It’s like Qmechanic is a web crawler for the physics portions of the internet physics.stackexchange.com/questions/11878/…

wot

@NovaliumCompany isn't it amazing how much mileage a joke can get supplemented with a quick google? :P

Gave a bounty of 200 ! and still in the chat

Feel like a boss

Dear chat feel free to weigh in on the bountied question I used to have 500% more points :P

Sigh ... Is there any procedure to remove a spam answer?

Nvrmind flagged it ...

Used a flag for the first time

Also any many world-ers in the chat?

Shit ... I can't comment no more :((((

@bolbteppa I referenced a paper defending your arguments (but I'm not sure if you would be happy with the paper)??? Either way I'm sure u can post an answer and get +200 points :))) physics.stackexchange.com/questions/502221/…

@enumaris \o Sir Amune

But seriously I can't even downvote atm

@ACuriousMind Slight style question

If I discuss faraway gravitational influence from celestial bodies, can I write "We're at the mercy of the stars"

arxiv.org/pdf/1802.02661.pdf

And me thinking I was doing a new idea :V

Bloody hell

And it wasn't even published one year ago

Sad!

Fortunately it's an obscure enough topic for me to just go on with new develpments

"Strictly speaking, no. Since there is one and only one Sun in the entire universe, there is one and only one star in the universe that has sunspots. That's rather pedantic."

and here is astro me

Hi all, what is the magnetic vector potential $\vec{A}$ for the QM $n$-electron case?

I would be tempted to define it as a sum of one-electron vector potentials $$\vec{A}=\sum_{j=1}^n \vec{A}_j$$ with $$\nabla_j \times \vec{A}_j= \vec{B}$$

But in that case I miss the meaning of $\nabla$ in a $$ \nabla \times \vec{A}=\vec{B}$$ type definition.

@EmilioPisanty, any idea?

25 messages deleted

@ACuriousMind I wasn't using satire (atleast) the last bit of the messages

@SirCumference How often do I have to remind you that you shouldn't discuss suspended users in here unless it's relevant for some other reason?

I really think he has good intentions but is seriously under-estimating the value of university eduvcation

@MoreAnonymous I did not delete the messages because of anything specific they said, but because it is not fair to users to discuss them here when they cannot respond.

ah ... Ohk ... fair enough ... In the mood to talk about GR (or equivalence principle not in SR)?

Also I'm not sure if u read the controversial question of mine has a 200 point bounty

@Rudi_Birnbaum What is the "n-electron case"? You'll need to be a bit more specific what you're thinking of here

@MoreAnonymous As you might have noticed I have just logged on. Give me a few minutes :P

time independent QM problem of n-electrons

in some (thus static) potential

@ACuriousMind Is there hopeeee for the "controversial question" :P

An unknown yet "curious" hero may put the world outta it's misery (especially mine) :P

@ACuriousMind non-relativistic as well.

@Rudi_Birnbaum So if the potential is external and static, where do the $A_i$ come from? Why is it not a single $A$?

doesn`t each electron need its own, just like it needs its own momentum?

No, why would it?

You don't get n different potentials in the classical case of this either, do you?

I see an interaction: Is it standard QM, advanced QFT or a measurement .. NOooooo Its a curious mod interacting with a post :P

because each electron has its own coordinates $\vec{r}_j$

@Rudi_Birnbaum Sure, so there's a term $\frac{1}{2m}(p_i - A(r_i))$ in the Hamiltonian for each of them

So when I construct an Coulom potential $\frac{1}{2} \r_j \times \vec{B}$

@ACuriousMind thats exactly what I mean

But there is only a single potential function $A(r)$ to which you feed the different $r_i$. There aren't $n$ potentials $A_i$.

So the "whole" QM-vector potential is the sum I wrote above, right?

No

I don't know what your $A_i$ are

There's only one $A$.

what is your $A(r_i)$?

can you write it down?

"There's only one A." But the hamiltonian contains a sum terms that derive from that one potential, right?

Well, for a constant magnetic field $B$, the vector potential is $A(r) = \frac{1}{2}r\times B$, so it's $\frac{1}{2} r_i \times B$

for electron $i$.

@Rudi_Birnbaum Sure, the Hamiltonian contains a sum $\sum_i (p_i - q_i A(r_i))$.

right

Now $\sum_i p_i = p$

the total momentum, right?

