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Exclusion principle

  1. Jul 7, 2004 #1
    I am confused!
    Fermi particles cannot occupy the same quantum state simultaneously. When considering quantum states in solid material, I was taught that in solid material, an electron occupies one quantum state and then many electrons occupy lower states making Fermi sea. Then, is this situation correct when considering quantum states in the universe, i.e., in vacuum? This seems to conclude that the number of electrons with a specific quantum state is only one in the universe!
     
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  3. Jul 7, 2004 #2

    vanesch

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    Yes, what's the problem with that ?

    Patrick.
     
  4. Jul 7, 2004 #3
    I hope I'm not misunderstanding your question, but if you are asking if that according to the Pauli Exclusion Principle that a certain electron's quantum numbers is the only electron in the universe with the exact same quantum numbers then the answer is no. I'm pretty sure the Exclusion principle was only refferring to electrons in a particular orbital, not the universe.

    Definition: there cannot exist an atom in such a quantum state that two electrons within it have the same set of quantum numbers.\

    well, that's the definition I have written in my notebook.
     
  5. Jul 7, 2004 #4

    vanesch

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    Nope, the answer is "yes" !
    Consider two hydrogen atoms at, say, 5 km apart. Strictly speaking, you HAVE to take into account the two protons in order to calculate the energy eigenstates of this system, and you will then find that you DO NOT have "twice" a state with energy 13.6...eV as a ground state, but that you will have an itsy pitsy tiny energy split with one energy level slightly above and another slightly under that value. Of course, when they are 5 km apart, for all practical purposes, you ALMOST have identical states around the first and the second atom. Even at 1 mm apart this is still almost true. But when the atoms get closer, this difference becomes bigger and this is in fact the origin of the chemical bond in a hydrogen atom.
    So indeed, each electron in the universe has its own UNIQUE orbital, with its own quantum numbers (but there are lots of these quantum numbers). Of course, once an atom is reasonably well isolated by distance from others, you can neglect all these extra quantum numbers, and just consider these that are locally relevant. In that restricted sense you can have "identical" quantum numbers (n, l, m, s) locally, but you should keep in mind that you've been neglecting others.

    cheers,
    Patrick.
     
  6. Jul 7, 2004 #5
    Thank you for your response.
    If the number of electrons with same quantum state in the universe is only one, they make a sort of Fermi sea in vacuum which should, I think, be called Dirac sea. But according to my memory, Dirac sea is denied by QFT. Does Dirac sea truely exist?
     
  7. Jul 7, 2004 #6

    vanesch

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    The "Dirac Sea" is a concept from what one could call "relativistic quantum mechanics", and is, as such a useful concept within that conceptual framework. The problem is that "relativistic quantum mechanics" doesn't really exist, because you get all sorts of weird problems if you try to keep the number of particles a constant in a relativistic context.
    Combining the notions of quantum theory (the superposition principle of states) and of special relativity (speed of light and causality), you almost naturally arrive at quantum field theory. Here, individual "electrons" don't make sense: you have ONE single quantum field which represents the "electrons". Electrons and positrons are nothing else but excited modes of this field. So the Dirac sea (almost) disappears, except for an infinite constant which appears in the hamiltonian and which is usually simply done away with, and could, if you want to, be identified with all the modes corresponding to the states in the dirac sea. But there's no real use of that concept anymore in QFT.

    cheers,
    Patrick.
     
  8. Jul 7, 2004 #7
    Let's turn to the case of hydrogen atoms.
    Supose that there are two hydrogen atoms here. The electrons in 1S orbit in each atom have same quantum numbers. Yes, each electron is in different system. For different sysytem, there is different Hilbert space. But what is the effect of being different system in physical term? There are many different systems in the universe. So there are many electrons with same quantum numbers. But in vacuum, there is no electron with same quantum numbers. WEIRD!
     
  9. Jul 7, 2004 #8

    ZapperZ

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    There is a major confusion, and misapplication of quantum statistics, here. One needs to be aware that quantum statistics kicks in when there is INDISTINGUISHIBILITY between the particles involved. This means that there is a significant overlap of the wavefunction between all the particles involved. If two electrons are far apart that they have no overlap, it really doesn't matter what quantum state those two are in. They are now classical, distinguishable particles and obey the typical Maxwell-Boltzmann statistics, rather than Fermi-Dirac statistics.

    I must admit that I am having a difficult time understanding what you learned about the Fermi sea in solid state physics. It sounds rather strange - not the concept of Fermi sea especially in conductors, but your understanding of it.

    Zz.
     
  10. Jul 7, 2004 #9

    reilly

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    ZZ -- Whether there is overlap or not between electron states does not matter, the exclusion Principle holds-- that's the current theory. In QFT this follows from the anticommutation rules for electron creation and destruction operators(One could argue chickens and eggs, but the denouement is a blanket Exclusion Principle). In a little less highbrow approach: a wave function must be antisymmetric in all fermion cordinates/variables (in, of course, a complete commuting set of operators). So, when two fermions have the same quantum number -- for illustrative purposes only -- the wave vector wil be zero.

