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Pauli's exclusion principle bosons

  1. Aug 27, 2004 #1
    I've heard quite abit of things about bosons and am quite confused. The biggest thing which distinguishes fermions from bosons, would be Pauli's exclusion principle. But I've also heard things about bosons having half- integer, while fermions have interger spin, among many others. I've also heard also that liquid helium can be considered a boson, but I once thought that bosons can be applied only to fundamental particles.

    What are the properties which distinguishes bosons and fermions?
  2. jcsd
  3. Aug 27, 2004 #2


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    Bosons, named after Satyendra Nath Bose, are particles which form totally-symmetric composite quantum states, they have integer quantum spin.

    All elementary particles are either bosons or fermions, so everything not a fermion is a boson

    Gauge bosons are elementary particles which act as the carriers of the fundamental forces/messenger particles


    Particles composed of a number of other particles (such as protons or nuclei), like atoms, can be either fermions or bosons, depending on their total spin. Hence, many nuclei are bosons. While fermions obey the Pauli exclusion principle(no more than one fermion can occupy a single quantum state) there is no exclusion property for bosons, which are free to crowd into the same quantum state. This explains the spectrum of black-body radiation and the operation of lasers, the properties of liquid Helium-4 and superconductors and the possibility of bosons to form Bose-Einstein condensates, a particular state of matter.

    Because bosons do not obey the Pauli exclusion principle, it is much harder to form stable structures with only bosons than with fermions. This difference accounts for the difference between what we think of as matter and things that confuse people between if it's matter and not sometimes, such as light.

    Fermions, named after Enrico Fermi, are particles that obey the Pauli exclusion principle, and Fermi-Dirac statistics. The spin-statistics theorem states that fermions have half-integer spin. One possible way of visualizing spin is that particles with a 1/2 spin, i.e. fermions have to be rotated by two full rotations to return them to their initial state.

    The elementary particles which make up matter are fermions, belonging to either the quarks (which form protons and neutrons) or the leptons (such as electrons). The Pauli exclusion of fermions is responsible for the stability of the electron shells of atoms, making complex chemistry possible. It also allows the stability of degenerate matter under extreme pressures.

    * electrons
    * quarks
    * protons
    * neutrons
    * neutrinos

  4. Jun 13, 2005 #3
    All identical particles, of whatever spin, obey the same general exclusion rule:

    "When all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

    It just happens that with spin 1/2 particles, the composite spin must be 0 and so the spins must be opposite. This results in the Pauli principle.

    So the notion that there are two types of particles, fermions and bosons, is misleading. It just happens that only those particles which obey the Pauli principle can give any properties to matter characterized by a specific spatial location. Other particles just form an undifferentiated soup.

    The generalized rule is a fundamental property of QM and can be proved assuming only the obvious:

    1. Unique state vectors require a uniquely described physical state.
    2. Particle permutation is an artifice of state descriptions and not physically observable.

    The rule then follows as an interference effect.
  5. Jun 13, 2005 #4
    Hmmm... Clebsch-Gordon might say different?

    Maybe I don't know what you mean by an "undifferentiated soup", but bosons are real and they exist in our every day lives- Read Mk's post for examples. I also have no idea what this undifferentiated soup has to do with the OP question about bosons and what they are. Maybe your ideas are better kept in Theory Development.

    misogynisticfeminist- I would stick with Mk's information- it is sound.
  6. Jun 13, 2005 #5


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    Then you will have a heck of a time explaining the existence of spin-triplet superconductors like the ruthenates and superfluidity of He3.

  7. Jun 15, 2005 #6
    Not for identical particles when all other quantum numbers are the same and "permutation" is not a physically observable operation. Please see the context of the remark you are questioning (only even eigenstates allowed) and you will understand it better. The generalized exclusion rule I quote is verifiable throughout physics. Identical particle scattering (p-p, pi-pi, d-d, alpha-alpha, etc) is the obvious example. Even when symmetry breaking is present, particles which belong to the same multiplets obey this rule approximately, whether bosons or fermions.

    Please inform me when you find a pair of indistinguishable electrons forming a composite spin triplet.

