The h Bar

danielunderwood

00:22

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

enumaris

03:11

wot

More Anonymous

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

@ACuriousMind slept and refreshed ... going back to https://chat.stackexchange.com/transcript/message/51706662#51706662

But let's say I'm in flat space time ... Both GR and SR give me the same answer?
If so what is the difference for an accelerated frame seeing a null geodesic (light) curve? (according to both theories?)

@PM2Ring I'm gonna be giving a massive bounty on: https://physics.stackexchange.com/questions/502221/does-the-copenhagen-interpretation-quasi-classical-apparatus-allow-one-to

(as we agreed) ... Dear all not sure if I'll be able to participate in chat after that ... :/

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

More Anonymous

03:59

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 :((((

More Anonymous

06:58

@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/…

Novalium Company

07:36

@enumaris \o Sir Amune

More Anonymous

Nooooo ... I got a downvote :((((
19 more downvotes and I wont be able to chat:P

But seriously I can't even downvote atm

Slereah

07:59

@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

Sir Cumference

08:27

-6

A: Can we see evidence of sunspots on other stars

So far, no, but Hubble has taken photos of the red supergiant Betelgeuse in Orion which show markings in the form of bands, and bright spots which might be flares, but no sunspots. A state-of-the-art super-telescope in the Chilean Andes has also taken photos of Betelgeuse which compare favourably...

Slereah

"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

Rudi_Birnbaum

09:30

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?

More Anonymous

10:02

I bet he didn't win another Nobel after that I suspect alcohol might be the reason :P

History is on my side!!

ACuriousMind

25 messages deleted

More Anonymous

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

ACuriousMind

@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?

More Anonymous

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

ACuriousMind

@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.

More Anonymous

10:10

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

ACuriousMind

@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

Rudi_Birnbaum

time independent QM problem of n-electrons

in some (thus static) potential

More Anonymous

@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

Rudi_Birnbaum

@ACuriousMind non-relativistic as well.

ACuriousMind

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

Rudi_Birnbaum

10:17

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

ACuriousMind

No, why would it?

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

More Anonymous

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

Rudi_Birnbaum

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

ACuriousMind

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

Rudi_Birnbaum

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

@ACuriousMind thats exactly what I mean

ACuriousMind

10:20

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

Rudi_Birnbaum

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

ACuriousMind

No

I don't know what your $A_i$ are

There's only one $A$.

Rudi_Birnbaum

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?

ACuriousMind

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

Rudi_Birnbaum

for electron $i$.

ACuriousMind

10:24

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

Rudi_Birnbaum

right

Now $\sum_i p_i = p$

the total momentum, right?

(operator)

ACuriousMind

you can define that, yes

Rudi_Birnbaum

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

ACuriousMind

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.

Rudi_Birnbaum

There is something called mechanical momentum density, right?

ACuriousMind

10:28

I've never heard of it.

Rudi_Birnbaum

sorry I mean mechanical momentum.

vs momentum.

momentum is $\sum p_j$

ACuriousMind

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

Rudi_Birnbaum

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

ACuriousMind

$p_i$ 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 $P = mv$.

Rudi_Birnbaum

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

ACuriousMind

10:31

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.

Rudi_Birnbaum

switching on and off the magnetic field?

Slereah

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

ACuriousMind

@Rudi_Birnbaum what about it?

Rudi_Birnbaum

in that way you get physical importance, I suppose

ACuriousMind

@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.

Rudi_Birnbaum

10:39

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

ACuriousMind

No, that's not what I mean

Slereah

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

(which is gauge-invariant)

Rudi_Birnbaum

the meaningful expectation values will be gauge independent

ACuriousMind

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

Rudi_Birnbaum

@ACuriousMind sure

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

More Anonymous

10:41

@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?

Slereah

@Rudi_Birnbaum this quantity is gauge invariant

ACuriousMind

@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.

Slereah

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

More Anonymous

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?

