The h Bar

Claudio Menchinelli

00:09

If I have a power series $\sum_n a_n\rho^n$ and the recursive relation for obtaining the subsequent coefficient is the one displayed in the image

How do I obtain the factorial

Oh I see hahaha. Brain is completely fried at this point

Semiclassical

03:11

thinking about some physics that i usually don't

specifically: what's the difference between the forces acting on someone submerged in water, vs those acting on someone engulfed in a granular medium?

the latter seemingly isn't as simple as "the weight of material displaced", owing to the high amount of friction between particles

naturallyInconsistent

03:25

@ClaudioMenchinelli you need to learn to recognise when pushing through is just counterproductive and just go get some rest.

@Semiclassical correct. But it might well be wonderfully approximated by fluid dynamics predictions and then adding an uncertainty band due to the friction being able to create pressure differential of either sign.

Semiclassical

right

mostly i want to be able to (somewhat) accurately communicate why jumping in a pool is safe while getting engulfed in a grain silo can be lethal, despite them being similar bulk densities

but that can wait until i get to the point of mentioning viscosity

Silly Goose

is the crossover of mathematical fields a new phenomena? for instance, using groups to characterize topological spaces or using lie algebras to characterize lie groups

consider a correlation amplitude $C(t)$. assume the spectrum of the system of interest is quasi-continous such that we can approximate $C(t)$ with an integral over the appropriate interval of energy values.

I don't understand the argument sakurai runs through to deduce behavior of the approximate $C(t)$.

In particular, it seems like Sakurai is saying that i) rapid oscillations $\implies$ cancellation, ii) we assume the energy spectrum is peaked with some width, so that the contributions to $C(t)$ are localized to some interval centered at the peak, iii) if the interval for which there are not rapid oscillations is much smaller than the interval of the width, then there will not be much contribution to $C(t)$

I am uncertain about 1) does saying the energy distribution is peaked mean that we only, qualitatively, care about what the distribution looks like at the peak and within the interval of the width of the peak? 2) does rapidly oscillated imply cancellation? 3) Is the condition $\lvert E - E_0 \lvert \approx \hbar /t$ the condition of not rapidly oscillating?

naturallyInconsistent

04:00

@SillyGoose (i) is true. i.e. (2) is true too. Read the wiki link at the bottom, but first let's talk things through. (ii) Yes, Sakurai is trying to get you to consider a case like this. (iii) is true too. (3) is true. (1) is false. Equation (2.72) is true regardless of which $E_0$ you pulled out; However, it is obvious that if you pulled out a $E_0$ far from the peak $E_0^{\text{best}}$ then you cannot see what the argument is about. en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma

After all, since by assumption the initial energy eigenstates constituting the state being scrutinised is peaked around $E_0^{\text{best}}$, the contribution from a small window near $E_0\neq E_0^{\text{best}}$ will be tiny, and then in the argument you cannot see why the Equation (2.72) will extract out by itself a $E_0^{\text{best}}-E_0$ and return you the same results. Hence why Sakurai's argument only starts by considering $E_0=E_0^{\text{best}}$

@SillyGoose Lie algebra and Lie groups are the same theory... it is not a new phenomena... However, why would you be expecting that mathematicians would not try to use existing tools and knowledge when exploring new realms?

Silly Goose

@naturallyInconsistent hm i guess im confused how the assumptions sakurai states tells us that a distribution with a peak with width has most contribution around the peak

for instance i can have this distribution

while perhaps what is in mind is somethign more like

naturallyInconsistent

04:17

@SillyGoose correct, think only of the latter

Silly Goose

hmmm interesting :P is there a reason the distribution should be considered as being so benign? do you know of another resource that discusses this approximation in more detail or perhaps an example of the realistic physical scenario sakurai alludes to

naturallyInconsistent

@SillyGoose I mean, if you had this kind of weird peaking initial state, then C(t) will still be near one if the "width of acceptance" $\hslash/t$ is wide enough to take all of them into account, i.e. short times. After a while, of course, this width of acceptance will shrink until it omits some of the other peaks, leaving you with $0<C(t)\ll1$

