Brian Cox and the Pauli Exclusion Principle

[atyy]

I am not sure the answer is yes, even on an immeasurable level. For stronger correlations, like those found in EPR or quantum eraser experiments, measuring/interfering with a particle has 0 effect on the expectation values for the 2nd observer measuring the other, entangled, particle.

Yes, indeed. I answered too hastily, I still had in mind your original question as to whether he changes the state of the other electron.

Edit: As well as confusing various meanings of effects as Ken G notes just above.

So let's see if this is correct, at least for the Bell experiments. There are two entangled particles (not identical for simplicity). Measuring one and getting a particular result causes the wave function to collapse across an entire spacelike hypersurface, so it does affect the state of both particles. However, no classical information is communicated faster than light, since only when the two experimentalists get together will they see the correlations in their results.

 

[Ken G]

So let's see if this is correct, at least for the Bell experiments. There are two entangled particles (not identical for simplicity). Measuring one and getting a particular result causes the wave function to collapse across an entire spacelike hypersurface, so it does affect the state of both particles. However, no classical information is communicated faster than light, since only when the two experimentalists get together will they see the correlations in their results.

I think to call that "correct" you would want to relax the "causes collapse" part, that's really just one interpretation (Copenhagen), which involves "causing the physicist to adopt a different wave function." In interpretations that treat the wave function as something real that evolves unitarily, all that is affected is the evolving correlations, no collapse is occurring. The key difference is that a physicist can globally change a wave function to reflect new information without any "signals" being propagated (and ensemble-type interpretations can insert their own language here), whereas in the treatments where the wave function only evolves unitarily, it does not suddenly change globally. Since all we ever check are the ultimate correlations, we don't get to know which picture is more correct!

 

[atyy]

I think to call that "correct" you would want to relax the "causes collapse" part, that's really just one interpretation (Copenhagen), which involves "causing the physicist to adopt a different wave function." In interpretations that treat the wave function as something real that evolves unitarily, all that is affected is the evolving correlations, no collapse is occurring. The key difference is that a physicist can globally change a wave function to reflect new information without any "signals" being propagated (and ensemble-type interpretations can insert their own language here), whereas in the treatments where the wave function only evolves unitarily, it does not suddenly change globally. Since all we ever check are the ultimate correlations, we don't get to know which picture is more correct!

I agree, only in Copenhagen, not dBB or many-worlds where there is no collapse.

One thing I don't quite understand is - is the anti-symmetrization of the wave function enough to enforce a Bell type nonlocality for two identical fermions?

 

[Morberticus]

Yes, indeed. I answered too hastily, I still had in mind your original question as to whether he changes the state of the other electron.

Edit: As well as confusing various meanings of effects as Ken G notes just above.

So let's see if this is correct, at least for the Bell experiments. There are two entangled particles (not identical for simplicity). Measuring one and getting a particular result causes the wave function to collapse across an entire spacelike hypersurface, so it does affect the state of both particles. However, no classical information is communicated faster than light, since only when the two experimentalists get together will they see the correlations in their results.

If the wavefunction is a subjective description of the system, its collapse or decoherence won't be a problem for relativity because nothing is physically being collapsed.

When observer 1 makes a measurement, his knowledge of the system is updated. Observer two, however, has not made a measurement, so he does not reduce the wavefunction. He still traces over all the degrees of freedom of observer two. The probability that he observes a spin state X has not changed. It is still:

P(observer 2 sees spin state X) = Sum_over_Y P(observer 2 sees spin state X | observer 1 sees spin state Y)

 

[Ken G]

Certainly yes if we have a total state that isn't a simple tensor product of component states but rather superpositions thereof then the composite system being described by the total state cannot be meaningfully decomposed as two separate subsystems in respective specific states.

Yes exactly, so if we are being completely precise we should not talk about the "states of the electrons within the diamonds" and the "states of the electrons halfway across the universe" for two separate reasons: there are not actually individual separate electron states here, and there are not even individual separate electrons here either! Of course we get away with not worrying about that, but since Cox is taking even the tiniest prediction of quantum mechanics extremely literally, he should be consistent with that attitude, although I do believe he feels that might be too technical. So the fundamental issue is, when should we "dumb down" certain aspects of quantum mechanics, while retaining other aspects completely literally? Does that merely create confusion?