Rudi_Birnbaum

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

More Anonymous

10:44

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

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

Slereah

You know I always wonder

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

More Anonymous

@Slereah whats "fake EM"?

Slereah

@MoreAnonymous up to gauge

Rudi_Birnbaum

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

Slereah

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

ACuriousMind

10:46

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

Answers should answer the question, and nothing else.

Slereah

You can reduce the bundle for point particles at least

More Anonymous

"Answers should answer the question, and nothing else."

Wait so if it's an unclear question (which I hope mine isn't but under the impression it is) ...
Then the answer not address the questioner's intended question ... even if that reflects in the comment section (I hope mine doesn't require that ... but who knows)

Slereah

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

'course it's a bit of a different case

ACuriousMind

@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.

More Anonymous

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

Slereah

10:53

An unclear question should be DESTROYED

ANNIHILATED

More Anonymous

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

Rudi_Birnbaum

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.

ACuriousMind

@MoreAnonymous You interrupted a pending close review by offering the bounty (bountied questions cannot be closed, which is a double-edged sword). We're currently discussing whether to refund the bounty so the review can proceed or not.

Slereah

Singularities can very much destroy informations

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

More Anonymous

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

ACuriousMind

10:56

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

Slereah

@ACuriousMind A length of lead pipe will do

More Anonymous

@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?

ACuriousMind

@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.

Slereah

@ACuriousMind very long pipe

Although the curvature of the earth might pose a problem then

More Anonymous

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

ACuriousMind

11:02

@Slereah Just ram the pipe through the mantle?

Slereah

@ACuriousMind lead doesn't do well with magma

More Anonymous

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

On a question?

Slereah

Click on the score

it will split into the up and down

ACuriousMind

That's a 1k rep privilege :P

Slereah

Oops

Forgot to check my privilege

More Anonymous

11:04

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

More Anonymous

11:20

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

ACuriousMind

@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.

More Anonymous

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 :/

Slereah

Science fashion

More Anonymous

11:37

@Slereah I bet you don't know about

https://www.youtube.com/watch?v=nDjz5qHIzsc

Its a secret episode

Slereah

Please

Rudi_Birnbaum

@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 $\nabla \vec{A}=\vec{B}$

Slereah

there's too much antivaxxer jokes around

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

More Anonymous

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

Point proven I am the science faashionn police

ACuriousMind

@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!

Slereah

11:39

What are you talking about

She's wearing the planet episode dress

Literally the first one

Rudi_Birnbaum

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?

ACuriousMind

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

Rudi_Birnbaum

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?

ACuriousMind

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

Rudi_Birnbaum

corrected.

ACuriousMind

11:44

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.

Rudi_Birnbaum

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

ACuriousMind

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

Rudi_Birnbaum

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

ACuriousMind

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.

Rudi_Birnbaum

such that nabla refers to that default

otherwise these equations contain more implicit convention than I thought.

ACuriousMind

11:48

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

Rudi_Birnbaum

@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

More Anonymous

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

Rudi_Birnbaum

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

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

ACuriousMind

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!

Rudi_Birnbaum

@ACuriousMind I see.

ACuriousMind

11:53

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

Rudi_Birnbaum

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)

ACuriousMind

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.

Rudi_Birnbaum

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.

Novalium Company

12:31

First day at school, tommorow... SUUUUUUUU

More Anonymous

13:41

I'm begging anyone to tell me to "stop"!

I really gotta stop editing this post:

https://physics.stackexchange.com/questions/502221/does-the-copenhagen-interpretation-quasi-classical-apparatus-allow-one-to

Please lemme know if you think it's sufficient?

C'mon chat help your friend?

PM 2Ring

@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...

More Anonymous

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

PM 2Ring

@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.

More Anonymous

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

PM 2Ring

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.