(operator)

you can define that, yes

Is $A = \sum_j A_j$ anything worthwhile defining?

There is no analogous "total potential", since summing the $A(r_i)$ doesn't make any sense as they are the values of a single potential, just at different places.

I mean, you can sum them, I just don't see why you would.

There is something called mechanical momentum density, right?

I've never heard of it.

sorry I mean mechanical momentum.

vs momentum.

momentum is $\sum p_j$

Oh, I'd call that kinematic vs. canonical momentum

mechanical momentum is $sum p_j + A(r_j)/c$

is just a name (=dummy index)

so there is something called kinematic momentum, right

and there is something called momentum

$p_i$ is the canonical momentum of a single electron, $(p_i - q_i A(r_i))$ is the kinematic momentum, i.e. the momentum actually related to velocity by $P = mv$.

an when we look at the difference of the two for the n-electron case

Yes, the difference between kinematic and canonical momentum is $\sum_i q_i A(r_i)$.

But the canonical momentum doesn't really have a physical importance, it's just a variable in the Hamiltonian formalism, not something we directly measure, so this difference is not really of physical importance, either.

switching on and off the magnetic field?

The Noether charge for EM is the electric charge

As it is known

And there is an EM momentum and energy, of course

But they are not the bare $A$

Also $A$ has a gauge associated to it so its sum wouldn't mean much

@Rudi_Birnbaum what about it?

in that way you get physical importance, I suppose

@Slereah Indeed, the specific value of $A$ (summed or not) isn't even well-defined, so I'm confused what Rudi's trying to do here.

@Rudi_Birnbaum You certainly don't - because, as we just said, $A$ has a gauge freedom so that value isn't even uniquely defined. Gauge-variant quantities are not physical observables.

Sure you have to calculate expectation values, as its just operators we talked about, thats self-understanding

No, that's not what I mean

(Momentum of the EM field is the Poynting vector $E \times B$)

(which is gauge-invariant)

the meaningful expectation values will be gauge independent

I mean that the value of $A$ - if you take its expectation value or not - is meaningless to begin with

@ACuriousMind sure

$\int \Psi^* (p+A) \Psi$ ..

@ACuriousMind ... I'm not sure if it's u or another mod ... But if u read the comments which have come outta the "moved to chat" ... They alone seem to indicate that the question is "still unclear"? Umm... Why were those comments moved outta the chat?

Maybe the vzn one is okay because I make a direct reference to it in the post?

@Rudi_Birnbaum this quantity is gauge invariant

@MoreAnonymous Because they are the start of the conversation that was moved to chat and therefore an indicated for other users what they're going to find in the chat.

since the gauge of the EM field is associated with a gauge of the matter fields

Ah .. .ohk .... Just checking ... Also what are the guidelines about answers being relevant to the comment section when the comment section has been moved to a chat?

@Slereah sure this is why I used it as an example to explain what I mean.

I mean I really try to take every comment and improve the question ... I swear

I've never edited a question this much before ...

You know I always wonder

What's the topology of the real EM configuration space?

@Slereah whats "fake EM"?

@MoreAnonymous up to gauge

@Slereah like how many holes does it have :-)

The EM field as it is is a trivial bundle $\mathbb{R}^4 \times \mathbb{R}^4$

What happens when you quotient out the $U(1)$ gauge

@MoreAnonymous What do you mean "answers being relevant to the comment section"?

Answers should answer the question, and nothing else.

You can reduce the bundle for point particles at least

From the $(n+1)$-dimensional bundle over $\mathbb{R}$ to an $n$-dimensional one

'course it's a bit of a different case

@MoreAnonymous Yes, an unclear question should not be answered until it has been edited to be clear.

In fact, an unclear question should be closed.

Hmmm ... Okay since mine has not been closed yet ... Maybe there's hope?

An unclear question should be DESTROYED

ANNIHILATED

Does the bounty get paused if closed (on vote to close due to lack of clarity)

@Slereah not even ur freaking singularity in GR can destroy information ... The tales of UNCLEAR questions will live on forever .. MUHAHAHA

I mean is it any different from a Coulomb potential term like say from an atomic nucleus. In sense each electron feels it and thus you have one term $(r_j-r_{nuc})^{-1}$ in the Hamiltonian. In way its for all electrons "the same" potential but then the potential energy for all electrons is a sum of these terms.

