The h Bar

user228700

13:07

@JohnRennie Ah, no, that does make sense. It isn't going to be released till next year! :-)

user228700

@JohnRennie :-(

user228700

@JohnRennie Did you find anything?

Emilio Pisanty

@Phase all I'm saying is that the text is probably not as clear to other people as it looks to you.

it may help to leave it alone for a couple of days and come back to it with fresh eyes

Phase

ok, but afaik every term i used is standard

Is it the grammar?

0ßelö7

I'm not sure what your point is

user228700

13:09

@JohnR: Ah, electrostatics again? :-)

0ßelö7

You're saying things that are correct, but to what end

Phase

Well it's to talk a person through eigenfunctions and TISE

0ßelö7

Oh wow that OP is awful

Phase

the question or the answer?

Semiclassical

Here's an intro quantum question that's on my mind.

0ßelö7

13:20

The question. That's what "original post" means

Phase

yeah I just wanted to be extra certain

0ßelö7

@Semiclassical shoot

Semiclassical

My students will be doing the case of an attractive delta-function potential next week for discussion. As a part of that, they'll show that the ground state of this potential is a bound state and that all other states are free.

0ßelö7

@Semiclassical I think you helped me with that one a year ago

Semiclassical

This illustrates a generic fact about bounded 1D potentials: they've got at least one bound state, but need not have any more than that. (This isn't true in 3D, but there's at least one answer on PSE that explains why.)

0ßelö7

13:23

@Semiclassical yeah there's a somewhat nice physicist proof of that fact

I've forgotten how it goes

Phase

If you remember, could you link me it?

Semiclassical

I might need to look that up. My question is related but slightly different

Why at least one?

0ßelö7

Because that's how the proof goes? I think you perturb a flat potential and do some magic

Semiclassical

Or, in terms of the delta-function case: What is it about the delta potential that forbids having a bound state with odd parity?

0ßelö7

Wait...how is a delta function bounded

Semiclassical

13:25

Hmm

Limiting case of a bounded potential

0ßelö7

It's a bounded linear operator

Semiclassical

gotcha

0ßelö7

Although I think bounded means literally not greater than something numerically

In this context at least

Semiclassical

My guess is that the physicist resolution of it revolves around the fact that, if the first excited state is bound, then the symmetry of the problem would force it to have a node at 0.

0ßelö7

@Semiclassical actually the space of test functions isn't normable so that might not be right

It's a continuous linear functional, not sure if that implies bounded for LF spaces

Semiclassical

13:27

no clue.

I'm trying to think in terms of a physics-level argument

0ßelö7

@Semiclassical doesn't the node theorem require a continuous potential?

I'm assuming it does...

Semiclassical

ugh.

probably

you can think of the delta function as the limiting case of a gaussian, though.

so it may hold in the case of the delta function regardless.

0ßelö7

Actually, I think this was an exam question of mine

Semiclassical

Looking at my copy of Shankar, they leave the proof of showing that an attractive potential in 1D has a bound state as an exercise , with the intention being for the student to use the variational principle

which I can buy

0ßelö7

@Semiclassical ah, yes

You show that the first EV is necessarily negative, by comparison with a step

Semiclassical

13:31

problem is, variational principle as I remember it gives bounds on the ground state

and it's the first excited state that's of interest here

John Rennie

@Kaumudi.H you around?

0ßelö7

@Semiclassical hmm

You said one bound state

Not two...

Semiclassical

yes, and my question is why there's no bound first excited state

there is only one bound state in the case of the delta potential. Why?

user228700

@JohnRennie Yes, sort of :-)

Semiclassical

what's special about the ground state vs. the first excited state?

user228700

13:33

user228700

As a C.S Engineering student, I am expected to learn the block diagram representation of a thermal generating system.

Semiclassical

my instinct is this. if there was a bound first excited state, by the symmetry of the potential it would be an odd function and so would have $\psi(0)=0$. (this is not so for the ground state, which is even)

user228700

I'm having the time of my life!

John Rennie

@Kaumudi.H I have several battered looking USB DACs in a box. I have no idea which if any of them work but I'd say there's a good chance some of them are working. When I've had a chance to test them I'll ping you. It'll be in the next couple of days.

user228700

AHA, OK! Thank you! :-)

Semiclassical

13:35

And since the delta potential corresponds to an interaction occuring only at $x=0$, the first excited state would not interact with this potential whereas the ground state would.

I think the difference should be coming from there, but I can't quite close the loop.