Semiclassical

@naturallyInconsistent one other meandering thought: would one still have something akin to Archimedes' principle in the grain case? my gut would be that it's still there, but much smaller than the frictional forces

naturallyInconsistent

Essentially, what this "self-correlation" argument implies, is that Schrödinger evolution automatically makes initial states into purer and purer energy eigenstates simply by you waiting longer and longer before using the states. That is, if you want to prepare a pure energy eigenstate for use in an experiment, you just have to wait longer for it to self-purify. Extremely realisitic.

@Semiclassical It must be there, and it has to work no matter what. Weak or strong friction, we still need to obey classical mechanics.

Semiclassical

yeah, the usual argument for archimedes' principle still holds

@naturallyInconsistent ah, just wait long enough for recurrence /s

(except the energy levels are incommensurate and thus too are the frequencies etc etc)

naturallyInconsistent

04:29

@Semiclassical Poincaré recurrence do not work here.

Semiclassical

psh

spoilsport

Sanjana

06:04

@Slereah Yes exactly...The problem is that I just found two references which seem to point out different motions which can be interpreted as dilations!

Silly Goose

06:46

i feel like the first chapter of sakurai is wonderful and the rest of the book is so opaquely written...

I do not quite understand why the propagator is not introduced immediately as the Green's function for the (wave function) Schrödinger equation. Instead, we follow a very circuitous path to the result...

Ryder Rude

what path does he follow?? @SillyGoose

Silly Goose

he begins with a general initial ket with time evolution operator applied to it

then supposes existence of an operator $A$ such that $[A, H] = 0$ where $H$ is the hamiltonian, so we have access to a basis $\{a'\}$

then uses the completeness relation to expand the time evolved initial state $\lvert \alpha (t, t_0) \rangle$. then, writes this in the position basis, i.e. as a wave function. then does a bunch of relabeling and calls something the propagator

i feel like the fact that what sakurai calls the propagator is the green's function for the (wave function) schrodinger equations should be obvious, but i am trying to see how it should be obvious

i think this is it

Ryder Rude

he shudve stated that the eigenvector stuff is just a computation technique to solve for the Green's function of Schrodinger eqn

Silly Goose

if we are interested in finding the green's function for the schrodinger, then we want to know how an individual position eigenstate $\lvert x \rangle$ evolves in time

this is $U(t, t_0) \lvert x \rangle$, which written as a wave function is $\langle x ' \lvert U(t,t_0) \lvert x \rangle$

@RyderRude i feel like i am missing something because the two sentence argument i wrote above is satisfactory to me

Ryder Rude

07:01

the argument u wrote does not help u actually compute $U(x,t, x',t')$. u have just written what u want to compute

the eigenvectors are a method to compute $\langle x'|U(t,t')|\langle x$

Silly Goose

hm i see but that is just application of completeness relation to the end of my argument

Ryder Rude

yeah

ACuriousMind

@SillyGoose Presumably because Sakurai generally takes a somewhat more algebraic approach and hence doesn't want to "spoil" this with the theory of Green's functions - here, the propagator is more seen as the kernel of the time evolution operator. Of course the two viewpoints are equivalent, but it's perfectly normal that one text does not consider all possible viewpoints

Ryder Rude

you can also use the easier technique : $U(t)=\sum _{E} |E\rangle \langle E | e^{-iEt}$. to get $U$ in the $x$ basis, just substiutr $|E\rangle =\psi _E(x)$ in this.

