12:26 AM
@vzn From you know where from today:

"As a student of John Wheeler's, Hugh Everett III wrote the thesis, The Theory of the Universal Wavefunction, in 1956. A year later, in 1957, Review of Modern Physics published the version Relative State Formulation of Quantum Mechanics.

These texts are mixtures of wrong statements and trivialities. The main wrong statements are that "the wave function is an objectively real, observer-independent, and therefore 'universal' set of classical degrees of freedom" and that "it doesn't ever collapse".

12:55 AM
bowchicawowow
got the go ahead to do my next project
now I can use TPU v3 resources hehehehe
now I gotta be careful to remember to shut down the instance when I'm not using it though...TPU v3 is pretty expensive...

4 hours later…
5:08 AM
Hwy, there's a rod hinged at the center with no friction. Suddenly an Impulse is applied on the end of the rod, then describe the motion of the rod?

For a bound state corresponding to a Hamiltonian $\hat{H}$, show :$\left<\Sigma q_i \frac{\partial v}{q_i}\right> = 2\left<T\right>$, where the terms have usual meanings.

Will it move with constant angular velocity around its axis

Is the question correct? I know a proof using Euler's theorem of homogenous functions if the $q_i$ were instead the derivative of $q_i$

5:33 AM
@SwapnilDas yes, the angular impulse (impulse times distance from the axis) is equal to the change in angular momentum. Just like in linear motion impulse is equal to the change in linear momentum.

@JohnRennie Thanks!
I had a question of coordinate transformation in 3d
Say a point is located at (3,4,5)
the observer is looking at the positive y axis and it at the origin
the coordinate axis is rotated 60 degree clockwise about Y axis
What's the new coordinate of the point?
How do I approach the problem? The answer is: (-2.9, 4, 5.1)
@JohnRennie

5:50 AM
@SwapnilDas I need to work now for half an hour or so ...

@JohnRennie Never mind, I could wait till 3 more hours

In the question I posted earlier, I think $v$ should be $V$.

@enumaris Phinneas and Ferb
@JohnRennie As it turns out, making the chemicals at home is not possible for me (it requires expensive lab machines and chemicals I have no idea how to work with). My last option before........ before..... giving up, is to find already made super cheap e-paper display (or something similar) and cut it in the desired size. Is it possible they sell just charged TiO2 in oil, or just the solution, or at least the display without any cables? If not, is there anything else I can do?

6:21 AM
@Forge hi
@NovaliumCompany I don't know to be honest ...

@JohnRennie I asked on Stackexchange, Reddit, Chemistry Forums... you're not the only one :P
@JohnRennie You warned me this will happen :\
I learned a lot though.

@NovaliumCompany it's always good fun trying things even though they may not work.
@SwapnilDas if you're rotating about the y axis than the y coordinate will be unchanged. It's just a rotation in the xz plane.

@NovaliumCompany what

@enumaris Practical experience, the one I value the most

@JohnRennie Ya correct, 4 remains the same

6:28 AM
I don't follow

@enumaris ?

@SwapnilDas so if you look in the xz plane you have the vector (x=3, z=5) and you are rotating this clockwise by 60°.

yup

You can do this geometrically by drawing a diagram and working out the rotated vector, or you can just use the rotation matrix directly.

@enumaris Oh, I think I get what you mean.

6:31 AM
I have no idea what you're talking about lol

@enumaris Me too xD

I got the answer, thanks! @JohnRennie

@SwapnilDas cool :-)

7:04 AM
@JohnRennie Hi, John. We were talking about what is meant by quantum fluctuation in Motls's article.
You said that it is the measurements that fluctuate not the field.

@Forge hi :-)
Yes, it is the measurements that fluctuate. With inflation the argument is a bit more subtle.
Inflation is driven by an inflaton field.

But then he wrote that “At the end, he says that there shouldn't be eternal inflation – in which regions of space repeatedly see a random jump of the inflaton field to a higher level which produces an exponential expansion which may start somewhere in the newly created regions again and again – because the eternal inflation requires quantum fluctuations and they just "proved" that quantum fluctuations don't exist. Or, in other words, eternal inflation is on par with the Boltzmann Brains so because they decided that they don't like the Boltzmann Brains,
“But all of this is just atrocious nonsense. There are numerous fundamental differences between eternal inflation and the Boltzmann Brains. One of them is that the eternal inflation is a credible picture in the cutting-edge cosmology discovered by some of the leading minds of the scientific community”

We actually don't know what the inflaton field is so we just assume it has some form. The key thing is the energy of this field because if the energy falls below some value the field decays and inflation stops.