Sid

@Kaumudi.H That is because, you are going to be an Engineer first, and then a major of your subject. You should know the basic things that an Engineer should know

And it isn't mandatory that you work in only what you specialised in

John Rennie

@Kaumudi.H And now I was going to cycle into town to get lunch, but it has just started raining. Hmm. How badly do I want to go into town?

Balarka Sen

Finally back home.

user228700

@JohnRennie Ah :-(

John Rennie

I think I might go to the corner shop instead. There's always tomorrow :-)

0ßelö7

13:39

@Semiclassical why is the ground state even?

And excited state odd?

user228700

There's a corner shop? That's convenient! :-)

Semiclassical

I guess I have the node theorem in mind for that. Ground state has no nodes, first excited state has one, etc

user228700

@Sid Well, I didn't sign up for this :-( Oh, well.

ACuriousMind

@EmilioPisanty No, I mean, what sort of stuff are you voting to delete that is not also flaggable as NAA/VLQ?

user228700

Oops, a bit of a silly typo there; continent, haha :-)

Sid

13:41

@Kaumudi.H Well, you get what you get. Life isn't always fair

Emilio Pisanty

@ACuriousMind flaggable according to whom?

Semiclassical

and since the potential is even in space, the eigenfunctions are also even/odd

user228700

@JohnR: Before you go, check Gchat.

Emilio Pisanty

I reckon this is flaggable as VLQ physics.stackexchange.com/a/358226/8563, or at least it should be

I guess it also makes NAA criteria

Semiclassical

but an odd function vanishes at zero and any other zeros comes in positive/negative pairs, so it's got an odd number of zeros.

so first excited state has one node and is therefore odd.

ACuriousMind

13:42

@EmilioPisanty Saying "The sun will not expand" as an answer to "If the sun expands,..." is not an answer.

0ßelö7

@Semiclassical What are the exact hypotheses for the node theorem?

@JohnRennie how do i get an ! in a print statement

the internet is very unhelpful

Emilio Pisanty

@ACuriousMind OK, take this one physics.stackexchange.com/a/328889/8563 as an example

it's not NAA, because it does attempt to answer the question

Semiclassical

this answer seems to have the formal statement as developed by Courant: chemistry.stackexchange.com/a/14825

Emilio Pisanty

is it VLQ?

0ßelö7

"hermitian differential operator"

Semiclassical

13:44

with the special case for 1D sturm-liouville problems being discussed a bit further into that answer

0ßelö7

@Semiclassical that's a red flag right there. No mathematician uses that phrase, so it could be physicist math

ACuriousMind

@EmilioPisanty I'd say so, because the it leaves entirely unclear how the statement in the answer actually relates to the question.

Semiclassical

He's quoting Courant-Hilbert.

0ßelö7

that's an old book

Semiclassical

Might just reflect old terminology

0ßelö7

13:45

they likely just mean symmetric then

Emilio Pisanty

@ACuriousMind well, there's at least one mod that disagrees with you

> not an answer – Emilio Pisanty Apr 25 at 10:50 declined - flags should not be used to indicate technical inaccuracies, or an altogether wrong answer

0ßelö7

@Semiclassical Oh, this is Courant's nodal domain theorem. There's a similar result in Riemannian geometry for eigenfunctions of the Laplacian.

ACuriousMind

Btw @EmilioPisanty if you are complaining that the VLQ criteria are...not very well delineated, then I completely agree

Semiclassical

sounds right

ACuriousMind

@EmilioPisanty Ah, and another (rob) agrees since he deleted that answer...

Semiclassical

13:47

@ACuriousMind the old line re: obscenity holds here, in that it's mostly decided on the basis of "I know it when I see it"

Emilio Pisanty

@ACuriousMind yes, well, my contention is that this would be best handled in a queue where there's a bigger voter pool, less discretion, and more transparency

@ACuriousMind yes, that's the core complaint

However

ACuriousMind

@EmilioPisanty Well, the queue is the VLQ queue. The VLQ flagging system is messed up because VLQ flags also show in the mod flag interface if they are not handled within 15 minutes

Emilio Pisanty

I don't think we're in a position to make that move at the moment

Balarka Sen

Seems like we're finally discussing physics and not memes. I am not the hero hbar needs right now.

