@MoreAnonymous Yes, I'm saying you're not allowed to take $\epsilon$ to 0. QM makes predictions about what pre-measurement states lead to which post-measurement states, and measurement itself is a black box. You're not allowed to make the black box vanish.
If we could do that, we would have solved the measurement problem :P
So then in that case is it okay to be unsatisfied with the "Heisenberg Equaitons of motion" within the context of the "black box" ... which joshphysics answer seems to rely on?
@MoreAnonymous Yes. That dissatisfaction is precisely what is called the "measurement problem" and the reason for the existence of the plethora of competing QM interpretations.
I see ... in that case I feel I have been unfairly downvoted :/// ... Maybe someone could post exactly what your saying? (I would but I fear misrepresentation and more downvotes)
As I said, you can model measurements e.g. with decoherence, by which you essentially shift the black box from the moment of the measurement apparatus interacting with the measured system to the moment of some experimenter interacting with the measurement apparatus
@bolbteppa They are not "wrong", but they simply do not apply to the measurement black box. Very simply: The Born rule is not a consequence of the Heisenberg equations of motion.
Let a quantum system with Hamiltonian $H$ be given. Suppose the system occupies a pure state $|\psi(t)\rangle$ determined by the Hamiltonian evolution. For any observable $\Omega$ we use the shorthand
$$
\langle \Omega \rangle = \langle \psi(t)|\Omega|\psi(t)\rangle.
$$
One can show that (s...
Great so now we have a top voted answer convincing people on the grounds of subtle ( and perhaps also unsatisfactory) grounds talking about an uncertainty principle which in this case people misleadingly think what he's talking about also applies to the measuremnt
@MoreAnonymous If you are talking about the downvotes on physics.stackexchange.com/q/501145/50583, then the problem with that question is: 1. It assumes that measurement is usually thought of as some "discontinuous process", which it isn't, and only after what you've written here I understand why you thought so.
2. The question focuses on some particular uncertainty principle when in fact you can't talk about the state "during" a measurement at all (more precisely: about the state during the period where you put the "black box", which as I said can be shifting depending on how you model everything)
The last few messages on here have convinced @MoreAnonymous that his suspicion that continuity and unitarity are illegitimate is okay, because 'the measurement problem', this is absurd haha
Time-energy involves two measurements at two separate times
Time is not an operator
It is still completely unexplained how 'discontinuity of measurements' even matters for measurements done at two separate times when all you care about is two separate ones
@MoreAnonymous So, because Wikipedia uses the word "discontinuous" once to describe von Neumann's collapse postulates, you have decided that there is someone out there who models measurement like in your $\epsilon\to 0$ approach? Ignoring that it directly after that talks about different interpretations not requiring such collapse?
I really think the problem with the question is that it asks something very specific when you really are confused about something much more general (see point 2. above).
@MoreAnonymous No one said that. You don't need different derivations. joshphysics answer is the derivation of the time-energy uncertainty in all interpretations.
This sidetrack about subsystems can be completely eliminated by modelling a measurement apparatus as a quasiclassical wave function and the system you're measuring as it's own system and then deriving the time-energy uncertainty relation for the total energy of the system (c.f. Landau)
I would love to ask that general thing you are referring to ... But after the down vote experience ... I would love to ask you what it is you think Im refering to so that I can ask it?
@bolbteppa "I don't know why pretty much every popular QM piece has to include this noble outsider vs. rigid establishmentarian nonsense" Is the quote not accurate?
@bolbteppa as far as I know the many worlders dont think so
@ACuriousMind then can you tell me why my objection assuming collpase of the wavefunciton is wrong when clearly "joshphysics answer is the derivation of the time-energy uncertainty in all interpretations."
@user76284 the quote is accurate but it's still ridiculous, framed to portray people espousing outsider QM interpretations (which the author is partial to, coincidentally) that are extremely contentious at best as noble future-seers, in a general public article
@MoreAnonymous The derivation simply doesn't apply to that, just like the Heisenberg equations of motion. Again, measurement is a black box - QM does not model what happens during it.
As josh defines, $\Delta t$ is simply an abstract measure for "how quickly the system's expectation values are changing", and the principle relates this to $\Delta E$.
If you understand how the Schrödinger equation works, the principle is really not surprising: It is saying that the smaller $\Delta E$ of a state is, the slower its expectation values are changing.
The states where expectation values aren't changing at all ($\Delta t \to \infty$) are the stationary states - eigenstates of the Hamiltonian, which have $\Delta E = 0$.