    Of course. They just don't give any structure to matter. That is why they are called "force" particles and fermions are called "matter" particles.

    There is nothing contentious or "unsound" about this difference. The differentiation of matter is critically dependent on the fact that "matter" particles have spin 1/2 and obey the Pauli rule. For example, without the Pauli rule we would have no chemistry and no life.
    Last edited: Jun 15, 2005
  8. Jun 15, 2005 #7
    Please explain why. Are you claiming a violation of the Pauli principle or, like Norman, ignoring the context of the remark you quote?
  9. Jun 15, 2005 #8


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    I'm sorry, but it is you who have to do the explanation. Nothing is being violated here, least of all the pauli exclusion principle (you seem to be forgetting that the total asymmetric wavefunction can be satisfied with either asymmetric spin OR asymmetric spatial part). All you need to do is look up the superfludity of He3 and the superconducticity of Sr2RuO4 as examples. They are well-verified spin-triplet pairings, or else they won't have given Tony Leggett his Nobel Prize a couple of years ago.

    Last edited: Jun 15, 2005
  10. Jun 16, 2005 #9
    mikeyork, you are clearly confusing the term 'boson' with the force carrying 'vector bosons'.
  11. Jun 16, 2005 #10


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    1.First of all,composing 2 spins 1/2 will give 2 irreducible spaces:eek:ne of weight 0 which is unidimensional and one irreducible space of weight 1 which is 3 dimensional.
    2.Bosons can be "matter particles".

  12. Jun 16, 2005 #11


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    I like that, good statement it helps me visualize the concept of intrinsic spin. :biggrin:
  13. Jun 16, 2005 #12


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    Try -



    And while at it try -

    Spin - one the distinguishing characteristics that differentiates fermions and bosons.
  14. Jun 17, 2005 #13
    Since the statement you questioned was none other than the Pauli principle, you seem to be contradicting yourself now. It's you that seems to have forgotten the import of the Pauli rule (no two electrons can be in the same state). If the spatial part is asymmetric, then the spatial quantum numbers cannot be the same and the Pauli rule does not apply.

    I cited a more general rule which I'll repeat here:

    "For all identical particles, of whatever spin,... when all other quantum numbers are the same, only even eigenstates of composite spin are allowed."

    For identical spin 1/2 particles, when all other quantum numbers are the same, as I said before, this implies a spin singlet. This, in turn, implies the Pauli rule

    When you refer to spatial asymmetry this means that not "all other quantum numbers are the same" and so the spin singlet rule does not apply.

    If this condition does not hold, the rule needs to be re-expressed. For example, when it comes to pairs of particles in their CM frame, since their momenta are opposite, the rule needs to be transformed to the CM frame. In fact it becomes:

    "For all pairs of identical particles, of whatever spin, in their CM frame, when all other quantum numbers are the same, only even sums of composite orbital and spin angular momentum are allowed."

    i.e. L+S must be even. This rule is the same for bosons and fermions and has been verified in identical particle scattering experiments.

    Even for non-identical particles belonging to the same isospin multiplet (e.g. p.n) it is possible to derive that L+S+I must be even as an approximation and this rule is used to help understand the structure of nuclei.
  15. Jun 17, 2005 #14
    In general, yes of course, but not when all other quantum numbers are the same and permutation is not observable.

    Some forms of matter are indeed bosons. But even they are fermion composites. Bosons cannot give the structure which differentiates matter. That is the point I am trying to make.
  16. Jun 17, 2005 #15


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    It doesn't matter whether the bosons are composed from 2 fermions.They're still described by a massive scalar field,electrically charged or not.

  17. Jun 17, 2005 #16
    Even though this is a theory-dependent statement, not one that is directly observable, let's assume it is true. I still have to say "so what?".