ACuriousMind

So we mods probably end up handling more of them than entirely necessary, i.e. you get one person reviewing the flag instead of the 2-7 in the VLQ queue

Emilio Pisanty

13:49

because if the queue messes up and deletes content that it shouldn't, we don't yet have an effective corrective mechanism

@ACuriousMind well, that's what this was about

24

Q: Flagging low quality answers: an experiment

Abstract: We are experimenting with expanding the scope of the very low quality (VLQ) flag. For two weeks, August 8-21, moderators will not decline any VLQ flags. We will allow these flags will sit in the queue as long as needed for the community to deal with them. We ask you, the community, t...

but it wasn't particularly effective

Semiclassical

@0ßelö7 I'm not too worried on this point, despite the sloppiness. If I replace the delta function potential with a sufficiently narrow Gaussian, then there's no issue with smoothness and it should still have only one bound state.

0ßelö7

@Semiclassical knowing that the first nontrivial eigenfunction doesn't change sign is actually surprisingly important

Emilio Pisanty

and I stopped pushing in that direction because of the lack of an effective corrective mechanism

ACuriousMind

@EmilioPisanty The main corrective mechanism is that the queue can't delete positively scored content, and that reviews that are strongly controvesial (more "delete" than "looks OK") raise a special mod flag

Semiclassical

ya

Emilio Pisanty

13:51

@ACuriousMind yeah, I don't think those are strong enough

ACuriousMind

@EmilioPisanty What would you like? Because I can't quite see a straightforward way to have stronger barriers while still keeping the VLQ queue effective at deleting unequivocal garbage.

Emilio Pisanty

@ACuriousMind we have the tools, we just don't have enough people yet with the capacity to use them and enough people using them regularly

ACuriousMind

Oh, you just want more reviewers? You and me both.

Emilio Pisanty

@ACuriousMind not just reviewers

I want a large pool of 10kers regularly auditing the Recently Deleted list

though some additional filtering on that list would be nice

Semiclassical

How big of a pool of 10k users do you have to draw from?

Emilio Pisanty

13:55

e.g. to be able to separate mod vs self vs queue deletions

ACuriousMind

@Semiclassical About 75, not all of them active anymore

Emilio Pisanty

and relatively few of them taking particularly active community moderation roles

so e.g. if you look at physics.stackexchange.com/help/badges/85/reviewer, there's relatively few 10k'ers

I think

though that's a subjective impression and might not bear out w.r.t. actual numbers

ACuriousMind

@EmilioPisanty I have a feeling that compared to the 3k - 10k bracket, they aren't doing less, there just are fewer of them.

Emilio Pisanty

@ACuriousMind yeah, that's entirely possible

but either way, there's just not enough at the moment, I should think

Phase

14:13

can I ask a QM question relevant to what we were talking about before @EmilioPisanty

Emilio Pisanty

@Phase don't ask if you can ask, just ask

Phase

oh right

ok

Can you actually do something like the infinite well in a matrix formalism?

Emilio Pisanty

@Phase if you want a finite matrix, then no

if you're OK with infinite-dimensional matrices, then you can do pretty much everything in QM that way

Phase

Ah ok so let me see if i understand this

Emilio Pisanty

20

Q: Separability axiom really necessary?

I know other people asked the same question time before, but I read a few posts and I didn't find a satisfactory answer to the question, probably because it is a foundational problem of quantum mechanics. I'm talking about the Hilbert space Separability Axiom of quantum mechanics. I'd like to u...

quantum-mechanics quantum-field-theory hilbert-space

Phase

14:19

Would you end up with an infinite dimensional vector space where each basis vector corresponds to a value of the function at a point?

Semiclassical

@EmilioPisanty If you're doing bound states, definitely. Not so sure about scattering states given that there's a continuum of energies.

Phase

So say the boundaries are 0 and L,

Semiclassical

No.

Phase

$\psi(0)$ would correspond to <a|$x_0$>?

oh ok rip

Emilio Pisanty

@Semiclassical well, there's always a discrete basis for you to do any calculation you want. Whether it's useful or not is another matter

Semiclassical

14:19

At least, that's not the story I know.

@EmilioPisanty Yeah, true.

Emilio Pisanty

@Phase what?

Phase

nvm

Semiclassical

The story I know would be to find (infinite) matrix representations of how operators act on the infinite-well energy eigenstates.

Phase

im not entirely sure what the vector space would represent

Semiclassical

e.g. $x\phi_0(x)=\sum_{n=0}^\infty c_n \phi_n(x)$. One more often sees some discussion of this when doing the simple harmonic oscillator.

