Brian Cox and the Pauli Exclusion Principle

[atyy]

Of course the many-electron wave functions are always superpositions of antisymmetrized products of one-body wave functions. The Hilbert space is just that, i.e., any N-body wave function fulfills
$$\psi(t,\vec{x}_{P(1)},\sigma_{P(1)};\vec{x}_{P(2)},\sigma_{P(2)};\ldots;\vec{x}_{P(N)},\sigma_{P(N)}) = \sigma(P) \psi(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2;\ldots;\vec{x}_N,\sigma_N)$$
for all $P \in S_N$. The wave function is always totally antisymmetric under permutation of electrons. It must be taken antisymmetrized at the initial time and then quantum-theoretical dynamics keeps it in the antisymmetrized state, because the Hamiltonian must commute with all permutation operators. Otherwise electrons weren't indistinguishable from each other.

This means in the dynamics is nothing which must antisymmetrize the wave function at any time step, but that's fulfilled automatically.

Further, in the relativistic realm, we only use local QFTs today, and they are fulfilling the linked-cluster principle, clearly contradicting Cox's statements!

Why isn't being fulfilled "automatically" the same as being fulfilled immediately? The anti-symmetrization seems like a global constraint on the wave function, which is not present if electrons are not identical.

 

[Morberticus]

Perhaps another way of resolving the apparent paradox is by insisting the quantum state is not a physical quantity, but rather a description of the system in the context of observer interactions.

The relevant question is not "Does Brian Cox alter the state of distant electrons by rubbing a diamond?" but rather, "If Brian Cox rubs a diamond, does it alter the expectation values for a distant observer performing measurements on his distant electrons?"

 

[atyy]

Perhaps another way of resolving the apparent paradox is by insisting the quantum state is not a physical quantity, but rather a description of the system in the context of observer interactions.

The relevant question is not "Does Brian Cox alter the state of distant electrons by rubbing a diamond?" but rather, "If Brian Cox rubs a diamond, does it alter the expectation values for a distant observer performing measurements on his distant electrons?"

My naive thought was - yes - he does, because of the requirement that the wave function of both electrons remains anti-symmetrized.

However, the effect is so small as to be immeasurable. I think Shankar's quantum mechanics textbook discusses why even though in principle we must include the pions on the moon for the wave function of pions on Earth for anti-symmetrization, in practice the error one makes for pions on Earth when the moon pions are neglected is too small to be measured.

Furthermore, even in the case of entangled non-identical particles, the collapse of the wave function across an entire spacelike hypersurface does not communicate any classical information faster than light.

So all is well with quantum mechanics and relativity.

 

[Ken G]

Why doesn't the wave function anti-symmetrization occur instantly? Naively, I'm thinking the wave function must always be antisymmetrized, so if one electron shifts, the entire wave function must immediately shift so that it remains antisymmetrized.

The total wave function must always be antisymmetric, so however it evolves in time under the Schroedinger equation, it will remain antisymmetric. That must include anything Cox does to the diamond, he's in that evolution too. This means the correlations between distant measurements on electrons might show some ultra tiny connection to what he is doing to the diamond, but again it must be interpreted as an evolution of a total wave function, not something that he is doing to one set of electrons that shows up on another set of electrons, because there just aren't "sets of electrons" there are "observations on electrons." Had he just said that distant observations may show tiny correlations on the outcomes of what he is doing with the diamond, I think he would have been correct, and perhaps in his mind, that distinction is too technical for his audience.

 

[Ken G]

Theory of electronic states of atoms. One can talk about individual electrons. There is definite natural number $N$of them in each atom of any common element.

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! 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.

They are different particles since they all contribute to mass and charge of the electronic cover of the atom and one can assume that they have definite positions without much difficulty (although apparently, this was not necessary to get many useful results from the model). There are no measuring apparatuses measuring their positions (or any other property of theirs) involved. Schroedinger's equation is just a mathematical model and $\psi(\mathbf r_1, \mathbf r_2, ...,\mathbf r_N)$ is an associated mathematical device useful for describing $N$ electrons in atoms and molecules.

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.

In the domain of atoms, it is not used as a device for describing what apparatuses will measure on the electrons. Rather the use of $\psi$ is to calculate expected average values of electronic quantities or probabilities of their configuration (momenta), irrespective of any apparatuses.

And how do we check those expected averages?

I do not see any reason to think that electrons cannot be treated as individual particles. Could you explain why you think that?

The theory of indistinguishability that leads to fermionic and bosonic statistics.

 

[Ken G]

Of course the many-electron wave functions are always superpositions of antisymmetrized products of one-body wave functions. The Hilbert space is just that, i.e., any N-body wave function fulfills
$$\psi(t,\vec{x}_{P(1)},\sigma_{P(1)};\vec{x}_{P(2)},\sigma_{P(2)};\ldots;\vec{x}_{P(N)},\sigma_{P(N)}) = \sigma(P) \psi(t,\vec{x}_1,\sigma_1;\vec{x}_2,\sigma_2;\ldots;\vec{x}_N,\sigma_N)$$
for all $P \in S_N$.

I'm sorry, I don't see where that is a product of one-body wave functions!

The wave function is always totally antisymmetric under permutation of electrons.

