The h Bar - 2019-11-09

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2:45 AM

@Student404Mus In seriousness if you're doing a doctoral thesis then presumably there's at least something you find interesting in physics

just look into that

if you're not knowledgeable at all about e.g. astro then that's probably not the best choice

@SirCumference hey

available for a question?

i'm busy studying but you can ask and see if someone knows

oh, good luck!

5 hours later…

7:52 AM

Good morning/afternoon/night for everyone!

So, tangent spaces $T_{p}(\mathcal{M}$) of a point $p$ of a given manifold $\mathcal{M}$, are vector spaces. Therefore they reemsables the notion of "the place where we can apply linear algebra".

Now, before tangent spaces, the primer structure which these spaces are defined are then the various (differentiable) manifolds structures $\mathcal{M}$ . A (intuitive) way to picture a mental concept of a manifold is something that are "flat" quite close to a given point $p \in \mathcal{M}$ (more precisely homeomorphic to a $\mathbb{R}^{n}$ vector space…

Is this a good intuitive discussion for explain Tanget spaces?

tangent*

4 hours later…

11:37 AM

Hi all! I think about the electronic angular momentum (non-relativistic QM) for bound states in some potential $V$ (like a molecule). In a linear molecule the Hamiltonian $H$ commutes with rotations around the say $z$ axis, the eigenstates $|n>$ of $H$ are also eigenstates of the angular momentum component $\hat{l}_z$ operator, so that $\hat{l}_z |n>= m \hbar |n>$ with some integer $m$.

When we now slightly non-adaiabatically "bend" the potential by an angle $\epsilon$ s.t. our new states become $|n(\epsilon)>$ I wonder what we have to expect for the expectation value $<n(\epsilon)|\hat{l}_z|n(\epsilon)>$? I suppose by perturbation theory we can show it will be close to $m\hbar$.

Apparently for most real world cases (unsymmetric molecules in electronic ground states) this expectation value is (close to?) zero. This is called "quenching of angular momentum" (as far as I understand now). I now wonder if the possibility to observe non-zero for that is somhow connected with degeneracy and symmetry?

@EmilioPisanty any idea?

In the sense of dicrete point group symmetry, I mean?

1 hour later…

1:03 PM

1

Q: Is it right to post the same answer twice?

This question was marked as a duplicate of this question. The first one was about the chain-shape of a stream of flowing milk and the second one was about the helical shape of a stream of flowing wine. It had an impressive answer that correctly explained the phenomena. The first question was hen...

support duplicates

1 hour later…

2:20 PM

@Rudi_Birnbaum you're using non-degenerate perturbation theory, which doesn't apply here

If you bend the molecule, you will break the degeneracy between the $\Pi_x$ and $\Pi_y$ states.

The eigenstates of the deformed molecule will follow those states instead.

@EmilioPisanty have you studied any quantum organic chemistry?

(just wondering :)

Abhas Kumar Sinha

@skullpetrol I tried, that's really f**y way of introduction to QM. I ended up without understanding anything.

yeah, they take a memorization first approach

Abhas Kumar Sinha

@skullpetrol that's what I do. Btw, I dun like OC

sure, for vocabulary I do also

Abhas Kumar Sinha

2:30 PM

@skullpetrol vocabularies?

yup

Abhas Kumar Sinha

A good source to learn Quantum Organic Chemistry - ncbi.nlm.nih.gov/pubmed/17799679

biochemistry is the worst

imho

Abhas Kumar Sinha

@skullpetrol a lot of names to remember, once I dared to ask why not using IUPAC names for Bio-Molecules, whole class ended with laughter.

:(

Abhas Kumar Sinha

2:32 PM

:(

i find that math likes to hide a lot of meaning in their vocabulary

Ehhhhh

So does physics

yup

but with physics, at least we've seen some of the words in math :)

What subject doesn't have a vocabulary that hides a lot of meaning? It seems almost necessary so it doesn't take forever to explain things, instead having to use the definitions of the subject-specific words every time.

Can you imagine if physicists called acceleration "the rate change of the rate change of position over time over time" every time.

2

@skullpetrol depends on how large of a molecule you mean

I'd probably say that I know molecular physics, and I've worked with quantum chemists, more than having a background in the latter myself

"quantum organic chemistry" is a weird way to phrase things, though

2:49 PM

would "quantum biochemistry" be a better phrase?

No

I've never seen qualifiers added to 'quantum chemistry' by a professional

(not saying it doesn't happen, just saying I've never seen it)

3:03 PM

@EmilioPisanty quantum electrochemistry appears to be a legitimate qualifier :-)

@EmilioPisanty OK, what does that mean for $<n(ϵ)|l_z|n(ϵ)>$?

