The h Bar - 2019-06-21 (page 1 of 2)

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3:41 AM

Problems of chat and real life conversations sometimes:

The social norm of "chats are just chats, nothing to be terribly serious about"

That means, if you cannot answer someone's heated question within some window, the person will lost interest in finding the answer to said question even when you finally managed to get back with the answer

People in conversations are thus "forgetful" in some sense, in that most of the time, they will not even remember what they have just said or want some moments ago

This is one of the reasons I do not like interruptions when I chat with people, because like any two trajectories in a chaotic system, once the interruption sets in, the chance of returning to the topic exponentially decreases to zero

Online chats meanwhile is more resistant, due to the asynchronous nature of chat conversations

but still, leaving behind a question for too long, and people will start forgetting about it, all the heated debates, passions, focus etc.

That perhaps explained why people tend to be surprised when I shared with them on some of my year old questions which I still continue research on in order to find the answers. I think it must be very odd for the social norm to have a person that hold onto things for that long

Friendship likewise has a similar phenomenon. Not meeting anyone who is not a close friend for too long, and they started to forget each other

What makes the social norm so short sighted, I have no idea. Perhaps modernisation and its convenience may be to blame

Captain Bohemian

4:27 AM

the air temperature is whopping 36 C now.

I feel hungry but don't really want to eat a hot meal. I just want to eat some ice but I am not sure if ice would stave off hunger. Hunger would hinder thinking and that's why I hate it.

5:05 AM

A nice cake is both cool and filling

try that

1 hour later…

6:32 AM

@JMac Oh brother! That's so bad. I assume he's claiming that his free energy scheme is somehow exempt from the laws of thermodynamics, but it's hard to tell. I like how he spells L.A.S.E.R. I bet he uses air quotes when he says it. :D

@PM2Ring Alright had a long think about how my calculation differs from Wigner's and it boils down to this ... He doesn't have an interaction hamiltonian ... I do ... Without the interaction term I'm pessimistic Wigner or his friend can measure anything ..

In my opinion the paradox stems from bad modelling than being an insight into the workings( or non-workings) of QM

Captain Bohemian

I have arrived the city library, but still feel so hot. How nice if the library opens 24 hours so that I can stay here all day long!

@MoreAnonymous Wigner's friend opens the Schrödinger's cat box, thereby collapsing the wavefunction and obtaining a measurement of the cat's state. But Wigner doesn't know what that state is until he "measures" his friend by asking him. So for Wigner, the (cat + friend) wavefunction hasn't collapsed yet. Hence Wigner claims that the wavefunction represents the observer's knowledge of the system, rather than being intrinsic to the system itself.

@MoreAnonymous I think you may be correct.

@CaptainBohemian Yoghurt is good on hot days.

6:49 AM

@PM2Ring Wow! I wonder if my model is worth publishing ... Time to knock on a QI (Quantum Information) prof's door :P

Morning

mourning (that I'm not in academia)

(hey)

Captain Bohemian

@PM2Ring I dare not eat yoghurt. I usually eat shaved ice to lower body temperature.

@MoreAnonymous you also have this kind of mourning!

if you aren't affiliated with an academic institute, you can still make publication.

I saw some of my ex-boss's papers show his affiliation to be with a company.

@CaptainBohemian I do ... While sometimes a beautiful calculation brings smile .. The fact of not having time for the other and being pressed for time makes me glum ...

I see ... interesting ... How seriously do people take non-academician papers? (and do all journals accept?)

Captain Bohemian

@MoreAnonymous what are you pressed time for?

7:00 AM

I'm currently (supposed to be) learning a big data and machine learning course

Captain Bohemian

@MoreAnonymous but his company is related to his research subject quantum computing, so I think his papers are justified to be accepted by journals.

Time to knock on a uni prof's door in that case ...

Captain Bohemian

but I find there are a lot of companies working on quantum calculation or quantum information.

when I went to conferences of quantum calculation or quantum information, there were often some participants who are not affiliated with academia. Then I realize there are a lot of companies working on quantum calculation or quantum information.

And now u have a third option being affiliated with noone! (Thats basically me :P )

the problem with option 3 is I doubt anyone will take you seriously let alone let you in a conference

Captain Bohemian

I used previous institute to publish to go to conference when having no affiliation.

7:08 AM

I wasn't aware one could do that .. What do mean by "used previous institute"

?

Captain Bohemian

but they won't find.