Singularities can very much destroy informations

QM in GR isn't unitary if it's not globally hyperbolic

@ACuriousMind I was under the impression I had sufficiently edited the question to address the concerns .. I swear it wasn't intentional

@Slereah Sadly, SE was not willing to fund my orbital laser array

@ACuriousMind A length of lead pipe will do

@ACuriousMind Maybe we can have a system where if u flag unclear you can atleast extend the time period to put a bounty? (that might help) ???

Can I participate in the meta of "We're currently discussing whether to refund the bounty so the review can proceed or not."???

Link?

@MoreAnonymous I know it wasn't intentional - there is no indication for <1k rep users that their question has pending close votes, and there is no mechanism that prevents offering a bounty on a question with pending reviews. SE does not seem to be interested in fixing this, as it has been that way for years (cf. meta.stackexchange.com/q/188916/263383 and linked questions)

@Slereah The laser array would have had the advantage of not having to get out of my chair.

@ACuriousMind very long pipe

Although the curvature of the earth might pose a problem then

"SE does not seem to be interested in fixing this" ... Im a victim of the system I tell u! :P

@Slereah Just ram the pipe through the mantle?

@ACuriousMind lead doesn't do well with magma

Also is there a nice way to just count number of upvotes versus downvotes?

On a question?

Click on the score

it will split into the up and down

That's a 1k rep privilege :P

Oops

Forgot to check my privilege

I feel the poor always pay the price in war! :P

But I suspect they are poor in the "war of the ideas" for a reason :P

Also is it possible to see who has flagged my post before all flags happen .. .Maybe they'd be willing to help?

@MoreAnonymous No, it is very much intentionally not possible to see that. If the close voter wants to help you clarify your question, they can always leave a comment.

Hmm ... okay ... I'm doomed

@PM2Ring @PM2Ring maybe we should have waited for the end of the universe before putting a bounty on this one :/

Science fashion

Its a secret episode

Please

@ACuriousMind: What bothers me if we say its the value of the potential $\vec{A}$ at position $r_j$ is that then $\vec{A}$ becomes a function from $\Bbb R^{3 n}\to\Bbb R^3$. But in this instance I miss the defining equation $\nabla \vec{A}=\vec{B}$

there's too much antivaxxer jokes around

I'm so sick of it I'm rooting for rubella

@Slereah my point was she doesnt wear anything in your science fashion wardrobe in this episode

Point proven I am the science faashionn police

@Rudi_Birnbaum No, $A$ is still a vector field on $\mathbb{R}^3$. Just because I plug $n$ different values into a function doesn't turn it into a different function!

What are you talking about

She's wearing the planet episode dress

Literally the first one

When I do $\nabla \vec{A}(r_j)$ I get 0, I have actually to do $\nabla_j \vec{A}(r_j)$ to get $\vec{B}$. See what I mean?

I think you're a bit confused about how functions work.

sorry I mean $\times$

$\vec{A}$ is defined to be something that gives $\nabla \times \vec{A}=\vec{B}$ for a given mag. field B.

right?

The definition is that the curl of $A$ is $B$, yes.

corrected.

Note that your equation doesn't give the dependent variable of $A$ a name. It doesn't matter whether you call the argument to $A$ $r$, $r_2$, $r_i$ or anything else - the curl is still the curl.

I cannot bring that definition in agreement with a $\nabla\times\vec{A}(r_j)=B$

I rather need $\nabla_j$.

@ACuriousMind hmm

@ACuriousMind not sure if I can understand that

If you give the dependent variable a name (say $r$), then you have to write it everywhere, i.e. $(\nabla_{r} \times A)(r) = B(r)$

@ACuriousMind yes and I thought the default is (x,y,z)=r

This is the same equation as $\nabla\times A = B$, just written with the variable named. You're now free to give it a different name, but the equation doesn't change.

such that nabla refers to that default

otherwise these equations contain more implicit convention than I thought.

There is no "default" when you write $\nabla \times A = B$. The variable simply isn't named there at all.

@ACuriousMind it sounds a bit like the "good old transformation problem". Where one must decide if you want to keep the object or the algebraic expression fixed.

both work

but only if not confused half-way

@ACuriousMind I gotta go ... Talk GR another day?

usually nabla is defined as (d/dx, d/dy, d/dz)^T

and r is defined as (x,y,z)^T.