 

[atyy]

One thing I don't quite understand is - is the anti-symmetrization of the wave function enough to enforce a Bell type nonlocality for two identical fermions?

Here's a paper that may be relevant http://arxiv.org/abs/quant-ph/0401065

 

[Ken G]

One thing I don't quite understand is - is the anti-symmetrization of the wave function enough to enforce a Bell type nonlocality for two identical fermions?

You already pointed out that the Bell state exists even in distinguishable particles, so we know Bell nonlocality does not necessarily involve indistinguishability. You are wondering about the converse-- does indistinguishability imply Bell nonlocality, even in a theory where all interactions are local and all signals are subluminal? I think the two issues are pretty orthogonal, since all Bell calculations I've seen effectively assume the particles are distinguishable. But is there some phenomenon we are overlooking by not accounting for indistinguishability? Perhaps so, because we cannot necessarily assume the individual state wavefunctions that we are superimposing are not themselves global. So there's always some tiny overlap, tracing back to the Big Bang, and always some tiny change when we account for indistinguishability. That must also be felt in the Bell-type correlations, though very tiny of course. This is all within the theory of quantum mechanics of course, quantum field theory might have something to add, and we should also remember that understanding the ramifications of any theory at an observationally untestable level is a dubious exercise to begin with!

ETA: yes, that paper you link would seem to be an excellent way to get to the heart of this matter.

 

[Morberticus]

This discussion has been very helpful. Thanks to all involved. I think the conclusion is, in a sentence:

The probability P(x) that you measure x in a distant part of the universe is not affected by Brian rubbing a diamond, because it is always

P(x) = P(x | Brian rubs diamond) + P(x | Brian doesn't run diamond)

[edit] - Correction (thanks to Ken G): Should be

P(x) = P(x \cap Brian rubs diamond) + P(x \cap Brian doesn't run diamond)

 

[Ken G]

This discussion has been very helpful. Thanks to all involved. I think the conclusion is, in a sentence:

The probability P(x) that you measure x in a distant part of the universe is not affected by Brian rubbing a diamond, because it is always

P(x) = P(x | Brian rubs diamond) + P(x | Brian doesn't run diamond)

If P(x|y) means "the probability of x given y", then I believe you mean, if you are out of the light cone, then P(x | Brian rubs diamond or Brian doesn't rub diamond) = P(x | Brian rubs diamond)*P(Brian rubs diamond) + P(x | Brian doesn't rub diamond)*P(Brian doesn't rub diamond) = P(x | Brian does whatever), since we must have P(x | Brian rubs diamond) = P(x | Brian doesn't rub diamond) = P(x | Brian does whatever).

 

[Jano L.]

That is not the same thing! That just says that if we do a measurement on the electrons of the universe, we will get that N are in that atom. That does not mean we have a set of N electrons there!

You' re almost incomprehensible to me here. There is no measurement of electrons involved. Disregarding that, if "we get that N are in that atom", that surely $does~mean$ we have a set of N electrons there, trivially. All results of theoretical chemistry for atom or molecule are, as far as I know, consistent with individual presence of definite number of electrons - that is the basic assumption behind all calculations and derived approximate methods like Hartree method or Hartree-Fock method. Adopting antisymmetric wave functions does not in any way imply that the electrons lost their individuality (it does not prove that they have it either). The simplest picture is, the atoms of chemical elements have definite number of electrons, and they are there, and after reflecting the results, they seem best described with anti-symmetric wave function.

The indistinguishability of electrons clearly disallows us to imagine we are dealing with a set of some N electrons, all we can say is that electrons will show up in groups of N, and we don't know what electrons those will be. Of course this distinction is never of any importance, yet it is there all the same, so we would only mention it if we were trying to amaze an audience about how strange the universe is.

