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Charge Independence and Pauli Exclusion Principle

  1. Apr 11, 2008 #1
    I remember reading about charge independence; about how the energy levels of mirror nuclei (correcting for differences in the colomb term) are identical… I think this suggests that the force between any two nucleons is the same, so the attraction of neutron-proton=proton-proton=neutron-neutron.

    One of my textbooks states the following:

    "The average neutron-proton attraction in a nucleus is greater than the average neutron-neutron or proton-proton attraction, essentially as a consequence of Pauli Exclusion Principle."

    questions:

    1)is the textbook right about the force between n-p not being equal to p-p or n-n?

    2)Can someone explain how Pauli EP is relavent here?

    I've tried reasoning it in my head, but I'm getting confused: I probably don't know what I'm talking about but n-p has an antisymetric isospin (right?), therefore the space wavefunction has to be symetric? Is this relavent?

    Thanks in advance:)
     
  2. jcsd
  3. Apr 11, 2008 #2

    nrqed

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    They are including the effect of the "exchange force" (which is not really a force in the conventional sense of the word). However I don't quite understand how they draw a conclusion without saying something about spin. Assuming that they are talking about ground state and that the pp (or nn) are in the triplet state (so symmetric) the spatial wavefunction must be antisymmetric which shows up as an apparent repulsive force.
    But again, this depends on the spin configuration. So it sounds a bit strange (or rather incomplete) to me.
     
  4. Apr 11, 2008 #3
    Thanks for your reply nrqed. I think I understand why the n-p force does not equal p-p or n-n (but I may be way off the mark).

    just to clarify can I write:

    [tex]\Psi=\psi(spacial)\psi(spin)\psi(isospin)[/tex]?

    I don't know much about isopin to be honest, so I'm not sure if I am applying it rightly.

    I would think a p-p or n-n pair must have an antisymettric spin wf because of Pauli EP ((I[tex]_{z}[/tex] must point up and down). Now, isospin has to be symettric (for p-p or n-n). So for the overall wf to be antisymettric (as they must be for fermions), the spacial wf must be antisymettric. So if the spacial wf is antisym, there's less overlap, so the binding energy is less.

    For n-p, isospin can be sym eg. (2^-0.5){|p>[tex]_{1}[/tex]|n>[tex]_{2}[/tex] +|n>[tex]_{1}[/tex] |p>[tex]_{2}[/tex] } OR antisym eg. |n>[tex]_{1}[/tex] |p>[tex]_{2}[/tex]; therefore the spacial wf can be EITHER sym or antisym. So if spacial wf is sym, the binding energy is more.

    Does this reasoning make sense?
     
  5. Apr 11, 2008 #4
    duplicate post - deleted.
     
  6. Apr 12, 2008 #5

    malawi_glenn

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    Yes the nuclear force is charge independent, but it is isospin dependent.

    "The NN-force is to a very large extent charge independent. This means that the
    interaction is the same for the same isospin T regardless of the T_3 value. This means
    also that the interaction is not the same when T = 0 and T = 1. For example, the np- interaction
    is different when the pair couples to T = 0 compared to when they couple
    to T = 1. In the T = 0 (S = 1) case the np-pair is bound to a deuteron, while in the T =
    1 (S = 0) case the np-pair is unbound. Note also that disregarding from the Coulomb
    interaction, the pp- and nn-interaction is the same since these states both have T = 1."

    From my lecture notes in Nuclear physics
     
  7. Apr 12, 2008 #6

    pam

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    1) Yes.
    2) The nucleon-nucleon force is a bit more attractive in the L=0, spin one state than in the spin zero state. Because of the Pauli principle, n-p can exist in the L=0, spin one state, while p-p and n-p can't. Incidentally that is why the bound deuteron state is only n-p.
     
  8. Apr 12, 2008 #7
    thanks malawi_glen. I think this has cleared it up!
     
  9. Apr 12, 2008 #8
    thanks pam.

    what do you mean by "spin one" the case of n-p? I mean, n and p have seperate nuclear wells do they not? Are you talking about isospin?

    And sorry for might be a stupid question but why is the nucleon-nucleon force more attractive when L=0?
     
  10. Apr 12, 2008 #9

    malawi_glenn

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    When L = 0 it means that the particles have no relative angular momenta and it should then correspond to the lowest lying state since higher angular momenta usally means that energy must go to give the particle relative motion (this was a very rough explanation altough)

    the (bound) n-p system have J = 1 and S = 1, these two can be deduced from experiment, and it is easy to work out the L from theory and then test it by comparing it with the magnetic moment of the deutron (which can be measured).

    Now the isospin follows the same algebra as ordinary spin, so according to Pauli you must couple L,S & T so that the total wave function is anti-symmetric, so it is straightforward to obtain the possible configurations of S,L and T. Then you perform experiments and find out what components you have in the N-N potential.

    So the force is charge independent, but due to Pauli - the n-p system (with J=1 & S = 1)is the only 2 nucleon system) etc.
     
  11. Apr 12, 2008 #10
    thanks glenn!
     
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