    Two fermions can be arbitrarily close in position or momentum, and the wave function might be arbitrarily close to zero, except the infinities of QED might raise havoc. Who knows?

    If two fermions are light years apart, the far distant part of the wave function will have little or no impact on the local wave function, or, more precisely, will have little or more impact on local expectation values.


    The "Dirac Sea" might be the most inspired idea Dirac ever had.

    Regards,
    Reilly Atkinson
     
  11. Jul 7, 2004 #10

    ZapperZ

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    1. I have three H atoms. One on earth, one on alpha centauri, and the other on the other side of the universe. In the ground state of each H atom, an electron occupies the 1S state. Can you tell me how the exclusion principle go to work here?

    2. Look at the FD and BE statistics. As I change the parameters (T maybe?), do they at some point converge to MB statistics? What does this imply? That a fermion can, under a range of parameter values, no longer need to be described by quantum statistics?

    3. I can "turn off" any overlap/interaction for the conduction electrons in a typical metal. When I do that, they behave like a CLASSICAL electron gas, and then I immediately obtain, via classical statistics, the Drude model of a metal which explain where Ohm's law (and other properties of a conductor) came from.

    In all of the above, at no point did the FD statistics come into play. This is because there are no indistinguishibility. The interaction/overlap are too weak that they are not indistinguishable. Classical (MB) statistics allows for distinguishable particles that in principle, you can tag and follow. And we all know there are no "exclusion principle" in classical statistics.

    Zz.

    1. http://rodin.hep.iastate.edu/jc/322-03/7
    2. http://www.chem.ufl.edu/~bowers/thesis/chapter2.pdf : "For a collection of identical quantum mechanical particles, however, the trajectories become clouded when the wave functions overlap."
     
    Last edited: Jul 7, 2004
  12. Jul 7, 2004 #11

    reilly

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    My response is that you reread Dr. Bower's thesis. He explains clearly the results of the exclusion principle, and thus illustrates the well known ortho and para-hydrogen states. His comment about overlap simply means that the time dependent electron position expectation values would yield highly intertwined "sorts of trajectories." There's nothing in his chapter that says the exclusion principle only applies in certain circumstances.l



    I would add to his list of references, Statistical Physics by Landau and Lifschitz. Here you can find very clear and elegant discussions of FD, BE, and MB statistics and their interrelationships. And, you will also find that your claim about the free electron gas is quite incorrect at low temperatures -- free, so called completely degenerate fermion gases actually exert a pressure at zero temperature -- proportional to the 5/3 power of the density.

    Spend some time catching up on quantum statistics, and you'll be able to answer your own questions. By the way, there is the so-called Spin and Statistics Thrm of Pauli which proves that state vectors must be either symmetric or antisymmetric, in the appropriate observables, for integer and 1/2 integer spins respectively. That is, the various statistics are hard-wired into the fundamentals of QM.
    Regards,
    Reilly Atkinson
     
  13. Jul 7, 2004 #12

    ZapperZ

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    And maybe you would like to read up on bonding and antibonding bands in H2, and compare that with 2 seperate H atoms. These two are distinct cases on when 2 H atoms are separated, and when the two of them are close enough that they have a significant overlap of the wavefunction to generate a hybridized state. Two H atoms uncorrelated to each other would have the identical ground state. Don't believe me, look at the spectroscopy results. This, by itself, would contradict the idea that the whole universe must obey the exclusion principle for ALL electrons being in a unique quantum state.

    When you look at a chapter on quantum statistics of fermions, and see the Slater determinant, what do you think THAT is?

    And you missed the point of that example. The point being that the indistinguishibility would be removed above certain conditions. Once it is removed, these are nothing more than classical particles. Open any Solid State physics text and the first chapter is typically on the free electron gas of the Drude model. I did NOT say that this model works under ALL parameters and circumstances. In fact, the 3rd chapter of Ashcroft and Mermin is a coverage of the failure of free electron model. So you don't have to inform me of the limitation of this model. However, the example here was to point out that if you can have indistinguishibility being removed to merge (as it should) into the classical limit, it clearly proves that there ARE cases where all electrons in the universe are not simultaneously govern by such exclusion principle.

    As a condensed matter physicist, I deal with strongly correlated system all the time. In fact, I deal with fractionalized system where an "electron" can fractionalize into separate charge and spin excitation as in a Luttinger liquid. Having a charge and spin currents with different dispersion will severely test one's understanding of quantum statistics. So you will understand if I find it rather amusing that you want me to "catch up" on it. I asked you those questions not because I have no answer to them, but rather I want you to apply what you understand to those situation. This will give a clear indication how what you understand is applied to something fundamentally simple.