Metal balls in a bearing are often indistinguishable in their properties to human as well. Yet there is definite number of them in the bearing. You may not be able to say which is which based on their photo, but nevertheless this is not a reason to deny their individuality.

Similarly, you may not be able to determine which electron is which from probability density generated by anti-symmetric wave function (cf. photo), but this does not in any way imply that there are not individual electrons there.

There are always measuring apparatuses involved, this is physics. The equations are intermediaries between initial and final observations, and demonstrably so, that's just how physics works. I know you realize this, I am clarifying my point.

Actually I do not agree. That's not necessarily the way how $theoretical~physics$ has to work. Measuring apparatuses came into theoretical physics in 20's and that was quite bad thing to happen to it. There is no satisfactory theory of measurement in quantum theory after almost a century.

Yet physics did evolve at marvelous pace for more than 300 years before that. Would you say the work of Kopernik, Newton, Maxwell, Boltzmann etc. was not physics because there were no apparatuses involved in their theory?

The theory of indistinguishability that leads to fermionic and bosonic statistics.~$is~a~reason~to~think~that~electrons~cannot~be~treated~as~individual~particles$

(Italic mine)
Could you please give a reference that shows that and could you please explain at which step exactly?

 

[Morberticus]

If P(x|y) means "the probability of x given y", then I believe you mean, if you are out of the light cone, then P(x | Brian rubs diamond or Brian doesn't rub diamond) = P(x | Brian rubs diamond)*P(Brian rubs diamond) + P(x | Brian doesn't rub diamond)*P(Brian doesn't rub diamond) = P(x | Brian does whatever), since we must have P(x | Brian rubs diamond) = P(x | Brian doesn't rub diamond) = P(x | Brian does whatever).

Sorry, I wasn't using correct notation.

By | I meant "and".

 

[Ken G]

Sorry, I wasn't using correct notation.

By | I meant "and".

OK, but even so, it is important that P(x) be determinable independently of what Brian does. Correlations between x and Brian's diamond can require that we know what Brian does, but P(x) by itself doesn't, even when we are being exact-- that is I believe the point you are making. But even that statement might require an interpretation of QM-- in some interpretations, Brian is not free to do whatever he wants, so it's a little unclear just what P(x | Brian does something) means!

 

[Jano L.]

... Certainly yes if we have a total state that isn't a simple tensor product of component states but rather superpositions thereof then the composite system being described by the total state cannot be meaningfully decomposed as two separate subsystems in respective specific states.

It depends on what you mean by "state" and by "meaningfull decomposition". If you just need to derive density operator for sub-system $\rho_1$ from density operator of composite system $R$, that can be done, although the equation of motion for the density operator of the sub-system will be formidably complicated. I believe one can even derive function $\psi$ for sub-system that will give adequate probabilities, although this is much more cumbersome to do practically.

Indeed, superposition of tensor products is often not $factorizable$ into tensor product, but beware, this $does~not~mean$ one cannot describe the subsystem by $\psi$ or by $\rho$. This happened for all successful applications of Schr. equation and quantum theory - one can often forget that the system interacts with environment and neglect the correlations.

 

[Morberticus]

OK, but even so, it is important that P(x) be determinable independently of what Brian does. Correlations between x and Brian's diamond can require that we know what Brian does, but P(x) by itself doesn't, even when we are being exact-- that is I believe the point you are making. But even that statement might require an interpretation of QM-- in some interpretations, Brian is not free to do whatever he wants, so it's a little unclear just what P(x | Brian does something) means!

I believe that is guaranteed if "Brian rubbing/not rubbing diamond" is a complete, orthogonal basis set.

 

[WannabeNewton]

Indeed, superposition of tensor products is often not $factorizable$ into tensor product, but beware, this $does~not~mean$ one cannot describe the subsystem by $\psi$ or by $\rho$. This happened for all successful applications of Schr. equation and quantum theory - one can often forget that the system interacts with environment and neglect the correlations.