So if it's the measurement that fluctuates not the field value, then how can quantum fluctuations cause the random jump of the inflation field to a higher level which leads to eternal inflation?
I grasp what his statement as : the field is fluctuating which results in random jump of the inflation field to a higher level, thus leads to eternal inflation. Is his statement about quantum fluctuation leading to eternal inflation wrong, or is it my misunderstanding?

@Forge because the inflaton field is not an eigenfunction of energy so it exists in a superposition of energies.
If it exists in a superposition of energies, and that superposition includes values below the critical energy then the field must exist in a superposition of having decayed and not having decayed. That is it exists in a superposition of still inflating and not inflating.
@Forge with me so far?

1 hour later…
8:40 AM
@JohnRennie Not really, I must admit that I'm just a layman who knows nothing about QM or QFT but like to read about cosmology.
The controversy is like this.
In this article, Motls said that the idea that quantum fluctuations can cause random jump of the inflaton field to a higher level, which leads to eternal inflation, is correct.
But in this article, Carroll said that the idea that quantum fluctuations can occasionally poke the field to go higher, which leads to eternal inflation, is wrong.

@Forge the problem is that Motl (like Matt Strassler) is writing for a popular audience, and as you're discovering trying to explain what really happens rapidly gets incomprehensible.

So the more I read about quantum fluctuations in eternal inflation, the more I get confused. Because many users on this site said that quantum fluctuation of the field is just the fluctuation of the measurement, or that quantum fluctuation of the field is just the fact that the value of the field at a point is just not determined because the field is distribution and has nothing to do with fluctuation.
My real question is to ask about the validity of the claim that quantum fluctuations can lead to eternal inflation.
@JohnRennie That's true :D

@Forge this is what I'm trying to get at, but unless you can grasp what it means to be in a superposition then I can't give you any better answer that the (over)simplified ones you've already read.

I guess a good step for the idea of getting a spacetime from measurements is
Given a 2D $\mathbb{R}^2$ spacetime, globally hyperbolic and hole free
Foliated by timelike geodesics
Each emitting null curves at every point
is the information vehiculed by all those null curves enough to get back the whole metric?

@JohnRennie Ok I will look up what is superposition and measurement. But can you please tell me first whether their claim that quantum fluctuations can lead to eternal inflation is valid. Thank you.

8:50 AM
@JohnRennie I'd say this isn't the only Motl problem
wink

@Forge eternal inflation is certainly possible, and it is made possible by the fact the inflaton field can exist in a superpositions of energies. If you want to call that quantum fluctuations then you can, and many do, but if you really want to understand what it going on you need to look deeper into it.

9:01 AM
@Slereah I wish Motl would get over his addiction to bile and spleen. When he can control it he writes really clear and informative articles.

Crush some antipsychotics into his borsht
could b of use
Why does Harvard still have that awful archive site
have they never heard of PDFs

9:56 AM
Does anyone know where I can buy microencapsulated electrophoretic imaging film?

I just buy it at my green grocer
They look suspisciously like cashews

Pff... I want to make sure I've done everything I can before I give up from the project

1 hour later…
11:01 AM
philosophy:
machine learning theories vs crank theories
We have machines like AlphaZero which can kept on playing against itself to find a new solution to a game
but they are not very robust because they lack input from the outside, despite they can came up with out of the box solutions
Likewise, some cranks is someone who came up with half baked models that are recursively pitted against themselves, and came up with a model that at best only match some of the experiments, but give absurd and contradictory results for others
For what reason we can place faith on the conclusions drawn from machines and not humans. What makes a human crank which a machine unable to?

@JohnRennie @PM2Ring I know I'm repeating myself, but I have to be sure - Is there anything else I can do or try before I can give up in peace?

@NovaliumCompany you never really explained what you were trying to do. You hinted that the display was just a means to an end without actually saying what the end goal was.
I'd be inclined to use a light emitting polymer, though I guess that would take more power than an eInk type display.