    The issue here is still clear:

    1. Quantum statistics becomes relevant when particles become indistinguishable.

    2. Distinguishable particles are governed by classical statistics (which does not contain any "exclusion principle").

    3. We know that kinetic theory works for ideal gass. Most real gasses under ordinary conditions can be accurately approximated by this. In fact, most engineers don't even learn about exclusion principles when they deal with gasses and liquids.

    4. From #3, it implies that two identical atoms of the gas can be in the same quantum state, meaning that inevitably, there is one electron in one atom in the SAME state as an electron in another atom. This does NOT violate the exclusion principle because these two electron are distinguishable!

    Now which part of these above are not in Landau and Lif****z, or any quantum statistics book?

    Zz.
     
  14. Jul 7, 2004 #13
    Another idea came to my mind. I hope this example inspires all of you futher.
    If the number of electrons with same quantum number is only one, same two scattering process don't occur in the universe. Because electrons in the universe have different quantum numbers, in state of scattering process is different process to process and it seems to be the case for out state. For electron in vacuum, electron is free particle so described by a plain wave with infinite spreading. So all electron's wavefunctions in vacuum overlap over much long distance. Yes, some QM book says a true wavefunction should be wave packet. But wave packet spreads as time passes by. At last, all electron's wavefunctions overlap in vacuum, i.e., in universe.
     
  15. Jul 7, 2004 #14

    vanesch

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    For all practical purposes, you are right of course. But in PRINCIPLE, there is a tiny difference (which is unmeasurable when the distances of the systems are macroscopic, hence the results of calculations when considering the particles to be distinguishable, and classical M-B statistics are highly accurate).

    If you have an atom here and another one at Alpha centauri, then the ground state of this two-electron system is given by an anti-symmetric wave function Psi_1(1,2) made up of an antisymmetrisation of essentially 1S(earth)(1) + 1S(alpha-centauri) (1) and 1S(earth)(2) - 1S(alpha_centauri)(2). The 1S+1S state has a slightly lower energy than the 1S-1S state (it is not fully the sum and the difference, there is a tiny correction to it because of the very small overlap of the 1S(earth) and the 1S(alpha-centauri) orbitals: indeed: these wave functions go in exp(-r/a) which is very small but not 0 at 4 lightyears).

    Now the difference in any expectation value between using these correct wave functions, and the "independent" wave functions where you consider a 1S(earth) and a 1S(alpha-centauri) state, will numerically only be influenced by the overlap integral 1S(earth) x 1S(alpha-centauri). This is not 0, but so small that you can neglect it, and so it SEEMS to you that you can treat the two electrons as independent. But orthodox quantum theory says that you should, in principle, ALWAYS have an antisymmetric wave function for all electrons in the universe. Practically, as I tried to show, it doesn't make a difference from the moment the overlap is neglegible.
    The only way to truely not to have to do this in principle is when there is an infinite potential barrier between the two systems. THAT is the only case where you do not have to deal with anti-symmetrical wave functions.

    cheers,
    Patrick.
     
  16. Jul 8, 2004 #15

    ZapperZ

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    But I think what you have just described is exactly what I have previously described in other postings of the H2 molecule. There is a difference in energy state of 2 H atom and 1 H2 molecule. The ground state of the bonding band of an H2 molecule has a LOWER energy than the ground state of 2 separate H atoms. And not only that, the existence of an addition antibonding band (and an energy gap between the two) clearly differentiates these two situations.

    Now the question is, is there ANY measureable differences to show that two separated H atoms, in widely different locations, would have some resemblance of an H2 molecule? One can argue ad nauseum that no matter how far apart they are, there's some "correlation", but then again, why not also pick on gravitational field, electric field, etc, which has an even longer range effect. Can one detect the gravitational influences of Alpha Centauri on earth?

    I think that issue is moot. The very fact that classical mechanics (and statistics) ARE approximations that actually work under "normal" conditions was the very fact that I've been trying to point out (obviously without success here). To say that "in principle", every electron in the universe must obey the exclusion principle no matter under what condition they are in is to deny pragmatism and to deny what already works. It seems to imply we MUST make use of those principles at all times or else things would not come out right. Those people working in particle beam physics would laugh if you make such statement - all their computer modelling routines to track the particle beams (especially electrons) treat them as classical particles with ZERO influences from the exclusion principles (example: PARMELLA routine from Los Alamos). And these darn models work.... under certain range of parameters (as does everything else that we know) or else all those synchrotron centers all over the world would not have been constructed.

    .. but then again, all this could simply be the ugly experimentalist in me who just doesn't have the patience with theoretical arguments that, "in principle", simply can't be measured.

    Zz.
     