That I agree with, except that there is no reason to imagine what you are permuting is electrons. You can (and I argue should) imagine that what you are permuting are the apparatuses with which you associate those coordinates. I cite two types of evidence for this claim:
1) you can permute apparatuses, as they are distinguishable, whereas you cannot do that to electrons, because they are not, and
2) what you mean by those coordinates is demonstrably the outcome of experiments, so permuting the coordinates means permuting the apparatuses.

Further, in the relativistic realm, we only use local QFTs today, and they are fulfilling the linked-cluster principle, clearly contradicting Cox's statements!

Yet not contradicting Bell's theorem. So clearly, nonlocal correlations must be taken into account. Ergo, had Cox simply said that things he does with the diamond will correlate with outcomes of experiments on distant electrons, and that this correlation is mediated by or related to the Pauli exclusion principle, he would have been completely right, as Morberticus said. Perhaps that is what he meant to say, just felt it was too technical a distinction. And we can certainly take issue with claims that unmeasurably small effects would "happen all the same", because we have no idea if things we can't measure actually happen or not, but that's kind of the separate issue of whether popularizers of physics theory have the right to imagine they finally have the "exact theory" after all these millennia!

 

[atyy]

The total wave function must always be antisymmetric, so however it evolves in time under the Schroedinger equation, it will remain antisymmetric. That must include anything Cox does to the diamond, he's in that evolution too. This means the correlations between distant measurements on electrons might show some ultra tiny connection to what he is doing to the diamond, but again it must be interpreted as an evolution of a total wave function, not something that he is doing to one set of electrons that shows up on another set of electrons, because there just aren't "sets of electrons" there are "observations on electrons." Had he just said that distant observations may show tiny correlations on the outcomes of what he is doing with the diamond, I think he would have been correct, and perhaps in his mind, that distinction is too technical for his audience.

OK, my naive understanding is the same. I'd grant Cox some latitude on the fine point, ie. I don't think he actually said this implies faster than light communication is possible.

Would you also agree that in the approximation in which the correlations between identical electrons are too small to be measured, then we do get spatially separated sets of electrons?

 

[Ken G]

Would you also agree that in the approximation in which the correlations between identical electrons are too small to be measured, then we do get spatially separated sets of electrons?

Yes, I agree completely-- I think physics is the way we use it, not a blind adherence to a set of equations well outside the realm they have ever been checked. Still, if we frame it as "this is what our current best theory suggests might be the case" rather than "this is what we experts know would happen if we could ever measure it", I'd be fine.

 

[atyy]

Yes, I agree completely-- I think physics is the way we use it, not a blind adherence to a set of equations well outside the realm they have ever been checked. Still, if we frame it as "this is what our current best theory suggests might be the case" rather than "this is what we experts know would happen if we could ever measure it", I'd be fine.

Yes, but again I'm willing to grant Cox some latitude there. Even for in the domain where the non-relativistic Schroedinger equation is supposed to hold, I don't think we can measure, for example, the discreteness between the 100th and 101st energy levels in a solid with 10²³ particles, although I think that is possible for the hydrogen atom.

 

[vanhees71]

@KenG: The wave function is in general not an antisymmetrized product of N single-particle wave functions (or Slater determinants) but are built by superpositions of those!

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!

 

[Ken G]

@KenG: The wave function is in general not an antisymmetrized product of N single-particle wave functions (or Slater determinants) but are built by superpositions of those!

All the same, a superposition of those is a very different animal-- it does not allow us to imagine that we have individual electron states any more. It's a bit like saying that a superposition of a photon going through various slits means we cannot say the photon went through a slit, the slit the photon went through becomes something indeterminate. That's what I'm saying about electrons: both their identity, and their state, is something indeterminate, so we should not use language that suggests "the universe knows the answer", unless we are adopting a specialized interpretation that is retrofit expressly to make that kind of language possible (which requires extra apparatus that is outside of quantum mechanics).

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!

I agree, I've tried hard to enforce that distinction. I believe it was even in the original post.

 

[vanhees71]

That I agree with. Indistinguishability of particles in quantum theory really means, well, indistinguishability. That's what's encoded in the antisymmetry of the N-body wave functions under permutations of their single-particle arguments (in my example position and spin projection).

 

[WannabeNewton]

All the same, a superposition of those is a very different animal-- it does not allow us to imagine that we have individual electron states any more.

For me at least, it's a lot clearer now what you're trying to convey. 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.

 

[Morberticus]

My naive thought was - yes - he does, because of the requirement that the wave function of both electrons remains anti-symmetrized.

However, the effect is so small as to be immeasurable. I think Shankar's quantum mechanics textbook discusses why even though in principle we must include the pions on the moon for the wave function of pions on Earth for anti-symmetrization, in practice the error one makes for pions on Earth when the moon pions are neglected is too small to be measured.

Furthermore, even in the case of entangled non-identical particles, the collapse of the wave function across an entire spacelike hypersurface does not communicate any classical information faster than light.

So all is well with quantum mechanics and relativity.

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.

 

[Ken G]

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.

The issue there is that two different meanings of "effect" are being used. atyy means an "effect of including or not including" various elements of the full wave function. You are talking about the effect of a cause, the effect on one measurement of the outcome of another. The latter can only appear when one looks at mutual correlations, not sets of outcomes of just one experiment, so that is the more stringent requirement to be an "effect" that you are talking about. The word "effect" is a bit overstretched, it seems!

 

[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