I saw my former classmates also did that.

hmmm ... seems dodgy (which makes it my kind of thing! :P)

Captain Bohemian

you just fill your previous institute in that column and tell your previous institute to the organizer of the conference you go to if they ask you.

Oh well .. I'll think about doing this when someone answers my model (the one with bounty on it ... Like I'm still uncertain if it's correct .. though that might stem from me never having confidence in my calculations and ideas)

Captain Bohemian

but I feel it very difficult to find an affiliation here.

7:14 AM

here = ??

ah ... do u mean in academia in general?

Captain Bohemian

no, in my country.

Can I ask which country that is? (I completely understand if u dont wanna give that informaiton on the internet)

Captain Bohemian

almost all research assistant positions here are in applied fields very far away from theoretical physics I have learnt.

sounds like india ....

Yea ... in india (and im guessing ur country 2) basically the government doesn't invest in the sciences that much

Captain Bohemian

But I hear that people having MSc in India can hold the position of assistant professor to teach math in universities in India.

in my country that's impossible.

7:28 AM

yea ... which is why what your suppose to do is go abroad and fulfil ur dreams (which is what I did except the fulfil ur dreams part) ... sigh ... I think it hasnt been a great mourning

Captain Bohemian

@MoreAnonymous indeed. holding teaching assistant position for physics experiments has far higher salary than research assistant position based on my previous experience.

but teaching those basic experiments is very boring.

@MoreAnonymous Good idea. I think you need some feedback from people who understand this stuff better than I do.

@PM2Ring I hope the bounty is enough to make someone go through all the calculations I did ...

@CaptainBohemian Oh, ok. There's no nutrition in shaved ice, though. Maybe have some fruit juice or salad.

@PM2Ring I hope the bounty is enough to make someone go through all the calculations I did ...

whoops somehow I double posted

7:36 AM

The chat software is a bit glitchy. It sometimes creates double posts when you edit too quickly, especially on mobile.

Captain Bohemian

@PM2Ring shaved ice has red beans, peanuts, mangoes, strawberries, green beans, etc. depending on the flavor one chooses.

@CaptainBohemian Oh, that sounds alright then. I thought you meant plain shaved ice, which is pretty boring. :)

The "Wigner's Friend" thought experiment had an interesting effect. It convinced some people that quantum measurement requires a conscious observer. It convinced others that wavefunction collapse is not objectively real, and gave impetus to interpretations that don't have collapse, primarily MWI.

@PM2Ring I think I love my interpretation of "interpretations of QM" which is they are unnecessary bagguage

If I'm right hopefully the whole intrepetational issue will b considered philosophy (rather than any form of physics)

Though it's too simple a point ... Id b very suprised if no-one thought of it before

which is why I ask in my post "Does anyone articulate this point in the literature?"

@MoreAnonymous That's basically what "shut up & calculate" says.

Well I wouldn't go that far ... Like if sure you can always ask questions like Wigner did which I think shut up and calculate discouraged ...

(in the sentence above ... remove if)

Im wondering if I shd post my question on quantum computing stackexchange then am I more/ less likely to get an answer

?

8:16 AM

Hm

If there's a timelike curve $\gamma$ going from $p$ to a singular point $q$ and a timelike curve going from $q$ to $r$

Is there a timelike curve going from every point of $\gamma$ to $r$

I'd say yes, but that is based on treating $q$ like a non-singular point

It's probably true by some limit process I suppose

2 hours later…

10:05 AM

Won't that scenario be the punctured spacetime explored by 0celo some years ago?

einsteinandtheevidence.wordpress.com/2016/03/27/test

2

Thus the future null ray to q would block any attempt to bypass q for points on or some infeintesimal distance away from $\gamma$ I think

so I think you cannot have a curve that go through every point in $\gamma$ except $q$

Wow! @Secret you also have your own blog?

So I got into a conversation with someone which basically had me rethink joshphysics's highly upvoted answer ... I'm thinking it only makes sense when there is unitary evolution which is not the case during a measurement .. Since he uses basically the Heisenberg equations of motion .. .

82

Q: What is $\Delta t$ in the time-energy uncertainty principle?

In non-relativistic QM, the $\Delta E$ in the time-energy uncertainty principle is the limiting standard deviation of the set of energy measurements of $n$ identically prepared systems as $n$ goes to infinity. What does the $\Delta t$ mean, since $t$ is not even an observable?

quantum-mechanics time heisenberg-uncertainty-principle

Anyone care to chime?