@JohnRennie Sure, but that would emit light, and would be too awkward for people to wear

I think an eInk display would also be very awkward. In fact anything requiring lots of wires is probably going to be a challenge.

Hi guys, goodmorning.
I have a problem here, anyone can help me?

https://physics.stackexchange.com/questions/501039/rotating-rope-with-a-ring

11:16 AM
@ABC what is trascurable mass?

negligible mass
I'm going to change

@JohnRennie The wires would have been under the shirt, not visible from the outside. The prototype will be awkward because the capsules are too big, but as we slowly work towards better resolution, it would look more and more natural from a distance, rather than glowing LEDs.

@ABC In the non-inertial frame you assume the ring is moving in a circle and then calculate the force pushing it outwards. If you are working in the inertial frame then you start with the assumption the ring is moving in a straight line and calculate the force deflecting it froma straight line.
I suspect the calculation is a bit messy in the inertial frame. I usually recommend students work only in inertial frames because there are hidden traps in non-inertial frames, but in this case I'd be inclined to work in the non-inertial frame.

Ok, perfect. But I don't understand why in the inerzial frame my ring go away from the bar

Random thing found when misreading the starred message as Even Physicists Don't understand Quantum Police

11:25 AM
@ABC Suppose there was no force between the bar and the ring. Then the ring would move in a straight line. It would be moving away from the bar because the bar moved in a curve away from the line of the ring.
So if you looked at it from the bar the ring would move away even though no forces were acting.
Now when you turn on the forces what they do is accelerate the ring inwards. It still moves away from the rod, but it moves away less rapidly than it would if no forces were acting.
The key difference is that in the non-inertial frame there is an outwards force accelerating the particle outwards, while in the inetrial frame there is an inwards force accelerating the ring inwards.

@JohnRennie I'm sorry I can't see what you say in the first few lines
I didn't quite understand what you mean by "It would be moving away from the bar because of the bar moved into a curve away from the line of the ring."

@ABC if no forces act on the ring it moves in a straight line. Yes?

yes sure

user280247
why there are 2high tides a day?Any good resource?

But the bar is rotating around the axis so any particular point on the bar is moving ina circle centred on the axis of rotation.

11:31 AM
ok but now we are supposing that the ring is on the bar or away from the bar?
In my mind I see ring moves in a straight line away from the bar

@ABC we started out saying there is no force between the ring and the bar. That means the ring can pass through the bar.

and the bar is rotaning

So the ring isn't attached to the bar. They just happen to start off at the same point with the same velocity.

ok perfect, ring can pass through the bar
My doubt was here
ok perfect
so now?

But obviously the ring can't pass through the bar so the bar exerts a force on the ring. That force deflects the ring away from a straight line.

11:33 AM
lol

ok yes bar deflect the ring

I'd have to draw a diagram to see exactly what direction the force acts in, and I need to go now so I don't have time.

in real world
Ok I will continue to think about it. In the meantime, thank you for the great help! Can we continue the discussion?

Yes, I'll be around later or failing that tomorrow.

thanks, have a good day!

user280247
11:42 AM
uhmm none of those worked for me

12:05 PM
@santimirandarp You read all of that stuff in twelve minutes? Wow! ;) Seriously, the answers by David Hammen to the top two questions in that list are very good, but I admit they are fairly technical. So, after having carefully studied that material, do you have any specific points that you don't understand?

Fuck me that synchronization problem in GR isn't trivial
Even assuming global hyperbolicity
I need to look up more on time functions and geodesic foliation shenanigans
Also redshifting in GR
You have to factor in like the motion of both curves, the metric and the foliation I think

@NovaliumCompany You want to make this thing wearable? I don't remember you ever saying that before.

It's easier in SR because you can more or less assume that your observers are all geodesics, so using the proper times and redshifts you can about synchonize all observers

user280247
@PM2Ring my question is simpler than that, so I didn't go through all the answer. And I do, but I won't discuss it with you.

12:49 PM
Hm
I think that instead of having Reichenbach synchronization with $\epsilon$ being roughly the "tilt" of the hypersurface, GR is going to have a set of synchronization that depends on the foliation
but the hard part is making the synchronization surface a Cauchy surface
Which makes sense since $\varepsilon$ basically corresponds to a linear foliation of Minkowski space

1:22 PM
@PM2Ring :P The whole idea behind this was always to make it into changing clothing. Like a chameleon. You choose from your phone what image to display, and the t-shirt changes.
@PM2Ring It was supposed to be a unique business idea, that's why I kept it in secret. But I won't be able to make it anyways.

