The h Bar

naturallyInconsistent

2:24 AM

B A H ~

naturallyInconsistent

3:58 AM

myow just got IQ and lifespan robbed by cold fusion stupidities, sigh

Obliv

@naturallyInconsistent how so

naturallyInconsistent

marketing people had to be nice to guests, guests suggested we hire this guy who is into cold fusion, and that means miao miao had to read his papers. Luckily, a cursory reading of the presented data is sufficient. No need to spend a lot of time on the details. For people who write papers incorporating sections on theoretical investigations, one would have hoped that they would be wise enough to estimate the "large energies" that they obtained.

Alas, they did not even try.

Obliv

Yikes. Yeah I felt that way except on the receiving end when I went to a job fair and the person there was talking to me about how I could contribute to stellarator fusion research im like brother I haven't even started learning physics yet

marketing people are indeed nice.

naturallyInconsistent

@Obliv that's not that bad. Actual research rabbit holes are so deep that, it only makes sense to learn on the job. i.e. it is not that bad to hire not-even-yet-starting-undergrads, if it means you can get fired up learning earlier.

Obliv

True, but I'd feel bad for not knowing how to do literally anything, including writing/debugging code meant for their systems. I don't foresee myself doing anything involved with the physics side of things in research so I should at least learn how to help with the codemonkey tasks :P

btw how is an operator defined on a dual space? Do they map functionals to other functionals?

naturallyInconsistent

4:14 AM

Dont feel bad that way. There are a lot of stupid shit that always can use intern labour.

Obliv

@naturallyInconsistent Okay, I'll keep that in mind.

keep an open mind*

naturallyInconsistent

@Obliv You don't have to. Since you already have Riesz rep theorem, you can just define the operator on the dual's dual, and that is more than enough.

Obliv

oh I guess I'm missing context in that img

eh I'll mull this over

I'm rushing too much

also this copy of ballentine has a lot of typos on some pages

and why do physicists insist on notational differences from mathematicians :(

like which argument of the inner product is linear and antilinear for example

naturallyInconsistent

@Obliv Look, the opposite is true. We physicists are extremely fixated on notation that is proven to be least likely to make mistakes. Operators work on the right in the normal sense, and it is the left entity that is getting the conjugation. It thus has to be the left that is antilinear and the right is linear. It is the mathematicians that simply don't care about physics convention.

Dannyu NDos

Does Dirac notation force physical quantities to be real numbers?

I mean, observable physical quantities?

naturallyInconsistent

4:27 AM

Why would that be?

Dannyu NDos

'Cause the operators are unitary.

Obliv

Okay my goal for tomorrow is finish the pre-reqs so I can finally start learning the material.

Meow

naturallyInconsistent

M I A O ~

@DannyuNDos You have made a mistake. The eigenvalues of unitary operators are not real.

Dannyu NDos

(Sorry, the cat in my profile picture doesn't belong to Schrödinger.)

naturallyInconsistent

miahahaha

Dannyu NDos

4:31 AM

@naturallyInconsistent I guess I mixed up unitary and Hermitian then.

naturallyInconsistent

@DannyuNDos And if the operators are not Hermitian, then those observables are not necessarily going to have real numbers as their eigenvalues. Like, what is it that you think Dirac notation is enforcing?

Dannyu NDos

Nevermind. I guess I realize that Dirac notation is just for convenience of notating vectors and covectors. It's operators that enforce anything.

naturallyInconsistent

I'd point out that you would not have needed an entire postulate asserting that physical observables come with Hermitian operators if Dirac notation is sufficient on its own.

Dannyu NDos

And I mixed up unitary and Hermitian because I thought the operator would need to be invertible. It really isn't when it involves observation.

naturallyInconsistent

Not sure how you made that mistake, but $\hat N=\hat a{}^\dagger\hat a$ and $\hat p_x=-i\hslash\partial_x$ are both examples of physical Hermitian observables that have zero as a possible eigenvalue and thus are not invertible in the matrix sense.