ACuriousMind

@SillyGoose 1. What do you mean by using groups for topological spaces? 2. I'd argue that the conceptual separation of math into so many somewhat distinct subfields is the modern phenomenon here, not the existence of connections between them

Ryder Rude

07:07

i must say Sakurai shud emphasize that this is is just a technique to compute $U(t)$ and is by no means general. it doesnt work when the Hamiltonian is time dependent

so defining $U(t)$ using eigenvector stuff is not the way to go

it's like when books call the eigenvalue problem as "time independent Schrodinger eqn", when it is just a computation technique from Linear algebra

Silly Goose

hm wait but aren't we interested in green's functions to solve inhomogenous diffeqs, not homogenous ones such as

the LHS here

@ACuriousMind to my understanding i would consider the following as an example for 1.: the fundamental group of based topological space will tell you about its topological properties. but the fundamental group is an algebraic object

@RyderRude i believe he mentions that the hamiltonian is assumed to be time independent. though i think the text is not as careful to talk about whether degenerate or nondegenerate spectrums are considered but i might be mistaken

for instance, in classical mechanics one resorts to green's functions not to solve the simple harmonic oscillator, but to solve the harmonic oscillator with an arbitrary driving or damping force

Silly Goose

07:31

okay i think this is my question. why should we find the green's function of the wave schrodinger equation? As opposed to straight up solving the wave schrodinger equation. To my understanding, solving for the green's function will let you solve the homogenous and inhomogenous schrodinger equation, but this is more than we are wanting to do.

ACuriousMind

@SillyGoose "Green's function" is indeed the inhomogeneous view, "propagator" is the IVP view. The two are equivalent by Duhamel's principle.

Silly Goose

in addition, why do we not solve for the green's function of $\mathcal{L} \equiv -\frac{\hbar}{2m}\nabla^2 + \frac{\partial}{\partial t}$? Then, we can introduce a potential $V(x)$ as the "driving term", solving the schrodinger equation in general using the same method but i guess making use of the ability to solve inhomogenous equations

ACuriousMind

@SillyGoose I would see this another way: Topologically, we just define the fundamental group as the set of non-contractible loops. It just so happens to also have a natural group structure, but there is nothing "fundamentally" algebraic about it from the Pov of elementary topology.

Silly Goose

07:52

@ACuriousMind okay i think i see maybe. so sakurai is not incorrect in calling the propagator a green's function due to duhamel's principle, but it seems incredibly misleading because i) we do not need to actually solve what green's functions exist to solve, ii) strictly speaking we want to solve the schrödinger equation, not for this "impulse schrödinger", iii) duhamel's principle is not really something to internalize as a general truth between what people might call propagators and what is a gf

Ryder Rude

sorry im.not sure

Ryder Rude

08:04

Science, too, is based on faith: The Problem of Induction

►

Slereah

The set of 0-homotopies does not have a group structure on it

Ryder Rude

the video is true. everything is on faith.

Slereah

08:17

@SillyGoose It's $L f = g$, not $Lf = gf$

Silly Goose

oh i see

Slereah

ACuriousMind

08:37

@SillyGoose at least there is an actual meaning to the usage of the word here compared to people calling n-point functions "Green's functions" which they definitely aren't for $n>2$ :P

Silly Goose

so what would one call the propagator (which is a collection of solutions paramterized by $x$) in the math world? is it just a "basis" for the homogenous solution space?

Sanjana

Who's spamming the starboard :p?

John Rennie

08:58

@Sanjana That happened yesterday as well.

Ryder Rude

09:25

@SillyGoose it's just the linear operator that evolves any initial condition forward or backward time. in $U(x,t;x',t')$, the final time can be smaller than the initial time. but if u consider $G=U(x,t;x't')\theta (t-t')$ so that G can only evolve a wavefunction forward in time, then $G$ is the Green's function of $i\frac{\partial}{\partial t}-H$

@SillyGoose it's in the first section en.m.wikipedia.org/wiki/Propagator

Slereah

The Chaos emeralds???

