If the state variable, x, actually contains all their is to possibly know, then yes, locality would be broken if these inequality were broken. But our history says it's far more likely that any experiment we could run would fail that precondition, we simply can't know 100% experimentally. Experimentation has entropy, even when theory doesn't. The hidden variable is very likely to be in the 'x' itself. We also can't assume c is the speed of causality in general, only in our frame of reference, and only if there's a not yet discovered way of breaking it. Entanglement shows that either there's a way to cause changes faster than c, or there's some still unknown measurable (hidden variable) that connects the two systems. That said, this is by far the best video series on this I've seen on KZbin, though I am still far from convinced that Bell's disproves hidden variable theory, but instead proves it. It does put a damper on EPR's argument though.
@DavidRoux-uv3bl15 күн бұрын
Enjoyable video. What is a "spin orbital"? Confusing terminology.
@DavidRoux-uv3bl15 күн бұрын
Very clear. Brilliant.
@sabamacx27 күн бұрын
Absolutely fascinating.
@wmstuckeyАй бұрын
It’s ironic that Michelson and Morley discovered the null result for the ether drift experiment revealing the observer-independence of the speed of light c, and the “fine structure” of the 656-nm hydrogen line revealing the existence of electron spin. Quantum information theorists have actually linked these two results en route to a “principle” understanding of what’s going on with spin. We explain this in our book, "Einstein's Entanglement: Bell Inequalities, Relativity, and the Qubit" Oxford UP (2024) and I’ll summarize it here. According to Einstein, special relativity (SR) is a "principle theory," i.e., a theory whose formalism follows from an empirically discovered fact. For SR that empirically discovered fact is the light postulate - everyone measures the same value for the speed of light c, regardless of their relative motions. Since c is a constant of Nature according to Maxwell's electromagnetism, the relativity principle (the laws of physics are the same in all inertial reference frames) says it must be the same in all inertial reference frames. And, since inertial reference frames are related by uniform relative motions (boosts), the relativity principle tells us the light postulate must obtain, whence the Lorentz transformations of SR. Likewise, quantum information theorists have rendered QM a principle theory and its empirically discovered fact is called Information Invariance & Continuity (Brukner & Zeilinger, 2009). In more physical terms, Information Invariance & Continuity entails that everyone measures the same value for Planck's constant h, regardless of their relative spatial orientations or locations (let me call that the "Planck postulate"). Since h is a constant of Nature according to Planck's radiation law, the relativity principle says it must be the same in all inertial reference frames. And, since inertial reference frames are related by relative orientations or locations in space (rotations or translations), the relativity principle tells us the Planck postulate must obtain, whence the finite-dimensional Hilbert space of QM. Quantum superposition is one consequence of the Planck postulate and here is how it makes perfect sense using spin as shown in this video. Suppose you send a vertical spin up electron to Stern-Gerlach (SG) magnets oriented at 60 deg relative to the vertical. Since spin is a form of angular momentum, classical mechanics says the amount of the vertical +1 angular momentum that you should measure at 60 deg is +1*cos(60) = 1/2 (in units of hbar/2). But the SG measurement of electron spin constitutes a measurement of h (Weinberg, 2017), so everyone has to get the same +/- 1 for a spin measurement in any SG spatial orientation, which means you can't get what you expect from common sense classical mechanics. Instead, QM says the measurement of a vertical spin up electron at 60 deg will produce +1 with a probability of 0.75 and it will produce -1 with a probability of 0.25, so the average is (+1 + 1 + 1 - 1)/4 = 1/2. In other words, QM says you get the common sense classical result on 'average only' because of the observer-independence of h. Give up your dynamical bias for QM (just as is done for SR) and the physics of spin makes perfect sense. And if you explore the Bell states for spin-entangled pairs using this approach, you find it solves the mystery of entanglement without violating locality (as in Bohm’s pilot wave), statistical independence (as in superdeterminism or retrocausality), intersubjective agreement (as in QBism), or the uniqueness of experimental outcomes (as in Many Worlds).