11:07 AM

That's not my blog. That's Ryan's

Captain Bohemian

@MoreAnonymous yesterday you said you want to work on quantum gravity. Which theory of quantum gravity do you want to work on the most?

good question ... I was more interested in quantum field theory in curved spacetime

to be fair

Captain Bohemian

oh

I don't know very much about quantum field theory. So I prefer to work on gravity.

Oh wow! u must be a boss at GR then .. Good for u

Captain Bohemian

and I see a lot of research groups in quantum field theory have a lot of papers regarding calculating scattering amplitudes or correlators. I don't understand why they do that. If it's not due to some novel mathematical structures appearing therein, I imagine doing that many calculations of that kind looks boring.

11:20 AM

Any particular example of which papers your talking about?

Captain Bohemian

I have not read any paper of that kind in detail because I have never got the impulse to do that. I would only read a paper in detail if its abstract attracts me enough. I just find a great proportion of publications of some researchers are calculations of scattering amplitudes or correlators. In the past, I considered they are like CERN collaboration or ALICE collaboration so just overlooked them. But afterwards I find there are some publications of that kind have pure theoretical contents

like some symmetrical structure or number theory

and start to suspect some of that kind of paper may be interesting, but I am still not very sure, probably because I have not read any of them in detail.

but I may not be able to understand them easily if I really read them because I don't know very much about quantum field theory.

11:56 AM

I'm honestly not the right guy to ask this ... Even though my dissertation is about QFT in cirved spacetime ... As I've claimed before I havent understood fundamentally GR or QFT

12:10 PM

Anyone knows if joshphysics is still an active user .. (Scroll up to see why)

/

What is your dissertation on

QFT with Robinson boundary conditions ... The QFT in curved spacetime actually came in when I had to make contact with Prof Jourma's original paper (which was about the firewall (in some sense))

In fact a lot of the results I use in this post come from that dissertation

7

Q: Using a time-like boundary as a computer?

Question and Summary Using classical calculations and the Robin boundary condition I show that one calculate the anti-derivative of a function within time $2X$ (I can compute an integral below) $$\frac{2 \alpha}{\beta} e^{ \frac{\alpha}{\beta}X} \int_0^{X} e^{- \frac{\alpha}{\beta}\tau} \ti...

quantum-field-theory boundary-conditions klein-gordon-equation algorithms computer

My supervisor wasnt keen on me putting this as part of the dissertation

though personally I found(/and still find) this result astounding

is it because he's not keen on having to read GR

I think it was more of him being a mathematician and found the result too speculative and irrelevant to main point we were trying to make contact with which was Jourma's Paper

Mathematicians are usually not keen on having to read physics indeed

Someday I should make a physicist to mathematician dictionary

12:16 PM

yea .. But it was sooo cool .... Basically the claim the calculations seemed to make was in QFT one can take the anti-derivative within finite time (infact its linear time)

user image

2

like thus

Any computer scientist would amazed by the result .. I tell u

@Slereah at first I thought u posted another meme :p

I'm afraid I don't know much about computing wrt GR

i know there's some fun results regarding CTCs and supertasks

Though it's not super useful :p

And the MONSTER SPACETIMES, of course

Hey semi-classical are you any good with QM?

(nvrmind seems to have left)

oh wait ur back

Obviously only good at semiclassical QM

12:25 PM

haha ... but he could have had that name as an undergrad ... and now wants to change the name to purely quantum :P

Honestly I don't know anything about semiclassical QM

I read about it in one of my more esoteric book

But I think it's one of those thing that hasn't really been popular in 60 years

12:50 PM

Is this comment too much? "I think a better analysis can be done? I mean the uncertainity principles are usually a statement about the measurement. But here you are using the Heisenberg picture (which relies on unitary evolution) ... Further your also integrating but we know the measurement is discontinuous so I'm not sure if that would be an intuitive definition .... Please update your answer in light of my comment"

1:24 PM

user image

@Slereah what a beast

the power!

@Slereah i think ur confusing Witten and one of these guys (nerdbastards.com/2012/02/23/…) ... Though ppl do the same with Elon Musk and Lex Luthor

user image

you can't use star trek because they believed in warp drive :P

1:35 PM

maybe that's why I never met captain Kirk .. He warp drived outside my lightcone :P

1:52 PM

did you get a reply from penrose @Slereah?