2

Ryder Rude

in lie theory terms, $U$ is the group element generated by $H$. so it is the exponential of the generator

but in QM, u also hav time dependent Hamiltonians for which this notion does not hold. u then hav to take a time ordered exponential to define $U$

Slereah

"By contrast, a chaotic space is so excessive in its cohesion that any point can be moved to any other point without any 'effort', that is, with no attention paid to the nature of the space-'time' which might be used to parameterize the motion."

lol

Ryder Rude

the general definition of $U(t,t')$ is the eqn $\psi (t)= U(t,t')\psi(t')$

Slereah

(Chaotic is the word Lawvere uses for the trivial topology)

Ryder Rude

09:35

the only metrics on rationals r trivial, subtraction or p-adics

why is it that in the subtraction topology, we get extra numbers in-between the rational numners (i.e. irrationals)

but in the p-adic topology , there's no extra numbers

real numbers are defined using Dedekind cuts on rationals

Sir Cumference

@JohnRennie It's been happening on a bunch of chats, it seems

math, music, etc

guess someone's bored or let a bot loose

Ryder Rude

10:00

apparently u do get extra numbers. not every p adic is rational

maybe only the terminating or repeating ones are

Slereah

Why does Lawvere claim that distributive categories are "like spaces"

Is it just the fact that every topos is distributive

I'm not sure I see the fundamentally spatial aspect of distributivity

Mr. Feynman

10:29

@ACuriousMind wait, what's wrong with calling n-point correlation functions "Green's functions"?

I mean, they're not fundamental solutions to a DE but that's pretty standard terminology ig

naturallyInconsistent

10:56

@SillyGoose 1) solving tiSE gives you energies and wavefunctions. Having the propagator is one way to get them, see Feynman Hibbs doing the opposite argument of the usual $U=\sum_E\left|E\right>e^{-iEt}\left<E\right|$ that combines energies and wavefunctions into the propagator. 2) solving tdSE for arbitrary wavefunctions, is not illuminating, unless we do it via propagators. Propagators are just useful in themselves, evolving any initial wavefunction to any final wavefunction as you like.

3) Sakurai is obviously trying to make the transition from QM to QFT smoother. That is also why scattering is treated in depth in the text.

ACuriousMind

11:39

@Mr.Feynman That's exactly what's wrong with it - this usage of "Green's function" has nothing to do with the mathematical definition of a Green's function. But as you say it's standard in physics, I have no hope of changing this anytime soon :P

fqq

@ACuriousMind while working in our masters thesis, a colleague and I solved this by writing them with a green pen and dropping the 's

naturallyInconsistent

@fqq $\color{green}{\text{Green}}$ functions everywhere?

John Rennie

12:08

Do we have a mod in the house? An answer is being edited to be obscene and every time I roll it back it gets edited again.

I have raised a flag.

2

A: Amplitude and frequency of a wave

𝕊𝕙𝕚𝕥, 𝕡𝕚𝕤𝕤, 𝕗𝕦𝕔𝕜, 𝕔𝕦𝕟𝕥, 𝕔𝕠𝕔𝕜𝕤𝕦𝕔𝕜𝕖𝕣, 𝕞𝕠𝕥𝕙𝕖𝕣𝕗𝕦𝕔𝕜𝕖𝕣, 𝕥𝕚𝕥𝕤, 𝕤𝕙𝕚𝕥, 𝕡𝕚𝕤𝕤, 𝕗𝕦𝕔𝕜, 𝕔𝕦𝕟𝕥, 𝕔𝕠𝕔𝕜𝕤𝕦𝕔𝕜𝕖𝕣, 𝕞𝕠𝕥𝕙𝕖𝕣𝕗𝕦𝕔𝕜𝕖𝕣, 𝕥𝕚𝕥𝕤, 𝕤𝕙𝕚𝕥, 𝕡𝕚𝕤𝕤, 𝕗𝕦𝕔𝕜, 𝕔𝕦𝕟𝕥, 𝕔𝕠𝕔𝕜𝕤𝕦𝕔𝕜𝕖𝕣, 𝕞𝕠𝕥𝕙𝕖𝕣𝕗𝕦𝕔𝕜𝕖𝕣, 𝕥𝕚𝕥𝕤, 𝕤�...

naturallyInconsistent

Ah, it seems to be the author, probably having a bout of depression

Ryder Rude

12:46

hi

if we have a group-type algebra with closure, unique identity and and unique inverses. does this imply associativity?

if we have the first three properties, then each row of the multiplication table is a permutation of elements

and permutations are associative

so do these three properties imply associativity??

then y do we hav associativity as a separate axiom??

