The h Bar

bolbteppa

12:10 AM

@vzn motls.blogspot.com/2018/10/… '3 page susy proof of RH'

One thing I can say is Das has written about 8 textbooks and they are pretty much awesome

Seems like the proof has issues with plane waves being normalized in the continuous spectrum vs the discrete spectrum

danielunderwood

12:53 AM

Can there ever be nonlinear operators in QM?

Al Nejati

5:35 AM

this guy has been posting a lot of relatively nonsense questions physics.stackexchange.com/users/209257/andassi

I googled some of the things he wrote and they seem to be largely copy-pasted from here kah-len.com/adrieiuous

It's some kind of new age cult thing

Avantgarde

@danielunderwood You mean something like $\hat{O} (| \alpha \rangle + | \beta \rangle) \neq \hat{O} | \alpha \rangle + \hat{O} | \beta \rangle$?

danielunderwood

@Avantgarde yeah or I suppose even $\hat{O}(c \lvert \alpha \rangle ) \ne c \hat{O} \lvert \alpha \rangle$

I'm pretty sure either would break the normal ways of doing anything though

Avantgarde

@danielunderwood That you can have, sure. Anti-linear operators act like this: $\hat{O} (c | \alpha \rangle) = c^* \hat{O}| \alpha \rangle$

The coefficients in front of the kets get complex-conjugated.

danielunderwood

5:51 AM

Ahh I don't think I've run into any of those. Are there common examples? And maybe a similar condition, but no systematic way to relate them

Avantgarde

6:24 AM

@danielunderwood Time reversal operator is anti-unitary

(and anti-linear)

Al Nejati

7:41 AM

Is it just me or are the quality of questions today abnormally low

John Rennie

7:57 AM

@AlNejati we sometimes get runs of abysmal questions. I doubt there's any significance to this. I'd guess it's just the sort of clustering you see in randomly distributed events.

Slereah

hello

you can have non-linear operators on Hilbert spaces, but they're rarely considered in QM

Well, except for the norm, I guess

The norm is nonlinear

$$| \psi + \xi | = |\psi| + |\xi| + \langle \psi, \xi \rangle + \langle \xi, \psi \rangle$$

and it's not with values in the Hilbert space

I recall that there's some non-linear symmetry operators that are sometimes used

Avantgarde

This is an interesting question.

Slereah

I also vaguely recall people considering nonlinear Schrodinger equations for QM

But I think that never went anywhere

Avantgarde

Yeah, I've come across that.

Slereah

Like I think people were hoping nonlinear QM would help with collapse

Avantgarde

8:11 AM

Nonlinear Schrödinger equation

In theoretical physics, the (one-dimensional) nonlinear Schrödinger equation (NLSE) is a nonlinear variation of the Schrödinger equation. It is a classical field equation whose principal applications are to the propagation of light in nonlinear optical fibers and planar waveguides and to Bose-Einstein condensates confined to highly anisotropic cigar-shaped traps, in the mean-field regime. Additionally, the equation appears in the studies of small-amplitude gravity waves on the surface of deep inviscid (zero-viscosity) water; the Langmuir waves in hot plasmas; the propagation of plane-diffracted...

I've never used non linear Schrodinger equation, but it seems to have some applications

Slereah

yeah

Avantgarde

The equation has a cubic term in $\Psi$.

But at the level of states, Hilbert spaces, etc. is it possible/realized anywhere?

Slereah

Well it can have a whatever term

@Avantgarde By definition observables are linear operators

So you'd have to change QM to make it well defined

Avantgarde

Yeah, so I'm asking if any such thing has been attempted?

Slereah

as I said, yeah

Not sure what properties it has, exactly

Secret

8:15 AM

journals.aps.org/prl/abstract/10.1103/PhysRevLett.119.110401

Avantgarde

I'd be happy to have a look at them, if you recall the references.

Secret

can't read which 3 articles have cited this thus not sure the most recent progress of that nonthermal noise which is suggested to be the collapse field

Slereah

don't have any paper in mind

Secret

researchgate.net/publication/…

Anonymous

8:42 AM

@Avantgarde I have a small doubt :P. From the article it seems that the $\Psi$ in the nonlinear SE has a very different meaning from what we call $\Psi$ in QM (i.e. the quantum state) isn't it? It says that it is a classical field equation and has applications in while studying fibre optics and water waves. It doesn't seem like it is incompatible with QM (also it says that "Unlike its linear counterpart, it never describes the time evolution of a quantum state" - the Hamiltonian)

Anonymous

Am I missing something?