Errr

Reply to what

I don't think I ever mailed Penrose

Wow! if u had ... U would have been a hero to me

I mailed Geroch a few times

He is quite nice

It's an honour to chat to a hero

To be fair I have once written to Scott aaronson ... and he replied

penrose use to be very active on twitter

1:57 PM

alas most famous GR people tend to be fairly old

really? dang any insightful tweets

or dead

I think Penrose got private funding so he doesnt need to do PR anymore ...

his account is still there

Doing GR for the rich and famous

2:00 PM

yea ... though some of the rich ppl have gr8 impact on physics

If Elon Musk wants to pay me to do GR he is free to do so

I'm guessing not tho

2 days ago, by Slereah

Did they find Hawking's paper on wormhole pair production

oh yeah that

I'm not sure who I could ask rly

thelifeofpsi.com/wp-content/uploads/2014/09/Hawking-1992.pdf

It's the ref. [8] paper

I guess I could ask Gibbons, if he's not dead

journals.aps.org/prl/abstract/10.1103/PhysRevLett.69.1719

Oh wait

It might be this paper

2:38 PM

Would someone answer a question real quick, is the electric field outside a polarized dielectric always 0?

2:52 PM

@knzhou Ah, I noticed you used a sneaky trick to keep a question off the HNQ. IIRC, the mods prefer that we don't use that loophole, and just raise a custom flag instead.

3:09 PM

@PM2Ring what? Link?

please^

Oh, ok. Latex in the title stops a post from being HNQ. physics.stackexchange.com/questions/482882/…

thnx

As you can see in the edit history, that question got Tweeted.

wahh .. did this question get so much view mileage becaue of tweets?

I gotta work on te PR of my questions

3:15 PM

i've lost interest in tweeter

because of trump

I didn't intend for knzhou to cop flak for this. But I don't think that deciding HNQ-worthiness is something that 1 non-mod user should take unilateral action on.

Has there ever been insightful physics tweets (never personally been there)?

I think u need a blog post for that

@MoreAnonymous Virtually any question about radioactivity gets lots of views ATM because of the Chernobyl doco.

Thats it im gonna accidently add that tag to all my questions

:P

I;m gonna rephrase the Wigner's friend question in terms of Schrodinger's cat with some radiation in it ...

lol

3:22 PM

BTW, pittsburghjoe, the negative quantum spacetime guy is back, with another non-mainstream personal theory post. physics.stackexchange.com/q/487291/123208 1 more vote needed.

By vote do u mean upvote?

:P

THis is gold! Can we have HNQ for these kind of questions

-10

Q: Would the first consciousness uploaded be able to cause a wave collapse?

If he/she had a camera connected to their local computer and he/she were the only ones with access to the double slit experiment results. If the results would show interference ..would that imply he/she didn't have a soul?

waves wavefunction-collapse

I think I shd run for being a mod ... my ideas would surely make stackexchange great again!

@MoreAnonymous No. Close vote. Asking for up or down votes is very bad etiquette on Stack Exchange sites!

@MoreAnonymous another pov, held not only by me, is that the numerous interpretations of QM are pointing to "something deeper" aka cracks in the edifice so to speak, and are hinting of a subquantum mechanics that is expressly denied at least by 1 interpretation: copenhagen. ie exactly as einstein asserted, QM is incomplete. this is not the same as being incorrect, a point sometimes mixed up even by experts...

@PM2Ring I love his opening line "You and I both know the idea of Dark Energy is preposterous" Yes, that's why it's such a prominent hypothesis. Just total nonsense someone made up; definitely not describing observed behaviour or anything...

yea .. I blame Wigner's friend for this fiasco and u can;t really blame einsiten since he had a unique way of physics which worked spectacularly well in some cases but this isnt one

@PM2Ring

Is askin to close a question ethical (on chat)

?

3:35 PM

I think it's bad to ask anyone to vote any particular way on any question. That said, drawing attention to questions for the purpose of getting closed votes is fairly different than trying to get attention to a question to try and get up/down votes I would say

@MoreAnonymous Asking people in chat to consider casting a close vote is generally ok, although some people don't like it. But asking for up / down votes is different because rep points are involved.

in theory salon, 25 secs ago, by vzn

PHYSICISTS HACK THE UNCERTAINTY PRINCIPLE TO SEE AN ION WIGGLE / wired

@vzn I'm curious whats the Bohemian take on Wigner's friend?