Slereah

Ah, Carlin's 7 words

A classic

But not of physics

Ryder Rude

13:01

sorry i meant only for finite and discrete algebras

algebra : = set with binary operator

do these three properties imply associativity?

Slereah

Quasigroup

In mathematics, especially in abstract algebra, a quasigroup is an algebraic structure resembling a group in the sense that "division" is always possible. Quasigroups differ from groups mainly in that the associative and identity element properties are optional. A quasigroup with an identity element is called a loop. == Definitions == There are at least two structurally equivalent formal definitions of quasigroup. One defines a quasigroup as a set with one binary operation, and the other, from universal algebra, defines a quasigroup as having three primitive operations. The homomorphic image of...

^it is a loop

Non-zero octonions are apparently an example

Ryder Rude

14:02

but im getting that the action of each element can be identified with a permutation matrix, and hence the elements must associate? @Slereah

Ryder Rude

14:13

but associativity shudnt be implied

but permutations r associative

Ryder Rude

14:25

ok so my theorem is incorrect. using the three properties, i may identify each element with a permutation of group elements but the multiplication of the elements is not identified with a multiplication of the permutation matrices

Mr. Feynman

14:36

I know the version of Frobenius theorem formulated in terms of involutive distributions. How is it related to the version $\mathrm{ker}\phi$ is integrable iff $\phi\wedge d\phi=0$ (i.e. the hypersurface version)

Is $\phi\wedge d\phi=0$ involutivity?

Slereah

perso.imj-prg.fr/vincent-minerbe/wp-content/uploads/minerbe-pub/…

Check p. 27

(p. 33 of the PDF)

Apparently stems from the formula $$d\alpha(X,Y) = \mathcal{L}_X(\alpha(Y)) - \mathcal{L}_Y(\alpha(X)) - \alpha([X,Y])$$

Mr. Feynman

14:56

God bless that formula

And you too :P

@Slereah I have the same habit of pointing out this lol

Slereah

I'm rly not good at math, I am just very good at bibliography

Let the math people sort it out I say

Mr. Feynman

Sure being good at bibliography is fundamental but no one believes you're not good at math lol

By the way I'm satisfied now

Instead of GR books throwing $\chi_{[a;b}\chi_{c]}=0$ at you because of "Frobenius theorem"

Slereah

15:11

If you want to see a lot of good Frobenius theorem stuff in GR you can check Straumann's book

it has a lot of nice applications

Mr. Feynman

Fun thing is that when I first studied in DG I thought it was useless

Slereah

It has fairly niche uses but they are good to have

Mr. Feynman

Well, I found it in elementary stuff such as surface gravity in stationary spacetimes

See, @ACuriousMind before fighting "Green's functions" we should cleanse this world from GR-style differential geometry :P

Slereah

It also has nice use for vague statements like "Symmetric spacetimes in so and so variables have a metric independent from those variables"

You get to decompose your metric into different terms if your Killing vectors are integrable

Basically splitting it along the foliation charts of your foliation

Mr. Feynman

Aren't vector fields integrable due to the flow theorem?

Locally

Slereah

15:18

One vector field is always locally integrable yeah

But what if you have multiple Killing vectors

Mr. Feynman

Oh so you mean that the distribution generated by them has to be integrable

Slereah

yes

It generates a whole submanifold where the metric is essentially the same everywhere

but only if integrable

Mr. Feynman

Oh yeah, thinking about it it's straightforward

The metric is Lie-transported along the foliation

*the submanifolds of the foliation

Now I gotta fight against the Kerr-metric

Slereah

Don't fall down the singularity

Mr. Feynman

I can't because I don't even remember the metric, too long

But if I somehow manage to get through the ring and make it outside without destabilizing the BH, I'll let the Slereah on the other side know

Slereah

15:29

A secret condition in Schwarzschild metrics that they do not tell you btw is that the Killing vectors are integrable

Which is usually implied by the rotational Killing vectors being orthogonal to the time/spacelike Killing vectors

Mr. Feynman

What are the Killing vectors there? The asymptotically timelike one and $\mathfrak{so}(3)$, right?