Indeed I was referring to the inability to write the total state (or density operator if mixed states are involved) as a factorized (uncorrelated) product of states of the subsystems. I have no disagreements with you regarding the ability to describe the subsystems themselves by specific states-I was referring to the composite system itself.

That being said, I think "indistinguishably" is being taken way too literally in the physical sense here.

P.S. if you want to italicize you can use

 

[Morberticus]

I believe that is guaranteed if "Brian rubbing/not rubbing diamond" is a complete, orthogonal basis set.

Nevermind. I see what you are saying. In non-relativistic qm, the conditional probabilities would have to be the same, which is not obvious. I am in danger of begging the question.

 

[Ken G]

Could you please give a reference that shows that and could you please explain at which step exactly?

People, indistinguishability is indistinguishability. If you cannot distinguish the electrons, you cannot say you have some here, and some other ones over there, that's perfectly clear even if you imagine you are exchanging the electrons rather than the measurement coordinates (despite my points about why that is essentially meaningless). So why do some of you seem to think electrons are somewhat indistinguishable and somewhat distinguishable? The fact is, they are formally indistinguishable, but we get away with distinguishing them in practice in our usual ways of doing that. Why should that seem controversial?

 

[bhobba]

People, indistinguishability is indistinguishability.

Yes.

But when electrons are bound to nucli in atoms I question that indistinguishably because its entangled with the nucleus which means its state is dependent on the state of the nucleus. Atoms can be bosons or fermions depending on their total spin even though the electrons they contain are fermions.

Bound electrons cloud the issue considerably IMHO.

Thanks
Bill

 

[Ken G]

Yes.

But when electrons are bound to nucli in atoms I question that indistinguishably because its entangled with the nucleus which means its state is dependent on the state of the nucleus.

Certainly true. Nevertheless, any such entanglement must be consistent with the basic indistinguishability of all the electrons in the universe, or it is simply incorrectly written. That's formal quantum mechanics talking, not me. Of course I realize we would never actually do that, we deviate from the rules of formal quantum mechanics whenever we do real calculations, we have just learned how to do that judiciously. I wouldn't even mention it except in a thread about injudicious connections between electrons in diamonds and everywhere else. Cox is trying to be formally correct, but he's only doing so with part of the issue.

Atoms can be bosons or fermions depending on their total spin even though the electrons they contain are fermions.

And that makes the electrons distinguishable how?

 

[bhobba]

And that makes the electrons distinguishable how?

It means when bound to an atom you can't really consider it as separate particles - you have to consider it as a system because they are entangled.

The reason electrons can't be in the same state is if electron 1 is in state |u1> and electron 2 in state |u2> the composite system is |u1>|u2>. If |u1> and |u2> are the same state then nothing happens on exchange in contradiction to the fact it must change sign. BUT if bound to an atom its entangled with the nucleus and you can't specify the state of the electron by itself and the argument breaks down. You must consider it as a whole and when that is done it could be a fermion on boson.

Added Later:
I think I may see Kens point. Yes, even in bound atoms you can't tell the difference between electrons. What I am saying is, even though that's still true, the consequences are different - it doesn't necessarily lead to they can't be in the same state. Or maybe another way of looking at it is being entangled with another system, its already in a different state anyway to anything else not also so entangled such as other electrons in the atom.

Thanks
Bill

 

[atyy]

No. It's called the Cluster Decomposition Principle.

I found this article which explains a bit why Bill K's answer is right. The confusing thing for me was that spatially separated entangled states of non-identical particles are nonlocal in the sense of Bell, because they produce distant correlations, and for them entanglement can also be defined by not being a product state. The anti-symmetrization of identical fermions does seem to make them "entangled" by the latter definition, yet is not enough to make them nonlocal.

http://arxiv.org/abs/1009.4147

 

[Ken G]

It means when bound to an atom you can't really consider it as separate particles - you have to consider it as a system because they are entangled.