Anonymous

I'm not sure non-linear SE has anything to do with non-linear "QM"

Secret

12:25 PM

Computing a Universe Simulation | Space Time

►

How to know the universe is not a simulation:

Check if every matter and energy have something in common

If everything can be made up by a bunch of objects of which there are only finitely many kinds of it, then the universe can be simulated as data

One way to envision a universe that is not a simulation is when not everything is reductive other than its existence

CooperCape

3:21 PM

If I'm given only the frequency of a photon emitted from the transition of an electron in a hydrogen atom, is there any clean way of finding out what the transition was?

Anonymous

@CooperCape Umm, estimate the frequency range it lies in - uv, visible, infrared, etc and then try to compare with the hydrogen series?

Anonymous

There are two variables involved - $n_1$ and $n_2$...so brute force solving is going to be a bit difficult by hand

CooperCape

@Blue Yeah I was looking here on wikipedia but it stops at $n'=6$ at a wavelength a lot shorter than mine. Might start at $n'=10$ and see what happens but all the question says is identify the transition

Novalium Company

@Blue Hi, I have a quick question for school: In the $E = \frac{F}{q}$, where q is the charge of the other particle, not the one we get the electric field for, right? And if we wanted to find the electric field of a particle, without knowing the charge of the other particle we use $E = \frac{kQ}{(r)^2}$, right?

Anonymous

Hmm, then brute force seems the only way. Try plugging in suitable integer values :P

Anonymous

3:26 PM

But this doesn't sound like a good question

CooperCape

@Blue part i we have to derive an approximation for $\nu$ for high levels of $n$ with small transitions $\Delta n$, and part ii says here is a frequency, find the transition and show that the approximation is not accurate enough for precise identification

Lozansky

@NovaliumCompany Looks correct

Novalium Company

Alright thanks.

CooperCape

So possibly could have to use the approximation and show that it doesn't give integer values of $n$ or maybe the $n$ value doesn't quite give the frequency back

Lozansky

Electric fields follow the superposition principle

Anonymous

3:29 PM

@NovaliumCompany Firstly, electric field of a particle is always independent of any other particle (you determine it by using a test charge)

Anonymous

Secondly, I think you misunderstood the first part

Anonymous

Lemme explain

Anonymous

3

A: How can the accurate value of electric field intensity be calculated?

The electric field intensity is usually defined in introductory textbooks as the limit $$\lim_{q\rightarrow0}\frac{\mathbf F}{q},$$ i.e. the force on a test charge, per unit charge, when that test charge goes to zero. However, you are right in noting that the limit $q\rightarrow 0 $ is impossible...

Anonymous

Okay, there's already an answer ^

Anonymous

The point to remember while using the formula $F/q$ is that if there is another particle having a charge $q$, it will produce it's own electric field too. So you cannot take a measurement of the first particle's charge correctly

Anonymous

3:35 PM

Which is why it is important to write the limit $q\to 0$ (at least while you're studying classical electrostatics) so that theoretically the other test charge's electric field won't affect your measurement. However, in reality, charges are quantized and so you cannot actually approach that limit

Anonymous

So yes, the formula you wrote for $E$ is correct. But I hope you got the reasoning now

Lozansky

What kind of definition is that?

Anonymous

@Lozansky Which one?

Lozansky

The limit one

Anonymous

That's the standard definition of electric field...

Novalium Company

3:38 PM

@Blue Thanks, got it.

Lozansky

I've only seen $\mathbf{E} = \dfrac{1}{4\pi \epsilon_0} \int \dfrac{\rho(\mathbf{r}')}{R^2} \hat{R} d\tau'$

Anonymous

@Lozansky Sure. They're equivalent

Lozansky

Seems circular

Anonymous

@Lozansky I don't see how

Anonymous

Emilio's answer covers it perfectly

CooperCape

3:42 PM

Okay I really don't get this q now. We have to show that $\nu\approx\frac{2Rc}{n^3}\Delta n$ which I did, then it says a radio telescope detects a hydrogen emission line with frequency $4877$MHz (seems really small). Then we're asked to identify the transition and show that the approximation isn't accurate enough for precise identification.

Novalium Company

@Blue "The point to remember while using the formula F/q is that if there is another particle having a charge q, it will produce it's own electric field too. So you cannot take a measurement of the first particle's charge correctly" - That kinda confuses me.

CooperCape

I'm just not sure what to do at this point

Semiclassical

@CooperCape well, what do you get for $\Delta n/n^3$ in that case?

Anonymous

@NovaliumCompany Basically, charged particles can have self-interactions with their own field

CooperCape

$\nu / 2Rc$ which is roughly $7.4$x$10^{-7}$

Anonymous

3:45 PM

Self-interactions are a bit complicated

CooperCape

But I don't see how that helps me?

Semiclassical

hmm

Novalium Company

@Blue Ok. I gotta go now, thanks for the help, although I didn't really understand what you were trying to point out.

