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Desintegration of particle into 2 fermions

  1. Jan 10, 2010 #1
    Why the spin part of wavefuncition of two particle of half-integer spin (fermions)
    which was created after desintegraton of spinless particle is always
    antisymmetric (let's assume that orbital angular momentum was 0
    before and after desintegration). This implies that spatial part
    of wavefunction is symmetric (maybe it the result of momentum
    Last edited: Jan 11, 2010
  2. jcsd
  3. Jan 11, 2010 #2


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    You "assume that orbital angular momentum was 0". This is a spatially symmetric state.
  4. Jan 11, 2010 #3
    Thanks for replay.
    I didn't mean spatially symmetric in that sense - sorry for being imprecise.
    I menat symmetry of spatial part of wavefunction under exchange of particles.
    I would like to prove that [tex]\psi(x_1,x_2)=\psi(x_2,x_1) [/tex].
  5. Jan 11, 2010 #4
    Conservation of angular momentum implies that you two fermions should be in the spin singlet state (S=0).

    [tex] \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right)[/tex]

    This state is anti-symmetric wrt interchange of the spin-indices of the two particles. As you said, this implies (since the total wavefunction must be antisymmetric wrt interchange of all particle coordinates (i.e. spin and position) the spatial part of the wavefunction must be symmetric.
  6. Jan 11, 2010 #5
    I would like to prove more general fact. If as a result of desintegration two particles
    with spin n/2 (with odd n) are created the spin part of wavefunction is antisymmetric
    (one can check that it's true by examining Clebsch-Gordon coefficients table but
    I'm looking for prove for all n).
  7. Jan 11, 2010 #6
    The actual phenomena are more complicated.

    The antisymmetric means that if we use the same functions in the 2 electrons, the whole wavefunction becomes zero.
    The way of dividing the wavefunction into the spin and spatial parts and using symmetric and antisymmetric relations is one of the one-electron approximations.
    This way uses the hydrogen wavefunction of each energy level and correct the charge value and so on.
    This is an approximation, so the result is different from the experimental value. (especially when there are many electrons).

    Both the spin and orbital have "the angular momentum" which names are the same, but these angular momentums have the entirely different properties. Because the spin g-factor is 2 and the orbital g-factor is 1 (As a result, both the magnetic moments are the same (2 x 1/2 = 1 x 1).

    So when we use the J (= spin + orbital angular momentum ), its g-factor becomes complex (See this thread).
    And there are some cases in which any coupling methods are difficult to express the experimental spectrum lines.

    (Oh, while I am writing this, I notice one thing. I heard that some people said the antisymmetric relation of the electrons appears only in the relativistic QFT. But the case you say in this thread are clearly non-relativistic. Is it OK?)
    Last edited: Jan 11, 2010
  8. Jan 12, 2010 #7


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    Spatially symmetric in "that sense" does mean symmetric under the interchange x1<--->x2.
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