A night with the stars (Brian Cox on telly)

Ken G

Exactly. I'm sympathetic of that problem-- we might not all agree with how Dr. Cox negotiates it, but we're all in glass houses on that score. If one person thinks Cox is doing more harm than good by stressing the more mystical elements, another can say he is doing more good than harm by simply getting people interested in some of the more fascinating new elements of what we have discovered. The fact is it might take centuries before we really understand what all this means, remember Feynman's wonderful words about quantum mechanics:
"We have always had a great deal of difficulty understanding the world view that quantum mechanics represents. At least I do, because I'm an old enough man that I haven't got to the point that this stuff is obvious to me. Okay, I still get nervous with it.... You know how it always is, every new idea, it takes a generation or two until it becomes obvious that there's no real problem. I cannot define the real problem, therefore I suspect there's no real problem, but I'm not sure there's no real problem."

unusualname

It's only a problem if our ability to "control" the material is inconsistent with global unitary evolution. ie is Brian's Cox's choice to rub the diamond any different from a diamond being shifted around underground by a natural process such as an earthquake?

When a supernova explodes it undoubtedly has a significant effect on the state vector of the universe, but it ought to be consistent with unitary evolution according to the Schrödinger Eqn.

Of course, this isn't an issue if you don't believe in macroscopic wavefunctions, especially one describing the entire universe, but in that case you need corrections to the current standard formulation of QM.

The no-communication theorem says a measurement in one place cannot change the probability distribution of any observable outside the future light-cone of the first measurement.

But science has no consensus on the nature of free-will, and such theorems may not apply. However, if free-will does break unitarity in a deterministic way then we may also need a reformulation of relativity since we would otherwise have the possibility of causal paradoxes.

Ken G

Yet that's a pretty small "but". It is a "but" that is more or less the defining quality of science!
Actually, I don't think the theorem can quite say that. A probability distribution is always contingent upon what you already regard as known, versus what unknowns you are simply averaging over. So changes in knowledge, here, can change probability distributions about distant events, reckoned here, without any causality violations (as in EPR type experiments). Hence, you can reckon that the probability distribution somewhere else, outside your light cone, can be changed by your measurement-- it is just the physicists outside your light cone that cannot know that. It's a question of what any probability distribution is contingent on.

unusualname

Yes, obviously I meant the probability distribution wrt to the observer observing the observable.

SecularSanity

Disclaimer: Pre-coffee

I thought that quantum entanglement had to be created by direct interactions between subatomic particles, but this guy says that the entire universe is in this entangled state. I don’t know but I don't like it.

Was Brian Cox Wrong?
However, I did find a poor quality video of John Bell stating, “You cannot get away with saying that there is no action at a distance. You cannot separate off from what happens in one place from what happens in another. They have to be described and explained jointly.”

Bell Himself Explaining the Implications of his Inequality

Does it prove that the entire universe is in an entangled state simply because there are methods of creating entanglement? Is quantum nonlocality equivalent to entanglement? Aren’t there limits to quantum nonlocality, e.g. Tsirelson's bound?

BTW, doesn’t he look like Johnny Depp as Willy Wonka?

“Oh, you should never, never doubt what nobody is sure about.”~ Willy Wonka

sheaf

As far as I know, the discussions on this issue are still ongoing. I thought I'd describe the situation from the viewpoint of my armchair.

Regardless of the discussions regarding whether Brian Cox should perhaps have said “quantum state”, rather than “energy level” in the TV show, this whole discussion has made me try to understand the applicability of the concept of entanglement to a situation such as this. Certainly Cox and Forshaw in their book did have entanglement in mind in connection with this issue, since they state:

Entanglement does indeed allow quantum measurements to display “instantaneous” influences, but no information can be transmitted using this mechanism. But, how would you go about applying entanglement to the scenarios they're discussing?

The model Cox and Forshaw are using is the double rectangular potential well. This model is described here. The energy eigenstates of a single rectangular well are split into pairs of energy eigenstates with very closely spaced energy eigenvalues. One member of a pair is a wavefunction with odd reflection symmetry about the origin and the other has even reflection symmetry.

We now populate the double well system with a pair of fermions. For simplicity, they could be spinless electrons, which would have to be in different states to respect their fermionic nature. As an example, they could be in each of the two lowest energy eigenstates, so the system state would be
[tex]|\Psi \rangle={1\over{\sqrt{2}}}(|E_1 \rangle |E_2 \rangle-|E_2 \rangle |E_1 \rangle) \ \ \ (0)[/tex]

The sort of question we would like to ask is whether or not there is entanglement between quantities measured in the left hand well and quantities measured in the right hand well?

