The h Bar

user400188

01:13

For the Schrodinger equation to hold, the wavefunctions second spatial derivative needs to be defined.
But are there any cases where the solution to the Schrodinger equation is not analytic? I.e. are there any cases where its third derivative or higher is not defined? I can't think of any off the top of my head.(Besides the infinite square well, but that doesn't count because it's unphysical, and also the Schrodinger equation isn't satisified everywhere anyway).

Krrish Dhaneja

06:54

Hi

John Rennie

07:05

@KrrishDhaneja Hi :-)

Krrish Dhaneja

07:46

remember me @JohnRennie :) ?

John Rennie

I don't. Sorry :-(

Were you one of the JEE students I chatted with?

Krrish Dhaneja

09:53

Ah no! Its been 2 years since i visited this site, anyways no probs

Slereah

11:28

@ACuriousMind Is the polarization of the quantum bundle somewhat equivalent to the choice of a Hilbert space for a quantum theory

The trivial examples tend to be mostly for pretty equivalent ones since they're for point particles, but what if you have a QFT?

Can you have non-equivalent Hilbert spaces based on the polarization

ACuriousMind

@Slereah the polarization is "choosing what your wavefunctions are functions of"

Slereah

Don't they lead to different Hilbert spaces tho

ACuriousMind

define "different Hilbert space"

Slereah

IIRC the basic polarization leads to the dumb old L² one while the complex one leads to the Bargmann one

ACuriousMind

the main "problem" with polarization is that it restricts the class of observables

Slereah

11:31

those ones are equivalent, but what if you're dealing with a case where they may not be

ACuriousMind

again, "equivalent" in what sense

all Hilbert spaces of the same cardinality are isomorphic

Slereah

In the Haag sense

ACuriousMind

@Slereah Haag is about isomorphisms as representations of the of CCR

but do you even get the CCR from geometric quantization?

there's no guarantee that $x$ and $p$ preserve the polarization

so you don't even have a guarantee that the resulting Hilbert space is a representation of the CCR

Slereah

It will lead to some CCR, no?

Or CR I guess :p

Not necessarily a canonical one

You still have your Poisson bracket and all

ACuriousMind

the problem with geometric quantization is that most classical observables don't turn into quantum observables

only the ones whose flow preserves the polarization

and so for two different polarizations you have, in general, no guarantee that they share even a single non-trivial observable

so it's very hard to "compare" the resulting Hilbert spaces in general

they carry a priori completely different and far too "small" algebras of observables

Slereah

11:42

can it be done for very basic ones?

ACuriousMind

can what be done?

Slereah

Compare them

ACuriousMind

again, to "compare" them you'd need to specify a notion of isomorphism that makes sense in general

usually we do that via "isomorphic as representations of the CCR" or "isomorphic as representations of the algebra of observables"

but in this case the algebra of observables itself changes when the polarization changes

so this notion isn't really applicable

Slereah

I guess I'm basically trying to figure out if you can do like Stone-von Neumann from the geometric quantization of a point particle

Show that all polarizations lead to equivalent Hilbert spaces in some sense

Hm

13 is one of the OG GQ book

Woodhouse

ACuriousMind

@Slereah I really think the problem here is that the algebra of observables isn't "stable"

in general you'd like your Hamiltonian to be in the algebra

but e.g. the free Hamiltonian $p^2$ isn't even quantizable for the vertical polarization on some $T^\ast Q$

the quantizable observables in that case are things that look like $f(q) + g(q)p$

if you take the horizontal polarization you get $f(p) + g(p)q$

Slereah

11:55

I wonder if it's related to my master thesis

My original master thesis was about quantization of QM on curved manifolds [using path integrals]

You did have pretty weird versions of the Hamiltonian due to various issues

ACuriousMind

in what sense could these two things ever be "equivalent"? In one of them we know which observable corresponds to the classical $p^2$, in the other we don't directly but we could form it by squaring the quantum equivalent of $p$, and if you choose the Kähler representation you get expressions with $z$ and $z^\ast$ instead and who knows how to get $p^2$ from that :P

the fundamental problem here is that geometric quantization circumvents Groenewold-van Howe by sacrificing the "functions of classical observables quantize to become functions of quantum observables" part of what we want from a quantization map

no one can actually do anything useful with an algebra of observables that's at most linear in momentum :P

Slereah

It is too bad

I much prefer to do geometry to analysis :p

ACuriousMind

I think the "intended" way of how this should work is that you should get your polarization from the Hamiltonian

then you can at least do normal QM in the sense that you have an equation of motion

but there's no guarantee you get an $x$ and a $p$ operator

Slereah

I mean that is fine

Surely in the spirit of geometry you shouldn't get bogged down in a specific set of coordinates

ACuriousMind

yes but the problem is we usually can measure position :P

Slereah

12:01

In a much more dire problem, my chat switched to the mobile version

I tried removing it in my preferences but it still looks weird

ACuriousMind

click on the hamburger in the left upper corner and choose "full site"

Slereah

there's no hamburger

An American tragedy

ACuriousMind

the mobile view should have a hamburger menu (three stripes) to the left of the room name

Slereah

looks like this

ah I managed to find where to switch back

Phew

Woodhouse's book barely costs $180

🏴‍☠️

user 726941

13:35

Do you guys think we can build a solf-question around this quote?