That's a very...concrete view of things. I'd say that $\nabla\times$ - the curl operator - is not defined in terms of any concrete variable names at all - derivatives are abstract notions that don't rely on particular names.

Take a function $f$ of a single real variable. I can talk about its derivative $f'$ without specifying what I want to call its input variable at all!

@ACuriousMind I see.

And there is no appreciable difference between then writing $f'(y)$, $f'(x)$ or $f'(a)$ or anything else

I.e. the derivative $()'$ doesn't refer to any name "by default". $\nabla \times$ is the same

It seems to me this is the "Physics decision" (I am no Physicsist) is to give the objects a meaning and not the expressions. OK.

So in a sense the terms get their definition in the equation by the use.

You want the curl of A to be B so the variables are implied to be such that this is fulfilled.

right

?

(the safe option here is to say you don't know what I mean .. lol)

I...wouldn't phrase it like that. As I said, I don't think there are any variables "implied" by writing $\nabla \times A = B$. "The curl of A is B" is a statement that doesn't need to refer to any variables at all.

Yes. My actual problem is to write down and understand the expressions for the n-electron probability current density.

And that involves writing down the Hamiltonian of the n-electrons in the mag. field.

But it also involves the definition of the n-electron probability current.

And there you have "both" cases of A, for each single electron and one for the effective one-particle density ...

especially the latter one confuses me.

First day at school, tommorow... SUUUUUUUU

Please lemme know if you think it's sufficient?

C'mon chat help your friend?

@ACuriousMind FWIW, I did mention to @MoreAnonymous that there were 3 pending "Unclear" votes on that question. I also advised him to wait & see before doing anything drastic, and that he couldn't aford to spend a lot of rep for a bounty. OTOH, I did also mention that bountied questions cannot be closed...

@PM2Ring I wasn't accusing you of anything ...

@MoreAnonymous I wasn't accusing you of accusing me of anything. ;) But you didn't take my advice to wait, and went ahead & posted a bounty that stripped you of useful privileges.

But also said "t looks ok to me, but I don't have enough expertise in quantum physics to respond to it adequately. – PM 2Ring 2 days ago"

over here: physics.meta.stackexchange.com/questions/12430/…

That with wait (to which I said 2 days) ... I was interpreted bounty time = today

FWIW, that question is so huge that many people won't bother to even read it, what to speak of contemplating creating an adequate answer.

Also I was forced to make it soooo huge

@MoreAnonymous Yes, I said to wait for some positive feedback / response, I didn't specify an exact time frame. But I guess it's up to you to decide if the response has been sufficiently positive.

Hmm ... well the max it went upto was $4$ upvotes .. and then $3$ and I;m under the impression those who mention this question in chat are considered some kind of dark lord worshippers ??

No one forced you to make your question an incomprehensible jumble of quotes (and not-quotes, which are nevertheless put into block quotes for no apparent reason).

@ACuriousMind what am I supposed to do when people downvote and don't tell me the reason ...

Do I take them seriously ?? Do I guess that its unclear due to the flags raised?

@MoreAnonymous There's no simple correlation between up / down votes, and close votes. About all you can say is that someone who close-votes a question won't also upvote it.

probably won't upvote it. I've seen counterexamples :P

If u see the chat I take everyone quite seriously ... In fact lemme know if I have done anyone injustice

@ACuriousMind Fair enough. I guess I've upvoted questions & also close-voted as dupe, if the question is good in its own right, and adequate dupe targets are hard to find if you don't already know the answer, or the magic keywords.

One might also upvote a question, then vote to migrate it

True

@PM2Ring great now I have to edit my question so that this doesnt happen to me ....

My real problem is how do I find a balance between "incomprehensible jumble of quotes " and a condensed version

@MoreAnonymous Well, those comments have been preserved, by being moved into a chat room. But be aware that chat rooms get frozen after 10 days or so of inactivity: people will still be able to read it, but not post to it.

@PM2Ring sure ... I've never been backed into a corner so badly before that the material I have to provide to make a question self contained goes up so high ... I mean at what point is it okay to link it off (which for way $4$ Im still doing)

I feel i had made some of my points clear that was obviously not clear

@MoreAnonymous I agree, it's not easy. And it's hard for people to advise you what to do when they aren't clear on what it is that you are asking, and aren't clear on what you misunderstand.