Phase

14:22

Since functions exist in a vector space, could the eigenfunction solutions to the TISE be the basis vectors?

Semiclassical

That's what I was getting at, yeah.

the action of $H$ on such basis vectors is trivial, of course: $\hat{H}\phi_n(x)=E_n \phi_n(x)$

Phase

Oh i see, I didnt recognise it

yeah

Im confused

Semiclassical

it's when you're doing other operators like $\hat{x}$ or $\hat{p}$ that things aren't nice

Emilio Pisanty

@Phase any set of linearly independent vectors can be a basis

Phase

Why is $x \phi_0 (x)$ equal to $c_0 \phi_0 (x)$ and an infinite sum of other functions

Emilio Pisanty

14:24

@Phase why wouldn't it?

$x\phi_0(x)$ is a function

Semiclassical

because $x\phi_0(x)$ is also in the Hilbert space, which is spanned by the infinite-well eigenfunctions

Emilio Pisanty

that's all you need

Phase

pardon my idiocy but

Semiclassical

If you want to actually find the coefficients, you'd need to start computing integrals

Phase

To me $x \phi _0 (x)$ looks like it would just lie along $\phi_0 (x)$

am I missing the purpose of the x?

Semiclassical

14:26

Then you're wrong.

Phase

i was expecting that

Semiclassical

$\phi_0(x)$ isn't a vector in R^n, it's a vector in Hilbert space

I mean, suppose $\phi_0(x)$ is a solution to an ODE. Do you really expect $x\phi_0(x)$ to be a solution as well?

Phase

I guess I've only ever considered $R^n$ when it comes to Hilbert spaces

Well Im confused by what $x$ does there

Semiclassical

?

Phase

Oh

ohhhhhhhhhhhhhhhhhhhhhh

Semiclassical

14:28

:)

Phase

Wait

Now im even MORE confused

How would you even begin to express a product of functions as a sum of functions?

Semiclassical

Same as you usually would. Think Fourier series

Phase

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh‌hhhhhhhhhhhhhhhhhhh

god penny drop moments are the best thing in life

What if the function isn't periodic?

0ßelö7

Compactificy on a torus

Semiclassical

Depends what you're doing.

But keep in mind that if you're working on an infinite square well, you don't actually care about what's happening outside the boundaries.

And if you were working on a ring, you'd insist that the functions you work with all be periodic

Phase

14:31

oh i think i realised something

tell me if its wrong

If you consider a hilbert space of square integrable functions, any product of two elements exists within the same space because of closure, and there's then a way to express it in terms of eigenfunctions in that space

now ive said it, provided that I havent misunderstood L^2 etc, it seems trivial and not worth saying

0ßelö7

@Phase what

Semiclassical

that seems right, yeah.

Phase

So when is this approach useful enough to justify switching from Schrodinger formalism?

Semiclassical

i think the usual place this can run into trouble is when you take boundary conditions into effect

Phase

i meant when does it become useful?

Semiclassical

14:34

not that examples are rushing to my head at the moment

Phase

Oh ok no worries

ill look it up i s'pose

Semiclassical

tbh it's mostly useful when you have finitely many states

Phase

Where the matrix algebra can't be trusted

0ßelö7

What you're saying isn't right. L^2 is not an algebra.

Semiclassical

i mean, being able to express $x\phi_0(x)$ in terms of $\{\phi_n(x)\}$ isn't a bad thing

but specifically working in terms of matrix representations is usually not necessary.

Phase

14:36

@0ßelö7 are you saying that Vector spaces aren't necessarily closed for operations like f(x)g(x)? or have i made a different mistake

Semiclassical

oh. yeah, that's definitely wrong

Phase

r i p

Semiclassical

f(x)+g(x) being in the vector space is a genuine example of closure.

f(x)g(x) isn't.

Phase

Oh

Duh

Because if you have two eigenfunction

s

x and x^2

the product doesnt exist in the space

right?

like, a space with two basis vectors x and x^2

0ßelö7

@Semiclassical thank you

@Phase if you want things like that you need an algebra structure

Semiclassical

14:38

usually one really doesn't worry about what f(x)g(x) per se looks like

Phase

But do I have the right understanding of why I was wrong

Semiclassical

rather, one asks how various operators act on a basis vector $\phi_n(x)$

Phase

Because you could just easily make something you cant express as a linear combination of eigenvectors

does it uh

Well

Is it still restricted by the postulate that it corresponds to a real observable?