> Richard Feynman, the late Nobel Laureate in physics, was once asked by a Caltech faculty member to explain why spin one-half particles obey Fermi Dirac statistics. Rising to the challenge, he said, "I'll prepare a freshman lecture on it." But a few days later he told the faculty member, "You know, I couldn't do it. I couldn't reduce it to the freshman level. That means we really don't understand it."

nickbros123

Hi people, if i have a coordinate system fixed onto a rigid body, (ie the axes are rotating along with the body, and also translating with the com preferably), is the angular momentum 0 about this coordinate system?

Slereah

yes

nickbros123

This comes straight from the definition of angular momentum, right?

I have some trouble wrapping my head around $\frac{dL}{dt} = \frac{\delta L}{\delta t} + \vec \omega \times L $ where $ \frac{dL}{dt} $ and $frac{\delta L}{\delta t}$ are the time derivatives of L with respect to an inertial frame and non inertial frame respectively

Non inertial frame here is simply rotating with the body

Slereah

13:58

I mean it's also true of regular momentum

in general you can find some coordinates that will cancel out the moments of a rigid body

Ryder Rude

14:54

@ACuriousMind I have one prototype idea of defining probabilistic knowledge of observers on spacetimes with time loops. Let's say you take two different allowed solutions of an observer's worldwide on that spacetime. These solutions are such that they agree with each other on a portion of the spacetime, but diverge beyond that. Then we assign probabilities to each of these two solutions.

ACuriousMind

I don't know what a "solution of an observer's worldwide" is

worldwide is obviously a typo but I can't figure out for what :P

Ryder Rude

worldline :P

ACuriousMind

why would there be "two different allowed solutions"?

why would they agree on some patch and then suddenly diverge?

Ryder Rude

so from the observer's PoV, this will feel like ending up in one of the two universes at the moment that the worldlines diverge

@ACuriousMind this is because circular spacetimes are not deterministic

so the solutions can agree on some portion while diverge beyond that

ACuriousMind

that's...not how it works

the statement is that the Cauchy problem for not globally hyperbolic spacetimes is in general ill-posed

Ryder Rude

14:58

i thought this meant that you can't get a unique future by evolving initial conditions

ACuriousMind

that does not necessarily imply that this ill-posedness manifests as some weird "non-deterministic solution" where a worldline splits in two

Ryder Rude

oh

but if you can have an initial condition give non unique solutions, is it not the same as diverging from the initial conditions?

ACuriousMind

no

or, rather, we're not exatly saying that a CTC means that an initial condition gives non-unique solutions

the usual idea of making the Cauchy problem well-posed is via the idea of a Cauchy surface: A timelike hypersurface - an "instant of time" - which is crossed by each inextendible timelike curve (=possible world line) exactly once

Ryder Rude

yeah

ACuriousMind

the idea here is that you can then predict everything deterministically from data on the Cauchy surface because you just have to let it "flow" along these worldlines

the existence of a CTC prevents the existence of a Cauchy surface because the closed loop will cross the surface twice

Ryder Rude

15:03

yeah

and this does not necessarily imply that we can have diverging solutions

ACuriousMind

the problem then is when you now try to let the data on the surface flow along the curves, the data flowing along the closed curve ends back up on the surface at finite time

i.e. your "initial condition" changes itself

Ryder Rude

yeah

our solutions must be consistent with that

ACuriousMind

and so you can't really consistently talk about evolving some initial condition

this is what we mean when we say CTC destroy causality or "determinism"

but there's nothing here like worldlines that split or whatever you were talking about

there's just the fact the Cauchy problem is ill-posed

Ryder Rude

the allowed solutions can be constrained by the fact that they must take the same value after they're evolved along the full loop

this constraint will remove the inconsistency problems

if we only talk about solutions that pass this consistency constraint, then diverging worldlines may be allowed in the theory?