  17. Jul 8, 2004 #16

    reilly

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    Once again lightly. The Spin StatisticsThrm requires all, repeat all, wave functions/states to be antisymmetric for Fermions, symmetric for Bosons. This stricture in no way is antipragmatic; it is just basic physics.

    As I pointed out in my first post(with a dumb typo, more for no), there are times when practically speaking, the antisymmetrization is of no consequence -- exchange currents and potentials and the like are of negligiable magnitude -- so neglecting them is a good first order approximation. The nucleons and electrons in a H2 molecule, and in two separate H atoms are, indeed,in different states, and clearly have quite different energy states.

    I've done considerable work on high energy electron scattering. That the exclusion stuff does not enter beam physics is indeed the case. No argument there. Does it apply for, say, electron-electron scattering? Yes. I agree, sometimes the exclusion principle is practically important, sometimes not.
    Regards,
    Reilly Atkinson
     
  18. Jul 9, 2004 #17

    vanesch

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    I'm an experimentalist too...
    I think we agree somehow. The only point I wanted to make is that it is easier to understand that the exclusion principle holds always and everywhere, just as newtonian gravity goes in 1/r even for big values of r. I think that was the problem of the original poster: when does and when apparently doesn't the exclusion principle work? Answer: it works always. But IN PRACTICE in certain cases you don't have to bother because it doesn't make a difference. I personally think of this situation as much more satisfying, than to have to say that in certain cases, it applies, and in others, it doesn't, as if there was a different physics at work in the two cases.

    cheers,
    Patrick.
     
  19. Jul 9, 2004 #18
    "as if there was a different physics at work in the two cases"
    O.K. I understand. Thank you, everyone!
     
    Last edited: Jul 9, 2004
  20. Jul 9, 2004 #19

    ZapperZ

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    Ok, I was going to let this thing come to its ugly end, but I think it is important to address what you said here.

    When something "doesn't make a difference" whether it is there or not, then we are going in some strange territory here in physics. You might as well say that everything and anything in the universe comes into play when one formulate any physical description.

    If it is true that IN PRACTICE, we can use classical statistics for a bunch of electrons under MANY situations, then I really do not see why there was a problem in the first place when I mentioned that FD statistics should not be used when there isn't any significant overlap between the electrons. Isn't this saying the EXACT same thing of what you guys are trying to drill into me? Isn't this saying that effects such as the exclusion principle is undetectable for those cases and for all practical purposes, is simply not there?

    So then what are we arguing about?

    It is confusing at best when you "agree" that yes, all the electrons in the universe obey the exclusion principle, and LEFT IT AT THAT! If that was the case, then it is meaningless to actually teach people distinguishable and indistinguishable particles, and even classical statistics! That was what I was objecting to. I have NOT seen anyone arguing, nor has anyone been silly enough to perform an experiment to detect the exclusion principle in 2 separated H atoms in two different containers, for example. If the possible effect is simply to small, even in principle, for any physical detection, then why are we insisting that it is even there??!! Have we degenerated into trying to count angels on a pinhead?

    There's also something that is severaly overlooked here. The biggest difficulty in any EPR-type experiment is to maintain coherence of the entangled properties, be it over time or over distance. Now look at the fermionic spin wavefunction for 2 electrons, for example - there certainly is a built-in entanglement there! Considering how DIFFICULT it is to maintain the coherence of the spin states when you split those two electrons apart, is anyone going to tell me that yes, the exclusion principle still works even for that 2-electron system even when they have coupled to numerous degrees of freedom and have decohere. One can already see how absurd this is, because if this were true, then it would be a piece of cake to detect entanglement properties and we would see it all the time!

    Now if it is THAT difficult to maintain coherence and to preserve the FD statistics when we STARTED from an already entangled pair of electrons, what hope do we have if we start from electrons that were never created together or have separate origins and interactions? It is a very dangerous slope to sit on if we think we can use with cavelier the explanation that "oh, it is simply too small that it is undetected, but it is still THERE!"

    Zz.
     
  21. Jul 9, 2004 #20

    vanesch

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    In the end it is a matter of taste. But I prefer the theoretical Occam's razor over the practical one: I prefer a simple, consistent theory that applies to EVERYTHING, even if that means that it gives rise to a more complicated and entangled situation in practice. Afterwards, if I have to calculate things, I can then neglect stuff or use "older" theories that I use as "approximations".
    It would be a strange world view - to me - that the basic principles of nature change as a function of the numerical value they take on as compared to other theories that are simpler in their practical application.

    So I prefer to say that you ALWAYS have to apply antisymmetry, because that is a SIMPLE statement. Saying that you have to apply antisymmetry when the wave functions have a certain overlap, and that you are forbidden to apply symmetry in other cases, depending on the precision of your calculation, is to me, a more complicated and ugly statement. However, you apply the last statement in practice to simplify calculations.

    cheers,
    Patrick.
     
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