The h Bar

user 85795

03:14

@SirCumference how much did it cost you?

Sir Cumference

03:37

@user85795 I found it in my parents' storage

it's the one i had as a kid

had to replace the battery though

user 85795

cool

Sir Cumference

you grow up in the gba/gamecube era too?

user 85795

sorta yeah

123

07:08

Hello Everyone...

Kenshin

07:31

sup

Mr. Feynman

10:00

@SirCumference good old shit, man. Miss those days :'(

Silly Goose

12:01

how do I gain some intuition about how jump operators work

Ryan Unger

14:05

Is this chat dead

ACuriousMind

@RyanUnger less alive than some times in the past, but not dead yet (and Sundays are always slow)

how have you been?

Ryan Unger

good, playing kotor II now

how are you?

ACuriousMind

I'm fine, preparing tomorrow's D&D session right now

@RyanUnger ooh, for the first time?

Ryan Unger

Yes, was on sale on the switch

I played the first one last month after defending my thesis

4

or two months ago, time flies

ACuriousMind

@RyanUnger that means...congrats Dr Unger?

Ryan Unger

14:14

yes, thanks!

ACuriousMind

it's a bit bizarre to me that old games like kotor are on the Switch, but I guess it works

the second part was peak Bioware/Obsidian, they don't give you companions like those anymore :'(

Ryan Unger

I think its great that they port old games to the switch

gives me a chance to play them

they're pretty expensive unfortunately

ACuriousMind

yeah, it's not bad, just weird if you remember playing them...oh god so long ago

Ryan Unger

kotor is before my time...

ACuriousMind

so what's next after recovering from the PhD stress? Going to stay in academia?

Ryan Unger

14:28

Yes, I'm a postdoc at stanford now

user 85795

👏👏👏

ACuriousMind

neat, hope you enjoy it!

naturallyInconsistent

congrats

Ryder Rude

Unreal: Mongolian Throat Singing

►

it starts at 2:25. it's completely different from any other singing

Sanjana

Congrats

Ryder Rude

14:35

congrats

Ryan Unger

thanks :)

Ryder Rude

what is ur field? are you interested in quantum gravity?

ACuriousMind

@RyderRude it's not so rare: We were talking about KoToR, a star wars game, and Mongolian singing was used in the Star Wars: Fallen Order theme song by Mongolian band The Hu.

Ryder Rude

@ACuriousMind oh

ACuriousMind

come on people, we don't need to star every "congrats", okay?

4

Ryan Unger

14:41

can you remove the stars from my messages pls

thanks

naturallyInconsistent

while you are at it, there is a star on "hi" and on "oh ok ok"

Ryan Unger

is it normal that I can't understand the droids

ACuriousMind

yes, that's a running gag in the star wars universe - the characters can understand them fine, but the audience only ever understand what they said by the way the characters respond to the droids

Ryder Rude

@ACuriousMind it sounds so epic

ACuriousMind

it's the same in the movies and tv series - the droids are never subtitled/translated

Ryan Unger

14:44

ah right, I didn't think it was so bad in the first kotor though

maybe there were fewer droids

well, there was T3 but he wasn't so central

seems like I am talking to him a lot more in this one

Ryder Rude

what movie did u like in the past month?

i mean, something u watched the past month

ACuriousMind

I think you can still get from the responses what the droid said, but I understand it's annoying if you don't find the bit itself interesting

Ryder Rude

see Blade Runner for a more dark themed Star Wars

ACuriousMind

it's probably something more interesting to the writer than to the player: From a writer's perspective, it's very interesting to write a character you can't actually understand in a way that the audience still understands something about the character and their personality, but to the player, it might just come off as unnecessarily annoying

Ryder Rude

it explores slavery and humanity

ACuriousMind

14:48

@RyderRude in what way do you think Blade Runner and Star Wars are at all similar?