Obliv

5:18 AM

Yay, miao might be TA-ing some physics lab sections this fall. Probably general physics 1&2, maybe analytical physics lab (I think this one is like 2nd year physics with some calculus sprinkled in)

Also the lab coordinator emailed me at like 12:30 AM so I decided to respond lol. I was going to wait until tomorrow but I guess he's up late..Ok now I slumber.

naturallyInconsistent

yay

sneeepppuuu

Ryder Rude

7:13 AM

hi

Silly Goose

7:25 AM

@ACuriousMind do you know of examples in quantum mechanics other than perhaps aharnov bohm where properties of the base manifold manifests in some physically interesting consequence? and also of a discussion of how $L^2(M)$ depends on the properties of $M$?

ACuriousMind

@SillyGoose The spectrum of the position and momentum operators (if you even know what it should be) changes depending on $M$. Famously, the spectrum of the momentum operator on the circle $S^1$ is discrete, which is why you have Fourier series for periodic functions instead of the continuous Fourier transform

I don't know what you mean by "how $L^2(M)$ depends on the properties of $M$". $L^2(M)$ is a separable Hilbert space, and all such spaces are non-canonically isomorphic. What properties do you think $L^2(M)$ has that can change?

Silly Goose

naively, if i puncture a hole in $\mathbb{R}^1$ at $x = 1$, then I shouldn't be able to have access to $\delta(1 - x)$. if we are to naively use such delta functions and relate them to quantum states or write quantum states in terms of them, then this seems to be changing property of the state space

also i was thinking that properties of linear operators over $L^2(M)$ are in a sense due to properties of $L^2(M)$ itself. but maybe this is a wrong line of thought.

I mean really I think I am interested in the following. Observation: Classical Chern-Simons theory is an example in which the base manifold plays an essential role in determining the "state space" (set of on-shell connections). Question: Is there an example in particle mechanics in which the base manifold plays an essential role in determining the state space. In particular, is there such an example in quantum mechanics.

ACuriousMind

@SillyGoose And I already gave the answer: The state space (naively in canonical quantization) is $L^2(M)$, so yes, the state space is determined by $M$. What more do you want?

Silly Goose

well what you said above suggests that $L^2(M)$ doesn't have properties to change

ACuriousMind

I mean I said that no Hilbert space really has "properties"

because they're all isomorphic

so the question is really what you mean at all by the quantum state space "changing"

and I also gave already the more meaningful answer that the spectrum of your observables may be different depending on $M$, which isn't a property of the Hilbert space in the strict sense, but nevertheless is what we care about in physics

Silly Goose

7:41 AM

i think in asking my question i am trying to map the field theory framework into the particle mechanics framework and i think there is not an exact analogy

well I guess I don't know what it can mean for the spectrum of observables to change depending on $M$.

ACuriousMind

@SillyGoose I already gave the example of the circle and the momentum operator, which part is unclear?

Silly Goose

well there seems to be more than one way that a spectrum of an operator can change. some include removing/changing eigenstates and others do not. so it is unclear what happens to the eigenstates in general

Silly Goose

7:55 AM

separately, are there analogous aharnov-bohm effects for the strong/weak interactions? there is the canonical electromagnetic and the gravitational.

perhaps Wu & Yang work out the theory for potential analogous aharnov-bohm effects here and some experimental realization can be found here

ACuriousMind

@SillyGoose I don't know what you mean by "removing" eigenstates

For two arbitrary $L^2(M)$ and $L^2(N)$, you have no embedding of one into the other and thus cannot really compare the states themselves

you can compare the spectra of two operators since they are subsets of $\mathbb{R}$ in both cases

Silly Goose

hm i see

ACuriousMind

@SillyGoose There are certainly topological effects, but there isn't really one like the Aharonov-Bohm effect since there is no such thing as a "classical background strong field"

the AB effect has a background electric field

but there are no classical strong or weak fields - even if you can write down their naive classical theory, they clearly do not describe our world, we do not see classical strong or weak fields anywhere in nature

the stuff you linked uses exotic systems where you get a kind of non-Abelian gauge symmetry from electric and magnetic fields, it has nothing to do with the strong or weak interaction

Silly Goose

oh

Ryder Rude

8:12 AM

@ACuriousMind to experimentally describe an electron restricted to a circular pipe, do we model it using QM with circular position topology, or using QM in $R^3$ with the pipe being a potential?