Like $\hat E \phi _n (x) = e \phi_n (x)$

0ßelö7

@Phase no. It doesn't even make sense to multiply in a general vector space.

Semiclassical

e.g. $\hat{x}\phi_n(x)=x\phi_n(x)$, and if you've got $\phi_n(x)$ vanishing sufficiently fast at infinity then $x\phi_n(x)$ is also in the Hilbert space

Phase

14:40

gah

hopefully when I'm actually meant to be doing this stuff It'll come a bit easier

What does "and if you've got ϕn(x)ϕn(x) vanishing sufficiently fast at infinity " mean?

whoa copy and paste doesnt work

0ßelö7

He's being sketchy

I can give you a reference if you really want to understand this

Semiclassical

as I am want to be

Phase

absolutely

when is this stuff anyway? 2nd year?

Semiclassical

For a more elementary case of this, suppose you've got infinite-well states on [0,L].

Phase

ok

Semiclassical

14:43

Then your basis states are square-integrable and vanish on the boundaries.

Phase

yeah

what does it mean though

to vanish sufficiently "fast"?

Semiclassical

The first condition is fairly simple since the basis states are moreover continuous.

0ßelö7

@Phase first or second year of mathematics graduate school

Semiclassical

The key point is that if you multiply one of these eigenfunctions by $x$, you get a new function which is nevertheless still square-integrable and vanishes on the boundary.

Phase

grad school? Is that american Uni?

0ßelö7

14:45

No, graduate school

Phase

wait

Masters?

Semiclassical

PhD, usually

Phase

aw

that makes me feel a little sad

Semiclassical

@Phase Another example may help. Suppose you were doing the simple harmonic oscillator potential.

Phase

wait hang on another dumb question

0ßelö7

14:46

@Phase to really understand this you need measure theory and linear functional analysis

Semiclassical

Then your basis states would be of the form $H_n(x)e^{-x^2/2}$ where $H_n(x)$ is a Hermite polynomial in $x$.

Phase

I thought in vector space you can't multiply elements together like that, is there an exception for $x$?

0ßelö7

Plus a tolerance for inequalities

Semiclassical

Key point is that if you multiply such a basis state by $x$, then you get something which is still polynomial times $e^{-x^2/2}$

Phase

I picked up a Func analysis textbook but I had a few questions and no-one in hbar or my friends helped me resolve my dumb confusions

Semiclassical

14:47

and whatever polynomial you pick, $e^{-x^2/2}\to 0$ at infinity sufficiently fast that polynomial * $e^{-x^2/2}$ goes to zero as well.

by contrast, if you had function which vanished only as fast as $\psi(x)\sim x^{-1}$ at infinity, then $x\psi(x)$ need not vanish at infinity. so it wouldn't be square integrable and wouldn't be in the Hilbert space

0ßelö7

@Phase why didn't you ask me?

Phase

i think i did but i didnt get a response

0ßelö7

Then I didn't see it

Phase

It was about cosets and quotient space and ages ago about Banach space

Emilio Pisanty

@Phase why do you think that approach is different to the Schrödinger formalism?

Semiclassical

14:49

So this is where boundary conditions can be a pain.

0ßelö7

No functional analysis question is stupid. It's a hard subject

Phase

@0ßelö7 remember who you're talking about

0ßelö7

Even for you

Phase

jesus

It's THAT hard?

Semiclassical

In a lot of the cases you deal with in physics, $\hat{x}\phi(x)$ will be in the Hilbert space if $\phi(x)$ is

but it's not guaranteed, and there are exceptions

so there are headaches to be had.

Phase

14:51

Hang on im sorry to keep pestering you but

0ßelö7

@Phase it get really hard

Phase

Could you explain like you would to a household animal the deal about multiplying a function with x

Semiclassical

In functional analysis, you have to question quite a few of the things you take for granted or sweep under the rug in physics QM.

So while both would certainly agree in terms of the conclusions you'd end up drawing about cases of common application, the mathematical side is far more attenuated to the various ways in which that can go wrong.

0ßelö7

@Phase please stop being self deprecatory

Phase

like the finite dimensional linear algebra and how that makes the commutation relation spooky?

@0ßelö7 I'm not, I was just asking him to explain it as simply as possible because I still havent picked it up

Semiclassical

14:55

@0ßelö7 I'm forgetting. What's an obvious example of a potential whose eigenfunctions wouldn't vanish at infinity sufficiently fast?

Phase

hm

maybe if the function is made of multiple continuous functions?

Idk just randomly suggesting

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