Ryder Rude

the villain in the movie is supposed to be understood as the good guy

@ACuriousMind Star Wars also had clone slaves as far as i remember

ACuriousMind

these are two totally different franchises playing with two completely different genres

Ryan Unger

It is indeed annoying to have to scroll through all of my dialogue options to piece together what the droid said

Ryder Rude

but yeah, they r completely different

Blade Runner is based on earth

ACuriousMind

"clones exist" is just a common sci-fi trope, it's not specific at all

Ryan Unger

14:50

it's ok as a gag at first, because Atton can't understand T3

but the player character can

Ryder Rude

Blade Runner 2049 is a detective story

ACuriousMind

@RyanUnger I think it might also be meant to drive home the point that the player character isn't a blank slate - they have skills the player hasn't, and they have a past that matters (and in that past they learned how to understand droids)

this is relevant to the story's overarching themes, but I can't explain that without spoilers :P

Ryder Rude

15:18

anyone see Girl with Dragon Tattoo? is it good?

Sanjana

15:43

Anyone studied and used twistor theory in physics? I just want to know how useful it is in physics before delving into the details, which seem like a lot. Yes, I have seen some motivations here and there, but if any of you guys with whom I am more familiar can say a word or two about it, then that would be great.

If the above sounds "twisted", I just wanna ask "why twistors?" and not just spinors?

More Anonymous

@RyderRude yes and it's a thing of personal preference. Personally no

ACuriousMind

@Sanjana I have never needed it for anything and from where I'm standing it seems mostly an unnecessarily complicated way to talk about something that from a modern viewpoint is just a bunch of differential geometry of projective spaces, but I haven't actually tried to learn any of the details. I think @Slereah might be more familiar with some of it given his GR interests.

Binky McSquigglebottom

16:12

Two masses $m_1=5kg$ and $m_2=10kg$ are connected as in the figure. The plane, inclined by $\alpha=30°$, is rough with static friction coefficients $\mu_s=0.5$ and dynamic $\mu_d =0.3$.Determine whether the two masses, initially at rest, move and, if so, with what acceleration.

Im stuck at the First point , pls help

Ryder Rude

@MoreAnonymous thanks

Sanjana

@ACuriousMind oh ok thanks.

Sir Cumference

17:45

@RyanUnger yoo you're back

congrats on the PhD

imbAF

19:53

Are all bound states, stationary states?

ACuriousMind

@imbAF why would they be?

imbAF

They should not

Unless the potential is time dependent right?

independent*

ACuriousMind

Why would there be a relation between a state being bound and a state being stationary at all?

what's your definition of "bound state"?

imbAF

Bound state, Is a state in which the system is confined in some region of space, the energy values are discrete and E<0

ACuriousMind

I understand the first point, but what has discreteness of energy values or "E<0" to do with being bound?

what does "E<0" even mean for a general state?

imbAF

20:00

It implies that the state, is a bound state

that's all I can think of

ACuriousMind

I don't understand what you mean

imbAF

That the wave function is localized

ACuriousMind

I asked you to define what you mean by "bound state", and it's simply not clear what "the energy values are discrete" or "E<0" means for an arbitrary quantum state

imbAF

Those are the characteristics

when the quantum state is classified as a bound state

ACuriousMind

"the energy values are discrete" is a property of the Hamiltonian, not of a state, and it's not clear what "E<0" is supposed to mean at all. Given a state $\lvert \psi\rangle$, what do you mean by $E$?

imbAF

20:01

The system can only take discrete energy values

Well the E<0 implies we have a binding potential

ACuriousMind

you're not getting the point

imbAF

I don't really know what else to say

ACuriousMind

what is $E$, mathematically, as a function of the state $\lvert \psi\rangle$?

imbAF

E is the energy, so it's a physical thing

ACuriousMind

this is not classical mechanics, you can't simply speak of an observable as a number

The quantum energy operator is the Hamiltonian $H$. We have a quantum state $\lvert \psi\rangle$. You claim that this is a bound state if "E<0", but what is the number $E$ in that claim? How do you compute it from $H$ and $\lvert \psi\rangle$?

imbAF

20:05

Well how I compute it it really depends on the situation considered

ACuriousMind

that's not really an answer

imbAF

H=T+V

then I solve the SE in the position space

Throught boundary conditions I find the correct expression

ACuriousMind

why are we solving the SE?

