would be a useful quantity to talk about even when it's not the conserved charge of time symmetry
this formula was derived using time translations.
but there's other derivations too, like in Newtonian mechanics. are we going back to Newton?
i should just read this section. thanks

@RyderRude Note that when you derive generalized energy (Hamiltonian in the lagrangian variables) you assume that there is no explicit time dependence
And as I said above, if your coordinates do not depend on time explicitly, the Hamiltonian is the total energy

yes
so the definition of total energy is just T+V

So do you mean to say that the fact that momentum space wavefunctions being defined via a FT is a matter of choice?

I always wondered this---there are two distinct notions of duality---fourier duality and Noether duality (I don't know if this has an official name...but it's that thing...translation in one quantity conserves it's Noether dual...Noether dual is also the thing you get when the Lagrangian is cyclic in that quantity...the pair of stuff you get in HUP, that appears in classical Poisson bracket or commutator relations etc)

@Sanjana if you think a bit about the "choice" here you'll realize it's just the usual ambiguity in the Fourier transform: Some people define it with $\mathrm{e}^{2\pi \mathrm{i}px}$, some with $\mathrm{e}^{\mathrm{i}px}$, etc.
so there's nothing new here, we just now can explain these differences in this framework as different choices of isomorphism between $G$ and $\hat{G}$
as for the "duality": What you mean by "Noether duality" is called being canonically conjugate variables. Two quantities $p$ and $q$ are classically canonically conjugate if they have Poisson bracket $\{q,p\} = 1$.
clearly, canonical quantization turns this into the CCR $[q,p] = \mathrm{i}$, so in quantum mechanics canonically conjugate quantities are Fourier duals of each other by the SvN theorem

3:43 PM
@ACuriousMind By the SvN theorem? Where in SvN theorem do we mention anything about Fourier transforms?

@Sanjana the SvN theorem means that (if the Weyl relations of these quantities hold) the only irrep of this is as multiplication and differentiation operators on $L^2(\mathbb{R})$
and multplication and differentiation on $L^2(\mathbb{R})$ are related by the Fourier transform

@ACuriousMind Is FT the unique transformation satisfying this property?

what property

that the multiplication becomes a derivative when transformed?

well for one it's obviously not unique as we just discussed you can choose e.g. a $2\pi$ factor in the exponent :P
on the other hand I'm not sure why uniqueness would matter here
I'm just saying this shows they're Fourier related

3:47 PM
@ACuriousMind Here it doesn't
Just having A Curious Mind :p

He will seek revenge for that

I think there's a whole host of weird integral transforms that all have similar properties e.g. the Laplace transform (which in some sense is just a more general Fourier transform)

So again, it amounts to saying that Fourier transform is a choice, right?

choice of what?
no one forces you to think about the Fourier transform at all
if you wanted to you could do all of QM just in position space, it would just be more awkward :P

no no not in that sense 😂
I meant...If I choose to use Schrodinger representation I can use *any* integral transformation which gives me a multiplicative $x$ operator and a derivative $p$ operator...right?

3:55 PM
I don't understand what you think you need the integral transform for there at all
you need the transform to interchange the two, i.e. turn the derivative operator into a multiplication operator and vice versa

@Mr.Feynman yes. it is a very good discussion. it emphasizes the dependence of the Hamiltonian (and its conservation) on the choice of generalised co ordinates when doing the Legendre transform

@ACuriousMind Yes and I am asking whether this transformation needs to be the Fourier transform or it can be something else also...that's it

@Sanjana I really don't understand the thought process here
it's not that we have the objective "let's interchange multiplication and derivative" it's that we look at the Fourier transform and notice it does that and then we go "neat"
but anyway, the Fourier transform here is a unitary linear operator and so it is completely defined by its image on a basis
so you can't have another unitary linear operator that does the same
but this really is too vague - remember the $2\pi$ ambiguity in the exponent, for instance
if you want to ask a question about uniqueness, you first need to be really sure what properties you're demanding of the thing

@ACuriousMind Oh no...I didn't mean that. I understand that you don't need a FT to go from $x$ to $p$...rather we need that for a change of basis. I think I was/am wording it wrong.
@ACuriousMind Yeah, how about we first eliminate all the trivial equivalents of FT like multiplication by a constant in the phase or amplitude...

@Sanjana hi. the eigenstates of id/dx are only plane waves, so there's no other choice than the Fourier transform if we want to switch from position to momentum representation
there's no other eigenstates

4:09 PM
@ACuriousMind But you said previously that Laplace transforms and other might have this property...What about those then?

@Sanjana Have you looked at them?

