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12:52 AM
Psychics is truly a complex domain, with all the related word salad of definitions and math equations, which often leave you with more questions than you initially had, and a journey through different stages of "what is that supposed to mean?"
3 hours later…
3:52 AM
Let us hope that the first word is a mishap from autoincorrect.
4:10 AM
6 hours later…
10:19 AM
what got u into physics
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
3 hours later…
1:26 PM
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?
1 hour later…
2:32 PM
In a cavity, are fock states, modes ?
2 hours later…
4:09 PM
@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?
@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)
@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
4:37 PM
@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?
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
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?
@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
@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
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
4:51 PM
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
so I don't understand why there would be "bounded but not normalizable" functions
no wait
can they just oscillated and remain bounded
instead of decaying to zero
what's your definition of "bounded"
that's what I mean
4:54 PM
wait, are you using "bounded" in the literal mathematical sense of "bounded function" and not in the sense of "bound state"?
maybe hahaha
no 100%
is it not correct
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
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
Ok, then I must'have misinterpreted my book's words
5:00 PM
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
@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
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
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?
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
5:33 PM
@Obliv I made it by hand mostly as a project to test my web skills
Probably easier just to open a wordpress blog
@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
@Slereah How much does it cost to run that server & domain? I don't wanna entrust stuff to 3rd parties :(
I think it's like 70 bucks a year?
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
just make some HTML files on your computer?
5:47 PM
ok I see, html is for static webpages which fits my purpose
@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 $V_\infty$.
1 hour later…
7:03 PM
@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
7:20 PM
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
yeah I read that
wait are you saying that boundedness $\nRightarrow$ non-normalizable ?
I'm really not sure what the claim about boundedness by $L$ is doing there
that's what is written in my notes
by my professor on the chalkboard
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
it means that the eigenfunctions remain bounded by $L$ for $|x| \to \infty$
we classified them in this way
7:30 PM
normalizable, non-normalizable but remaining bounded, and unbounded
or, equivalently, proper, improper, non-physical
I don't know what this distinction between "improper" and "non-physical" is :P
based on vibes
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
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
7:41 PM
@ACuriousMind honestly, I don't even think any of the QM professors know what that is
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
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
@ClaudioMenchinelli that they might not know this doesn't mean it's not what they're doing :P
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
8:01 PM
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?
Luigi Picasso, lezioni di meccanica quantistica
chapter 7
@ACuriousMind no they did not prove this obv
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
I wasn't going to since I can't read Italian :P
lol I am dumb
I'm looking at the papers and I cant stop laughing
8:24 PM
And thanks for the helpful discussion as always. I tend to forget, but you already know I'm grateful at this point :)
no worries, this was interesting to figure out
8:51 PM
Does the phase space of the electromagnetic field exists?
@imbAF what do you mean by that?
How could a phase space fail to exist?
In our lecture today, we considered the phase space of the light field, and the axis where the real and imaginary part. Which is not how I know the phase space looks like.
It usually has momentum and position components
but here we had real and imaginary part of a field
And the following was written:
For the light field in a coherent state, the time evolution of the electric field (written in a very convoluted way) was:
$\lang \alpha|E(t)|\alpha\rangle=E_0|\alpha|cos(\omega t + \Phi)$.
Through the number and amplitude uncertainty, we obtain a phase uncertainty.
I don't see the amplitude uncertainty
sounds like you're talking about optical phase space, where we have coherent states $\alpha$ with $\alpha = q + \mathrm{i}p$ and $p,q$ behave like what we would ordinarily use as phase space coordinates
I still don't understand the question about whether or not this space exists :P
For mechanical systems, the phase space usually consists of all possible values of position and momentum variables.
This phase space had as axis Re(E) and Im(E). And even if we had position and momentum, how can one decide what the position of the field is. The field is spread in space
it's not like a quantum system i.e a particle and QHO
so yeah
@ACuriousMind Btw are you familiar with different concepts in quantum optics?
9:07 PM
ACM is actually Ed Witten in disguise lmaooo, he has all the physics knowledge of this world currently
Than perhaps tomorrow, I could ask you some questions. I have been attending lectures in quantum optics on the nano scale, and as I am continuing with new lectures, more and more questions pile up. I would probably ask you tonight, if it wasn't for the fact that I am at work tmrw. So, if you'll be online tmrw and have time to help me, I'd very much appreciate that.
I dont think I've ever seen you using a gif
my life is complete
@ClaudioMenchinelli wait until Witten logs and asks ACM a couple of questions
I wouldnt be surprised
9:16 PM
Hello ladies, gentlemen and ACM. What's up?
ACM may also mean A Clever Machine
@ACuriousMind Just one brief question, should there a distinction between a light field a mode and a state be made? Is it correct to say:

he light field is a superposition of modes. Now it also depends where we consider the light field right? You can have the light field in vacuum, so you have free space modes, or the light can be in a waveguide, in this case you'd consider waveguide modes or optical resonators and its own modes. And now a mode is characterized by a state, you say. This state can be pure i.e coherent or fock states, or a mixed state i.e a thermal state. But for the state o
Would the state of the light field be a tensor product ?
@ACuriousMind unrelated note: along with the improper usage of "phase space", during my years as a student I've also heard the expression "there is no phase space" :P
@imbAF All these things are different things. 1. I would suggesting talking about the electromagnetic field, not the "light field". Quantum fields are operators, not states. 2. "A mode" is a somewhat vague term that usually refers either a creation operator or to a state made from applying that creation operator (perhaps repeatedly) to the vacuum. 3. A state is just a state.
@Mr.Feynman who says that? what do they mean by it?
I don't have a book in mind, a Prof of mine (hep-ph/nuc-th) of mine said it but it was related to the behaviour of the cross section+the improper usage of "phase space" to mean the $d^4p\delta^4(\sum_i p_i)$
oh, that
9:23 PM
a state is a state of which system?
I forgot the details on purpose after the exam
My mind had been polluted
@imbAF whatever system you're looking at (e.g. "electromagnetism in vacuum" or "electromagnetism in a medium")
you'll also get people talking about the state "of a field" but I don't like that way of speaking since it tends to confuse people about the nature of the field itself as an operator, not a state
Ok but when we talk about a state, it needs to be associated to a system
At least that's how I have been taught
@ACuriousMind This was written in the lecture regarding thermal states:

And the commentary was :

For a thermal state the population distribution follows the Boltzmann law
These probabilities need to be normalised, which leads to the thermal photon number distribution


which describes the probability to find n photons in a certain mode, if we have overall only a single excitation.
Here clearly a distinction is made between the mode and the state. So if we are talking about a state, what is the corresponding system
Additionally, what if we don't have a single excitation, but more than one?
Why is there put an emphasis on this : "..if we have overall only a single excitation"
so "find in a mode" here just means that there is the creation operator $a_\lambda^\dagger$ for photons of a certain wavelength $\lambda$ and the state of $n$ photons "in" that mode is $\lvert n\rangle = (a_\lambda^\dagger)^n\lvert 0\rangle$
"only a single excitation" means you're considering only a single $\lambda$ and not the possibility of different $\lambda$
the "system" is whatever you're looking at - vacuum, a waveguide, a cavity, it doesn't really matter
I need to study this answer as it has plenty of info
but I assume since we are talking about thermal states, here we can have photons of different wavelengths
9:33 PM
@imbAF no, that's what the text is explicitly excluding
and how many photos= single excitation?
For example in a fock state a singe excitation of the state = the nr. of photos of this state
or am I wrong?
they're saying there is a "thermal state" of temperature $\beta$: $\lvert \beta\rangle$ and then $p(n)$ is the overlap $\lvert \langle \beta\vert n\rangle\rvert^2$, i.e. the probability to measure the thermal state to consist of exactly $n$ photons
@imbAF again, "single excitation" here means just that you're considering only a single wavelength, and not that there might be photons of different wavelength at the same time; this is the case e.g. when you have monochromatic lasers
Ok, thanks for the answers. Unfortunately I need to go, but if it happens, let's continue tomorrow
Thanks and goodnight
good night

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