I agree you can't consider the electrons separately from the rest of the atom, it's a bound system. What I'm doing is further adding that you also can't consider "the electrons in the atom" separately from the electrons in the rest of the universe, because that would be to imply they are distinguishable, and they are not. Of course in practice we do allow ourselves, though formally incorrectly, to use the language "the electrons in the atom", as long as we realize we really mean "whatever electrons show up in our measurement on the atom". They could be any electrons in the universe, and indeed quantum mechanics is quite explicit on this point, that's why the "total wave function" of the universe (if you are the type to believe there is meaning in such a thing, which Brian Cox clearly is) does not say which electrons are in that atom and which electrons aren't. The identity of the electrons in the atom never appears anywhere in the wavefunction of the universe, and importantly so-- that's where the PEP comes from. Granted, the PEP really only cares about electrons with overlapping wavefunctions, but here we are in Brian Cox-land where there is just one wavefunction for the whole universe, which contains all the information in the universe, and since that wavefunction lacks information about the identity of the electrons in that atom, there is, in formal quantum mechanics, no such thing as "the electrons in that atom." Of course this only matters for the tiniest of correlations, but that's what this thread is all about.

The reason electrons can't be in the same state is if electron 1 is in state |u1> and electron 2 in state |u2> the composite system is |u1>|u2>. If |u1> and |u2> are the same state then nothing happens on exchange in contradiction to the fact it must change sign. BUT if bound to an atom its entangled with the nucleus and you can't specify the state of the electron by itself and the argument breaks down. You must consider it as a whole and when that is done it could be a fermion on boson.

All the same, atoms have the structure they have because of the PEP. The entanglements with the nucleus do not change that, though I admit I've never thought about how it might subtly alter the structure of an atom to think about entanglements and not just interaction energies and exchange terms. Has anyone?

Added Later:
I think I may see Kens point. Yes, even in bound atoms you can't tell the difference between electrons. What I am saying is, even though that's still true, the consequences are different - it doesn't necessarily lead to they can't be in the same state.

But it generally does-- we have the level structure of atoms as a result. I presume you are talking about tiny deviations from the simple application of the PEP to electron states that are unentangled with the nucleus, and I've never even thought about those possible repercussions, though we know they have to be small. So I've not disputed your point about entanglements, it was the language about an atom taking its electrons with it that can't be right because "the atom's electrons" is not a formally correct concept. Instead, what the atom "takes with it" are the coordinates of whatever measurement we are doing on the atom, not "its electrons," because the latter require distinguishing that which cannot, by any experiment, be distinguished.

So I'm saying that formally, if we are being more precise than is generally necessary, all we can say is that the total wavefunction of all the electrons in the universe must contain aspects that guarantee we'll usually find some N electrons in that atom if we look. Entanglements with the nucleus, and the interaction energies, are what ensure that, but they are still buried in the evolution of the total wavefunction in ways that does not allow the electrons in that atom to be formally recognized as individual entities, and there's always some probability they will tunnel out of the atom and exchange with other electrons that tunnel in. Not only can no observation rule that out, occasionally we find an observation that requires it. The rest of the time, we'd be nuts to try to take that into account, and a lot of what we do in quantum mechanics is actually a "manual" approximation that we know will work. But even if we are being unnecessarily precise, we agree the universal wavefunction does not propagate signals halfway across the universe. But it certainly could swap electrons halfway across the universe, and not only couldn't we tell the difference, it is one of the central tenets of quantum mechanics that we couldn't tell the difference.

 

[Ken G]

I found this article which explains a bit why Bill K's answer is right. The confusing thing for me was that spatially separated entangled states of non-identical particles are nonlocal in the sense of Bell, because they produce distant correlations, and for them entanglement can also be defined by not being a product state. The anti-symmetrization of identical fermions does seem to make them "entangled" by the latter definition, yet is not enough to make them nonlocal.

http://arxiv.org/abs/1009.4147

Sure, but that's not what makes them nonlocal. What makes them nonlocal, formally and if we are being unnecessarily precise and following a probably improper allegiance to the literal details of quantum mechanics, is that their states have evolved within the universal wavefunction from the time of the Big Bang. Put differently, no wavefunctions are completely nonoverlapping, so the cluster decomposition principle is an approximation, whereas formal quantum mechanics is not (if we believe that, anyway). Brian Cox can only be held accountable to formal quantum mechanics, not to approximations like the cluster decomposition principle or other practical matters that allow us to disregard the law of formal quantum mechanics that all electrons are fundamentally and completely indistinguishable.