@davidhuo6902Ай бұрын
clearly and concisely
@tictacX12 ай бұрын
I watched a few videos on this subject, many of them jump a few steps. With this one, i finally get how they prove entanglement is real! Great video, thanks!
@alexgoldhaber17862 ай бұрын
3:52 - - The second term in the equation may be hard to understand but not to the point that you call it 'jerk'!
@mrslave412 ай бұрын
6:57 bohr: "renunciation of the classical ideal of causality..." so bell proved epr are wrong because epr is right??? lol :) this is absolute non-sense BY DEFINITION of NON-SENSE.
@mrslave412 ай бұрын
0:51 "can quantum-mechanical... be considered complete?" - good question. bell's theorem literally proves EPR to be CORRECT. proof by contradiction: assume QM is complete. QM leads to contradiction with our reality. Therefore, QM is not complete. DONE. 🥂
@mrslave412 ай бұрын
2:42 quantum particles are waves. so this discussion needs rephrasing.
@mrslave412 ай бұрын
0:43 there’s nothing wrong with particle wave duality. what classic physics is violated?
@timvw013 ай бұрын
What happens to the light after it has transferred momentum to the charged particle? Is its wavelength longer?
@GH-li3wj3 ай бұрын
If the correlations between the measures violate the Bell 's inequalities there is the possibility that you don't measure a , b or c but you measure the correlations C(a,b) between a and b C(a,c) and C(b,c) , measuring a value is not the same as computing a value, that 'all. Physic is local. If C(a,b), C(a,c) and C(b,c) were realy computed you could not violate Bell 's inequelities unless Physics is not local. This is not the case then the hypothesis that the correlation are computed is wrong. It's what happened If you looked closely to the Aspect ' experiment. looks here kzbin.info/www/bejne/Z4bQnZKcob-nbcU
@_dp3po_8813 ай бұрын
I don’t understand why the three probabilities aren’t: A and not B, B and not C, and C and not A… why would the last one be A and not C? I havnt watched part 2 yet but the explanations of this always seem to purposefully tweaked to ensure a specific outcome. Your explanation has been very clear and just enough in depth. I really appreciate it. I just always struggle to grasp the reasoning at this point. Any help? Or should I just move on to part 2?
@davidcarr22163 ай бұрын
Right there is the most stunning sequence of videos on YT. And a fabulous introduction to Quantum Chemistry for anyone considering studying it formally.
@davidcarr22163 ай бұрын
I'm just seeing your videos for the first time (starting with Q.M.). I'm both deeply impressed and deeply thankful. Most explanations are either too simplistic or they bury the viewer in maths.
@AnnoyingCritic-is7rp3 ай бұрын
Floathead physics also does this.
@willbarron12593 ай бұрын
This is by far the best explanation of Bell’s inequality, and i thoroughly enjoyed the presenters detailed and balanced description of classical and quantum mechanical positions on the subject. I do disagree with the analysis of the available data. Superposition and wave function as currently defined by QM seems irrational to me. Most if not all correlations that violate Bell’s Inequality are a product of misapplied testing methodology, and the failure to account for the angular potential of measurement acceptance being reduced to binary pass/fail, up/down spin. If entanglement can to be considered a local hidden variable, then that combined with the wide angular acceptance vector being reduced to a binary pass/fail outcome, it can be understood that correlation seemingly in violation of Bell’s Inequality is actually classically mandated.
@NicholsonNeisler-fz3gi4 ай бұрын
I see what you did there
@justanotherguy4694 ай бұрын
Thank you for this clear and concise lecture.
@fxy174 ай бұрын
not "invisible" but stationary. still makes more sense than "imaginary particles"
@Underduress20005 ай бұрын
Best video I have ever seen on this topic.