Conventionally, entanglement questions would be treated by decomposing the full Hilbert space in the form
[tex]{\mathcal{H}=\mathcal{H_{L}}\otimes\mathcal{H_{R}}}[/tex]
For example, in the "classic" EPR entanglement scenario, this sort of decomposition is clear - [itex]\mathcal{H_{L}}[/itex] is the two dimensional Hilbert space of spin states of a LH-travelling spin 1/2 decay product of a spin 0 singlet state, and [itex]\mathcal{H_{R}}[/itex] the RH-travelling equivalent.
For any pure state [itex]|\Psi\rangle\in\mathcal{H}[/itex]I can choose an orthonormal basis [itex]\{|\Psi^L_{i}\rangle\}[/itex]for [itex]\mathcal{H_{L}}[/itex] and [itex]\{|\Psi^R_{i}\rangle\}[/itex]for [itex]\mathcal{H_{R}}[/itex] such that
[tex]|\Psi\rangle=\sum\limits_{i}\alpha_{i}|\Psi^L_{i} \rangle \otimes|\Psi^R_{i}\rangle \ \ \ (1)[/tex]
here [itex]\alpha_{i}[/itex] are a bunch of coefficients (which can be chosen to be real and positive). This is the Schmidt decomposition. Given this, a good measure of entanglement - namely the entanglement entropy - can be defined as
[tex]S_{A}=-\sum\limits_{i} \alpha_{i}^2log \alpha_{i}^2[/tex]
The higher the entropy of a state, the more entangled it is.

Now trying to apply this to the double well scenario, we immediately run into trouble, because it is not clear how to perform the decomposition [itex]\mathcal{H}=\mathcal{H_{L}}\otimes\mathcal{H_{R}}[/itex].

If we want to ask the question "is there any entanglement in the double well model?" a key problem is that the two electrons in the system are indistinguishable fermions, so when one tries to construct a two particle state, it must be antisymmetric in the two electron identities. For example, ignoring spins, the position wavefunction representation of a two particle state might be constructed from single particle wavefunctions as:
[tex]\Psi(x_1,x_2)={1\over{\sqrt{2}}}(\psi(x_1)\phi(x_2)-\psi(x_2)\phi(x_1)) \ \ \ (2)[/tex]
An n-particle state would be the same, except it would be a normalised sum over all the even permutations of [itex]x_1,x_2,...x_n[/itex] minus all the odd permutations. Such states/wavefuctions are sometimes called Slater determinants.

Now, there is a fairly large body of literature around which discusses entanglement in multi-fermion systems. However, much of it is concerned with treating entanglement in systems appropriate to quantum computing - for example entanglement between quantum dots. In these cases, the mere fact that you cannot express a two particle state as a product state, but rather a difference of such, like in (2), is deemed *not* to constitute entanglement. For example Shi defines entanglement in a multi-fermion system to be the inability to express the state (by choosing a suitable single particle basis) as a single Slater determinant (like (2) for the case of 2 particles). In other words, a state is *not* entangled if you *can* express it as a single Slater determinant.

Adopting this definition would immediately rule out the double well energy eigenstate (0) as being entangled – it's a single Slater determinant. But is this criterion really appropriate for the double well discussions? As far as I can tell, the reasoning behind considering (2) as unentangled has immediately made an assumption regarding remote exchange correlations, namely that they can be ignored due to the large separation. Schliemann, whilst arguing the case for using Slater rank as the entanglement criterion states ( where I've substitued the wavefunctions in (2) for his notation) states:

So by construction the double-well electrons will be unentangled if we use Slater rank as the entanglement criterion, so this doesn't really help.

There are other approaches to entanglement of fermions, such as the one discussed by Zanardi et al(http://arxiv.org/abs/quant-ph/0308043). They state that it is meaningless to discuss entanglement of a state


This approach avoids the need to perform the decomposition (1) and instead focuses on the properties of various observables on the state being checked for entanglement. The criterion of Zanardi et al seems quite complex, but its essence is captured in a simpler formulation described in a reference by Kaplan, to which I was referred by PF user Morberticus. Basically the question of whether or not a state is entangled is asked *with respect to a pair of observables* [itex]A[/itex], and [itex]B[/itex]. A state [itex]\Psi[/itex] is deemed entangled with respect to [itex]A[/itex], and [itex]B[/itex] if the covariance function
[tex]C_{AB}\equiv \langle\Psi|AB|\Psi\rangle-\langle\Psi|A|\Psi\rangle\langle\Psi|B|\Psi\rangle \ \ \ (3)[/tex]
is non zero.

