bound states and potential

Claudio Menchinelli

May 28, 2024 10:23

Regarding the hydrogen atom hamiltonian, the effective potential potential energy is given by $V_{\text{eff}}(r)=-e^2/r + \frac{l(l+1)\hbar^2}{2mr^2}$ and $V(0,\infty)=+\infty$ so that, for every energy value $E$ the eigenfunctions remain bounded and normalizable

If a perturbation $V(r) = 1/2m\omega r^2$ is added but, as we can see, it does not break the rotational invariance, the picture does not change and again the eigenfunctions are in $L^2(\mathbb{R}^3)$ for every value of $E$

if I add a perturbation $V_z = 1/2 m\omega z^2$ then $V_{eff}(r)=+\infty \text{ only for } z=\pm \infty$

but what can I say about the eigenfunctions?

@RyderRude sorry didn't notice your message, hi to you too my guy

imbAF

May 28, 2024 13:26

Regarding the 2nd quantization the following is said in Wikipedia:
"In this approach, the quantum many-body states are represented in the Fock state basis, which are constructed by filling up each single-particle state with a certain number of identical particles."

What does it mean to fill up single-particle states?

imbAF

May 28, 2024 14:32

In a cavity, are fock states, modes ?

ACuriousMind

May 28, 2024 16:09

@ClaudioMenchinelli I'm not sure I understand what you're saying here - the $V_\text{eff}$ you wrote down goes to 0 as $r\to \infty$, no?

bolbteppa

@ClaudioMenchinelli The eigenfunctions of the hydrogen atom are not all normalizable, there is both a discrete spectrum of normalizable states (bound state), and a continuous spectrum of non-normalizable states (corresponding to scattering states), to take the usual waffle interpretation you have to believe a particle has no energy, no momentum, no angular momentum when passing by another atom and scatters off, however if the particle gets sucked into orbit around the atom it then does

(Just one says for a free particle which has a continuous spectrum, where e.g. Griffiths talks about a free particle having no energy or whatever he says)

Claudio Menchinelli

@ACuriousMind yeah I miswrote, it does indeed go to zero, I mixed up the two. The eigenfunctions for the normal effective potential are \textbf{not} normalizable but remain bounded when $E > 0$

and when u add the $V_r$ potential, then the asymptotic values are $V(0,\infty)=+\infty$ as I incorrectly stated before

@bolbteppa yeah I know, I copied my notes wrong hahah

yeah Bound states for $V_{min} <E<0$ which is the case I usually solve my exercises for, namely discrete energy spectrum

I just wanted to know what happens to the eigenfunctions when The perturbation is just along a preferred axis

ACuriousMind

May 28, 2024 16:37

@ClaudioMenchinelli what problem do you have in applying perturbation theory to find that out?

also what kind of physical situation produces such a $1/z^2$ term?

Claudio Menchinelli

no no no no

it's a simple harmonic potential $\frac{1}{2}m\omega z^2$

I did the computations already hahah

no the last question of the exercise is purely about a simple discussion what happens to the eigenfunctions upon the addition of either of the two perturbations

ACuriousMind

on a qualitative level all you could say is that the $r^2$ will preserve rotational symmetry (and hence degeneracy among states with different $m$) while the $z^2$ will not, are you looking for something else?

Claudio Menchinelli

@ACuriousMind out of curiosity, have you read my messages from Sunday or only the ones from today?

if u scroll up you'll find the hamiltonian

ACuriousMind

@ClaudioMenchinelli after you said your question was not well-written I assumed you'd come back with a better written version if it still bothered you :P

Claudio Menchinelli

no basically I went to my professor

and he said that there are ulterior degeneracies to take into account

namely with respect to a reflection wrt the $yz$ plane

and also the commutator $[H,J_{\pm}]$ is null inside the subspace

that's why the degeneracy was not removed by the perturbation

it was a long interesting discussion that made me realize how much more to an exercise there is to think about than just what is requested in an exam scenario

@ACuriousMind I just wanted to know if there could be normalizable eigenfunctions after adding $V_z$

and not just bounded

ACuriousMind

May 28, 2024 16:51

I'm a bit confused about your terminology here

the eigenfunctions for bounded states (i.e. the eigenvalues in the discrete part of the spectrum) are the normalizable ones

it's the unbounded/continuous spectrum eigenfunctions that are not normalizable

Claudio Menchinelli

exactly

ACuriousMind

so I don't understand why there would be "bounded but not normalizable" functions

Claudio Menchinelli

no wait

can they just oscillated and remain bounded

instead of decaying to zero

ACuriousMind

what's your definition of "bounded"

Claudio Menchinelli

that's what I mean

ACuriousMind

May 28, 2024 16:54

wait, are you using "bounded" in the literal mathematical sense of "bounded function" and not in the sense of "bound state"?