ACuriousMind

The particle on the ring is a popular and common QM exercise.

Ryder Rude

Or maybe it can be modelled as both, but they have different experimental accuracies?

@ACuriousMind yes, but I've never seen a comparison of these two models

The electron has interactions with the pipe, so i think we should model it as a potential

ACuriousMind

you don't model the interaction of the electron with the walls of the double slit, either

it's just a needless complication that's not relevant in any case I know of

Ryder Rude

oh

this means non trivial topological space theories r physically useful too

thanks

i knew this becuz the quantum rotor already has circular position

But it's nice to know that we can replace pipes and barriers with topologies

Silly Goose

@ACuriousMind I think I am looking for an analogous statement for quantum mechanics to the following statement (even if perhaps not true, I mean the spirit of the following statement). In abelian classical Chern-Simons theory, the set of on-shell connections is precisely the first de Rham cohomology class of the base manifold.

ACuriousMind

8:22 AM

@SillyGoose There will be no analogous statement because - if we look at QM through a QFT lense - the base manifold is always $\mathbb{R}$, i.e. time.

the $M$ above plays the role of the target space of a $\sigma$-model (with the positions being the fields), not of the base manifold of a QFT

Silly Goose

blebs

well let's suppose that I want to allow punctured $\mathbb{R}$ base manifolds for time.

ACuriousMind

and what physical situation will be modeled by that?

if you puncture $\mathbb{R}$, what you get are two disjoint copies of it

Silly Goose

well I don't see why for a system with non-reversible dynamics I can't just treat the section of the time manifold that has passed in whatever way

ACuriousMind

what happens when time evolution runs into the gap? This makes no physical sense, all frameworks I know always consider a connected spacetime

@SillyGoose When you model a physical system, you are not starting at a particular time - the "time that has passed" only makes sense along a particular trajectory of the system, not for its state space

Silly Goose

what about for a scattering event in which you only really suppose you know the state far before and far after. is there maybe some wiggle room in between to treat time strangely?

alternatively, is the assumption for a "universal" time well-motivated?

Ryder Rude

8:34 AM

in the big bang model, for e.g., time has a beginning

but idk how punctured time could be justified

Silly Goose

@ACuriousMind i had in mind the following. 1) initialize initial state $\lvert \psi \rangle$. 2) set initial time $t_0 = 0$. 3) free evolution for $\delta t$. 4) suppose $\mathbb{R}$ is now punctured somewhere in $[0, \delta t)$. 5) perhaps now the state space is changed (becase we have changed the topology of the base manifold). hence, perhaps the future dynamics are affected.

but puncturing in the interval of passed time is not inconsistent with anything (to my understanding).

ACuriousMind

@SillyGoose Scattering theory also rests on the assumption that time is $\mathbb{R}$, with an asymptotic future and past

@SillyGoose But what kind of physical operation is this "puncture"?

again, the definition of the spacetime is something you do before looking at specific states

since by definition the trajectory in time is a function $\mathbb{R}\to S$ for whatever state space $S$ we have

There simply, in our standard models of classical or quantum mechanics, is no such thing as saying that the underlying (space)time changes during time evolution

note that this doesn't even make conceptual sense: What does it mean that time changed "after" some time $\delta t$? How can time change after some time? How could we go back to "before" (and we should always in principle be able to run in reverse) if time itself has changed?