However, to apply this to our double well system, we need to be able to define the operators A and B appropriate to "making an energy measurement in the LH well" and "making an energy measurement in the RH well".

The only energy operator I can think of that would be consistent in the two-fermion system would be the total energy operator [itex]E_1+E_2[/itex]. This is symmetric in permutation of the electron identities 1 and 2 as it should be. However, to evaluate (3) to check for entanglement, I'm still left with the job of defining a "left hand well energy operator" [itex]E^{A}_1+E^{A}_2[/itex] and a "right hand well energy operator" [itex]E^{B}_1+E^{B}_2[/itex].

I've no idea how to do such a thing, and I'm inclined to agree with the conclusion of Arnold Neumaier in his answer to my question on physics stackexchange (http://physics.stackexchange.com/que...le-well-system), namely that there is no simple way to progress this discussion !

Q-reeus

Apart from the no doubt important observations made in #108, as a QM outsider it strikes me there is something fishy about bcox's example. Correct me if wrong, but within say a chunk of semiconductor, where electronic wavefunctions appreciably overlap, 'instantaneous' adjustments in energy level are part of the overall energy budget book-keeping - within that notionally closed system. The bcox example of rubbed diamond is merely one part of an energy exchange system, the other being bcox (the diamond rubber). Isn't it the case the proper perspective here is one of a closed system diamond/bcox having zero net energy-mommentum change? So what basis is there for anything outside this net constant energy system to care about? Another example to my mind illustrating this might be an elastic solid bar of non-uniform cross section. Struck with equal impulses at both ends and set into vibratory motion, there will be periodically varying energy levels present. Bar momentum and energy are time invariant overall, yet owing to non-uniformity, at any given instant peak upper and lower energy density excursions will not be symmetrical. But again, will 'the rest of the universe' care at all, as long as net energy of that system is constant? Sorry if these points may have been raised before.

Ken G

The issue, as investigated is detail in #108, is whether or not indistinguishability of particles counts as "entanglement" in quantum mechanics, to the extent that we can say that changing the energy of "one electron" affects them all. But note that the indistinguishability is crucial-- so if we only look at energy conservation issues, and imagine that the "electrons in the diamond" are separate from any with whom they do not overlap, then we are begging the question. The diamond is not a set of electrons, it is whatever is happening that correlates with the coordinates of the diamond's location. There is no "set of electrons" inside a diamond, there is only a number of electrons there-- with no implied connection to which electrons that refers to.

So I think the problem is in the basic language of saying we rub a diamond and it "affects an electron"-- there is strictly no such thing as "an electron" inside that diamond, versus outside of it. Instead there is a state of all the electrons everywhere, and since they are indistinguishable, we can never say "which electron" we affected. We can't say we only affected the electrons "within the diamond" because there's no such thing, so we have to speak in terms of how we affected the dependence of the electron's state with respect to coordinates within the diamond. So we are only changing how the state vector depends on those coordinates, and we are not changing how it depends on causally unrelated coordinates. But that language does not carry over into a statement about "which electrons" we affected, because there is no such thing. So I would say we err equally in saying that we did not affect any "other electrons" as if we said we did affect any "other electrons." The language is just basically incorrect, it doesn't matter whether we claim there is or is not such an effect.

Next we can ask if it's really such a problem that the language is basically incorrect. The language is intended for a broad audience, not for quantum experts. So we should expect it will be basically incorrect. So the real question is, is it basically incorrect in a disastrous way, or does the incorrect language still manage to carry the flavor of, and convey what is amazing or profound about, the correct language? The answer to that can easily vary from person to person, but I personally am not terrible exercised about that particular way of conveying the surprising aspects of particle indistinguishability in quantum mechanics.