 

[atyy]

Sure, but that's not what makes them nonlocal. What makes them nonlocal, formally and if we are being unnecessarily precise and following a probably improper allegiance to the literal details of quantum mechanics, is that their states have evolved within the universal wavefunction from the time of the Big Bang. Put differently, no wavefunctions are completely nonoverlapping, so the cluster decomposition principle is an approximation, whereas formal quantum mechanics is not (if we believe that, anyway). Brian Cox can only be held accountable to formal quantum mechanics, not to approximations like the cluster decomposition principle or other practical matters that allow us to disregard the law of formal quantum mechanics that all electrons are fundamentally and completely indistinguishable.

I don't understand this overlapping part - is that something that would hold also for distinguishable particles?

 

[Ken G]

I don't understand this overlapping part - is that something that would hold also for distinguishable particles?

Sure, but the ramifications would be different. For one thing, overlap of indistinguishable fermions produces an "exchange energy" term in the interaction energy. But for the PEP, we don't need any interaction energy, so no exchange energy, we can just look at counting the states. So a quantum statistical mechanical application would be most appropriate. Two overlapping states, once expanded on eigenstates, would normally have a nonzero amplitude of being found in the same eigenstate. But the antisymmetrized wave function would cancel that amplitude. If there is no overlap, there is no such cancellation, because there are no amplitudes for having the same quantum numbers for two completely separated wavefunctions. However, Cox might have been referring to the idea that there is always some tiny amplitude of being anywhere, so there's always overlap, so there's always a need for the antisymmetrized wavefunction to cancel amplitudes of the same quantum numbers, when observations of those quantum numbers are being carried out.

 

[Morberticus]

Here's a decent answer on stack exchange (This time in the context of two election "energy levels" in wells with large separation (ignoring other quantum numbers for the moment))

http://physics.stackexchange.com/qu...-a-two-fermion-double-well-system/22304#22304

"one can't measure the energy "in one well only" with the accuracy needed to distinguish E1 and E2"

"If your measurement apparatus is confined to the vicinity of one well, the error in your energy measurement can't be smaller than E1−E2 so you won't be able to say "which of the two nearby states" the electron is in. The same holds for the vicinity of the other well which is why the measurement in one well can't influence anything detectable near the other well."

So it seems this type of correlation is permitted because of the fundamental uncertainty present in quantum mechanics. It's not just imperceptible, it's formally imperceptible.

 

[Ken G]

I think that post is actually saying something similar to what I said, just in another way, and I can show this. Their key point is that you can't measure the energy "in one well only" with the accuracy needed to distinguish E1 and E2, but why would you want to measure the energy "in one well only" in the first place? That would be a strange interpretation of a quantum mechanical energy measurement, we should actually be looking for energy eigenstates of the joint wavefunction! If we do that, we get no confusion, the two-electron system with two widely separated wells has just a single ground state, and a first excited state that is imperceptibly close. In the limit that the two wells are infinitely separated, that looks like two separate and independent ground states of just one electron each, but remember, that is not the situation if we are not taking that limit-- if we are trying to be absurdly precise, as Brian Cox is essentially doing.

If we stick to the absurdly precise non-limit, the presence of two slightly different energy states of a two-electron system, not two separate ground states of two one-electron systems, is a manifestation of the indistinguishability of the electrons, and the need to write a joint wave function that is not only antisymmetric under exchange of the observing coordinates, but also respects that basic indistinguishability. The failure to respect that indistinguishability is what motivates language like "measure the energy of one of the electrons", which translates into "measure the energy of one well only" if you think you can distinguish the electrons. But neither of those statements make formal sense in quantum mechanics, an energy measurement is an energy eigenvalue of the whole system, pure and simple.

