Tim Maudlin - What theories qualify as quantum theories without observers?

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  • เผยแพร่เมื่อ 4 ธ.ค. 2024

ความคิดเห็น • 41

  • @atmanbrahman1872
    @atmanbrahman1872 หลายเดือนก่อน +1

    Can't be done.

  • @CarlosElio82
    @CarlosElio82 2 ปีที่แล้ว +1

    Bell's postulate resembles a human being and the cells he is made of. "Macroscopic objects are made of local beables at microscopic scale." If beable is the property of being an individual, then Bohr position is not bizarre at all. My pencil is not the sum of the molecules contained in it. My pencil has the individuality of being my pencil. Silly is my example, but coherent and in agreement with Bohr. As a mathematical beable, the open set (0

  • @Mentat1231
    @Mentat1231 10 ปีที่แล้ว +3

    Finished. Fantastic lecture and q-and-a. Thank you so much for uploading this. I thought the one gentleman was a bit too insistent that mentioning superluminal particles was not only irrelevant but harmful. Sure, there are non-relativistic quantum mechanics where we can explain the violation of Bell's inequality without worrying about propogation speeds, but why shouldn't Maudlin just mention how it works in the relativistic scenarios as well, and deal with the whole issue of superluminal transmission? Indeed, I wish he had spent *more* time on it.

    • @david203
      @david203 2 ปีที่แล้ว +1

      But superluminal transmission has little to do with relativity. It has to do with entanglement. Two entangled spins are always opposite in direction, and no information is ever transmitted. For example, you can't send a spin from one place to another superluminally. But you can measure spins in two different places simultaneously or in either order. It is the nonlocal characteristic of QM, and does not involve any transmission of information at any speed. This error in understanding the EPR experiment keeps happening again and again no matter how often the physicists correct it.

  • @CarlosElio82
    @CarlosElio82 2 ปีที่แล้ว

    Seen beable as the property of individuality, helped me understand Bell's program. A photon is a beable, so is the electron it surrounds. I am a beable.

  • @darrellee8194
    @darrellee8194 ปีที่แล้ว

    OMG. I only have two years of college physics, but I feel like I’m understanding this better than some of the guys in the audience asking questions. Who are these people? 1:03:47

  • @KipIngram
    @KipIngram 2 ปีที่แล้ว

    33:50 - Wait a minute - how can the wave function have any impact on the beables? You don't GET anything tangible from the wave function without applying an observable operator. If beables are functions of the wave function, then there *is* a way to observe aspects of the wave function without going through the usual process and, one would presume, without collapsing it.

    • @zemm9003
      @zemm9003 6 หลายเดือนก่อน

      The Wave Function is the only thing that exists in Quantum Theory. Nothing else even exists. The Born Rule is something introduced externally to explain experiments but it is an ad-hoc addition that has no relation to the basic principles.

  • @boliussa
    @boliussa ปีที่แล้ว

    Should have much longer QnA, for a few hours.

  • @david203
    @david203 2 ปีที่แล้ว

    Okay. I have to say: yes, we seem to make only local measurements in QM experiments, and infer QM nonlocality using them, granted. But why does that enforce the creation of two types of beables in the new theory, local and nonlocal? Why can't Bohm's idea that all wavefunctions are subsets of the Universe's wavefunction, specifically that the experiment must include the measuring device in its wave function, be joined by the idea that all beables are nonlocal? Yes, in practical terms one might measure a nonlocal beable locally, but why did Maudlin not propose that beables only be of one type: nonlocal? I don't get it.

  • @Mentat1231
    @Mentat1231 10 ปีที่แล้ว +1

    I'm only 45 minutes into this excellent lecture, but I'm curious about the reasons why superluminal particles are being discounted. Maudlin seems to dismiss them off-hand and implies that we should already know the reasons why such couldn't exist. I'm not versed in this part of physics well enough to know why that's the case. Can anyone name a reason?

    • @ChrisSchiebelbein
      @ChrisSchiebelbein 10 ปีที่แล้ว +1

      I noticed at the beginning it says Quantum Theory without Observers III.
      Do you know if there is a I & II that I'm missing? I'm also somewhat an amateur with physics, but this does look compelling enough to watch.