Captain Bohemian

I wonder what the following means. "an opinion of a research fellow holding a degree or the academic degree of doctor habilitated on the possibility of supervising the doctoral student in the event of admission to the Doctoral School"

@PM2Ring I see this a gr8 meta post question ... which will definately bring out different perspectives

3:39 PM

It's also considered ok to educate new OPs about the process of accepting answers, with a link to a relevant meta FAQ page, but even then it's better to ask them to consider accepting an answer, rather than point-blank begging them to accept your answer.

this point is less controversial than the other one

(to me atleast)

@MoreAnonymous not an expert on this, dont think bohmian mechanics addresses it directly, but think some of the confusion comes from the language. the words "observation" and "observer" are not so simple in advanced/ "emergent" physics. there are classical systems that exhibit "observer-dependent" effects. etc but they are somewhat low visibility in teaching QM theory, and that needs to be reexamined. the whole thing is ripe for a major revision... its coming, on horizon. the signs are all there.

@PM2Ring TBH I've never even liked doing that when I have an answer. I usually just let it slide if people don't use the accept feature. I should start doing it when I read questions that other people answered though

@vzn "its coming, on horizon. the signs are all there." My personal take is its just bad modelling ... However if that means fails to address the problem then I think then I'll have to admit there is more meat in interpretations of QM

Monster spacetimes must be cool to the touch due to its enormous entropy

meaning they should be ridiculously efficient heat sinks

3:46 PM

@JMac On SO, I post a comment about accepting on questions that have good answers. Old questions without an accepted answer are slightly more likely to attract low-quality new answers. But more importantly, a question without an accepted answer gives the impression that none of the answers were good enough to satisfy the OP, so future readers are inclined to skip them, and look for a question that does have an accepted answer.

Have u done a calculation (or are u using physical intuition)? @Secret

intuition, my GR still too suck to do any calculation, at least not on the back of an envelope

I will need to develop some kind of GR intuition in order to e.g. guess what some component of the riemanian tensor be like at semi quantitative levels

I only successful at developing QM intuition due to the maths of QM being easier to grasp (and my huge love for quantum)

Meanwhile, GR directly conflict with the notion of distance I use to build my memory palaces, making it hard to link GR concepts together without some adjustments

Oh gr8! ur a boss at QM ... any takes on Wigner's friend?

Of course, often the OP is the least qualified contributor to a page to determine the quality of the answers. But the green checkmark is a fundamental part of how Stack Exchange works, and we're stuck with it. ;) At least now accepted answers on self-answered questions no longer automatically get pinned to the top of the page.

@PM2Ring Yeah, I just always feel really uncomfortable with the idea of doing it for questions with my own answer. That's why it's a good reminder that I should start doing it when I notice it for other questions. I've always assumed a lot of the answers that go unchecked for long periods of time are "drive-by" questions, where OP may not come back to SE on that account

Captain Bohemian

3:51 PM

@Secret but I feel math in GR easier than math in QFT.

@MoreAnonymous I take the view where the subsystems get into entanglement as soon the measurement happened, which is why the results of wigner and his friend can correlate

@MoreAnonymous what is "bad modelling"?

Captain Bohemian

But I can't be completely sure. Maybe that's just because I haven't learnt a lot of math used in QFT.

1

Q: Why doesn't Wigner's friend interact with the system?

So I was recently modelling something that turned out to be basically Wigner's friend. I saw there were some differences (in the Wiki page) in how it was modelled: Namely, that Wigner's friend somehow managed to perform a measurement without interacting with the system! (I mean if there was an ...

quantum-mechanics quantum-information resource-recommendations hamiltonian

@CaptainBohemian I am note dared to said anything about QFT, because I am still very beginner on it so I don't know enough details

3:53 PM

Wigner ignores the interactino term which enables him to perform the measurement in the first place

Ignoring it only makes sense before and after the measurment

@Secret I think you can only think of it as entaglement when there is no interaction hamiltonian .. See my post above

Captain Bohemian

QFT uses a lot of functional analysis and operator theory and I haven't known very much about them.

@vzn modelling the measurment globally without an interaction is bad modelling in my opinion

@MoreAnonymous Wigner's friend's measurement is local to the system inside the box, I don't see why there will be an interaction hamiltonian that can shift the state of the combined system

You need a hamiltonian that can affect the combined system in order to affect Wigner's outcome. The measurement also cannot affect the outcome Wigner measured because the two systems (the inside and the outside of the box) are not entangled to start with

@MoreAnonymous have read some about wigner friend experiment. it seems to relate highly (like much of QM) to conditional probability which can be a slippery concept. which is probably still not fully understood in QM. anyway my main idea is that introducing consciousness to QM was something of a mistake, a red herring, but somewhat understandable given the language of "observation + observer" etc

@Secret Can I ever measure a system without interacting with it? Also once u include the interaction then this is no simple relation (tensor product) between the wavefunction of system 1 aand that of 1+2+interaction between 1 and 2

@vzn I think conditoinal probabilty is understood see the comment

8

Q: How is conditional probability handled in quantum mechanics?