@ExistenceUniversity5 ай бұрын
This was the most helpful video I have found thus far for helping me understand the thought experiment Einstein formulated in his 1907 paper in the 18th section titled "Space and Time in a Uniformly Accelerated Reference System". However, Einstein finds the local time of the Accelerated reference frame at high numbers becomes t'*[1+(gamma x'/c^2)] I was curious if this was for similar reasoning
@JohnFairstein5 ай бұрын
Thank you!
@AntiCSFC5 ай бұрын
Uploaded 10 years ago? I find this now? Thats crazy. The way you are showing ACTUAL experiment results leading to every new fact about reality is great and unparalleled.
@detlevtiszauer94196 ай бұрын
I like the thorough treatment with references to the original papers on the major concepts. Would have liked to hear though that the electron wave function is not a scaler but a spinor.
@ragnaarminnesota67036 ай бұрын
Reality is non-local. For that to be true, the speed of light limit is violated. And violated by something we can't explain. You end by saying we don't know. I say the speed of light remains the undefeated champion. I am with Sabine on this. Measurement independence. What does that mean when in the light cone of the experiment is you deciding how the experiment will be conducted? When they did the Quaser experiment for the Nobel Prize on this subject, that Quaser was in a sufficient light cone too.
@ryandupuis58606 ай бұрын
I'm not sure why Rutherford, with his gold foil experiment, assumed that the Thomson model was completely wrong and replaced the charge clouds with orbiting points. If the nucleus is found to be much smaller and heavier than the electron and its orbit, then simply switch the sizes of the clouds in the Thomson model! Instead of the positively charged cloud being large, shrink it to be the appropriate size and keep the electron's orbit as a negatively charged cloud around it. Maybe I'm speaking in retrospect, but I think people hinged too much on particles being point objects since they had no reason to assume that at all (their shape had never been seen).
@ryandupuis58606 ай бұрын
Thomson had it more right than Rutherford, thinking of the particles not as discrete points, but clouds
@gst93256 ай бұрын
there are no particles. particle is more or less frozen energy wave
@geraldinejasnin73787 ай бұрын
possible en francais !!
@benjaminkaufmann24827 ай бұрын
I find the way you explain the "quantum jump" misleading. Every quantum mechanical system (even in quantum field theory) is described by continuous equations, and so every quantum system changes continuously. In other words, the transition from one electron state to another occurs through a continuous change in the wave function. The wave function never changes abruptly, as indicated in the video. Only the macroscopic measurement result seems to have this jumpy behavior. This means that if we ask the quantum system what state it is in, we will get a "discrete answer" - this state or that state. The wave function, however, changes continuously. Edit: And also in QFT, the creation of a photon is a continous process! (Only the macroscopic oberservation shows "jumpy nature".)
@viascience6 ай бұрын
Wave functions can undergo discontinuous changes. See en.wikipedia.org/wiki/Wave_function_collapse
@benjaminkaufmann24826 ай бұрын
@viascience I was referring to 4:37 where it says that “radiated transitions must occur in quantum jumps”. This is not true. In all quantum mechanical calculations involving radiation, the time evolution is continuous. There is no sudden change in wave functions. The sudden change in wave functions only occurs during a measurement. And what really happens there is still not clear.
@DavidRoux-uv3bl15 күн бұрын
@@benjaminkaufmann2482 this makes me ask, what is meant by "continuous" and "sudden".
@KetanSingh7 ай бұрын
This series is really good. I come back to it for revision every few years. Btw, how can I see more of those visualisations similar to the occilating charge radiating?
@benjaminkaufmann24827 ай бұрын
The series is really good! But in this episode I see a mistake. The description of incoherent light is incorrect. Incoherent does not mean that the average field is zero.
@viascience6 ай бұрын
For incoherent light the expected value of the field is zero. This is true in both the quantum description and the classical description. If the field is cos(omega*t+phi) with phi random, then the expected value <cos(omega*t+phi)> = 0 at any time t.