    • @metatron5199
      @metatron5199 8 ปีที่แล้ว

      Relativity is the reason, and also a quantum theory without observers would make the discernment that Bell's theorem in fact proves the universe to be non local specifically in regards to the experiment that proved bell's theorem throw showing spatial non locality . Which than implies ER=EPR on some level.

    • @Mentat1231
      @Mentat1231 8 ปีที่แล้ว +2

      Meta Tron
      I was hoping someone would say "Relativity" since I would then respond (in the spirit of Bell himself) that a Neo-Lorentzian version of Relativity allows superluminal propogation and so technically the most parsimonious way to look at non-locality is to say it shows Neo-Lorentzianism to be superior to Einstein's theory. Quentin Smith has written a paper in which he defends this view and then goes further to show how a Bohmian view of QM couples with a Neo-Lorentzian view of Relativity and yields an alternative to GTR which is equally good but has none of the problems that GTR has getting along with QM.

    • @david203
      @david203 2 ปีที่แล้ว

      But superluminal information transmission has little to do with relativity. It has to do with entanglement. Two entangled spins are always opposite in direction, and no information is ever transmitted. For example, you can't send a spin from one place to another superluminally. But you can measure spins in two different places simultaneously or in either order. It is the nonlocal characteristic of QM, and does not involve any transmission of information at any speed. This error in understanding the EPR experiment keeps happening again and again no matter how often the physicists correct it.

  • @david203
    @david203 2 ปีที่แล้ว +1

    Why is the uncertainty principle often presented as part of quantum mechanics, when it is entirely a classical law? So long as time exists and is relevant to experiments, the position of a particle and its momentum or energy are going to be dependent on each other simply because of the way they are defined! Specifically, speed is the derivative of position with respect to time. That is its classical AND quantum definition! Dependency on each other means that there could be a lower bound on the accuracy of knowing/measuring both at the same time.
    Jean-Baptiste Joseph Fourier showed that measurements in the time and frequency domains are always going to have a kind of reciprocal precision: you only need one measurement to determine position in time, but you need multiple measurements to determine speed in time. It is that simple. So, no matter what position or frequency/momentum values you choose, the error of their product will have a lower bound, right?
    But that's just the Heisenberg uncertainty principle. It has nothing to do with quantum mechanics, just the simultaneous measurement of two dependent functions.

    • @alexanderkohler6439
      @alexanderkohler6439 2 ปีที่แล้ว +1

      The Heisenberg uncertainty principle is not a classical law, despite of Fourier's findings.
      Classical theories always attribute precise/single theoretical numerical values for ANY observable to ANY physical system (leaving thermodynamical systems aside) at ANY time. This includes observables like position and momentum. This attribution is done independently from any real measurement of observables. Additionally, classical theories themselves do not put any fundamental limitations on the precision obtainable by a real measurement of observables. Hence, you don't have a Heisenberg uncertainty principle in classical theories. One reason why you actually won't get arbitrary precision by real measurements even in a classical description, is due to uncountable many uncontrollable environmental disturbances during a measurement. Another reason why you never could get arbitrary precision by real measurements, even if you could control all of the environmental disturbances, is because classical theories are actually not capable of fully describing physical reality.
      Quantum theories attribute precise/single theoretical numerical values for A given observable to physical systems only in very exceptional circumstances, namely in case of physical systems that are in an eigenstate of the operator corresponding to that observable. In all other cases (the overwhelming majority of non-eigenstates), quantum theories attribute theoretical distributions of numerical values for that observable. Again, this attribution is done independently from any real measurement of observables. In case of eigenstates, quantum theories themselves do not put any fundamental limitation on the precision obtainable by a real measurement of the observable. In case of non-eigenstates, quantum theories however put a fundamental limitation on the obtainable precision, because they do not attribute a single numerical value for that observable in the first place, but already a whole distribution of values.
      Most physical systems are in a state, that is simultaneously a non-eigenstate of the operator corresponding to the observable for position and a non-eigenstate of the operator corresponding to the observable for momentum. For these states, quantum theories simultaneously attribute a theoretical distribution of position values and a theoretical distribution of momentum values. Again, this attribution is done independently from any real measurement. Heisenberg's uncertainty principle now states, that the product of the width of these two theoretical distributions is always bigger than a global nonzero constant. This is a pure quantum mechanical result that you don't get in classical theories. In classical theories there are simply no states that attribute distributions of values for observables. It is only single numerical values that are being attributed by classical theories, hence no uncertainties the likes of Heisenberg's.