@skullpetrol you are free to invent whatever names you like.

question will be what they address, and who will agree.

quantum chemistry is a synonym for computational chemistry

usually

and they are both kind of the applied branch of theoretical chemistry

thee also exist computational and theoretical chemistry that is not quantum mechanics based, though but that usually is called mathematical chemistry or chemical informatics, latter is stuff like chemical libraries

@skullpetrol if you're not going to listen to what I say then I don't see how it makes sense for me to continue with this conversation.

and former contains some graph theory and stuff

@Rudi_Birnbaum also molecular dynamics.

yep

and I agree "quantum organic chemistry" sounds weird

3:14 PM

@Rudi_Birnbaum that your ket isn't too well defined ;-)

@EmilioPisanty I'm listening, pal...I just wanted to show you a word that you've never seen.

Well the bent molecule's electron surely has a state function or a space of state functions

, no?

@skullpetrol sorry, but 🙄. If you want my attention on a Saturday, put something real on the table.

@Rudi_Birnbaum yes

But what do you mean by $|n(ϵ)\rangle$? Particularly at epsilon=0.

the n-th state unbent

linear case

It's degenerate

How do you break that degeneracy?

3:19 PM

yes, is it a problem?

@Rudi_Birnbaum yes

Basically

maybe this is something I never really fully understood.

If you do things right

@Rudi_Birnbaum it's zero throughout

cant one choose just any nomalized linear comb.?

Including at zero

3:20 PM

why?

I'm at a small keyboard, I can't explain in depth right now

:-(

:(

It sounds like you need to take a deeper look at degenerate perturbation theory, though

Your confusion hinges on the differences from the non-degenerate theory

but cant one choose a basis of the space?

the deg. space I mean?

and then look at the exp. values for all the basis (vectors) separately?

@EmilioPisanty insert "again", I usually all knew that once upon a time but meanwhile I forgot a lot.

its usually enough if you give me just the important key word ...

3:37 PM

I mean isn't it that if the perturbation breaks the symmetry such that the degnerate symmetry species branches then you also obtain your (new) basis? I just can't see at the moment how that affects the angular momentum expectation value.

Abhas Kumar Sinha

another cyclone today night

@EmilioPisanty now I got it, thanks for the hint, you were again brilliant!

deg. pert theory: (youtube.com/watch?v=9JhX_UNcQvE) its a good one I think ...

@JMac Some subjects use vocabulary to hide a lack of meaning ;)

lol

@ACuriousMind You say that now; but when my zero-point magneto-mechanical resonator starts generating free energy due to the quantum subposition of the magnetic field, you'll see.

Abhas Kumar Sinha

3:50 PM

@JMac I use resonators for time travel

@AbhasKumarSinha What anti-particle base do you use to stabilize the temporal resonance field?

Abhas Kumar Sinha

@JMac that's a @Secret

:P

<_<

I tape a bunch of fridge magnets to a battery, it seems to work really well.

Abhas Kumar Sinha

quora.com/profile/Abhas-Sinha-8

@JMac lol, me toooooooooooo :)

3:52 PM

You held the o key a bit too long there.

Abhas Kumar Sinha

@JMac no that was just time lag due to slow time travelling processors

"I'm Resonator, the time traveller, from 2030..." - some legend!

@ACuriousMind would existential philosophy be one of them?

Abhas Kumar Sinha

@ACuriousMind some subjects are meaningless....

@skullpetrol Do you mean like Existentialism, or the philosophy of existence in general?

@skullpetrol Existentialism doesn't really hide a lack of meaning, does it? It's one of its core tenets that we must create meaning ourselves!

3:58 PM

::scrambles to Wikipedia for a crash course::

:-)

Abhas Kumar Sinha

Long ago, I wrote this answer, - qr.ae/TWwe8l Only God and I knew how it works, now only god knows... :( XD

@skullpetrol this may help you - en.wikipedia.org/wiki/Wikipedia:Crash_Course_in_Wikipedianism

Wikipedia:Crash Course in Wikipedianism

The Essence of Wikipedia: A Crash Course in Wikipedianism Wikipedia is one and at the same time an encyclopedia and a community. The encyclopedia is a central repository of knowledge made available to the public (the entire world), and as such is one version of society's memory. The community is everyone who makes use of or takes care of the encyclopedia, and includes readers, contributors, editors, and the admins of Wikipedia. Everyone has the right to use Wikipedia, and therefore it is the responsibility of everyone to build, maintain, and defend Wikipedia. Anyone who participates in Wikipedia...

@AbhasKumarSinha thanks pal

Abhas Kumar Sinha

@skullpetrol hehehe XDD

user image

Abhas Kumar Sinha

k

4:16 PM

user image

@ACuriousMind is Winterfest coming again this year?

What's Winterfest?

HATS!

Oh, you mean Winterbash!

2

pardon my shouting

ooh ya, oops

I don't have any more information about that than you

4:19 PM

did you enjoy Octoberfest?

