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Exclusion principle question

  1. Jul 14, 2008 #1
    Imagine a system consisting of two hydrogen atoms with the electrons sitting in the ground state with the same spin. Surely there is nothing stopping us from constructing such a system. The system is completely symmetric - each electron experiences the same potential. The wavefunction of the system is thus symmetric under interchange of electron labels, but this violates the exclusion principle, which suggests that we could not contruct such a system. But this doesnt make sense, surely there can be more than one hydrogen atom in the universe with an electron in the ground state.

    Help please.
  2. jcsd
  3. Jul 14, 2008 #2


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    Remeber you have a spatial wavefunction and a spin wave function, where their product gives the total wavefunction. I think you're confused by just considering the spatial wavefunction which is symmetric

    Pauli Exclusion Principle:
    "For two identical fermions, the total wave function is anti-symmetric w.r.t their exchange."

    i.e. All fermions have antisymmetric total wavefunctions. Therefore 2 electrons, say, in the same spatial state within a Hygrogen molecule must have different spin states to remain antisymmetric.

    What do you mean? A single electron is not restricted by the PEP
  4. Jul 14, 2008 #3
    In the [tex]H_2[/tex] molecule, the 1s orbitals are separated in space because of coulomb repulsion, so 1s_a and 1s_b are not exactly the same orbitals! Therefore you could fill two electrons with equal spin in these.
  5. Jul 14, 2008 #4
    In your example the potential experienced by the two electrons actually differ, they are identical in structure but not equal. There is V(r) and V(r'), as long as the protons do not sit on top of eachother (which is not possible) the total wavefunction of the system is not symmetric under the interchange of electrons.
  6. Jul 14, 2008 #5
    ok im not convinced you guys understood my question so Ill put it in a different way. Say we take a uuu baryon, the pauli exclusion principle implies that we must have another quantum number, colour, to separate the states of the up quarks. But there is nothing stopping us from having another uuu baryon somewhere else in the universe in the exact same state - why does the exlusion principle apply between quarks confined in a single baryon but not to quarks in different baryons?
  7. Jul 14, 2008 #6


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    Then they must have a negative number of angular momenta vs. each other:

    [tex] |\Delta ^{++},\Delta ^{++}; L = \text{odd} > [/tex]

    since the parity of a state with angular momentum L is (-1)^L

    Same holds for example when two pions are emitted in a decay - but pions are bosons - so their L must be even in order to have a symmetric WF w.r.t particle exchange.
  8. Jul 14, 2008 #7
    You are answering yourself, think slowly. You say SOMEWHERE ELSE. It is allowed because it is somewhere else.

    The total function goes to zero for quarks in a particular baryon all these three quarks are affected by the same potential, their wave functions are identical and all depend on the same spatial parameter r. Thus Pauli Exclusion Principle. But for a quark in this baryon and another one their wave functions may be identical but one depends on spatial parameter r, the other on r'. Thus 2 quarks on different baryons are not subject to Pauli Exclusion. Because you can have a nonzero antisymmetric spatial wavefunction which is not 0 for two quarks on different baryons. Thus they can have symmetric or antisymmetric spin wave functions.
    Last edited: Jul 14, 2008
  9. Jul 14, 2008 #8


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    Just think of them as localised and isolated systems. PEP aplies to indentical particles in that isolated system, and the same occurs in an x number of other identical but seperate other isolated systems
  10. Jul 15, 2008 #9
    The three u quarks in the baryon arent in the same position, why does PEP apply to them? How far away do two particles have to be before PEP doesnt apply?

    The way I see it, if the universe consisted of two identical fermions, they would both experience the exact same potential due to the other (lets say they have a long range interaction). If I wacked that into the schrodinger equation, the problem would be completely symmetric under interchange of particle labels, thus violating PEP. The particles could be at either side of the universe or they could be a planck length away from each other - it doesnt seem that it would affect the solution of the SE.

    Sorry if Im beating a dead horse, I cant quite get my head around this. Also thanks for all the replies.
  11. Jul 15, 2008 #10
    We are talking about the ground state baryon here where all u's are in the same state in terms of position.
    the problem is indeed symmetric and the two fermions of yours are undistinguishable, so the wavefunction for the composite system of your two fermions should be anti-symmetric with respect to the interchange of them. This is the PEP.
  12. Jul 15, 2008 #11
    I don't think people have answered your question. I posted a question about two electrons in a helium atom that led to similar problems (maybe your question was a spinoff from mine?).
    What I am understanding in the case of the helium atom is that the function must be antisymmetric with respect to exchange of electrons. That is why the combination
    +-> - -+>
    is allowed, and the combination
    +-> + -+>
    is forbidden. It's not so much forbidden as the fact that the sign changes when you reverse electrons, so the two terms add up to zero. Your question is analogous.

    You have the combination of electron spin up at A and also at B. The combination
    (A+)(B+)> - (B+)(A+)>
    is OK, because when you switch locations the phase reverses sign, so the combination
    adds up to a net value. But oddly, the other combination:
    (A+)(B+)> + (B+)(A+)>
    appears to add up to zero, so should not be allowed. This does not appear to depend
    on how far apart the atoms are, which seems to be an odd restriction on what is allowed physically.

    While my analysis does appear to allow two hydrogen atoms to have the same spin,
    I think that nevertheless you've posted an interesting paradox which ought to have a decent explanation.
  13. Jul 15, 2008 #12
    As I said, the PEP is usually defined as the fact that the wavefunction for a composite system with several undistinguishable fermions must be antisymmetric with respect to the exchange of the fermions. There are deeper reasons for this spin-statistics connection, but at the level of QM it is taken as a postulate. If you don't take that as a postulate, what do you take as postulate ? If you accept this formulation of the PEP as of postulate for QM, then you realize that
    is misleading. This sign change is extremely important, it is not anecdotical. And
    I don't know what remains to be clarified. Here you have three identical quarks in the fundamental state of the hadron, the are all in the same position state, they cannot have different spins since the hadron is spin 3/2, where are you getting antisymmetry from if you don't get it from the color d.o.f. ?
  14. Jul 16, 2008 #13
    So the PEP applies when the fermions are in the same position state, but otherwise doesnt hold? I think I understand now.
  15. Jul 16, 2008 #14


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    The electron orbiting around hydrogen nucleus #A in orbit n=1, L=0, S=+1/2 is not in the same state as the electron orbiting around hydrogen nucleus #B in orbit n=1, L=0, S=+1/2
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