Silly Goose

well i was thinking that puncturing a hole in time could be a way of modeling time-irreversibility within subsets of time or something like this. at least in this sense it is related to something conceptual

but i think there is a problem with having "time depend on time"

ACuriousMind

"time-irreversibility" is a concept entirely disjoint from what we're discussing here

Silly Goose

but i mean it's not really different than how we operationally talk about "wave function collapse"

our rule to update a state post-measurement depends on the pre-measurement state

Ryder Rude

8:48 AM

how r u defining time irreversibility here? i think time irreversibility means "non time symmetric dynamics"

Silly Goose

so can our rule to update the time manifold depend on the original manifold

ACuriousMind

time-irreversibility just means the dynamics are not symmetric under time reversal, not that time itself is different

Silly Goose

oh

well this is like as time-irreversible as one can get I would think: the past is no longer even accessible

Ryder Rude

but yeah, there is a broader meaning in terms of wave function collapse which is also considered time irreversible

ACuriousMind

@SillyGoose that's not what anyone thinks happens in reality

Silly Goose

8:49 AM

well I guess I don't get why someone would think that time is literally reversible

ACuriousMind

it's just how all our physical theories work

Silly Goose

i mean maybe it makes things easier

but i don't see why it is justified

ACuriousMind

the whole point of the e.o.m. being an initial value problem is that we can choose any point in time at all for the initial data and then we predict the physics for all time

@SillyGoose It has worked in 100% of cases so far

pretty good justification if you ask me :P

Silly Goose

well in any case I am not proposing a modification to all physical theory. i just mean to explore what could be the case if one allows this small change

Ryder Rude

only wavefunction collapse is irreversible

@SillyGoose i think ur idea is not clearly defined yet

are u saying that, once we reach some moment in the universe, the manifold of spacetime changes to have a puncture in the past?

"moments" are already not a well defined thing because of the relativity of simultaneity

ACuriousMind

8:54 AM

@SillyGoose it is not a small change

you are trying to change a fundamental assumption of all of mechanics, namely that time is connected (and almost always $\mathbb{R}$)

I think you really underestimate how fundamental this change is, and how ill-defined it is to just say it in one sentence without any other concurrent changes to our mechanics

and again: What is the physical system that is supposed to me modeled by this? For which cases is time being $\mathbb{R}$ insufficient?

Ryder Rude

instead of adding a puncture in the past (where "the past" is not well defined in relativity), u can instead explore theories which always have a puncture

i personally find CTC theories interesting. Punctures can be cool too

but the manifold shouldn't change in an ad hoc way...

it should always have a pre defined topology

ACuriousMind

again, "puncturing" time just makes the time manifold disconnected

but we always assume connected (space)time manifolds, since what would happen in one part of a disconnected spacetime could not possibly influence anything in the other part via standard accepted physics

Ryder Rude

@ACuriousMind would you say spacetimes with black hole singularities are similar to SillyGoose's idea?

ACuriousMind

Not at all

Dannyu NDos

You might want to define "connected" rigorously tho. In topology terms, "connected" means absence of proper nonempty clopen subsets.

ACuriousMind

9:00 AM

@DannyuNDos I am using the proper mathematical definition.

Dannyu NDos

I mean, If you puncture one hole in a two-dimensional plane, the remaining space is still connected. It loses simple connectivity, tho.

Ryder Rude

let's say we take SR in 2D and remove a point. Does this make for a valid "thought-experiment-theory"?

ACuriousMind

@DannyuNDos The point is that time is the 1-dimensional manifold $\mathbb{R}$

Ryder Rude

In the above, the spacetime is still connected... So it's not uninteresting

Dannyu NDos

Oh.

