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Pauli exclusion principle, fermions, bosons

  1. Apr 14, 2006 #1
    Hi, im having trouble understanding these concepts. I checked out some websites but it still doesnt help. First of all what's the main postulate? That there exist 2 different kind of particles: bosons and fermions? What are their fundamental definitions which lead to the fact that an integer number of fermions can occupy ther same energy level, only 0 or 1 bosons can occupy same energy level. In what situations does this apply? Must it be a closed system or a system of particles so close together that the heizenberg uncertainty priciple dominates? Also what is meant by "the same energy" level (same numerical value of energy?) (because im sure there's one fermion here with energy = a and if you look thousands of km away youll find another one with the same energy...)

    Im sure this is an easy question for you guys..
  2. jcsd
  3. Apr 15, 2006 #2

    Meir Achuz

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    You have many questions that are all related. The overall answer is not webbable. You should take a course or try to read a textbook yourself.
    Griffiths would be a good start.
  4. Apr 15, 2006 #3


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    Way oversimplified, there turn out to be two kinds of wave function: those that add when you combine them and those that subtract.

    So if you combine two of the same kind you get in the one case twice the original, and in the other case, zero. The fist kind describe bosons and the combining rule says you can stick a bunch of them together and get one big boson; this is called a bose-einstein condensate. Bose had the idea of treating photons as a classical gas, and worked out that the classical statistics of such a gas did not depend on the order, or permutations of the photons, but only on their number, or combinations. A rough translation is that "identities" of photons don't matter. Einstein was more or less just Bose's front man in Europe as far as this phenomenon is considered.

    The subtractive kind of wave function says you don't get anything if you try to stick two wave functions together. So these describe fermions and the combining rule says you don't find two fermions in close bondage. That is Pauli's principle. Pauli was the first person to attempt a proof of the "Spin-Statistics Theorem" which is what my oversimplification described.
  5. Apr 15, 2006 #4
    Well i read this webpage: http://hyperphysics.phy-astr.gsu.edu/Hbase/pauli.html

    Its pretty useful. So basically the rule is: if the particle has half integer spin the wave function has a minus sign when you combine two indistinguisable states (called a fermion) and if the particle has integer spin the wave function has a positive sign (a boson). So is there a deeper way to explain this correlation between spin and wavefunction? All i know about spin is that its some sort of internal angular momentum.
  6. Apr 15, 2006 #5

    Meir Achuz

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    There is more to it than that. The sign has to do with the action on the WF when the particles are exchanged. Your questions show that you know too little to start web page by page.
  7. Apr 15, 2006 #6
    Suppose you have two uncorrelated particles. They are described by the tensor product |A>|B> of their individual states. Now suppose the letters A and B represent all the properties these particles have. Then our formalism contains one spurious property: being the "left particle" (namely |A>) and being the "right particle" (namely |B>). We don’t want spurious properties so we symmetrize and write


    Now exchanging A and B won't change anything, and this is how it should be. There is however another possibility: we can antisymmetrize:


    The particles that need to be symmetrized are the boson; those that need to be antisymmetrized are the fermions. Why we have these two possibilities is explained http://thisquantumworld.com/scat.htm" [Broken] (read the fine print).

    The famous spin & statistics theorem was proved by Pauli: if you symmetrize particles with half-integral spin and if you antisymmetrize particles with integral spin you get nonsense. Therefore particles with half-integral spin have to be fermions and particles with integral spin have to be bosons. HTH.
    Last edited by a moderator: May 2, 2017
  8. Apr 18, 2006 #7


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    "In the case of a quantum system of N identical particles not every vector from the tensor product space

    [tex] \mathcal{H}^{(N)}=:\otimes_{i=1}^{N} \mathcal{H}_{i} [/tex]

    -where [itex] \mathcal{H}_{i} [/itex] is the separable uniparticle space- leads to the description of a physical state of the system, but only the vectors completely symmetric from the subspace

    [tex] \mathcal{H}_{\hat{S}} =:\hat{S}\mathcal{H}^{(N)} [/tex]

    or vectors completely antisymmetric from the subspace

    [tex] \mathcal{H}_{\hat{A}} =:\hat{A}\mathcal{H}^{(N)} [/tex]

    . The particles described by completely symmetric vectors are called bosons and particles described by completely antisymmetric vectors are called fermions".

    That's how the VI-th postulate in Dirac-von Neumann's formulation is usually stated.

  9. May 10, 2006 #8

    If you want to know why particles with integer spin obey the Bose-Einstein statistics, and why particles with half-odd integer spin obey the Fermi-Dirac statistics, then I have two different reasons for you:

    - Non-relativistic Quantum Mechanics (ordinary QM): This must be put into quantum mechanics by hand. That is, it is a postulate only to match the theory with the experiments.

    - (Relativistic) Quantum Field Theory: It can actually be shown from "first principles," that bosons obey Bose-Einstein statistics and that fermions obey Fermi-Dirac statistics. It essentially comes from the fact that when it comes to quantizing various fields, one must use either the cannonical commutation relations or cannonical anticommutation relations to arrive at the proper relationships between fields, themselves. As selfAdjoint mentioned earlier, the realization of the statistics of a particular type of particle is known as the Spin Statistics Theorem.
    Last edited: May 10, 2006
  10. May 15, 2006 #9
    The usual postulate is known as the "symmetrization postulate" and is the basis for some of the replies here. But it is actually wrong. The correct postulates are both obvious and essential. They are:
    1. To compute interference effects, state vectors must be single-valued.
    2. Particle permutation is not physically observable.

    These are all you need to derive the observable effects due to particle identity. You can find a more detailed discussion of how this differs from the "symmetrization postulate" by reading my posts in the following threads in the Quantum Physics forum:

    Why Pauli's Exclusion Principle?
    Pauli Exclution Principles (sic)

    Not really. All identical particles obey the same generalized exclusion rules that limit allowed states of orbital angular momentum and composite spin. This happens to lead to different statistical behavior depending on whether or not the particles have integer or half-integer spin.

    ???? I think you've misunderstood something here. It looks like you've got fermion and boson the wrong way around. But the postulates above lead to the observable generalized exclusion rules.
    The exclusion rules apply to allowed composite eigenstates. Any situation in which this is meaningful qualifies.

    I doubt it. The "spin-statistics theorem" as it is commonly known has been misunderstood for about 80 years. Most physicists still are not even aware of the problems that abound in understanding it. Sudarshan and Duck wrote a whole book about it ("Pauli And The Spin-Statistics Theorem") and said, very appropriately, "Everyone knows the Spin-Statistics Theorem but no one understands it".
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