Claudio Menchinelli

maybe hahaha

no 100%

is it not correct

ACuriousMind

in that sense, all the eigenfunctions are always bounded

even the unbound/non-normalizable ones, they're just waves after all

their non-normalizability does not come from being not bounded, but from not decaying at infinity

Claudio Menchinelli

awesome

ACuriousMind

that's the intuition for why the unbound states are non-normalizable, after all: if the function was concentrated inside some compact $S\subset\mathbb{R}^3$ (i.e. decayed at infinity), then it would be bound (physically) since it's confined to $S$

this does not quite work mathematically since there are $L^2$-functions that do not decay at infinity, but as a heuristic it's fine

Claudio Menchinelli

Ok, then I must'have misinterpreted my book's words

ACuriousMind

May 28, 2024 17:00

lol, if it's a physics book it's normal if it claims that all $L^2$ functions decay at infinity :P

common sleight of hand because physicists absolutely don't want to do actual functional analysis

it's not a terrible lie since the smooth compactly supported functions are dense in $L^2$, so every $L^2$ function is in fact the limit of functions that decay at infinity

Obliv

@Slereah Do you have any advice on how to set up a blog such as your own? I am considering writing notes across different topics I'm reviewing/studying right now and putting them online so I don't have to worry about losing it physically

Claudio Menchinelli

Therefore, when I have a potential whose asymptotic values are $+\infty$ only along one axis, but along the x and y axis it's still the usual effective potential for the hydrogen atom, I cannot have bound states?

namely bounded and normalizable eigenfunctions

ACuriousMind

what does the behaviour at infinity of the potential have to do with the existence of bound states?

certainly the harmonic oscillator on its own has bound states and goes to $\infty$ at $\infty$, no?

Claudio Menchinelli

no I mean when the energy is much bigger than $V_{min}<0$

I need to go grocery shopping unfortunately, I'll think about this in the meantime, I think my ideas are a bit confused by what my textbook used to say about this topic, I need to read that chapter again to make things clear. @ACuriousMind Thanks for the help, I'll come back later and update you :P

Slereah

May 28, 2024 17:33

@Obliv I made it by hand mostly as a project to test my web skills

Probably easier just to open a wordpress blog

ACuriousMind

@ClaudioMenchinelli Okay, I think I know what's confusing you - tell me if the following is an accurate summary of your question(s): 1. The harmonic oscillator alone has a potential that goes to $\infty$ at $\infty$ and has no unbound states. 2. The hydrogen atom alone has a potential that goes to $0$ at $\infty$ and has unbound states.

3. For some reason (?) you have become convinced that the value of the potential at infinity determines the existence of unbound states, and now you wonder whether your potential that goes "partly" to $\infty$ at $\infty$ and partly to 0 has unbound states.

unfortunately, there is no such simple and general relation between the existence of (un)bound states and the value of the potential at infinity - in fact you cannot even be sure in general that the bound and unbound states are disjoint in terms of their energy values, since there are bound states in the continuum

Obliv

@Slereah How much does it cost to run that server & domain? I don't wanna entrust stuff to 3rd parties :(

Slereah

I think it's like 70 bucks a year?

Obliv

Okay, well maybe I'll just run an offline version for a theoretical website blog since it's just for myself anyway

any idea how I could do that? :P

Slereah

just make some HTML files on your computer?

Obliv

May 28, 2024 17:47

ok I see, html is for static webpages which fits my purpose

ACuriousMind

@ClaudioMenchinelli The only thing about the value of the potential at infinity that is absolutely true always is this: For $\liminf_{\lvert x\rvert\to\infty} V(x) = V_\infty$, the spectrum of $H = \Delta + V(x)$ below $V_\infty$ consists purely of normalizable bound states. There is no general statement about the spectrum above $V_\infty$.