i mean, not necessarily allowed, but in special cases

ACuriousMind

I don't know what you mean by diverging worldlines

again, there's no diverging worldline here, the problem is that the CTC is a worldline that will cross certain instants of time twice

locally the geodesic equation is still uniquely solveable, I don't know where this idea of a worldline that splits comes from

Ryder Rude

15:17

i meant trajectories on the loop spacetime that obey the geodesic equation but that the trajectories only agree with each other on a portion of the manifold

@ACuriousMind So this diverging wordline idea isn't allowed by the geodesic equation

then i will have to think of some other idea :P

ACuriousMind

I don't know how you came up with this idea of a "diverging worldline" in the first place :P

Ryder Rude

this makes sense. the parallel transport of the portion of the trajectory is unique

@ACuriousMind I just thought that if the theory allowed these trajectories that only agreed on a portion, then I could assign probabilities to those trajectories to define that observer's probabilistic knowledge

ACuriousMind

yeah but there is no part of the theory that contains such weird trajectories

I think you just heard "non-deterministic" and started speculating instead of looking at what the math actually does

Ryder Rude

yeah, i will abandon this idea becuz the geodesic equation does not allow it

@ACuriousMind Yeah, I am in dire need of learning GR:P

ACuriousMind

that's the problem with natural language: it is rarely precise enough to talk about such technical issues

Ryder Rude

15:23

yeah, i need to learn the math too

i've been postponing it becuz last time i tried it, it was overkill for me

too many proofs regard tensor indices stuff. lots of indices

ACuriousMind

but that's what it is

it's not a shame not to learn it, but then you shouldn't expect to make any sort of meaningful statements about extensions to this theory :P

Ryder Rude

unfortunately :P

no, i really want to make those statement :P. It will be great entertainment

i will try to learn it

i was using Wald's book. I'll try to make it thru Carrol's this time.

Relativisticcucumber

hello -- there is a part of sakurai where he says that $J_+J_+^{\dagger}$ must have nonnegative expectation values because $J^{\dagger}_+ \vert a, b \rangle$ is in dual correspondence with $\langle a, b \vert J_+$

why does this imply that the expectation values must be nonnegative?

ACuriousMind

@Relativisticcucumber what do you mean by "expectation value"? Usually we only talk about expectation values for observables, i.e. self-adjoint operators

oh, you mean the expectation value of the product, sorry

in that case, just observe that $\langle \psi \vert J_+ J_+^\dagger \vert \psi\rangle$ is the same as the (squared) norm of the state $J_+^\dagger\lvert \psi\rangle$ and norms of states are non-negative

Relativisticcucumber

15:42

oh no, i see but what does this have to do with them being in dual correspondence?

ACuriousMind

"dual correspondence" is a bit of a weird phrase for exactly what I just said

Relativisticcucumber

and one more thing -- so i know that using raising and lowering operators, we can show that there is a highest value and lowest value for an observable, then we iterate from some value to the bottom and top using these operators. but to show that these are, indeed, all possible values, do we somehow show that the corresponding eigenket spectrums pans the space? i think in the argument im looking at this isnt treated but i might just not have seen it?

i imagine we can just show that this basis formed is complete, but im not sure how easy or not easy this is

@ACuriousMind ok

ACuriousMind

@Relativisticcucumber do you know what "irreducible" means?

Relativisticcucumber

i dont think so

ACuriousMind

ok

in that case, you are right that, in the fully general situation, just applying the raising and lowering operators to one state does not produce a basis for the full space

however, the set of vectors $\lvert a_1,b_{1,\text{min}}\rangle,\dots \lvert a_1,b_{1,\text{max}}\rangle$ that you get from one such "string" of vectors you can get by raising/lowering spans a subspace of the full space on which the rotation operators "close", i.e. applying any of the $J_i$ to this space will never produce a vector that lies outside it

one can show that, in general, every space on which the rotation operators act decomposes as the direct sum of such closed subspaces

so in order to understand how rotation operators act, it suffices to understand these subspaces

(mathematicians call these subspaces irreducible representations of the rotation algebra)

Relativisticcucumber

15:54

interesting -- so this doesnt come back to bite us later at any point? like if we try to express certain solutions as sums of other solutions, we somehow stay in the subspace in question? what i mean is what if we set up our situation such that we are in one of these subspaces, but then we are trying to express a state that requires another subspace to express it? im not sure if this is possible or makes sense

god i have the physics skills of a preschooler

ACuriousMind

I don't think preschoolers know what linear operators even are :P

but indeed I'm not quite sure what your question is getting at

it can happen that you need to actually care about more than one of these representations at once

Relativisticcucumber

ok yes thats what i was wondering is basically when this picture of looking at just one representation starts to fail

or more like how we know when it's ok

ACuriousMind

e.g. the usual space of wavefunctions in 3 dimensions carries naturally the orbital angular momentum operators from $x\times p$ and via the spherical harmonics all the representations with full-integer angular momentum occur

@Relativisticcucumber What physical situation/problem are you imagining here where it could be unclear whether or not there is more than one irrep?