@RyderRude Seems intuitive...but how exactly do I show it's relation with F.T.?

again, in order to say that something here is "unique", you first need to properly define unique in what sense
what properties exactly is this object supposed to have
then you can show (or not show) that it is the only object with those properties

when we say "position and momentum are related by an FT", our goal isnt "FT is the only way to interchage x and id/dx", our goal is "FT is the way to switch from the eigenbasis of x to that of id/dx" @Sanjana

I already covered the case where we look at the bases above

4:13 PM
@Sanjana the eigenvectors are $e^{ipx}$. the FT is just the eqn for the projection onto these vectors

so if you want an operator that does the same to a basis it's trivially unique by general linear algebra. if you want something else, you need to properly define it

So I will just go and search for transforms with those properties?!
(If you say that there's no uniqueness theorem like that/not known if there's any---then I would definitely look...but it's like searching for a closed form solution in radicals to a generic quintic polynomial without knowing Abel Ruffini theorem!). And the other integral transforms I know of don't satisfy that property...so I was asking that.

@Sanjana My whole point here is that you haven't so far given a proper definition of the properties you want this unique object to have
maybe looking at other transforms would make you go "oh no this isn't what I want" or "oh sure, that works, too"
but I cannot answer your question because you keep evading my requests to make it precise enough to be answerable without guessing what you actually want :P

Thing is..I understand the notion that a theorem has it's assumptions...But It's just that I am flexible with them...not too flexible to include trivial alternate versions of F.T.
...Do you "feel" it?

nope
I am unfeeling

4:25 PM
Come on 😭
I want an integral transform which takes a position space wavefunction and gives me a momentum space one where in Schrodinger (position) rep, position is a multiplicative operator and momentum is $-id/dx$...

Fourier transformation with or without all those extra factors are what I will call "trivial"...I want a non-trivial one...possibly with a verrrrry different kernel, not an exponential---that could count as an answer...

but i just told u that the very meaning of "going from position to momentum space" is to "change the basis to the eigenvectors of the momentum operator"

@Sanjana So what about the basis argument? Any time you require that a unitary map maps one operator to another, you are also demanding that it maps all the eigenvectors of that operator to the eigenvectors of the other.

there's no other eigenvectors of id/dx than plane waves

But a linear operator is defined by its action on a basis, so this already completely defines the operator
this is the third time I'm trying to say this in different words but so far you've ignored it :P

Somehow...I missed it...Now I am seeing that both you and Ryder Rude have been telling me this so many times...so sorry.
Which operator are you talking about here? The FT one?
But aren't the eigenvectors of F.T. Hermite polynomials?

4:30 PM
@Sanjana The Fourier transform is the unitary operator. The multplication and differentiation operators are the ones it maps to each other. I didn't say anything about the eigenvectors of the Fourier transform tiself.

@RyderRude feels like sanjana blocked you too

@naturallyInconsistent hmm i dont think so
@naturallyInconsistent what do u mean "too"?

@RyderRude the list of people blocking you is not at all short...

@naturallyInconsistent can u please list them?

@RyderRude i dont think we can ever know all of them.

4:42 PM
@naturallyInconsistent then pls stop claiming that u know the cardinality of an unknown set :P

please don't speculate about who might have whom blocked, this is unconstructive gossip; if people want someone to know they've blocked them, they'll say so

btw it is 1, becauze I know
and i have blocked them too

it is not 1; I already know a few.
@ACuriousMind they said so

Meanwhile, me knowing that one can even block people here :o

@naturallyInconsistent pls give proof

4:44 PM
@RyderRude no

@naturallyInconsistent hmmm then we will just dismiss ur claims. thanks
@Sanjana u click at their name and click "ignore user" if u want :)
btw pls try to tell when u block someone, or it makes for counter productive discussions

the block universe lol

@RyderRude and :64501521 Can you explain this once more---then I will officially call myself dumb and move on to some other topic ðŸ™‚
I get the fact that FT is a unitary linear map which takes in a differential operator and gives a multiplicative operator...So they map the eigenvectors of the $x$ operator and $d/dx$ why is this last statement equivalent to the fact that FT is the unique unitary linear map? (unique upto the trivialities I mentioned above)

In other words, following Ryder Rude if I started with a generic unitary linear map which takes in $d/dx$ and gives $p$ then via what logic would I solve for the eigenfunctions of $d/dx$ and call the result the kernel in the unitary map?
@RyderRude Good to know but I can't even think why would anyone want to block anyone else in a physics chat site...there's so much to learn from each other! :)

@Sanjana 1. Do you know that a linear map is completely define by its action on basis vectors? I.e. instead of specifying an "entire map", I just need to pick a basis and say what the map does to the basis?

@ACuriousMind Yes...it's like mentioning the components of the vector instead of the abstract vector itself.

4:54 PM
2. Do you understand that requiring a unitary linear map to map one operator to another requires that map to map the eigenvectors of one operator to the eigenvectors of the other?
(this is the flip side of the statement that unitarily equivalent operators have the same spectrum, a statement we already invoked earlier today)

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