@user76284 the article implies these outsider 'interpretations', that mainly live in philosophy classes and legitimately have all sorts of problems and very debatably are just flat-out wrong and illegitimate, are legitimate ways of approaching QM and worse the scary establishment physicists actually "don’t seem to want to understand it" as if these alternative approaches somehow offer insight that wasn't addressed (and very debatably invalidated) at least over 50 years ago
and it gives the usual history examples which magically paint the outsiders as oppressed victims without any indication that these alternatives were directly debated decades ago and discarded
The farther a state is from being a stationary state, the more it is changing. Truly a much less mysterious statement than many try to pretend the time-energy uncertainty principle is, but it has the advantage of being true.
It is silly, in my opinion, to blithely dismiss questions of interpretation as only relevant in philosophy classes. But it is also silly to suppose that the physics establishment didn't have compelling reasons to adopt the interpretation of the formalism that they did.
So basically in the joshphysics answer of the uncertainity principle your talking about how you can never do 2 measurements immediately one after the other becasue we assume that there has to be a unitarity evolution between these measurements ... Which is the essence what the time energy uncertainity princple is about ... In which case just assume it?? @ACuriousMind
What I've always found amusing, though, is how different people 'won' the interpretation question. on the one hand, when it comes to physics textbooks, you very much have the Schrodinger equation front and center
@bolbteppa Let's go step by step: "that mainly live in philosophy classes" Are you implying they don't live in physics classes? Are you implying they're of no physical concern?
@bolbteppa "legitimately have all sorts of problems" The standard model has all sorts of problems. General relativity has all sorts of problems. Does this imply they're not legitimate subjects of physics, in your view?
@ACuriousMind if the joshphysics answer only refers to unitarity then I hand it to you it is correct ... But most people try to make uncertainity princples connnect to the measruemtns
@bolbteppa "very debatably are just flat-out wrong and illegitimate" Which interpretation are you referring to, and why do you claim it's "flat-out wrong and illegitimate"?
@Semiclassical Sure, if you want to measure uncertainty, you need to get infinitely many identically prepared states, measure them, compute the standard deviation, yadda, yadda, yadda, but that's all got nothing to do with stating or deriving the principle.
@bolbteppa "as if these alternative approaches somehow offer insight that wasn't addressed (and very debatably invalidated) at least over 50 years ago" Wait, are you serious? You're saying no interpretations provide ANY new insights?
@bolbteppa "magically paint the outsiders as oppressed victims without any indication that these alternatives were directly debated decades ago and discarded" No idea what the word "magically" is supposed to mean here, but regarding "these alternatives were directly debated decades ago and discarded", how precisely were they "discarded"? That's news to me, and I'm sure it's news to all the people working in this area.
@bolbteppa @Semiclassical Let's do this slowly: 1. The expectation value of a single state for any observable is well-defined (mathematically, I do not claim you can measure it on a single state). 2. I can define $\Delta E$ and $\Delta t$ purely in terms of expectation values of observables (and their time derivatives). 3. Therefore, the uncertainty principle is a purely formal statement about the (mathematical) properties of a single state.
I think what I've got in mind is that a frequent -application- of the time-uncertainty relation is to situations where two measurements are made and $\Delta t$ is the time between them. (whether that is a -sound- application is what I'm wondering)
@Semiclassical Well, the "time-energy uncertainty" joshphysics' answer is about is not about that, it's about something completely different (the "the farther you are from being an energy eigenstate, the faster you change your exp. values" I talked about above).
@ACuriousMind "It tells you the approximate amount of time it takes for the expectation value of an observable to change by a standard deviation provided the system is in a pure state" which involves two measurements of energy at two times!
@Semiclassical Admittedly, josh's (and Griffith's and most into QM texts that actually derive a "time-energy" relation, lest it seem this is some idiosyncratic notion of josh's) notion of $\Delta t$ is not easily experimentally accessible
An expectation value exists whether you decide to measure its random variable or not. I did not expect this to a controversial or at all difficult notion.
@ACuriousMind well the formality is confusing you about the meaning of time-energy, it literally makes no sense to be using $\Delta t$ and $\Delta E$ talking about one measurement at one time
@ACuriousMind one fun bit of pilot wave stuff: The delta-function may not make sense as a wavefunction, but you can nevertheless derive a sensible set of trajectories from it
@ACuriousMind can you atlest explictly state which experiement joshphysics derivation applies to in the lab .. Its okay to say the set of experiements is the null case?
From section 44 of Landau, the $\Delta E$ in the time-uncertainty relation is the difference "between two exactly measured values of the energy $E +\varepsilon$ at two different instants, and not the uncertainty in the value of the energy at a given instant" (for a system of two parts with energies $E$ and measuring apparatus energy $\varepsilon$)
@bolbteppa do they use the joshphysics derivation as well?