Claudio Menchinelli

May 28, 2024 19:03

@ACuriousMind you've perfectly summarized my situation unfortunately :P

@ACuriousMind I see

I've read my class notes

and the summary is (using your notation): when $E>\text{lim inf}_{|x|\to \infty}V(x)=V_{\infty}$, then $|\psi(x)|<L, L \text{ is a finite constant} $ and the energy spectrum is continuous

in other words, the eigenfunctions are non-normalizable

ACuriousMind

May 28, 2024 19:20

I think we need to separate a few claims here: Boundedness is really a separate property from being non-normalizable or the energy being "in the continuous spectrum"

what's true is that non-normalizable eigenfunctions are always in the continuous spectrum, and that they have to be above $V_\infty$ (by the contrapositive of my statement about $V_\infty$ above)

however, it can happen that there are normalizable eigenfunctions above $V_\infty$ (see bound states in the continuum above), meaning the discrete spectrum and the continuous spectrum are not disjoint

Claudio Menchinelli

yeah I read that

wait are you saying that boundedness $\nRightarrow$ non-normalizable ?

ACuriousMind

yes

I'm really not sure what the claim about boundedness by $L$ is doing there

Claudio Menchinelli

that's what is written in my notes

by my professor on the chalkboard

ACuriousMind

I mean, what you've written there does not claim that boundedness implies non-normalizability

it's just saying that the eigenfunctions are a) non-normalizable and b) bounded

while I don't know of such a result for boundedness of Sturm-Liouville solutions in this generality, it might be true, I just don't know why it's relevant

Claudio Menchinelli

it means that the eigenfunctions remain bounded by $L$ for $|x| \to \infty$

@ACuriousMind

exactly

we classified them in this way

ACuriousMind

May 28, 2024 19:30

hm?

Claudio Menchinelli

normalizable, non-normalizable but remaining bounded, and unbounded

or, equivalently, proper, improper, non-physical

ACuriousMind

I don't know what this distinction between "improper" and "non-physical" is :P

Slereah

based on vibes

Claudio Menchinelli

don't worry about it, I don't know why the second one does not sit right with you hahaha

@Slereah Looking at my notes, at this point, I'd say you're pretty much right :P

I get the situation

@ACuriousMind you're far too careful and rigorous for what my professors have discussed

anyways, I agree with your previous final statement

and I've completed my exercise finally

@ACuriousMind I limited myself to discuss normalizability according to what you wrote, so I should be safe hahaha

@ACuriousMind if you want to reads something, I'll translate the professor's solution to this last question so that you can see if you agree or not :P

ACuriousMind

I think what your notes are trying to do is the Lebesgue decomposition theorem into point, absolutely continuous and singularly continuous spectrum

this decomposition of the eigenfunctions into three subsets really can't be anything else

Claudio Menchinelli

May 28, 2024 19:41

@ACuriousMind honestly, I don't even think any of the QM professors know what that is

ACuriousMind

but I don't think I know a general proof that the absolutely and singularly continuous parts correspond to the (un)boundedness of the eigenfunctions

Claudio Menchinelli

what if the guy who wrote the book knew this, but it didn't include this since it's physics book, and my professors took what it's written there as it is

ACuriousMind

@ClaudioMenchinelli that they might not know this doesn't mean it's not what they're doing :P

Claudio Menchinelli

They use a particular book for this specific part of the course

I'm pretty sure it is since it fits perfectly

how deep are we going into the mathematical foundations of QM? That's my question

we barely scratched the surface in that sense, we didn't even introduced what an hilbert space is

ACuriousMind

May 28, 2024 20:01

okay, so this is Gilbert-Pearson theory (core.ac.uk/download/pdf/81962078.pdf, math.caltech.edu/SimonPapers/253.pdf) in disguise

and indeed absolutely continuous = bounded but not in $L^2$ and singulary continuous = not bounded and not in $L^2$ in terms of eigenfunctions

@ClaudioMenchinelli which one?

@ClaudioMenchinelli ???

this seems to me to be rather advanced for a course that doesn't even state what a Hilbert space is - how did they prove (or "prove") this statement about boundedness?

Claudio Menchinelli

Luigi Picasso, lezioni di meccanica quantistica

chapter 7

@ACuriousMind no they did not prove this obv

ACuriousMind

lol

Claudio Menchinelli

That's what I've been trying to say to you hahah

don't look into the book

nothing in it that talks about these things

ACuriousMind

I wasn't going to since I can't read Italian :P

Claudio Menchinelli

lol I am dumb

I'm looking at the papers and I cant stop laughing

Claudio Menchinelli

May 28, 2024 20:24

And thanks for the helpful discussion as always. I tend to forget, but you already know I'm grateful at this point :)

The h Bar

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