Actually maybe a good first step would be to find synchronizations for Minkowski space with arbitrary foliations
@RyanUnger Riddle me this, Batman
Is there an obvious space of foliations of 2D Minkowski space by Cauchy surfaces?
I guess it would be a space of curves with a bounded derivatives
But of course there's the usual counterexample of a spacelike achronal hypersurface that isn't a Cauchy surface

1:39 PM
Yo, topologists

@NovaliumCompany Maybe you should take a look at the kits these guys sell: eink.com/flexible-technology.html

There's this claim that humans are topological pretzels
Is it true?
If so, is there a good reference?

Pretzels are topologically equivalent to a ball
they're just a cylinder twisted around

@Slereah what?
They have genus 3

Well humans are of genus... 2 at least?
Digestive tract and nostrils

1:42 PM
@Slereah two nostrils

@JohnRennie Well yes but the nostrils are connected via the sinuses

@PM2Ring Is there something that can help me on that link? I cannot just buy 100 displays and glue them to a t-shirt, too expensive.

they don't extend all the way to another hole

@Slereah doesn't matter, it's still four holes

@Slereah but the nostrils connect to the mouth

1:43 PM
The nostrils, the mouth and the anus all join up.
3

@PM2Ring Those transparent coin holders that you see in my videos, I was going to glue them individually as separate pixels.

Add a few piercings and the genus gets remarkably large :-)

@Slereah I think Emilio is thinking of pretzels shaped like this:

@PM2Ring So you have no idea how I can continue the project? (now that you know what I'm actually doing)

@NovaliumCompany It's a large site, and they have lots of different products. They have some large displays, intended for signage, but they aren't cheap. And those ones probably aren't flexible.

1:58 PM
@JohnRennie An indent isn't the same as a hole
it needs to come out the other side
@EmilioPisanty mouth connects to anus, though
So this is one of those handle on a handle situation

@NovaliumCompany Sorry, I don't. I posted that link to give you an idea of the state of the art in flexible e-ink technology.

by diffeomorphism this is roughly a surface of genus 2

@JohnRennie @PM2Ring Welp, I guess I'm giving up with this project. I've learned a lot and I'm satisfied I spent my time with it. Thank you both so much for the help.

Oh wait
I see it now
Yes it is indeed a surface of genus 3

2:06 PM
@PM2Ring Cool, but they should cover whole t-shirt.
My only option is to buy large e-paper displays, cut them into 1cm x 1cm pieces and glue them to the t-shirt and the wires will be on the inside.
Ah... I hate to give up. I even told myself that I want to finish this no matter what. ;(

2:46 PM
I'm so glad I didn't post an answer when I saw this question about invariant mass: physics.stackexchange.com/q/501116/123208 I suppose it's just another example of confusion caused by relativistic mass... but I don't like the OP's attitude.

@PM2Ring oof

I'm so tempted to downvote the question, but I won't, since "OP is being a dick" isn't a valid downvote reason.
But it looks like a few people don't agree with that. ;)

no, but it -is- a reason to flag their comments :3

True, but they've already had a couple of comments deleted by tpg2114, and given a warning. Perhaps they need a sterner warning...

3:58 PM
lol! Lumo "get over his addiction to bile and spleen"... would there be anything )( left? :P

4:34 PM
Can someone tell me if my question is rubbish and if so why?
https://physics.stackexchange.com/questions/501145/validity-of-the-derivation-of-time-energy-uncertainty-principle?noredirect=1#comment1130199_501145

@MoreAnonymous what's the gist of the issue
What are the "discontinuous measurement interpretations of quantum mechanics"

something something collapse, maybe?