Q-reeus

Been a while since this thread started, and had forgotten many prior inputs. But I did recall this from the OP's #1:
Ken G, I take your point in #110 this is not just about energy levels, but Cox did say electron energy levels everywhere else would need to adjust, and that's what #109 was trying to address. OK so the state of all electrons is where it's at. Is there experimental confirmation though of truly instantaneous state/energy adjustments of the type discussed here? #107 expresses things pretty well imo (but how many will wade through that third link?!).

guillefix

I was thinking about this again (now that i have had free time after exams :P) and just wanted to know if the effect described by Brain is just entanglement. I mean the system of all electrons can be in several different states, and in all of them electrons have slightly different energies. Then measuring*one electron would make all ellectrons "collapse" to certain energies. Is it this? because I can't think of any other spooky action in qm apart from entanglement. And anyway why do all the electrons have to have different energies in each possible state anyway? The only reson I can think is what people have said that at each position the potential energy due to relative distance to protons etc is different. Please is this right? Because if so, noone here makes it clear!
Thanks!

Ken G

I get uncomfortable with the word "instantaneous", because it doesn't really mean anything. States reflect knowledge of a system, and when your knowledge changes, the state changes "instantaneously" (or at least as fast as your brain works), but that's a statement about how you regard the system's state, nothing has to "happen to" the system itself. In particular, the "state of the system" might not change at all for someone else, especially if they are outside the light cone of the measurement that changed your opinion of the particle's state. I would say that a "state" really just means "knowledge about the preparation of a system", so thinking about it that way strips the term "instantaneous adjustments" of its mystery.

Ken G

Entanglement is a somewhat different concept usually, because it is often applied to particles that are distinguishable but are well separated. When particles are indistinguishable and not well separated, they exhibit different types of correlations (like the Pauli exclusion principle) that are generally not what is meant by "entanglement" (but I suppose one could call it that also). However, when you have indistinguishable particles that are well separated, it is no longer clear if their indistinguishability matters any more-- to the extent that they can be distinguished by their separation, they have become distinguishable, but to the extent that their wave functions still overlap a tiny bit, they may yet exhibit the kinds of effects that are special to indistinguishable particles. Entanglement is a term that is often reserved for correlations that have nothing to do with distinguishability and does not require wave function overlap. So I'd say the semantics become a bit unclear here.

The force that keeps a white dwarf star from collapsing under its own gravity is pretty "spooky", but it is not action "at a distance", because the particles are very tightly packed together. It is called "degeneracy pressure", is related to the Pauli exclusion principle, requires indistinguishability of the electrons, and is generally not considered in the same breath as "entanglement."

I think Brian is imagining that perfect degeneracy (precisely the same energy in two different states) is formally impossible, just because nothing is ever perfectly equal. Of course, we should not expect quantum mechanics to be perfect either, so the distinction is a bit forced, and energy degeneracy is often a useful concept in practice.

guillefix

Ok, so then Brian's effect comes from the indistinguishability of electrons and the fact that the wavefunctions overlap. So I guess that means that if there are two protons A and B with two electrons A and B say, if I measure the electron in proton A, it might be either electron A or B, thus the only way for it to there be two electrons is if they have sligthly distinct energies.

Furhtermore, I think that what Brian said that if you heat this diamond it will "change" the electrons in that star is refering that the electrons are in a pure entangled state of the two slightly different energy levels, and if I collapse this one to one state by heating it or something, then it will ""change"" the electron in the star, really it has just collapsed it too. I just read wikipedia for indistinguishability (http://en.wikipedia.org/wiki/Identical_particles) and although some parts I can't follow I do get that indistinguishable fermions are in the entangled state I just described. Also I think that the state that distinguishable fermions can have is just an approximation for infinitivelly separated fermions or ones which wavefunctons don't overlap for some reason. As you say, for these ones you get "normal" entanglement.

So as I understand it, both types of correlations are a kind of entanglement, they are just caused for different reasons (one is caused by wavefunction overlapping and PEP; and the other by all other causes of entanglement like electron spin interaction etc)


What you mean? I thought that quantum mechanics predictions were exact as far as we know. So for distinguishable particles you could get perfect degeneracy, even in the real world, wouldn't you? Say in the BEC, you get bosons that are perfectly degenerate don't you? Unless uncertainty principle does something..

Lastly, for the case of the two protons, or any case really, if a new electron just popped out somewhere it will be distingishable until its wavefunction spreaded and reached other wavefunctions, won't it? And this wavefunctions don't travel faster than c, dont't they?

Ken G

Yes, personally I wouldn't have any issue with calling them both forms of "entanglement", as long as the important distinctions are made clear, though it might not be standard lexicon.