    • @david203
      @david203 2 ปีที่แล้ว

      @@alexanderkohler6439 I didn't have time to read your long posting, but you are quite wrong in claiming that Fourier mutual precision is not identical to Heisenberg mutual precision. In Fourier, the two measurements, time and frequency, are related in the same way as in Heisenberg the two measurements, location and momentum, are related. I'm really sorry you don't understand this, but Heisenberg uncertainty is not magic and is not special quantum mysticism. It is just due to the inverse precision relationship between any function x and its related function dx/dt.

    • @alexanderkohler6439
      @alexanderkohler6439 2 ปีที่แล้ว +1

      @@david203 I did not claim that. I said the Heisenberg principle was not a classical law.
      It is not a classical law, because you simply don't have that type of uncertainty relationship between position and momentum in classical theories. In addition, classical measurements of positions and momenta have nothing to do with distributions of signals in the time and frequency domain that Fourier cared about and that you try to reference.
      Sure, for the latter you have the same sort of uncertainty relationship between the variances of the respective signal distributions in time and frequency as you have it for the variances of the respective distributions in position and momentum in quantum (not classical) theory. And yes, the mathematical reason for this similarity is the same: It is the Fourier transform that relates the two distributions in either of these pairs. But that does not change anything with respect to the fact, that you don't have that uncertainty relationship in classical theories which makes the Heisenberg principle a result of quantum mechanics and not of classical mechanics.
      In classical mechanics positions and momenta are always well defined without any uncertainties. When you describe a pendulum in terms of classical mechanics you can always specify the position and the momentum of this pendulum at any time with arbitrary precision, i.e. no uncertainties. Fourier's result about signal distributions in time and frequency does not change that.

    • @david203
      @david203 2 ปีที่แล้ว

      @@alexanderkohler6439 I'm sorry, none of this is correct, and I'm under time pressure to complete a project. The reason you can measure speed and momentum simultaneously in classical mechanics is that the "uncertainty" in precision is too small to measure with instruments of our size. It only becomes apparent at dimensions of the atom, which is unthinkably small. We'll have to agree to disagree and I hope you get a chance to learn this stuff someday.

    • @alexanderkohler6439
      @alexanderkohler6439 2 ปีที่แล้ว +1

      ​@@david203 Classical mechanics is a mathematical theory that exists independently from any measurements one might or might not perfom in the real world. This theory uses mathematical objects like symplectic manifolds (phase spaces) and precise coordinates (position and momentum) on these manifolds in its goal to describe the state of systems that you find in the real world. These mathematical objects are not aware of any uncertainties in position and momentum that you actually observe in the real world when you "look" very closely.
      This is why there is no Heisenberg principle in classical mechanics as a mathematical theory.
      This is why classical mechanics as a mathematical theory ultimately fails in its goal to give a correct description of the real world.
      This is why you need to have another mathematical theory that uses other mathematical objects than classical mechanics does in order to achieve the goal of getting a correct description of the real world.
      That theory has to use mathematical objects that are inherently aware of these uncertainties that you observe in the real world. There actually is such a mathematical theory, called quantum mechanics which incorporates all this. The Heisenberg principle is a result of that mathematical theory called quantum mechanics. The Heisenberg principle is not a result of the mathematical theory called classical mechanics.

  • @robbie_
    @robbie_ 3 ปีที่แล้ว

    Tim, concerning tachyons....

    • @timothycoffman3436
      @timothycoffman3436 2 ปีที่แล้ว

      Well, do they exist?