I often feel people encountering QM for the first time tend to get lost in artificial questions where they imagine weird situations of incomplete information that will never happen in practice (but I suspect this is likely produced via some sort of subconscious training where we expect exam questions to just randomly remove information from a situation)

Relativisticcucumber

well my example is not so specific. i am thinking we say physical states can be represented as a sum of eigenstates and in my last qm class we were taught these raising and lowering generate the entire basis, but there must be some physical state that involves another representation so how could be know how many representations we need to write the decomposition of this state

well i dont think this is an exam thing im just thinking how can i apply this concept to do things

but i suppose that would come from a more legitimate exploration of the task at hand should it ever be at hand

ACuriousMind

okay, so "physical states can be represented as a sum of eigenstates" is true but it is really important to understand this isn't a statement that actually has a lot of content. The following things are true:
1. A quantum state is a vector/ray in a Hilbert space.
2. The eigenstates of a self-adjoint linear operator on a Hilbert space form a basis for that Hilbert space (hence every vector in the space can be expressed as a sum of these basis element)
3. The operators $L^2, L_z$ are self-adjoint and commute so their eigenstates will form a basis.

now, what you have learned via the raising/lowering operators is that you can start with some eigenstate of $L^2,L_z$ and produce a bunch of other eigenstates of it

that doesn't imply that starting from any single eigenstate produces all other eigenstates; this is only true when the representation is irreducible as discussed above

but the point of the raising/lowering operators isn't really that you apply them to any specific space to "find eigenstates" or something like that

it's more that this technique with the ladder operators enables us to say very general things about what can happen on a space on which the rotation operators act

for instance, you can show that $L^2$ can only have eigenvalues of the form $l(l+1)$ where $l$ is a half-integer

but you usually don't have to "discover" the space of states of a system. You start with some Hilbert space (e.g. wavefunctions in 3d) and then you can ask "Hey, into how many irreps with different values for $l$ does this thing decompose?"

that's a question with a clear answer, you don't have to "know" how many representations there are from the start, it's something you can derive

ACuriousMind

16:16

not sure if any of that rambling was useful to you :P

Relativisticcucumber

16:27

sorry i am back !! i will read now

ACuriousMind

no worries, as a wise user once said: chat is an asynchronous communication protocol

Relativisticcucumber

ok i see my confusion now. because in my previous class actually this was the "derivation" for the system's eigenstates. and now i have come to the conclusion that this is misleading but maybe pedagogically useful.

@ACuriousMind this is helpful !! i see

ty vm

ACuriousMind

np

nickbros123

@Slereah the angular momentum of a rotating body about it's principal axes is non zero no? Is the principal axis fixed in space at a particular instant?

Could someone walk me through what exactly does Euler equations mean

(in rigid body dynamics)

Mr. Feynman

20:29

I need a reality check about DoS. When one deals with emission of a photon by an atom there is a photon in the final state, i.e. the spectrum is continuum and the density of final states is $\rho(\hbar\omega_k)=\frac{V}{2\pi^2c^2}\omega_k^2$. Then in Fermi Golden rule one uses $\rho(\hbar\omega_k)\bigg\rvert_{\omega_k=\Delta E/\hbar}$ (energy conservation).

In the case of *absorption* I should have the same DoS, but we have no photon in the *final* state, so $\omega_k=0$ above. This is surely wrong and I'm drowning in an inch of water, so where am I getting confused?

*continuous

In other words, the continuous part is given by radiation as atoms have discrete spectra. We are concerned with the final states, but in the case of absorption there is no photon left (I'm assuming a process with a single photon $\lvert 1_{\vec{k},\sigma}\rangle\otimes\lvert \alpha_{atom}\rangle\rightarrow\lvert\boldsymbol{0}\rangle\otimes\lvert \beta_{atom}\rangle$

ACuriousMind

@Mr.Feynman Fermi's Golden rule looks a bit different for discrete final states, see en.wikipedia.org/wiki/…

Mr. Feynman

But before that consideration, shouldn't I get the same DoS for absorption and emission? As far as I'm concerned the only difference should come from the photonic amplitude

In this case the discrete part (atomic) should be trivial, only the continuum spectrum of photons plays a role in the DoS...

user 726941

21:45

Btw, is drowning in an inch of water the opposite of being in over your head :P

Mr. Feynman

It means you are getting lost over something stupid

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