Silly Goose

9:04 AM

@ACuriousMind i mean small as in directly changing very few things (namely, just one thing). but i agree possibly very consequential

but i think the point is that it is a fundamental change and that it is very consequential and i see no reason not to be able to do it (in that i cannot justify why one shouldn't do it)

ACuriousMind

if you poke a hole into a higher-dimensional spacetime, it remains connected, but you haven't really "punctured time" - you still have time-like curves that go from every point to every point (that are time-like separated), they just can't pass through that particular point anymore

Ryder Rude

oh

ACuriousMind

@SillyGoose I really don't know what you know by "i see no reason not to be able to do it" - if you read proper analytical mechanics texts (like Arnold), they always assume that time is $\mathbb{R}$ - and Arnold even contains a careful approach where he for a long time only talks about affine spaces etc. before essentially concluding that time has to be $\mathbb{R}$.

Silly Goose

i mean naively i see no reason. but if there is genuine justification--like certain important results are changed or invalidated--then i can concede

ACuriousMind

and I apparently cannot repeat often enough that the physical picture here is entirely unclear: How is anything in "one half" of this punctured time supposed to influence anything in the other half?

@SillyGoose also, again: How is this supposed to work? How does time evolution work here? What happens when $x(t)$ runs into the gap? You can't just say "let's have a hole in time" and then act as if the burden of proof is on the people who say it doesn't work

the burden is on you to actually present this as a well-defined theory that one can analyze

Ryder Rude

9:14 AM

ok so if we don't puncture time, but just remove a point in higher dimensional spacetime, then the Hilbert space of states and the spectrum of observables becomes time dependent, right?

like the spectrum of position is time dependent now

but Hilbert spaces r still isomorphic in the math sense

Slereah

If you just remove a random point from a well behaved spacetime you're gonna get a big singularity

and then you will have Troubles

Ryder Rude

@Slereah yes. Do differential eqns no longer work?

Slereah

Time evolution will fail to be unitary

Ryder Rude

yeah..

Slereah

unless you specify it by hand

Which is basically just saying what happens at the singularity

Ryder Rude

9:16 AM

so it is still an ugly thought experiment theory

Slereah

I mean what are you aiming for

Ryder Rude

beauty.. :

Slereah

not off to a good start

Ryder Rude

i find CTCs much more aesthetic, but unitarity should be weird again

Slereah

Thinking circles to be beautiful is the classic aristotelian trap

It doesn't work that well

Ryder Rude

9:20 AM

QM postulates require the most boring time imaginable...

I saw this movie called Mirage where a storm at two different points of time connects a room at the two different times

so the future room can make changes to the past room

and the changes also get reflected in the future world

So it's like a feedback loop

Highly recommended

It's a SciFi+psychology+romance+thriller movie

Claudio

this was an interesting discussion to read. Puncturing time is something I've never thought of. And now I wanna read Arnold as well :P

@Slereah could you explain in very simple terms why we lose unitarity?

I know nothing of GR lol

Slereah

@Claudio What happens to a particle if it goes in the hole

Claudio

Are we talking about $\mathcal{U}(t) = \exp\left(\frac{-iHt}{\hbar}\right)$

@Slereah oh I see

there's no t to plug in lol

wait then we don't only loose unitarity, the operator itself is not well defined

Slereah

9:35 AM

More importantly, if your particle disappears in the singularity, it goes from having a probability of 100% of being somewhere to 0%

or something in between because QM

If probabilities aren't conserved then there is a break of unitarity

Claudio

Oh yep I get that as well

Claudio

9:46 AM

@ACuriousMind This need to get starred for future physicists to see ngl

no wait maybe not. It lacks a bit of context imo. Dang, we missed a great opportunity

anyways, I found an interesting article. I don't know if it's famous but just the premise is really funny: arxiv.org/abs/physics/0110060v3

Slereah

"This startup is building the country’s most powerful quantum computer on Chicago’s South Side"

That's some very specific qualifier

Not even the biggest quantum computer in Chicago

ACuriousMind

is it just a qbit? :P

Slereah

Only half, it can just have one state

Biggest quantum computer on the southern side of chicago next to the bakery near the bus station

Ryder Rude

i would say the loss of unitarity is not that severe. It's just two disconnected universes like ACuriousMind said

QM can work with connected sub-intervals of $R$ as time

Claudio

@RyderRude this does seem severe to me :P

lol

Ryder Rude

9:58 AM

@Claudio it's just uninteresting...