In ordinary probability theory the conditional probability/likelihood is defined in terms of the join and marginal likelihoods. Specifically, if the joint probability of two variables is $\mathcal{L}(x,y)$, the marginal likelihood of $x$ is $\mathcal{L}(x) \equiv \int \mathcal{L}(x,y) \operatorna...

quantum-mechanics probability

@Secret as I say in the comments of the post the contradiction only comes when assumes there is no interaction term ..

Also i agree introducing consciousness = massive mistake

4:05 PM

But the interaction takes place before 1+2 becomes entangled by the measurement, thus the measurement cannot project 1+2 into some correlated pair of outcomes. But really, the answer to this question depends on whether there is a time separation between the measurement and the establishment of the interaction term (which for the sake of argument, assume that it somehow can reach far enough to affect system 2). In that case, I don't know

If the interaction term develops before the measurement, then the system will be entangled in the way you suggested, and there will be no disagreement in outcomes

Otherwise, you get the usual wigner friend outcome because the measurement will be on non entangled systems

@CaptainBohemian lol what functional analysis or operator theory comes up in e.g. Peskin and Schroeder

Wait can u repeat urself this time only with the case my analysis does not apply to?

"If the interaction term develops after the measurement, you get the usual wigner friend outcome because the measurement will be on non entangled systems"

Wait Wigner's friend will have an interaction term and so will wigner which one are u talking about?

I dont think u can ignore the interaction term during the measurement ...

@vzn cracks in the edificie of people's understanding of what QM says more likely

4:10 PM

Both. I am assuming that Wigner's friend's measurement produces an interaction term that affect 1+2 (when wigner's friend is inside the box that is 1)

@PM2Ring Nah, I just like LaTeX in titles. That one was briefly on the HNQ several weeks ago, it has long since fallen off.

I don't know what happens when the interaction term is produced at the same time as the measurement, since the time of measurement is not generally a well defined thing in quantum experiments

@knzhou Ah, ok. I guess. :D

Because the issue of the simuaneousity is that: You have two operations that were done to the quantum state:

1. A unitary evolution of the joint system 1+2

2. A projection of 1+2 into the eigenstate of the measurement observables of both Wigner and Wigner's friend

And these two do not commute

Captain Bohemian

@bolbteppa you mean their book or their theories? I am not sure I have read their book.

4:14 PM

thus it is hard to model what happens when the interaction term is produced at the same time as the measurement

I agree ... So to be specific let Wigner's friend measure at time $t$ and Wigner at time $t'$ ... Before $t'$ the sytem obeys unitary evolution

t <t'

@CaptainBohemian P&S is one of the most standard QFT books, I don't remember a single theorem from functional analysis even being invoked in the parts I've gone over, off-hand anyway

Captain Bohemian

but I can't usually remember the authors of a book I have read unless I have paid special attention to them.

Let's talk about this in the Heisenberg picture ... Let's say Wigner's friend measures an observable $O$ which obeys the usual time evolution (unitary) .. However to Wigner the Unitary operator of O will be the system 1+2+ interaction of 1 and 2 ... note both these methods only differ when the Hamiltonian's interaction 1 and 2's is small

which is true in all cases expect during the measurment

*only differ = do not differ

Captain Bohemian

@bolbteppa the QFT books I have read partly seem not to use functional analysis, either, or maybe they have used somewhere but I didn't pay attention or know. I sometimes don't know what some math terms exactly mean. For example, the term "algebra" is used in different places and I don't know why they are used there.