    • @david203
      @david203 2 ปีที่แล้ว

      @@timothycoffman3436 No. In spite of the fact that people want to believe weird things!

  • @mylittleelectron6606
    @mylittleelectron6606 ปีที่แล้ว

    Great job as always Dr Maudlin. As a side note, it's amazing how people insist on supplying justification for people saying that some people just don't listen...I guess it's full employment for Dr Maudlin though. Does the questioner ruffled by tachyons not understand that the goal is to produce a theory with an ontological description that correlates with reality, not just a tool for predicting experimental outcomes. Who cares that non-relativistic quantum mechanics explains the data by assigning no propagation speed. Is the gravitational force communicated instantaneously? How about the electromagnetic force? So we can go back to Newtonian gravitational theory? If not, then what characteristics of a particle aside from being massless and chargeless allows particles at the quantum level to travel faster than photons? If there are no restrictions on propagation, then why do photons have a constant finite speed? The point is, the questioner seems unconcerned by describing our shared reality using theories that contradict each other. The whole point is to avoid this embarrassment.

  • @david_porthouse
    @david_porthouse 2 ปีที่แล้ว +1

    According to Bell we need a nonlocal theory. Five minutes spent playing around with the Minkowski formalism will suggest that there is more than one way to travel faster than light. After another five minutes we might think about associating one of these ways with wavelike behaviour, and the other way with particle-like behaviour. If someone can show that I have the associations the wrong way round, I would be delighted. Otherwise it's a matter of working out how to make use of tachyonic Brownian motion in a computer simulation which will allow us to do numerical experiments with detectors made of antimatter.

    • @david203
      @david203 2 ปีที่แล้ว

      There is absolutely no way to travel or send information faster than light. This was proven by Einstein and has never been disproven. There is no such thing as "tachyonic Brownian motion" and no tachyons have ever been discovered.

  • @davidwilkie9551
    @davidwilkie9551 2 ปีที่แล้ว

    ONE-INFINITY Singularity Apature of entangled wave-packaging formation is the simplest way to recognise Euler's e-Pi-i infinitesimal relative-timing by holographic AM-FM relative-timing modulation shaping, and all wave cause-effect is totally interpenetrating at this CentreofLogarithmicTime, projection-drawing Communication = . dt axial-tangential zero-infinity Entanglement limit.
    Looking through the Observer's nothing in No-thing vertex "hole in Time" vortex of fractal point-line-circle conic-cyclonic coherence-cohesion i-reflection, instantaneous reference-framing of transverse trancendental Pi-bifurcation containment.

  • @darrellee8194
    @darrellee8194 ปีที่แล้ว

    Funny. I’ve listened two or three lectures now on Bell, and up until seeing it written out on the slide 15:12 . I though they had been talking about ‘vehicles’ not ‘beables’ Vehicles made sense to me because they were things that could carry an observation to an observable. Beables is an awful term.

    • @ricomajestic
      @ricomajestic 7 หลายเดือนก่อน

      Vehicles makes no sense in this context! Nothing is being carried or transported.

  • @mitesh8utube
    @mitesh8utube 3 ปีที่แล้ว

    WTF does that all mean?

    • @david203
      @david203 2 ปีที่แล้ว

      It all means that Bohm and Bell were right and Bohr was and still is misleading. But shush! It's a secret because few physicists know enough to believe it; we don't want to insult them.

  • @JohnVKaravitis
    @JohnVKaravitis 5 ปีที่แล้ว +2

    He's REPEATEDLY violated the "no more than 12 words in a powerpoint slide" rule. For shame!

    • @davecurry8305
      @davecurry8305 2 ปีที่แล้ว +1

      I have an old Pickett with lots of log log scales. Seriously, though, the slides are a distraction. I just close my eyes and listen intently.

  • @krigs_1434
    @krigs_1434 2 ปีที่แล้ว

    De la grosse cam 🔥

  • @5ty717
    @5ty717 8 หลายเดือนก่อน

    Can’t listen with hysterical overhead. Just tone down a little TM