Also I feel someone should take the moral responsibility and post an answer in mine or the joshphysics answered question (atleast to clarify somethings) ...
@MoreAnonymous It doesn't "apply" to an experiment. As I read it, it is merely a generic formal statement like the Schrödinger equation (the Schrödinger equation also applies to "no experiment", as we cannot measure the value of the wavefunction).
@MoreAnonymous joshphysics answer is technical, detailed, and correct. It does not mention any "measurements" or otherwise mislead the user. I see little room to improve it, and I have trouble understanding how it led to the impressions you seem to have gotten.
@bolbteppa The fact that they talk about the energy of the measurement apparatus already tells you this must be a different "time-energy uncertainty" than in joshphysics' answer since there's no measurement apparatus there at all
to be fair if i only recently learned that the Hesenberg equaitons of motion dont apply during the measurement .. I would be pretty shaken and start researching ..
@Semiclassical and yet we can all agree (yes including me) that it is valid ... But here Im not sure about joshphysics derivation
If someone can contact joshphysics directly that might be the best so we can ask him what experiment had in mind
In fact I just saw this comment of Joshphysics where he seems to know what Im talking about:
However, I've never delved deeply enough into the literature to be convinced that what people mean by this principle can in general be formalized such that it follows mathematically from Shrodinger evolution.
@MoreAnonymous I don't think he had any particular experiment in mind. But here's one: Take an ensemble of identically prepared quantum states, and split it into three: On the first ensemble, measure energy (now). On the second ensemble, measure some other observable (now). On the third, wait (for some $t$) and measure the same observable again.
@ACuriousMind maybe you if this is where it's applicabble then I agree but the mileage people seem to think it has seems dubious and should be clarified ... And this is fairly commmon which is what I think Cosmos Zach answer my question with (assuming he has responsibly read and understood whtat joshphysics is on about)
Compute the standard deviation of energy and the mean values of the other observable at both time. Do this often enough and with varying $t$s to find a $t$ where the difference between the two mean values is the standard deviation of the first measurement of that observable.
That $t$ is your approximation of $\Delta t$, and the deviation of energy is your approximation of $\Delta E$.
@ACuriousMind Please take my comments to clarify what joshphysics derivation is about seriously ... Because people are not under that impression and nor do they have the level you seem to have ...
@ACuriousMind the derivation in Griffiths is talking about the time derivative of the expected value of an operator, i.e. how fast the expected measurement is changing, this is where it involves two measurements, the whole derivation is looking for some way to measure this comparison between two measurements, just because they define some random thing and call it a definition doesn't somehow throw away the basic meaning of $\Delta E$ and $\Delta t$ involving two values at two times...
I admit I'm a little frustrated. long ago I asked questions from Physic that I didn't know the answer and these questions were ignored. Now I know a lot of answers. So I ask questions and give them answers showing that the questions made sense and are correct.
example
How to calculate the intern...
@MoreAnonymous Do you mean that there is a danger that people think that joshphysics' answer somehow is meant to be a proof of the pop-sci version of the "time-energy uncertainty" where it is regularly used to support such nonsense as "borrowing energy for a short time" and so on?
@bolbteppa If you look at my proposed toy measurement above I'm perfectly aware that attempting to measure the quantities involved here necessitates measurements at different times.
@ACuriousMind no ... But they assume they can get waaaaay more milage from it than they can (not as scary as the pop-sci version) but still waaaay more than your example
@MoreAnonymous And how does josh's answer support that? He's explicit about what $\Delta t$ and $\Delta E$ mean, I honestly do not see how one could misinterpret that unless one wanted to.
"In the system to which the quantum mechanical formalism is applied, it is of course possible to include any intermediate auxiliary agency employed in the measuring processes. . . . The only significant point is that in each case some ultimate measuring instruments, like the scales and clocks which determine the frame of space-time coordination— on which, in the last resort, even the definition of momentum and energy quantities rest—must always be described entirely on classical lines, and consequently be kept outside the system subject to quantum
@ACuriousMind obviously your too smart to relate to the rest of our misgivings ... But I feel that throughout this conversation you have been the only one to make sense to me ... Because I was under the assumption of its broad use ... I mean through this chat history and Zach's answer ... How can you not be under impression that people are having misgivings?
@MoreAnonymous Oh, I understand that there are many misunderstandings surrounding QM, no question about that! I just don't think that josh's answer is the source of that...
@MoreAnonymous and a few minutes ago you led people into convincing you the measurement problem implied that the Heisenberg equations sometimes did not apply, need to be careful about agreeing on points so fast!