you wanted to talk about bound states, so we need a definition for any quantum state to be a bound state or not

imbAF

Yes correcy

ACuriousMind

if you a priori restrict your definition to apply to solutions of the time-independent SE, then saying bound states are stationary states becomes a tautology

imbAF

20:07

Well the main thing is that you need to be able to do separation of variable for the solution to the TDSE

ACuriousMind

(and spoiler: No, bound states are not necessarily stationary states - the superposition of two bound states is again a bound state)

imbAF

which can only be done once you establish that the potential is time independent

ACuriousMind

@imbAF why are we solving the TDSE at all? I just asked you to define the number $E$ is your proposed definition for "bound state"?

either write down a definition for that $E$ or realize that it makes no sense

imbAF

To me $E$ is just energy, and we solve the TDSE to know about of the evolution of the state of the system

ACuriousMind

I don't know how that's an answer to the question

imbAF

20:10

as to why we solve the TDSE?

ACuriousMind

A quantum state is just a vector in Hilbert space (or a wavefunction, if we're doing the position representation)

if you want to make a definition of what "bound state" means, you need to write down a definition that you can in principle check for any such state as to whether it applies to it or not

so if I give you a random wavefunction $\psi(x)$, how would you check whether it is a bound state or not?

imbAF

I am not sure. Probably by measuring the energy of the system in that state

Or

sorry

ACuriousMind

What I'm trying to point out here is that when you ask a question of the form "Are all As also Bs?", you first need to have clear definitions of what A and B are, otherwise the question just doesn't mean anything

imbAF

You are correct

ACuriousMind

when you asked "Are all bound states stationary states?", that implies that you have separate definitions of "bound state" and "stationary state"

imbAF

20:15

I am currently reading Introduction to Quantum mechanics by Griffith, and I gotta say, the book is just lackluster and quite generic. Really disappointed with the way the book is presented

ACuriousMind

but so far you haven't produced an actual definition of "bound state", so perhaps the question you really want to ask is "What is a bound state?"

imbAF

And it confuses the hell out of me. with his, we take this thing as a fact and no explanation. WHy? we will see that it helps us later on. Great argument

ACuriousMind

note that usually intro texts don't really define this notion in general and just use the term in examples like the hydrogen atom where it's reasonably intuitive what the (un)bound states are

imbAF

Yes

that's why I cannot, apprently put into words what it is

Because, the bound state was introduced when we considered specific cases

So, naturally, I would say, that a system in a bound state, is localized in space

ACuriousMind

that's because the technical definition is a bit involved and the interesting stuff - such as proving that all bound states are indeed in the "discrete part" of the energy spectrum, known as the RAGE theorem - requires considerably more mathematical machinery than these texts want to introduce

imbAF

20:18

ahaa

so, half assed explanation from the book...sorry for my language

ACuriousMind

and indeed the correct definition of "bound state" is just "a state that remains localized for all times", but the technical definition of localized is also not that straightforward

imbAF

@ACuriousMind I agree. how can you put it into words what localized means?

the pdf is non-zero in some regions and zero elsewhere

that's a weak description of localization

ACuriousMind

I mean, I could write down the technical definition but again, that's beyond the machinery an intro text provides you with

imbAF

The book is, just bad

ACuriousMind

no, this is not a unique failing of this particular book, QM is just much more subtle and complicated than we want to make it at an intro level

imbAF

20:20

I agree but still the book is bad

Right now I am studying the delta like potential

And griffiths says the following:

The first derivative of the wavefunction in the boundary, tells nothing (doesn't show that). And then, without no logical explanation as to why, he does what he does

ACuriousMind

so an intro text will point out the curious fact that the discrete spectrum of the hydrogen atom is the bound states and the continuous part the unbound scattering states, heavily imply that this is a general phenomenon (which it is), but not prove it at all

imbAF

he says that we integrate the SE from -$\epsilon$ to $\epsilon$..........