I wasn't there, so not really

does the ankle that you broke bother you at all?

Not really, it's fine. But I haven't tried to run a marathon since ;)

:D

@ACuriousMind you're a marathon runner?

4:30 PM

@RyanUnger No.

I didn't try to run marathons before either :P

as of 6th November 2019...In light of recent events…

last year there was no 0celo and now this^

1 hour later…

6:10 PM

@EmilioPisanty but now lets say for the particle on the circle if we have a perturbation that does not break the state degeneracy but does break the continuous symmetry e.g. keeping a 4-fold (or even 3-fold) axis. I see then two cases, either we start from an $L^2$ eigenstate or an $L_z$ eigenstate.

The former will leave us with a degenerate state but not the latter, right?

One can actually write down this example pretty straightforwardly: The potential $V(\phi) = \alpha \cos(2n\pi)$ has $n$-fold discrete rotation symmetry. So we're basically considering the Schrodinger equation $-\psi''(\phi)+V(\phi)\psi = E\psi$ subject to $\psi(\phi)$ being $2\pi$-periodic.

exactly!

(Hi @Semiclassical)

hi

My original question was, what the expectation value of $l_z$ can become in non-radially symmetric cases

hmm

6:23 PM

and i was first wondering if a small perturbation to C1 symmetry would cause a big deviation from the original angular momentum quantum number.

I could see it being different for n even vs. n odd, but I don't have a clear sense either way

Emilio then brought up the point that one must obey deg. perturbation theory after all.

relieved :-)

yeah

so I got the two cases of either breaking the twofold degeneracy or not.

But still I do not see the full picture

I guess what I mostly notice is that the Lz-symmetry is a matter of the potential having parity symmetry or not

hence why even n vs. odd n could make a difference

6:26 PM

Well actually no

the twofold degeneracy in radially symmetric cases is maintained in any symmetries with rotational symmetry of order 3 or larger

But I see your idea of $\pm m$

Maybe lets start with a more basic thing. What happens if I prepare a state with $m=+1$ in the full radial symmetry?

Is that a proper possible eigenstate?

as in, with no potential?

yes (of $H$ I mean)

In that case the hamiltonian would be H=Lz^2, so yeah

So it is. Fine, then I perturb the system slightly to square shape.

What happens?

I don't know how you'd describe that. What I have in mind is that of a ring with a periodic potential applied to it

6:31 PM

Exactly like that

I don't see how a square shape corresponds to that. In particular I don't know what makes the corners different

adding some $V=\lambda sin(8\pi\phi)$

Well I meant just that there should be a $C_4$ or whatever symmetry with a fourfole axis

forget to corners

okay. so some kind of quadropole field

yep

I feel like that's actually a standard problem

Quadrupole ion trap

A quadrupole ion trap is a type of ion trap that uses dynamic electric fields to trap charged particles. They are also called radio frequency (RF) traps or Paul traps in honor of Wolfgang Paul, who invented the device and shared the Nobel Prize in Physics in 1989 for this work. It is used as a component of a mass spectrometer or a trapped ion quantum computer. == Overview == A charged particle, such as an atomic or molecular ion, feels a force from an electric field. It is not possible to create a static configuration of electric fields that traps the charged particle in all three directions (this...

6:35 PM

The gist of it would be?

the electric field is changing in that case, though. not totally sure that's the right analogous

But do you see my question? We start with an $m=+1$ state and perturb it such that the principle degeneracy between $m=\pm 1$ is not broken.

the usual way to deal with this quantitatively is to take $-\psi''(\phi)+\alpha \cos(4\phi)\psi=E\psi$ and expand $\psi(\phi)$ in a Fourier basis

Well the question is: What the expectation value of $l_z$ in that case.

writing $\psi(\phi)=\sum_{m=-\infty}^\infty c_m e^{i m\phi}$, I think one gets $m^2 c_m + \alpha c_{m-4}+\alpha c_{m+4}=Ec_m$

not totally sure if it's $\pm 4$ there

6:41 PM

OK.

note that c_0 is influenced by c_\pm 4, c_\pm 8, etc

whereas c_1 is only influenced by c_5,c_9,.., c_-3,c_-7,...

Hmmm

which in particular doesn't touch c_-1.

I feel like something is going wrong though

right the question is finally if the $c_{\pm m}$ terms cancel or not and when

my suspicion (based on some quick manipulations) is that the degeneracy between the even-m states would be lifted but not the odd-m states.

logic being that it seems like the equations for c_1 and c_-1, for instance, are identical (and so too for c_5,c_-5, etc) but with no dependence of c_1 on c_-1

so whatever solution holds for c_1,c_5,c_9,...,c_-3,c_-7, ... will also work for c_-1, c_-5, c_-9,...,c_3,c_7,...

The better approach, though, would be to show that there's still some symmetry operator which commutes with H=Lz^2+cos(4phi)

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