I'm pretty sure this derivation
5

In addition to Joshphysics' precise answer, let's mention another interpretation (the one I think Ben Crowell is referring to in his comment to the same answer). There's a formula from time dependent perturbation theory which gives the probability of an induced transition from an initial state $... generalizes to qft and is fine despite looking like it might be special I feel joshphysics answer can be used in a broader sense and I would like a version of his derivation (if possible) I don't think my question is soooo bad that it warrants a down-vote :( What does "the measurement is considered a discontinuous process" mean Also "But here you are using the Heisenberg equations of motion (which relies on unitary evolution)" implies that the Heisenberg formalism is somehow special, which is ridiculous 4:44 PM I do not say the Heisenberg formalism is special And for your first question: en.wikipedia.org/wiki/Wave_function_collapse IIRC you can do roughly uncertainty relations by integrating the operators over some volume Visser does it This idea of measurement being discontinuous is irrelevant For$\phi_\Omega = \frac{1}{\text{Vol}(\Omega)} \int_\Omega \phi $and$\pi_\Omega = \frac{1}{\text{Vol}(\Omega)} \int_\Omega \pi $, $$[\phi_\Omega, \pi_\Omega] = \frac{i\hbar}{\text{Vol}(\Omega)}$$ @Slereah in that case can you atleast shed light how one can differentiate and integrate as he does? This idea about not being able to use basic calculus is therefore also completely irrelevant The issue with time-energy uncertainty is that time is not an operator 4:48 PM I am aware u can use calculus in discontinous situaitons as well But the derivation he uses does not enable that The basic idea of the time-energy uncertainty is that if you could measure positions sufficiently close in time, you would be able to infer the momentum from it ie$p/m \approx (x_2 - x_1)/ \Delta t$You cannot quantify this idea that measurement is discontinuous and it's irrelevant, the issue is that time is not an operator, and the time-energy uncertainty relation involves two measurements at two separate times, which is different to the uncertainty principle measuring two quantities at the same time Can you quantify the idea the measurment is continuous? (I mean thats why we have the interpretations in the first place) I'm after the derivation not the intuition ... I know what you guys are talking about What does "measurements are discontinuous" mean anyway You can't measure things continuously Measurements are discrete Therefore they cannot be discontinuous .... I'm sure if it was easy as that the whole Quantum information community would be with you by now ... 4:53 PM non sequitor Im under the impression they say the measurement is discontinous becasue 'As Lev Landau once joked ”To violate the time-energy uncertainty relation all I have to do is measure the energy very precisely and then look at my watch!”' 5 Special relativity has four-vectors$(\Delta t,\Delta \mathbf{r})$and$(E,\mathbf{p})$, so we would like there to be a direct analogy between these two uncertainty relations. In fact the analogy fails, because position is an operator in quantum mechanics, but time isn't. Peierls has a nice discu... If we have a wavefunciton and I happen to measure it and assume the measurement is an instanteous process then due to the before and after of the measurement being significantly different it is discontinous '...time is not an observable. A measurement of time in itself does not convey any information about a physical system, and a statement about any other physical quantity usually implies that we are talking about its value at some particular time. In the case of a conserved quantity, such as the energy of an isolated system, the result then also gives the energy at any time. ' @MoreAnonymous and? How is that relevant So basically I can have the wavefunction to be a dirac delta operator at t - epsilon and a exponential function at t + epsilon ... And take epsilon to 0 ... This is the definition of discontinuity ... Did I suddenly make sense??? Or have I been deemed a lost cause?? I'm guessing a lost cause (with the way the dowvoting is going) ... ://// 5:08 PM Basically it is not clear how this idea of 'non-unitarity or discontinuity of measurement' is even relevant to your question Because in the josh physics answer he makes use of unitary and continuity (during the measurement) which I think some interpretations will be unhappy with Hey Jude, don't make it bad Take a sad song and make it better Remember to let her into your heart Then you can start to make it better Well your question doesn't make it clear you are questioning the validity of a standard quantum mechanics principle in non-standard "interpretations" of qm, veering to you know what territory basically The title is Im questioning the derivation ... Please do not misrepresent my views Yeah by questioning the most basic principles of qm like unitarity How does "The probabilistic, non-unitary, non-local, discontinuous change brought about by observation and measurement, as outlined above" even relate to the Hamiltonian or Heisenberg equations being unitary 5:16 PM I never said non-local ... Please stop making me sound like a crank You linked to a section of a wiki page with this quote explaining where you're getting this "non-unitary" and "discontinuous" stuff and implying somehow it means the Heisenberg equations are non-unitary somehow in some unstated (crank?) theories And also it is not a crazy opinion: 3 It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation. Why? The wave function is$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$where$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$The probabilities... and then saying this might affect a standard derivation using these properties, without explaining how 'non-unitarity of measurement' even relates to the Heisenberg equestions I never said that "implying somehow it means the Heisenberg equations are non-unitary somehow in some unstated (crank?) theories" ... Please clarify before you accuse me crackpottery " then saying this might affect a standard derivation using these properties," I am saying Im unsatisfied with a derivation ... Is it that hard to understand ... ofcourse it doesnt ... this is experimentally verified .. I am highly uncomfortable with the jump in logic you are doing to accuse me crackpottery 5:21 PM You did say that, you said "Because in the josh physics answer he makes use of unitary and continuity (during the measurement)" above and in the comment to his answer you said "But here you are using the Heisenberg equations of motion (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" this clearly means you either think non-unitarity of measurement = Heisenberg equations are non-unitary, or worse are not explaining how measurement being non-unitary even applies to this question but also questioning a standard derivation because it doesn't apply to non-standard theories of QM I'm not accusing you of anything, just trying to help you figure this out haha Alright .. Maybe you should answer this question as well where the measurement is not a unitary process? 3 It is said, that when measured, a quantum system undergoes "wave function collapse", which is a non-unitary transformation. Why? The wave function is$\Psi = \alpha \left|0\right\rangle + \beta \left|1\right\rangle$where$\left|\alpha\right|^2 + \left|\beta\right|^2 = 1$The probabilities... Look, it's completely irrelevant whether measurement is a non-unitary process, you need to explain how that even matters here, and you also need to clearly state whether measurement being non-unitary implies that Heisenberg's equations do not apply or give a non-unitary time evolution for some reason Because the derivation uses unitarity (indirectly through as you say Heisenbergs say) ... I haven't seen a derivatoin of the Hesenberg equations of motion which does not rely on unitarity ... Maybe there is one which Im not aware of See, you are saying that Heisenberg's equations must be wrong because 'measurement is non-unitary' Ofcourse not They obviously hold during the unitary evolution 5:27 PM The wiki page you linked to explicitly says that QM assumes wave function collapse is non-unitary, but also claims that time evolution of a system is unitary under the Schrodinger equation (equivalent to Heisenberg) But if ur calling velocity p/m during the measurement thats only true in a dimensional sense Well maybe theres a derivation Im not aware of either way in that case this seems to a well spread misconception at least making it worthwhile answering? it's worth noting that time evolution of a subsystem can be non-unitary even as the entire system evolves unitarily Heisenberg's equations are unitary at every step, not indirectly, but directly 'Unitary time evolution' @Semiclassical That would be a way out ... But then again joshphysics answer does allude to that so it's okay to be unsatisfied @bolbteppa yes directly would be a better choice of words I get worried when the long pauses come in the conversation and I wonder: Did I suddenly make sense??? Or have I been deemed a lost cause?? 'A physical note: unitary => invertable. Physically we know measurements are not always invertable.' is a good comment You're still not explaining how non-unitarity of measurement even applies So far you've tried to say that it applies because the Heisenberg equations rely on unitarity This is comparing apples and oranges, or do you disagree (if so, explain how) 5:37 PM well if joshphysics derivation uses unitary evolution (due to heisenberg equation) ... But if you want it to apply during the measurment then I'm skeptical ... He hasnt made the case for that ... Even more, you've tried to say that discontinuity of measurements somehow means we can't use calculus Atleast in collapse model No if you scroll up I also say that you can use calculus in discontinous cases as well but joshphysics's answer does not make that case He only sticks to the unitary evolution of quantum mehanics Great, so you are saying that a basic quantum mechanics principle, non-unitarity of measurement (point 1 of the wiki you linked to, see: en.wikipedia.org/wiki/… ), somehow implies that the unitary time evolution of a system (point 2 of the wiki you linked to) fails between measurements, surely you see this is ridiculous and would mean QM makes no sense Tbf I say I am aware you can use calculus in discontinous cases what it it sounds like is that you're trying to apply the time-energy uncertainty relation to a situation where$t_1$is before a measurement and$t_2$is after And that seems highly dubious. 