So what I'm saying is, Cox's error is in using language that suggests you can talk about "the energy of the electrons in a diamond", in the same breath that he says all the electrons in the universe obey the PEP. If you use the former language, you are treating the electrons as distinguishable, so you get no PEP. If you use the PEP, you don't have "electrons in the diamond", you just have the joint wavefunction of all the electrons, and their eigenstates represent the Hamiltonian of the whole universe, not just the diamond. That is tantamount to saying there is some tiny probability they can tunnel to other places and change places with the electrons that are over there, or if you prefer, the simple fact that you never know which electrons you are really experimenting on, you only know various probabilities and expectation values that you will get when you set up your apparatus and define your measuring coordinates.

Finally, I note that I characterize Cox's language as an "error" because of its inconsistency, but he might hold that the inconsistency is justified if part of the truth is too technical, and the other part isn't. Is inconsistency OK when we "dumb down" one part but not the rest? I don't think so, I think we need to be consistent even when we are simplifying for a nonexpert audience, but that is a different kind of complaint than saying he doesn't understand quantum mechanics.

 

[DevilsAvocado]

I am more worried for the ones who believe they truly understand it all well enough(now i am going into hiding).

:thumbs:

"I think I can safely say that nobody understands quantum mechanics." -- Richard Feynman​

 

[DevilsAvocado]

Then again: The linked-cluster principle does not contradict Bell's findings. There are non-local correlations, but there are no actions at a distance. That's an important difference! Local relativistic QFTs of course admit the Bell correlations (vulgo entanglement), but interactions are local, and the linked-cluster principle holds, i.e., local observables do not depend on interactions/experiments at far-distant places. See Weinberg's Quantum Theory of Fields, Vol. I on this!

Okay, local [Bell] observables do not depend on interactions at far-distant places. So how does the linked-cluster principle explain this?

Real-Time Imaging of Quantum Entanglement

https://www.youtube.com/watch?v=wGkx1MUw2TU
http://www.youtube.com/embed/wGkx1MUw2TU

[URL]http://www.nature.com/srep/2013/130529/srep01914/images_article/srep01914-f4.jpg[/PLAIN]

http://www.nature.com/srep/2013/130529/srep01914/full/srep01914.html

 

[DevilsAvocado]

[...] I think we need to be consistent even when we are simplifying for a nonexpert audience, but that is a different kind of complaint than saying he doesn't understand quantum mechanics.

I think we have to remember that "A Night with the Stars" runs on BBC 2 (watched by 1.6 million viewers), where maybe only a tiny fraction has any deeper understanding of QM, and the majority knows absolutely nothing. I don't want to be rude or something, but if Brian Cox were to talk like you (or me or any other "nerd" on PF), everybody except the already "baptized" will have zapped over to "Dame Edna & Lady Gaga –– Go-go Dancing with the Stars", or something, within a minute.

There's just not any room for Brian to talk about spin, angular momentum, quantum numbers, etc, even if this is what he should have done (to make it technically correct). It just becomes too dense for Average Joe. And still there are comments like this:

"Thank goodness for Simon Pegg. And Jonathan Ross, James May and Sarah Millican. That Brian Cox lecture would have been a struggle without them, that's for sure, writes Sian Brewis. Imagine, a whole programme without a single celebrity in there. How on Earth would we poor viewers be expected to concentrate?"​

Also note that the he did not talk about FTL communication, all he said was - Everything is connected to everything else...

https://www.youtube.com/watch?v=Mn4I-f34cTI
http://www.youtube.com/embed/Mn4I-f34cTI

But this was apparently enough for "Bhagwan-Brahmaputra-Guru", and followers, to light up the "hippie-räucherkegel" and start talking about "interconnected minds" etc, sigh...

Not directly Brian's fault, is it?

My guess is that he wanted to talk about the non-local nature of QM, and he preferred something that could be 'explained' in a few minutes before the audience had lost interest. It's probably impossible to be 100% technically correct all the time in a show like this, and if Brian can make some viewers interested enough to learn more about the subject, that can't be a bad thing, can it? Who knows, maybe the next "Feynman/Schrödinger/Einstein" becomes interested in QM (instead of Lady Gaga) because of a show like this. That is just great, IMHO!