ACuriousMind

see e.g. this answer for a high-level explanation of the RAGE theorem

imbAF

@ACuriousMind Yes

@ACuriousMind But again, what I fail to understand is that: Since the H atom exhibits discrete energy values, isn't that enough of an argument to say what bound states are?

ACuriousMind

I don't know what you mean

sure, you could define a bound state to be a superposition of eigenstates from the discrete part of the spectrum. But that's a silly definition, because what, a priori, does the discreteness of the energy have to do with remaining localized in space (which is what "bound" colloquially ought to mean)?

imbAF

20:26

One observes the H-atom or anything similar to it, and sees that the energy values of a system are discrete. So he labels the states, in these considered cases as bound. And since all these cases have to do with localized systems, we can simply say that a localized system's state, is called a bound state

ACuriousMind

what do you mean "the energy values of a system are discrete"?

the H-atom has a bunch of non-discrete energies in the spectrum above the binding energy, corresponding precisely to the unbound states

imbAF

that the system only exhibits certain energy values and not a continues spectrum of them

@ACuriousMind by unbound states you mean scattering states?

ACuriousMind

yes

and the hydrogen atom has those

imbAF

well then, that throws everything I know out of the window

Discrete energy values of a system that is described via scattering states

But I guess I will read the link you gave me

And also, Griffith says that, scattering states are not stationary states

Claim made by him in the book

ACuriousMind

depends on what exactly one means by "stationary states" and whether we allow eigenstates for continuous spectrum value or not, but it's not exactly wrong

again, the topic of bound vs. scattering states is rather subtle and unless you explicitly do scattering theory you won't explore it in any detail

imbAF

20:32

whether we allow eigenstates of what?

of what observable ?

ACuriousMind

of the Hamilton operator

the subtlety is that while the values from the discrete spectrum have actual eigenvectors, the continuous spectrum values don't have proper eigenvectors - this is the same as the "problem" with the non-renormalizability of e.g. position "eigenstates" that you should've already encountered - position has a purely continuous spectrum

imbAF

Yes I know

similarly the momentum operator

as well

Therefore, a scattering state is not normalized ?

ACuriousMind

yes, so the same happens for the continuous part of the energy operator - so the scattering states, which are in that continuous part, can't really be stationary states if you insist them to be normalized, just like real state can't be a position eigenstate if you insist on it to be normalized

imbAF

With what you just said, you are implying:
1. Stationary states= always normalized

Also : "just like real state can't be a position eigenstate if you insist on it to be normalized" I don't understand this part

what you mean by real state?

by what you say I can say

ACuriousMind

"real" as opposed to whatever you want to call the non-normalizable $\lvert x\rangle$ we sometimes write down

imbAF

20:40

a real state can't be a Hamiltonian eigenstate, when the spectrum of the Hamiltonian is continues

ACuriousMind

careful: The spectrum of the Hamiltonian, outside of the free particle, usually is not fully continuous - it has a discrete part and a continuous part.

that's the part where we already leave the machinery an intro to QM equips you with, because to properly talk about what I meant above we would need spectral theorems that tell is that just as the spectrum splits into a discrete and a continuous part, the Hilbert space the operator acts on also splits into corresponding parts, where the discrete part is spanned by the eigenvectors from the discrete spectrum, etc.

imbAF

I mean why think of a free particle and not of an unconfined system?

@ACuriousMind If this is the case, then I don't believe the reason my knowledge about qm is fragmented has to do witht he fact that I mostly studied with notes, because Griffiths book for example, doesn't say nothing at all about what you are saying

ACuriousMind

that's why I keep saying this particular question of bound vs. scattering states is beyond typical intros!

imbAF

Yeah I guess, mathematical physics is the direction I'd be inclined to study

@ACuriousMind According to Griffith, a stationary state is one that can be written via the separation of variables, which is possible only if the potential doesn't depend on time. WHich means stationary state = time independent potential

ACuriousMind

@imbAF that's not what Griffith is saying

he's saying that the separable states are stationary, not that the definition of a stationary state is being separable

the definition of "stationary state" is eqs. (2.6) and (2.7) and makes no reference to separability

they're called stationary because they're stationary in time

imbAF

20:54

Ok

In the link you provided

what is a d-dimensional SE? Like, where can one see the dimensionality?