Slereah

yeah

it is one of the common condition used to define the "standard" Schwarzschild metric

Mr. Feynman

Oh, actually yes I remember. That's how you define the coordinates themselves

The Schwarzschild coordinates I mean

Carroll had a sketch of how you foliate a spherically simmetric spacetime

I was talking about that a couple of weeks ago here

Dec 29, 2023 at 10:59, by Mr. Feynman

Do you have a clearer explanation of Carroll'a argument (concerning Birkhoff theorem and foliations) quoted in this question? Basically I'm asking the same thing as OP since I don't consider the accepted answer compelling enough

Claudio Menchinelli

15:44

The fact that they commute is due to the fact they are the trivial extensions of operators acting non trivially only in one subspace

is this what is being said here implicitly? I don’t see any other possible argument

Mr. Feynman

@ClaudioMenchinelli using the notation you like more, they just saying that these are operators of the form $T_1\otimes\mathbb{I}$ and $\mathbb{I}\otimes T_2$ that act on $\lvert r_1\rangle\otimes \lvert r_2\rangle$

Slereah

15:59

Oh wait lie bracket of two killing vectors is always killing

What was that condition

Claudio Menchinelli

@Mr.Feynman yeah that’s exactly the translation into math notation of my previous comment. Thanks

Ryder Rude

what do u do in free time

Mr. Feynman

@Slereah I mean, Killing vectors form Lie algebras so what about it?

@RyderRude I disguise and fight crime at night

Ryder Rude

so it was u

i was doing that crime

:)

"in an abelian grp, every element is in its own class"

oh this one's intuitive

there's soo much to learn

number theory sounds interesting

one can never learn all the interesting things. one wouldve to be an AI

Claudio Menchinelli

16:47

Regarding the two-body problem, I know that the potential depends upon the distance between the two bodies $r = |\mathbf{r_1} -\mathbf{r_2}|$ only. If I wanna prove that the Hamiltonian is translationally invariant then I need to show that $ [\hat{T}_1(\mathbf{u}),\hat{T}_2(\mathbf{u}),\hat{V}(r) ]$. The Problem being: how do I determine how the potential operator acts on a generic position ket $|\mathbf{r_1,r_2}\rangle$?

ACuriousMind

@Mr.Feynman I can fight more than one terminology war at once ;)

Claudio Menchinelli

just wrote potential operator lol

I meant $V$ as a function of position operators

I realize now that the absolute value is not a problem at all lol. Well, another embarrassing moment from me. Overthinking at its finest

Mr. Feynman

@ACuriousMind and that's why you're the Batman of physics

Why Batman? I dunno, in that context Flash didn't sound as cool

Ryder Rude

17:03

". However, the theorem does not hold for algebraic integers.This failure of unique factorization is one of the reasons for the difficulty of the proof of Fermat's Last Theorem. "

here en.wikipedia.org/wiki/Fundamental_theorem_of_arithmetic

Claudio Menchinelli

23:14

position and momentum operators do not commute, but, regarding the two-particle problem, if I consider the position operator of a single particle $x_{i1}$ and the momentum of the second one $p_{j2}$ then those two must commute

Claudio Menchinelli

23:25

I’ve tried summing up the commutation relations for the two-body problem, could someone assure me that those are correct before I go on, I hate the notation adopted by the book:$$ [r_{i}^{I},r_j^{J}]=0, [p_{i}^{I},p_j^{J}]=0, [r_{i}^{I},p_j^{J}]=i\hbar\delta_{ij}\delta_{IJ}, I,J=1,2 \text{ and } i,j=1,2,3$$

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