I just mean the "as far as we know" part. Been there, done that, 2000 years of science and so forth. Quantum mechanics is a mathematical structure that applies to idealized versions of the real world and gives incredible accuracy under certain rather special conditions. It isn't consistent with general relativity at the Planck scale, and so forth, but more than likely it will break down at much larger scales than that, given the many orders of magnitude of untested parameter regimes. Also, it reduces to classical mechanics in the limit of large quantum numbers, so in that limit, it can only be as exact as classical mechanics, so encounters the same need to idealize complex systems in order to make progress. What gets thrown out when we idealize complex systems? That's unknown, because the more complex the system gets, the more things we choose not to try and know about that system.

BEC are for indistinguishable particles, so they all sample the same states and can find the same one, so there is no degeneracy in the states there (degeneracy doesn't mean multiple particles in the same state, it means multiple states at the same energy).

But even if there were for some reason multiple states at the same energy, there'd be other quantum numbers to distinguish those states, and Brian Cox could have referred to them instead of talking about energy. I don't know exactly what he meant, but to me the point is that when the particles are indistinguishable, you have to focus on the differences in the states rather than the differences in the particles. If you do something to change those states, or change the accessibility of the states, then all the indistinguishable particles sample that change, expressly because they are all indistinguishable.
Yes, I think one could make a case that completely non-overlapping wavefunctions create some form of distinguishability among otherwise indistinguishable particles. So I would tend to think that only the other indistinguishable particles within the light cone of rubbing the diamond would gain access to new states, possibly contradicting his claim that they would all respond instantly. But who knows, maybe Brian Cox would ask how do you know the states are completely non-overlapping, do you know the entire history of the universe and can assure that no correlations persist from the Big Bang? This is a tricky area, involving both indistinguishability and relativity, so it's a hard problem. I think Brian is in effect deciding to frame his comments in the context of nonrelativistic quantum mechanics, which is like taking c to be infinite.

Bear in mind that we don't really make statements about reality unless we are giving specific experimental outcomes-- instead we give interpretations of theories about reality, and this requires that we choose a theory to begin with, and try to convey the essence of that theory to nonspecialists. Conveying the essence of some theory to nonspecialists should really be distinguished from making claims on how reality works "behind the curtain" where we never see.

guillefix

Well my point is that he might be right that it's an instantaneous effect if this correlation between particles behaves the same way as entanglement, which I think it does, that's why i like calling it so.. In this way possibly all particles in the observable universe are entangled in this way. Anyway, when talking about universal wavefunctions, we can just say some nice comment about the essence of our theory as you say, because trying to make any kind of predicition about reality taking into accoun the whole universe is ridicously hilarious. Still, I'm sure the ideas that come from these thought experiments will be of use in the future of both physics and technology. LOL in fact I just remember what Cox said that it is necessary to explain covalent bonds and transistors (of course, as you say we shall distinguish these kinds of things from giving essences of ideas to people). I just gotta read his book!!

Q-reeus

At the risk of more going around in circles on this one, way I see it there either is or isn't physically real 'instantaneous adjustments' a la bcox's claim there is. If there is, how can that not imply instantaneous signalling? Example - a single fibre of say diamond is subject at one end to rapid mechanical or electrical stimulation such that energy states there are modulated. If wavefunction overlap meaningfully extends universe wide, how much more within a continuous crystaline fibre of say a meter long! Hence there aught to be an appreciable instantaneous effect the other end. Which can be experimentally directly compared to e.g. conventional c limited signalling via optical fibre. My guess - nothing here worth rushing out to patent! If bcox still looks in on this thread, maybe he might care to comment.

guillefix

I think that FTL signaling is not possible because of the same reason that in Bell's experiment. Imagine you have the two electrons and protons. You increase the energy of the electron in proton A by a certain amount. What is the energy of the electron? You don't know, the electron in proton A was in a entangled state of having either energy 1 or 2, and when you add a bit, it will just be in an entangled state of energy 1' or 2'. You can then measure it to find out, but then the outcome is random, so not signaling possible. Finally, once you have measured it, you know the energy of the other electron, so you could say: aha I'll move my electron to that energy and so the other electron'll have to move. However, I think that because the wavefunction has 'collapsed' the two wavefunctions are now non-overlapping and the particles distinguishable (Ken G refers to collapse as to "your opinion of the particle's state", but i just used collapse now because it's shorter). Furhtermore when the wavefuntions eventually overlap (wether they do FTL or not) the two particles will become entangled againm, so you find yourself in the same situation. So in reality you are not chaging anything about the state of the universe while heating the diamond, the closest thing you are doing is changing your perspective of it, and you can't use this to signal, because the perspective is random.