ACuriousMind

it's in d spatial dimensions

imbAF

ah, ok

What is $L^2(R^d)$?

ACuriousMind

that's a sign that that answer is not yet for you

imbAF

So, I shouldn't read it?

ACuriousMind

you can read it, but you probably won't understand enough of it to be useful

as I keep saying, this topic is really beyond intros to QM

just table the issue of bound states as something you can't fully understand yet and move on

imbAF

21:03

Yeah, but you'd think that with a bachelors degree, one should be able to

wtf was I taught for 3 years..

ACuriousMind

I wouldn't think that :P

imbAF

But I am curious, could you help me understand the first paragraph, with all the math signs etc?

or describe it with words

and maybe I can do research on my own

ACuriousMind

really you'd need to read a book/lecture on mathematical quantum mechanics and/or scattering theory to get this

imbAF

book such as ?

Can you recommand any book that it actually, gives a proper explanation why we are doing whatever we are doing

not like the current book I am reading

ACuriousMind

the classical reference for rigorous scattering theory would be the multi-volume series Methods of modern mathematical physics by Reed and Simon

imbAF

21:09

where things are considered in a certain way....because it will prove useful later

ACuriousMind

it's not a particularly quick or easy read :P

imbAF

The way I see it, the need for mathematics in physics keeps increasing but I can't keep up. Can't read a book every month

ACuriousMind

I mean, plenty of physicists are fine without ever seeing or proving things like the RAGE theorem :P

imbAF

How can you operate

when you can't even put into words what a bound state is

which leads to asking dumb questions

with dumb implications

such as the one I did

ACuriousMind

just like in many legal systems, "I know it when I see it" works fine in many situations :P

imbAF

21:12

That's horrible

ACuriousMind

I mean in many situations you don't really care about what a "bound state" is - you just have your system and you do QM with it

you can do a lot of physics without ever caring about whether some state should be called "bound" or not and what the general theorems about bound states are

imbAF

Is mathematical physics the branch that deals with these problems ?

ACuriousMind

I mean, for this particular question, I don't know of an answer that does not involve mathematical physics

imbAF

such as a precise defition of what a bound state is

ACuriousMind

it is not, in general, the case that it resolves all issues you might have with QM :P

imbAF

21:14

What issue do I have other then

from what I can tell

mostly mathematical physics

ACuriousMind

the "you" there is generic, I don't mean you, specifically :P

imbAF

Ah

I have one additional question

Give me a bit of time to formulate it

ACuriousMind

you have all the time in the world

imbAF

Before that, just to confirm

Can I say that a stationary state is one, such as when the system is represented by it, the expectation value of any measurable quantity doesn't change over time

ACuriousMind

yes

imbAF

21:28

Le'ts consider a discrete spectrum of the Hamiltonian of some system. A solution of the TDSE, can be a superposition of eigenstates of H, and can be written :
$\Psi(\vec r,t)=\sum_n c_n \phi_n(\vec r)e^{-i\frac{E_n}{\hbar}t}$.
This expression was the result of the consideration that the potential was time independent, which then allowed me to use separation of variables.
But, there is also this expression:

$\Psi(\vec r,t)=\sum_n c_n(t_0)e^{-i\frac{E_n}{\hbar}(t-t_0)}\phi_n(\vec r)$

Can you explain me what is the difference in what we consider that these two expressions are different?

In one, the coef. do not carry a time dependency

while in the 2nd case, they do

Which I believe, this time dependency will be reflected in the probability of the system being in one of the eigenstates, and it's a changing one

ACuriousMind

your first case is just the second case for $t_0=0$

imbAF

Ah

One more thing. In Griffiths book, when he speaks about the delta like potential, he says that 2nd boundary condition tells us nothing. but then he says that the delta function must determine the discontinuity in the derivative of $\psi$

So, the 2nd boundary condition does tells us something. That we have a discontinuity in the boundary

or am I getting it wrong ?

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