@PM2Ring I dont think thered be a quesiton if I knew what I was misunderstanding ...

@ACuriousMind Is there any existing method to award say +50 points to someone who edited a question?? I'd really love to incentive that one

@MoreAnonymous No

Hmm ... I'm guessing there's a relevant meta where someone asked for this before and the answer was no?

@MoreAnonymous I'm not making any claims about the actual topic. I'm only discussing the "meta" aspects: eg, that the question is too large and complicated for most readers. If it were a programming question on Stack Overflow, I'd advise you to try & break it up your problem into a bunch of sub-problems, and to ask those sub-problems in separate questions. That may not be so applicable here. But if you can figure out a way to do that, you're more likely to get useful responses.

@MoreAnonymous I don't know of any specific post, but in general SE does not want people to gain much rep with anything but questions and answers.

You gain +2 rep for every edit you propose that is accepted but that stops at 2k, not coincidentally below the 3k threshold where users gain close votes.

@MoreAnonymous: "the derivation makes use of the Heisenberg picture" what do you mean here?

@Rudi_Birnbaum I accidently use the wrong word its in the chat: Heisenberg equation of motion is what I meant to say

@MoreAnonymous so what should be the problem with that?

Also in the post he was referring to it was always Heisenberg equations of motion ...

@Rudi_Birnbaum the objection section is the relevant part of the post u want

If ur okay with it I'm willing to copy paste

I am reading it

dont paste it

What do you mean, by rely on unitarity? Unitarity of what?

@PM2Ring Maybe I should ask something (I think) I know the answer to ... Something along the lines of: We know the measurement and often associate it with the Schrodinger picture .. How does it look in the Heisenberg picture?? (If this is possible at all)

@Rudi_Birnbaum Some lines later in the same section: I am not aware of a derivation of Heisenberg's equation of motion which is applicable during a non-unitary process (for example the measurement).

This should give you an idea about what I'm talking about

The measurement is not a quantum mechanical process. Its the interaction of microscopic (=quantum) objects with macroscopic (=classic) objects.

Its not clear why that what you describe should exist.

@Rudi_Birnbaum depends which interpretation ur onboard with .. .Either way maybe u should just skip along to the section "The fourth way out "

@Rudi_Birnbaum Maybe you should also see Basic Premise before that section? (Its quite small)

@MoreAnonymous hard to imagine. All conclusions of "alternative interpretations of QM" are afaik completely equivalent. So nothing in predictions should depend on the QM-interpretation.

@Rudi_Birnbaum yes that's what I was under the impression of too but somehow this question seems to challenge that ... either way go ahead to "More about way 4"

@Rudi_Birnbaum Also it could also be the case in different interpretations you have to do more derivations ...

To deal with more systems .. That too would be really interesting

@MoreAnonymous one should alway use Landaus interpretation, of course (in my humble op).

@MoreAnonymous Its exactly Landaus interpretation.

You cite it yourself

"A measurement is an interaction between a measuring apparatus which is (quasi-)classical and a quantum system"

Ah ... quasi-classical means semi-classical not necessarily macroscopic

Well thats how I would see it.

Still its no QM process

the modelling is quasi-classical but NOT QM

To be fair if you click one of the links:

quasi-classical

The link is this one: johnboccio.com/research/quantum/notes/bell.pdf

Also the question says: Copenhagen interpretation (+ “quasi-classical apparatus”)

Copenhagen interpretation does in some sense allow for the measuring apparatus to be special

There are interpretations where they just dont go to the apparatus

A consideration is pure QM iff it can be done using QM axioms alone.

So when you go about modelling the measurement I think by definition you exceed QM.

But either way I was hoping to accept a heuristic answer

Where the person says you know this derivation also implies u can't take $\delta t \to 0$

between subsequent energy measurments of the same system

when u combine it with Copenhagen interpretation (+ “quasi-classical apparatus”)

@MoreAnonymous It seems that is not in contradiction to what I wrote

?

@MoreAnonymous Yes there is no contradiction. This paper just says that

measurement postulates x,y,z can be deduced from postulates a,b,c

So that x,y,z are no postulates anymore.

yes but a,b,c are the same postulates describing unitary physics ...