@benjaminkaufmann24826 ай бұрын
@@viascience I reffered to 5:27 where it's said: "At a given time in place the total field being the sum of a large number of random contributions have an expected value of zero. That's the nature of incoherent light." This is not true. At 5:53 it's said about coherent light: "At a given time in place the total field ideally will be the sum of n identical contributions and have a defined expected value." Yes, but this is also true for incoherent light! In your comment you write "the expected value <cos(omega*t+phi)> = 0 at any time t". Now you mix up the average in time and the average at a certain point in space. The average in time is zero for coherent and incoherent light, but the average of n coherent contributions at a certain point in space is not necessarily zero in classical physics - neither for coherent nor for incoherent light. BUT in quantum mechanics, for the single photon field, the expectation value of the field at a certain point in space is always zero! This is completely different from the classical field theory.
@benjaminkaufmann24826 ай бұрын
@@viascience I reffered to 5:27 where it's said: "At a given time in place the total field being the sum of a large number of random contributions have an expected value of zero. That's the nature of incoherent light." This is not true. At 5:53 it's said about coherent light: "At a given time in place the total field ideally will be the sum of n identical contributions and have a defined expected value." Yes, but this is also true for incoherent light! In your comment you write "the expected value <cos(omega*t+phi)> = 0 at any time t". Now you mix up the average in time and the average at a certain point in space. The average in time is zero for coherent and incoherent light, but the average of n coherent contributions at a certain point in space is not necessarily zero in classical physics - neither for coherent nor for incoherent light. BUT in quantum mechanics, for the single photon field, the expectation value of the field at a certain point in space is always zero! This is completely different from the classical field theory.
@sylviachung25677 ай бұрын
life saver. thank you for the most precise and understandable explanation!!!!
@adriangheorghe23277 ай бұрын
De la anihilarea electronului cu pozitronul avem: h=(k.Qe^2)/(Re.Ffae)=(Qe/Re).(Qe/Tfae).Tfae.(k.Tfae)=(Uf.If.Tf).Df; rezulta k=adimensiunal, ;rezulta epsilon0=1/4.pi.k=adimensional; rezulta Farad/metru=adimensional; rezulta farad=lungime L. From the annihilation of the electron with the positron we have: h=(k.Qe^2)/(Re.Ffae)=(Qe/Re).(Qe/Tfae).Tfae.(k.Tfae)=(Uf.If.Tf ).Df; result k=dimensionless, ; result epsilon0=1/4.pi.k=dimensionless; results Farad/meter=dimensionless; results farad = length L.
@II-th7bn7 ай бұрын
😂😂 you tell the meaning of vector, as if it is very hard . But skipked many hard terms
@tajulislamkhan10627 ай бұрын
Simply brilliant.
@adriangheorghe23277 ай бұрын
De la anihilarea electronului cu pozitronul avem: h=(k.Qe^2)/(Re.Ffae)=(Qe/Re).(Qe/Tfae).Tfae.(k.Tfae)=(Uf.If.Tfae).(k.Tfae)=Wlu.Df. rezulta; k=adimensional; rezulta epsilon0=adimensional; rezulta (Farad/metru)=adimensional; rezulta Farad=lungime. From the annihilation of the electron with the positron we have: h=(k.Qe^2)/(Re.Ffae)=(Qe/Re).(Qe/Tfae).Tfae.(k.Tfae)=(Uf.If.Tfae ).(k.Tfae)=Wlu.Df. result; k=dimensionless; results epsilon0=dimensionless; result (Farad/meter)=dimensionless; results Farad = length.
@TheDavidlloydjones7 ай бұрын
Kwannamechanics. Who knew?
@blockhead18998 ай бұрын
I tried my absolute hardest to follow the math but there is so much I don’t understand and this is supposed to be the basics??? Do you know any videos that explain the math more indeph for beginners pls?