4:24 PM

@Secret I hope I'm talking sense rather than bullocks ... Feel free to answer the question if you think u've got my point but I havent got urs

So, at $t$ wigner friend made the measurement, which generates some $O_{WF}$ that acts on 1+2 and then at $t'$ wigner made the measurement, which in this order or the reverse, first project out the joint system and then evolve it under some $O_{W}$

If both $O_W$ and $O_{WF}$ are small, then we have the usual outcome of the Wigner friend experiment. Otherwise, there are multiple possibilities depending on the order of the interactions $O_W,O_{WF},M_W,M_{WF}$ where $M$ is the projection done to the state due to the measurement

if $O_{W}$ is earlier than $M_W$, Wigner may or may not register a disagreement depending on whether $O_WO_{WF}=\text{id}$

if $O_W$ is later than $M_W$, then there will be a correlation between Wigner and Wigner's friend measurement, which is different from the disagreement case

both of these assumes $O_{WF}$ is later than $M_{WF}$

Taken from the wiki page:

"The thought experiment posits a friend of Wigner in a laboratory and lets him perform a quantum measurement on a physical system (this could be a spin system or also Schrödinger's cat). This system is assumed to be in a superposition of two distinct states, say, state 0 and state 1 (or "dead" and "alive", in the case of Schrödinger's cat). When Wigner's friend measures the system in the 0/1-basis, according to quantum mechanics, he will get one of the two possible outcomes (0 or 1) and the system collapses into the corresponding state.

DOesn't this assume wigner does his measurment after his friend

I don't think Wigner has a way to know

thus he should be modelling the aforementioned 4 basis cases, along with their superpositions (one can put time ordering of operations into superposition due to something called indefinite causal order)

and only the final measurement outcome will tell Wigner whether he or his friend does the measurement first

Hmm ... So basically u choose Wigner;s reality as more fundamental than his friends for some reason

?

nah, I am just talking about how Wigner will assess the situation. You can do the same thing for wigner's friend because $O_{WF}$ affects both systems

Wigner's friend also have no way to know what Wigner will get from his measurement, because he is inside the box

Thus Wigner's friend would have made the measurement after Wigner did and he will not know until they compare results together

4:40 PM

Wait the one in the box is Wigners friend right?

yup

Maybe u shd answer the question I pose ... On why the interaction Hamiltonaian does not intefere with the problem ...

I'm not getting it like this

I don't see an easy answer. all I can think of is that the box is designed to be isolated until it is opened, meaning that there is no interaction between systems inside the box and the outside. It also means whatever $O_{WF}$ that is produced from the measurement can only unitarily evolve Wigner's friend subsystem and for Wigner it will be the identity operator

That however does not resolve anything about the ordering of $O_WM_W$, so I don't know how experimentally, $O_W$ is avoided

I know its a bit late in the conversation to ask this but what do you exactly mean by $O_{WF}$ $O_{WF}$ (I just wanna reconfirm ...)

$O$ is the interaction term you said will arise during measurement. I think I should have used the letter $H$ and that will be clearer

4:48 PM

So u think the interaction Hamiltonaian is valid objection?

I don't see any problem with that. You already showed there is energy consumption during measurement in one aother question of yours

I had said "Let's say Wigner's friend measures an observable $O$ "

yup

: / ... plz dont tell me this is the source of my confusion

I don't think there is a confusion here. The whole process has to go through one of these as wigner and Wigner's friend made the measurements

4:53 PM

I think Wigner cannot think of his friend + system as a tensor product

because he has to include the interaction term

which will destroy the tensoe product but give the right unitary evolution of the system

Then this will mean they will all be entangled in some way already before the measurement is done

An interaction term acting on a tensor product is more likely to give a partially entangled state than anything else

Thus when you destroy the tensor product, you end up with something that is more or less entangled

and that of course can unitarily evolve and avoid the disagreement

I'm not sure wat u mean by the first 2 lines can u supplement it with some math ...

So from wiki: "An entangled system is defined to be one whose quantum state cannot be factored as a product of states of its local constituents; that is to say, they are not individual particles but are an inseparable whole."

generally when things are entangled if I learn somthing about system 1 I learn something about system 2 ... This is not true in this case (atleast thats what I think)

You claimed that the modelling $\lvert 0\rangle_S \otimes \lvert 0\rangle_F$ is wrong because they forgot the interaction term, right?

yes ,,,

Now, suppose we have some interaction term $H_{SF}$. Then we have the state should be described as:

$\lvert \psi \rangle_{SF} = H_{SF} (\lvert 0\rangle_S \otimes \lvert 0\rangle_F)$. Such $\lvert\psi\rangle_{SF}$ is likely to be not factorisable thus suggesting $S$ and $F$ are entangled

5:06 PM

No I disagree ... entaglemtn means something else basically $H_1 \otimes 1 + 1 \otimes H_2$ ... Now the Hamiltonian with interaction its slef it $H_1 \otimes 1 + 1 \otimes H_2 + H_{12}$ ... I dont think u can say they are entagled

if they were when U measure S ud know something about F (not true during superposition)

Something acted by $H_1 \otimes 1 + 1 \otimes H_2 + H_{12}$ does not look like to be a product state. If it is not an entangled state either, then what kind of state is that?