    Let me remind you what my original post was about:

    The notion that bosons and fermions are two different types of particles is misleading. They both obey the same generalised exclusion rule (which follows from basic QM when you require unique state vectors for uniquely described states and unobservability of particle permutation). The essential difference between fermions and bosons is that, as a result of the exclusion rule, only fermions can give the differentiated structure to matter that we see in the periodic table, atomic nuclei and the tables of sub-nuclear hadrons. Without fermions there would be no chemistry or biology or nuclear energy.
    Last edited: Jun 17, 2005
  18. Jun 17, 2005 #17


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    And you seem to be IGNORING an important fact. If YOUR interpretation of what the Pauli Exclusion principle is is TRUE, then it has been severely and SPECTACULARLY violated by SEVERAL experimental observations! Go open Leggett's paper on superfluidity of He3, the paring state of the ruthenates, how the d-orbital is filled, and a whole zoo of other observations.

    Pauli exclusion principle requires an antisymmetric TOTAL wavefunction. This means that you can have an asymmetry spatial*symmetric spin OR symmetric spatial*asymmetric spin! A spin-triplet state requires that the spatial part is asymmetric, meaning the two spins cannot be very close to each other! It means that any pairing of the two will be considerably over a larger extent. This is why He3 becomes a superfluid at a significantly lower temperature than He4! It requires a much lower temperature to reduce the thermal fluctuation to be able to maintain coherence between two spins being paired over a LARGER distance!

    Hey, don't take my word for it, read it yourself! If you still maintain your postion, then the next posting you give on here better be to wiggle your way out of these whole bunch of experimental results that clearly have demolished your version of the exclusion principle.

  19. Jun 17, 2005 #18
    Not me -- although I understand that a selective reading of my posts in this thread, focussed only on where I used the very loose terms "force particle" and "matter particle" could lead you to that confusion.

    I'd also suggest that pions (which are scalar bosons) are also considered "force particles". This useage pre-dates the standard model and unified QFT. Back in the days of Yukawa, pions were considered to be the unique carriers of the strong force.
  20. Jun 17, 2005 #19


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    OK, so where is the "selective reading" here that will cause a confusion? You clearly are claiming that pions, just because it was MISTAKENLY taught to be the strong force carrier, is STILL a force particle? Whoa!!!

    And what force is it carrier of? And while you're at it, what force is the kaon and the J/psi particle a carrier of?

  21. Jun 17, 2005 #20
    I am well aware of the fact that the whole wavefunction needs to be taken into account. This is precisely why I talked about the necessity of equality of "all other quantum numbers" in what I wrote. Spatial asymmetry implies that there are quantum numbers which are not the same. So the Pauli rule does not apply. Please read it again.

    Although there is a slight technical difference between the observable rule I give and the conventional expression via the Symmetrization Postulate (which actually needs a technical qualification before it is true, see below) this difference has no observable effects. The difference you are trying to create is non-existent.
    My version of the Pauli exclusion principle is identical to Pauli's (no two identical spin 1/2 particles can be in the same state). No wiggling is necessary. You have just misunderstood what I wrote.

    BTW the technical problem with the conventional Symmetrization Postulate is that it requires an additional qualification regarding the construction of the wavefunctions. It is trivially easy to construct wavefunctions for fermions that are permutation symmetric using a different method, but giving the same observable results. They are related to the conventional wavefunctions by an order-dependent phase so that exchange results in a 2pi relative rotation of one particle's state with respect to the other in one case but not in the other. This has been known for more than 30 years. My first paper on this (unpublished pre-print from Istituto di Fisica, Rome) was written in 1975. The next year Broyles published his attempted proof of the spin-statistic theorem (Am. J. Phys. 44 (4), 340-343, (1976)) using essentially the same method. Berry and Robbins published a variation using configuration space (Proc. R. Soc. London Ser. A 453, 1771-1790 (1997)). However those proofs neglected to provide a definitive theoretical basis for deciding which wave functions are necessarily symmetric and which anti-symmetric. The definitive theoretical basis was spelled out in my own paper presented at the Spin2000 conference in May 2000 and published in the proceedings (AIP Proceedings 545, pp 104-110). AFAIK this is the only published paper which gives a complete proof of the spin-statistics theorem. You can view it here:


    All the standard field theory proofs and variations thereof, ignore the additional qualification needed regarding the way the wavefunctions (or creation operators) are constructed and are therefore incomplete.
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