Maybe I should mention that there are other "time-energy uncertainty relations", as alluded to in the answer by Nikos M. on the same question, or as mentioned by Martin in physics.stackexchange.com/a/259350/50583
Time-energy is very interesting, but I am really shocked at some of these claims e.g. that calculus doesn't apply or that unitarity of the Heisenberg equations can magically fail
That is, you shouldn't walk away from here with the impression that every application of time-energy uncertainty that is not covered by the Mandelstam-Tamm relation is bogus
@ACuriousMind yes ... But people are clearly thinking of josh's answer in contexts it does apply in ... And despite his best efforts his answer seems to enable them to do so ...
Going from M-T back to the other point, though: I do not think it 'obvious' that measurement is a unitary evolution. at a pragmatic level, at least, it is certainly not
@bolbteppa You're being, once again, extremely uncharitable. MoreAnonymous was certainly a bit confused, but I see no evidence of the malice you seem to imply.
@ACuriousMind I seriously think this answer needs more sentences explictly stating what it can and can't do ... I think bolbteppa thinks otherwise and hence the malice and downvotes
@MoreAnonymous I think maybe you think josh's answer is answering a different question? The question he is answering is not "What are the implications of the time-energy uncertainty principle?", but "What is $\Delta t$?"
@Semiclassical to be fair it was you who seemed to allude that this answer which I was specifically refering to seemed to imply a statement about 2 measurements ...
Whether or not this definition of $\Delta t$ supports any particular application of "time-energy uncertainty" is simply outside the scope of the question
@Semiclassical maybe then we can agree that this answer didnt exactly help you ...
@ACuriousMind Can we atleast agree this answer as is is not really helping most people understand where the implications of it are? (in fact Id b curious as to what the distribution of helps vs not help is like)
(or this :p ) Definitely not implying malice here, apologies if it comes off that way, but these statements are asking about non-unitarity of Heisenberg's equations, even asking for a derivation of them in the non-unitary case, the whole issue is questioning josh's use of unitarity (and even continuity) and whether this is legitimate
@MoreAnonymous that's like complaining that the instructions for how to build a doghouse don't tell you what the implications of owning a doghouse are
the question he was answering was, as ACM said, not "what does the time-energy uncertainty relation tell us about the world". it was what the $\Delta t$ was
$\Delta t$ has a definition in terms of time, the idea that we need to re-define what $\Delta t$ is is ridiculous, the question is if we can constrain the other quantities arising in the inequality in terms of the known thing called time...
Historically people tried to define a time operator, and the Griffith's derivation is basically a more correct form of the historically wrong definition of time-energy which actually used a time operator
@Semiclassical ... While I agree with the premise that when we use electricity ... automatically we dont know its gonna be commercially successful ... I feel it is a different case for trying to use (or waste) time using classical EM for uncharged particles
The time-energy uncertainty relation literally implies that position measurements in relativistic theories are impossible, it nearly destroyed all of quantum mechanics in the 30's, it completely invalidates the concept of a position-space wave function, something these 'alternative' theories still seem to have no idea about
@bolbteppa yes in fact in Grienier I read how to derive that dirac delta function in QM is essentially a gaussian approximate ... But Im not sure about the relevance of any of this?
@bolbteppa I think that's a strange statement, and it is rather the non-existence of a time-operator for systems with bounded energy that already poses considerable problems for "proper" relativstic position.
So let me get the main stream position on why josh's answer is gonna stay the same:
"Yes basically now we have a top voted derivation in physics where people are uncertain with the physical system it applies to in fact its okay for them to even misuse it"
And
"In physicsstackexchange were okay with the above quote ... And even if brought to our attention ... It okay!"
Of course it's bad when people misuse "time-energy uncertainty", but frankly people misuse half-understood QM concepts all the time, and it's not that answer's fault. We don't need to right all that's wrong with the world, sometimes it's enough to just give a well-defined answer to a well-defined question
@bolbteppa Oh, historically you may well be right! I'm honestly not very interested in history and one of my pet peeves is teaching physics as history when hindsight could suggest much better structures.
I mean .. Im of the opinion a well defined answer should not be able to confuse people who have degrees (I suspect masters included) ... I feel that alone should be a sufficient definition of well defined
Let a quantum system with Hamiltonian $H$ be given. Suppose the system occupies a pure state $|\psi(t)\rangle$ determined by the Hamiltonian evolution. For any observable $\Omega$ we use the shorthand
$$
\langle \Omega \rangle = \langle \psi(t)|\Omega|\psi(t)\rangle.
$$
One can show that (s...