More Anonymous

13:48

There is some promise if that's what u think wittingly mentioned in the beginning:

Note: The below is a labyrinth of positions one can take on this argument. And I feel there are 2 ways out (way 3 or way 4)of this maze. I'm asking for which way is better? (See Questions for what I mean by "better")

Also I was forced to make it soooo huge

And I end it with

P.S: I feel this post was self-contained but to do justice to way 4 I had to link it off. Please do not hesitate to contact me if you feel this question can be improved? Also vzn I feel your comment is beyond the scope of this question.

PM 2Ring

@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.

More Anonymous

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 ??

ACuriousMind

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).

More Anonymous

@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?

PM 2Ring

@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.

ACuriousMind

13:55

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

More Anonymous

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

Discussion on question by More Anonym…

Imported from a comment discussion on physics.stackexchange.co...

@PM2Ring or @ACuriousMind

Lemme know which sections or sentences are redundant?

PM 2Ring

@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.

ACuriousMind

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

PM 2Ring

True

@MoreAnonymous A little while ago you asked about answers taking into account the comments on a question. On the Stack Exchange network,
answerers are under absolutely no obligation to read *any* of the comments. The OP is expected to respond to comments that request clarification by editing the question itself. Comments are for temporary conversations that help you get the question itself into shape. They are sometimes called "second-class citizens", and compared to Post-It notes. And they may be deleted at any time.

More Anonymous

@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

PM 2Ring

14:11

@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.

More Anonymous

@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

PM 2Ring

@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.

More Anonymous

This is one of my comments:

o the derivation makes use of the Heisenberg picture (which in turn makes use of unitary - see wiki derivation). During the measurement some people say it undergoes "non-unitary" evolution. - I feel I have repeated the "Objection" section. Maybe I should ask why is the objection unclear?

its a response to this:

So are you claiming that the calculations in this answer are somehow invalid when measurement is taken into consideration?

@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

ACuriousMind

@MoreAnonymous No

More Anonymous

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

PM 2Ring

14:23

@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.

ACuriousMind

@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.

Rudi_Birnbaum

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

More Anonymous

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

Rudi_Birnbaum

@MoreAnonymous so what should be the problem with that?

More Anonymous

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

Rudi_Birnbaum

14:28

I am reading it

dont paste it

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

More Anonymous

@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

Rudi_Birnbaum

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.

More Anonymous

@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)

Rudi_Birnbaum

@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.

More Anonymous

@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

Rudi_Birnbaum

14:38

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

More Anonymous

@Rudi_Birnbaum I'm pretty sure Landau doesn't advocate:

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

Rudi_Birnbaum

@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"

More Anonymous

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

Rudi_Birnbaum

Well thats how I would see it.

Still its no QM process

the modelling is quasi-classical but NOT QM

More Anonymous

To be fair if you click one of the links:

quasi-classical

The LL formulation, with vaguely defined wavefunction collapse, when used
with good taste and discretion, is adequate FAPP. It remains that the theory
is ambiguous in principle, about exactly when and exactly how the collapse
occurs, about what is microscopic and what is macroscopic, what quantum
and what 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

Rudi_Birnbaum

14:46

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.

More Anonymous

@Rudi_Birnbaum this is quite a technical discussion see:

https://arxiv.org/abs/1811.11060

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”)

@Rudi_Birnbaum

Also don't take that paper to seriously as far as I know it hasn't been debunked yet ..
But there is a heuristic reason for both to believe the measurement is(or is not) unitary .. I do not ask for a proof of this .. I ask for something completely else

Rudi_Birnbaum

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

More Anonymous

How is:

the formula for assigning outcome probabilities (Born's rule) and the post-measurement state-update rule, can be deduced from the other quantum postulates

=

modelling the measurement I think by definition you exceed QM.

?

Rudi_Birnbaum

@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.

More Anonymous

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

In which case I think it would be more accurate to say:

modelling the measurement I think by definition you (don't) exceed QM.

Discussion on question by More Anonym…

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