Claudio

@RyderRude connected?

Ryder Rude

Unitarity would be harder to patch in a circular time @Claudio

@Claudio yes, like QM on [0,1] as time

It's just that time evolution is only defined on [0,1]

Claudio

I was thinking how do you connect something on a line like $$[a,b[ \text{ and } ]b,c] $$

Ryder Rude

It's just uninteresting...nothing mathematically troublesome

@Claudio I'm not getting u...

ACuriousMind

no, this is not how physics works

you can't just say "after one second there's no time evolution"

Claudio

10:01 AM

Time as a "1d manifold" as they were saying

like $\mathbb{R}$

ACuriousMind

would everyone please stop making unwarrantedly confidently assertions about their entirely non-mainstream idea of how time evolution works :P

Ryder Rude

@ACuriousMind yeah, but this is a "thought experiment universe". One can say that this universe is modelled by a bounded time. I'm not saying it's interestin, but only that it's not troublesome

ACuriousMind

This is nonsense. The time evolution operator is $\mathrm{e}^{\mathrm{i}Ht}$. You can always plug in the entirety of $\mathbb{R}$ in there

Claudio

@ACuriousMind sorry

ACuriousMind

@Claudio I wasn't talking about you

Claudio

10:04 AM

@ACuriousMind oh my bad I thought I was saying something weird.

Ryder Rude

@ACuriousMind yeah... Even as a thought experiment, it doesn't make sense to restrict the domain...this is y i say it's uninteresting

ACuriousMind

it's not "uninteresting" it's nonsense, inconsistent, not thought through at all

let us not pretend that there's a consistent idea of physics in these ideas; we need to be able to call out things that make no physical sense as what they are

Ryder Rude

yeah.. it is important to call out wrong ideas

user 20458579510081670432

1:50 PM

I'm willing to bet that Wacken '25 breaks their sellout record of 4.5 hours.

ACuriousMind

2:32 PM

@user20458579510081670432 you've lost the bet already - sale started Sunday, 8pm and it's still not sold out

last year there were a lot more "big" bands announced at the start (in particular stuff like the Scorpions, Blind Guardian, Amon Amarth) so there were more people buying the tickets quickly just for one particular band; the lineup they announced this time is a bit more niche

user 20458579510081670432

:(

did you get a t-shirt with your ticket :^)

Silly Goose

3:06 PM

@ACuriousMind I am specifically considering the case in which such a case could not happen

operationally, i picture the following:
1) start with an initial state $\lvert \psi \rangle \in \mathcal{H}$ at $t_0$.
2) evolve via closed dynamics for a time step $\delta t$, $\lvert \psi \rangle \mapsto U(t_0+\delta t, t_0)$, i.e. solve the Schrödinger equation over $\mathcal{H}$ and over time interval $t \in [t_0, t_0 + \delta t]$.
3) puncture a hole in $[t_0, t_0 + \delta t]$.
4) update the time interval to be $[t_0 + t_0 + \delta t]$ with a puncture unioned with the "next" time interval $[t_0 + \delta t, t_0 + 2\delta t]$.

with the hope to study how puncturing passed time affects the (appropriately adjusted) usual time evolution of a closed quantum system

ah blebs the typos

i mean really the situation i am interested in is perhaps just Schrödinger picture quantum mechanics with a time-dependent Hilbert space

naturallyInconsistent

3:44 PM

@SillyGoose and ACM had kept telling you that Steps 3, 4 and 5 make no sense + needs much more fleshing out before it can stop being vague.

ACuriousMind

6:57 PM

@SillyGoose what do you mean "just"? There is no such thing as a "time-dependent Hilbert space" in ordinary QM

there is, in some sense, in certain approaches to QFT where we talk about a QFT assigning a space of states to each Cauchy surface

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