So, how 'wrong' was Brian Cox?

Well, as in all cases where people talk about QM – it's a matter of interpretations. Matt Leifer sums it up pretty well:

My interpretation of what Brian Cox was trying to say is slightly different, but not necessarily more likely. Since the Pauli exclusion principle comes from the requirement that the global wavefunction of a bunch of fermions has to be antisymmetric under exchange, you can argue that there is a sense in which all the electrons in the universe are entangled, e.g. in a universe of only 2 electrons with two possible positions “here” and “far away”, the wavefunction would have to be of the form:

psi(1,here) psi(2,far) – psi(1,far)psi(2,here)

This looks like it is entangled, but there is nothing you could do to prove it by local measurements, since you can’t do a measurement that distinguishes 1 from 2 in any way.

Of course, anyone who has read Sakurai knows that, if we only have access to observables in the “here” region, then this state will be indistinguishable from a single particle state psi(here), which is manifestly local. If you are one of those people who think of wavefunctions as being physically real, then there is one description in which Brian is correct and one in which he is incorrect and since they are operationally equivalent it is impossible to adjudicate. Of course, if you insist on only regarding measurement outcomes as physically real, then you will never observe any nonlocality this way, and I don’t think anyone would argue with that.

There are obviously always different views:

https://www.youtube.com/watch?v=ASZWediSfTU
http://www.youtube.com/embed/ASZWediSfTU

 

[Ken G]

My guess is that he wanted to talk about the non-local nature of QM, and he preferred something that could be 'explained' in a few minutes before the audience had lost interest. It's probably impossible to be 100% technically correct all the time in a show like this, and if Brian can make some viewers interested enough to learn more about the subject, that can't be a bad thing, can it?

I basically agree, and I think he provides more of a service by getting people jazzed about it than he does a disservice by feeding the lunatic fringe. I'm just saying that if we are to critique the formal correctness of his language, the main problem would be failing to enforce the indistinguishability within a set of electrons that he is also invoking the PEP to talk about. I think you can make the same points he is making, and get the same people jazzed about them, without feeding the misconception that we have one set of distinguishable electrons within a diamond, and a different set halfway across the universe, and they affect each other via the PEP-- when the PEP is demonstrably a principle relating to particles that you cannot possibly distinguish. How important is it to get that right? I don't know, I think it's important when it's not any harder than getting it wrong, but everyone has to kind of draw their own line there.

Well, as in all cases where people talk about QM – it's a matter of interpretations. Matt Leifer sums it up pretty well:

Indeed I said something very similar earlier-- we have never tested the kinds of effects that Cox is talking about, so we can't really say "the universe is like this", because some don't accept that there "really is" any such thing as a universal wave function. But Cox does believe in that concept, and he could use language that is consistent with that concept, but he is only partly doing that, and partly not doing that. It is the inconsistency that is my issue.

ETA: let me put it this way. Sometimes when we say something that sounds a little too shocking, it is because we said it wrong. This is probably one of those cases. If one wants to invoke the PEP to explain energy shifts of electrons halfway across the universe, one needs to recognize that in the context of formal QM, and in particular the PEP, there is no such thing as electrons halfway across the universe, because they have to be indistinguishable to obey the PEP. Hence, to see the kinds of energy shifts he is talking about, one would need to do energy measurements on the entire universe of electrons. If one is doing energy measurements on the entire universe of electrons, it is a whole lot less surprising, but perhaps just as interesting, that what you do to a diamond can effect the energies of the electrons of the universe, expressly because the electrons of the universe are all indistinguishable from those in the diamond.

 

[Morberticus]

some don't accept that there "really is" any such thing as a universal wave function. But Cox does believe in that concept

I think Cox's position is one I sympathise with, but even non-local realism is under attack.

http://www.nature.com/nature/journal/v446/n7138/full/nature05677.html

 

[Dale]

Closed for moderation