5:41 PM This idea of measurement being discontinuous is really irrelevant "surely you see this is ridiculous and would mean QM makes no sense" ... Nope ... I'm saying Im unsatified with a derivation ... I'm okay with a sub system undergoing non-unitary evolution But the logic of what you're saying makes so little sense, it implies QM can never apply and calling that a measurment This subsystem stuff is irrelevant as well System 1 + sytem 2 undergo untiary evolution ... System 2 alone does not ... Im saying that is a possiblility ... @bolbteppa I say discontinous because if it were continous then u can get away with the derivation joshphysics uses 5:45 PM As I said, it sounds like you're asking whether$(\Delta t)(\Delta E)\geq \hbar/2$applies in a situation where the system is measured during the$\Delta t$interval @MoreAnonymous have no idea what you mean by discontinuous I haven't read the rest of this conversation but it is well-known that subsystems need not evolve unitarily. In fact, that is the case for most interacting systems since unitary evolution can't turn a pure state into a mixed state, which is what evolving a non-entangled state into an entangled does on the subsystems. @Semiclassical ... There is a difference between being unsatisfied with a derivation and questioning what it intends to proive if you measure the system before$t$and after$t+\Delta t$, then the system evolves unitarily in between those moments and hence there's no sense in which anything discontinuous happens @bolbteppa .. Lets stick with this: So basically I can have the wavefunction to be a dirac delta operator at t - epsilon and a exponential function at t + epsilon ... And take epsilon to 0 ... This is the definition of discontinuity ... 5:48 PM @ACuriousMind the question is about a general derivation of the time-energy uncertainty relation (literally no comment on subsystems, this sidetrack is a hoped way of validating a non-unitary Hamiltonian, which is desired because 'measurement is non-unitary' and so the question hopes that measurement being non-unitary somehow implies Heisenberg's equations are non-unitary sometimes) Given that I saw the name "joshphysics" mentioned, I'm almost certain the correct derivation has already been linked :P so if you want to have a scenario where anything 'discontinuous' happens, then the only one I can envision is when the system is measured in the midst of the$\Delta t$interval Well it's only correct if you can use calculus, but apparently we can't :p @bolbteppa ... I feel after this time I spent talking to you ... your misrepresenting my views more and more @MoreAnonymous haha 5:50 PM @bolbteppa Im not even sure if your actively trolling me at this point Oct 14 '16 at 20:44, by ACuriousMind I don't even know what calculus is @MoreAnonymous I have been trying to help you for a while now, I still haven't gotten an answer on how discontinuity of measurement even relates to the Heisenberg equations, nor how non-unitarity of measurement even relates to the Heisenberg equations of motion, I have asked in multiple ways at this stage Here is a more physically motivated example. Suppose you start with a particle in the ground state of a 1D box of length$a$, and then suddenly double the size of the box. Within the sudden approximation, you wouldn't suppose that the wavefunction suddenly changes at that moment indeed, what you'd assume is that the wavefunction starts out in the (old) ground state and unitarily evolves subject to the new potential Even more you have brought up non-standard QM theories while questioning a standard derivation Unitarity => Heisenberg Equations of motion ... Non-Unitarity of measurment => Take whole system into account to make sense Joshphysics does not do this I think I dont have to elaborate about this (as I have done before) 5:55 PM You do, what does " Non-Unitarity of measurment => Take whole system into account to make sense" even mean @MoreAnonymous If you want to describe the measurement process itself you cannot assume that it happens instantaneously. QM does not unambiguously say that the wavefunction instantaneously changes when a magic "measurement" happens. It doesn't even say that there is a unique wavefunction you can assign to a system ($\psi$-ontic vs.$\psi$-epistemic interpretations are both viable). If you are interested in describing the measurement process as an actual quantum mechanical process, look up decoherence theory. As semi-classical said: it's worth noting that time evolution of a subsystem can be non-unitary even as the entire system evolves unitarily And I think the measurement can be modelled as such (I'm guessing) The pure formalism is silent about both what a measurement is and what happens during it. It merely gives you a recipe for computing the outcome of the measurements you do in a lab - everything else is interpretation. It seems, to say it again, like you are considering a scenario where a measurement occurs during your interval$(t-\epsilon,t+\epsilon)\$
and asking whether or not the time-uncertainty relation is relevant for that time interval

@ACuriousMind I'm a bit confused ... Your saying there is no interpretation where this is right?

"I can have the wavefunction to be a dirac delta operator at t - epsilon and a exponential function at t + epsilon ... And take epsilon to 0 ... This is the definition of discontinuity"

5:59 PM
That's a pretty arbitrary and unphysical example