Entanglement is a property of states, not of Hamiltonians.

2

yea .. but for entanglement dont u need the Hamiltonain to be something like $H_1 \otimes 1 + 1 \otimes H_2$

?

There are Hamiltonians that can never produce an entangled state from an unentangled one (those of the form $H_1 \otimes 1 + 1\otimes H_2$, because this just gives a time evolution $U_1\otimes U_2$ under which all non-entangled states remain non-entangled), but all others can. Whether they actually do that for any given initial state and time frame depends on the specific Hamiltonians and states.

But u obviously can't think of the usual $\psi_1 \otimes \psi_2$ when we have $ $H_1 \otimes 1 + 1 \otimes H_2 + H_{12}$

5:14 PM

I don't know what that's supposed to mean.

As Acuriousmind said, it depends on the details of $H_{12}$ and the state you start with, it may be entangled or it might stay unentangled

So let us use some coulomb repulsion between two electrons (as an example) ... I dont think u can write the state as a tensor prodcut

What state?

The state of this Hamltonaian

That's not a thing

5:16 PM

U can start with the ground state (if u want)

A repulsive Hamiltonian has no bound states, therefore no ground state.

Any eneergy eigenstate?

Without tensor product ... I;m not sure how to express entanglement

And in this case I'm certainone can't express the wavefunction as a tensor product

I don't know what you're talking about. Given two subsystems with spaces of state $H_1,H_2$, the space of states of the combined system is always $H_1\otimes H_2$ and all states of the combined system can be written as $\sum_{ij} \psi_i\otimes \psi_j$. Those states that can be expressed such that the sum has only a single summand are not entangled, all others are entangled.

The Hamiltonian nowhere enters into this.

(the $H_i$ up there are Hilbert spaces, not Hamiltonians. Maybe not the best choice of notation in this context, but too late... :P )

How does one "combine" $H_1$ and $H_2$ ? I'm not aware

What is the combination of tensor product states in the example I give

$H_1 \otimes 1 + 1\otimes H_2 = H_1 \oplus H_2$ if I recall

which is very different from $H_1 \otimes H_2$

5:24 PM

yea ... I suspect ACuriousMind is either talking about some (more) advanced stuff or we're not on the same page

I think you should just restate your entire problem more clearly, as I don't think he is aware of the full context on what this whole discussion is about

@MoreAnonymous "combine" simply means that you consider both systems has having one overarching state, it's just a change in the description. As long as you don't introduce interaction between the systems, the combined state will never be entangled, i.e. basically just be the pair of the state of system 1 and the state of system 2.

I can't tell you how to express energy eigenstates of two electrons repelling each other since that would require me to solve the Schrödinger equation for that system and doing that even for two-bodies is famously hard :P

Alright but can u reference a proof that the solution (the states) are of the form of a tensor product ...

The proof is that that is how the space of the combined system is defined.

What do you think the solution space is, if not the tensor product?

Yes .. but the combined system in this case has an interaction ...

5:31 PM

The Hamiltonian is irrelevant for finding the space of states!

The Hamiltonian is an operator on the space of states - to even specify that there "is an interaction", you first need to have the space of states

joshphysics gives a nice explanation here why the combined space of states is the tensor product

I'm still reading and just hit the first comment to the answer which is basically how I think of this stuff

So does this mean secret was right when he said: "Now, suppose we have some interaction term $H_{SF}$. Then we have the state should be described as:
$\lvert \psi \rangle_{SF} = H_{SF} (\lvert 0\rangle_S \otimes \lvert 0\rangle_F)$. Such $\lvert\psi\rangle_{SF}$ is likely to be not factorisable thus suggesting $S$ and $F$ are entangled"

?

@ACuriousMind

Perhaps in spirit, but one does not obtain the resultant state of an interaction by applying the Hamiltonian to it! The time evolution operator is $U(t) = \mathrm{e}^{-\mathrm{i}Ht}$!

oh yeah, I made that U,H conflation mistake again....:P

5:52 PM

@ACuriousMind So adding an interaction term does not void Wigner's friend ?

Relevant post that started this discussion: physics.stackexchange.com/questions/487171/…

between me and secret

Honestly I still think the argument holds ... while it doesn't void the tensor product ... One guys unitary evolution operator doesn't include the interaction while Wigner's does

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