@rossholst53158 ай бұрын
How do you know if the particles you are testing are entangled? It would seem that entangled particles would always have some anti correlation if measured along a specific direction. So you could measure two entangled particles and get a spin up and spin down when measured along the same axis. But if we change it up and measure 1 particles along 1 axis and another particle along a different axis. We will get 1 measurement along the 1st axis, if it was entangled we would get the anti correlation. But instead we measure at a different angle. Now we get an orientation along this new axis and if it was entangled we would hypothetically know what its orientation along another axis. However can we still assume the expected result of a measurement we didn’t preform? What if the particles were not actually entangled? We would be able to tell it wasn’t entangled by measuring along the same axis as the first, but we used a different setting. So would it now be possible that they were not originally entangled? Also if we assume that information travels faster than the speed of light between these entangled photons, could the information about if they are entangled also travel faster than the speed of light. In some sense could we generate entangled photon pairs that don’t share a creation event? Where they are measured before enough time had elapsed that they could influence each other? I would be much more convinced of quantum “weirdness” and less supporting of hidden variables, if entangled photons did not have some shared location in space and time that was shared between the two. Because unfortunately we can never take entangled pairs and get them outside of their temporal cones to rule out a hidden variable theory. I would also think that any hidden variable theory should also make the same predictions as quantum mechanics. I don’t think a hidden variable theory should predict linear changes on an oscillation. Waves are products of acceleration, and thus predictions should be exponential. I would think they would be highly related to exponential trig functions sine and cosine. Having a linear prediction for probabilities seems like expecting a linear velocity of the x directional component for an object rotating at a constant angular velocity. Which would seem impossible if rotation was occurring. So I still don’t see why “classical physics” should make predictions that are linear when describing an exponential process?
@rossholst53158 ай бұрын
Also it would seem that the probability equation would only hold true if we repeatedly made observations on single photon, or multiple photons that shared the same states as the original photon. If the starting states of the photons are changing all of the time, it would not make sense to expect this inequality to necessarily hold. If we are measuring kids with hats, gloves, and scarves, would we expect this inequality to be true if for every measurement we used a new group of kids, who all had different proportions of each garments? The only way for the inequality to hold true would be to have a way in which the starting conditions remained exactly the same for each test. But if the starting conditions can change between tests these probabilities might actually be independent of each other. Also if we flip a coin, we expect the probability to get heads or tails is 50%. But what are the chances that 99 coin flips ends with an equal number of heads and tails. If we just started flipping a coin, we would only assume the average would trend to 50% as the number of attempts increased. However in all likelihood if we just looked at the difference between heads and tails, we would probably observe an oscillating pattern where heads went up by a few and then tails went up by a few. But every other throw one of the outcomes must be greater than the other as on an odd number of attempts the totals cannot be equal.
@sivaprasadkodukula79998 ай бұрын
Very good
@sivaprasadkodukula79998 ай бұрын
Very good
@dylanledermann86298 ай бұрын
6:25 Unfortunately, you seem to be confusing amplitude with displacement. We're talking about the probabilities of the current displacement being in some range dx for all the individual modes, amplitude is the max displacement for all the modes, these are NOT the same things!
@RooftopDuvet8 ай бұрын
The way I usually see this stuff taught tends to avoid langrange multipliers etc, as those tools are probably alien to a new audience (as they were to me when I first learned about it). However, now that I'm more comfortable with those tools, the approach here is so much more elegant... It's the first time I've appreciated how we can think about general systems as an abstract gas in phase space, and how things like Boltzman's law are actually just things that fall out of the geometric constraints on that 'gas'. Makes you see the 'mechanics' in 'statistical mechanics' much more clearly. Thanks a lot
@TheAlexfrascari9 ай бұрын
10 years and Just found this Channel. Still ond of the best
@ricardovencio9 ай бұрын
Clear and focused explanation. Chanels use lots of visual aids (or handsome faces) but here it is simple clean drawings and even better speach to convey the messg. Awesome