Semiclassical

I guess one can start playing around with cases. If $\Delta n=1$, then this gives $n^3\approx 1/7.4\times 10^{-7}\implies n\approx 110.558.$

And I guess the problem is that this about midway between $n=110$ and $n=111$?

CooperCape

Ahh I see

Do you think I should only consider the $\Delta n = 1$ case then?

I could potentially justify that with $\Delta n << n $ being in the question statement for part 1

Semiclassical

3:49 PM

Well, it might be worth checking higher values to see what changes

Anonymous

@NovaliumCompany Tbh, I think I'll need to read up a bit about self-interactions myself

CooperCape

$\Delta n=2$ gave me $221.12$

Semiclassical

for $\Delta n=2$ it looks to instead be 87.7

CooperCape

Oh oops

Anonymous

I'm not aware of the details

CooperCape

3:50 PM

of course I did that wrongly

Semiclassical

no, you're right

n^3/Delta n is constant in this problem

Novalium Company

@Blue Ok, how is uni going?

Anonymous

11

Q: Influence of charged particle's own electric field on itself

I read this in my textbook: A charged particle or object is not affected by its own electric field. Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an ...

Semiclassical

so if you increase Delta n then n should go up

Anonymous

Maybe this covers it a bit ^

CooperCape

3:51 PM

Yeah but only to 139.29

I did the calculation wrong

Semiclassical

ah, so we were both wrong

CooperCape

Yep

I guess I might write out the first few $\Delta n$ and show that none approach a reasonable integer value to certainly state that it was the transition detected

Semiclassical

I think one does have to make some assumption about $\Delta n$

for instance, when $\Delta n=5$ you get $n\approx 188.95$

Novalium Company

@Blue No, I mean, I didn't understand what you were trying to point out at all. I understood that the electric field is the limit as the charge goes to 0... but that's not really possible because, quantization and the charge can't go below e, so we use some other weird formula?

CooperCape

@Semiclassical Is that justifiably close? I'm still kinda stuck on the fact it says "identify the transition"

Semiclassical

3:54 PM

and it's not really surprising that, if you try enough values of $\Delta n$ in $n\approx 110.56(\Delta n)^{1/3}$, you'll eventually land near an integer

hmm

yeah, I dunno what they intend you to do there

Anonymous

@NovaliumCompany To determine the electric field at a point in space you need to place a test charge at that point and find the force on it (acceleration). Now, if during accelerating since the test charge $q$ has its own electric field, it can exert a force on itself too! So the net force on the test charge would become $\vec{F}_{\text{BY the particle whose electric field you want to measure ON the test charge}}+ \vec{F}_{\text{DUE to the test charge on itself}}$.

Anonymous

That extra second force will cause an error in your measurement of the electric field.

Anonymous

To make that self-force zero, you take $q \to 0$

Semiclassical

I guess you could also note that when $\Delta n=8$ you get $n\approx 220.994$

Novalium Company

@Blue I got it, thanks, but those are quite advanced. Why do we have to introduce time?

danielunderwood

3:57 PM

@Avantgarde that looks like a nonlinear extension to the position-basis equation. Nonlinear operators wouldn't necessarily imply such a form would it?

CooperCape

I might just stick a few small values of $\Delta n$ down and identify one's close to an integer and write a small note. Thanks for your help

Semiclassical

so both $n=189\to n=184$ and $n=221\to n=213$ should give approximately that same frequency

yeah, I'm really not sure what they expect you to do

or would it be 189->194? I'm getting myself mixed up

Anonymous

@NovaliumCompany Mathematically you can take the limit $q\to 0$

Anonymous

That's how they get the other formula from Coulumb's force

Anonymous

There's no experimental/physical aspect involved in that one

Anonymous

4:01 PM

$\lim_{q_2\to 0}\frac{k q_1 q_2}{r^2 q_2} = \frac{k q_1}{r^2} $

Anonymous

After that they're defining electric field using $kq_1/r^2$ instead of the other way round i.e. placing a test charge method and finding force on it method

Novalium Company

@Blue Ok, thanks but let's stop here. It's getting really messy, for some reason my mind won't concentrate. See you :)

Anonymous

No problem. I guess someone else can give you a better overview of the self-interaction thing. I never actually took a formal electrodynamics course

Anonymous

4:17 PM

Anyway, I think for your school physics you'll just need to state that you take $q\to 0$ because the test charge will otherwise alter the external electric field (this is in the context when you're trying to find the electric field due to an extended/many-particle body)

Avantgarde

@Blue I don't know. I didn't read the full article. But yeah, seems like it's just a Schrodinger-like equation.

@danielunderwood I'm not sure. Antilinear operators are the only non-linear operators I know.

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