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Exchange Forces and Symmetrization of Fermions/Bosons

  1. Apr 22, 2012 #1
    Im reading through Griffiths' Intro to QM book right now, and I can't quite understand a statement he makes about the symmetrization of the states of identical fermions and bosons explaining the covalent bonding between two hydrogen to make an H2 molecule.

    The book says that (based on the spatial wave function alone) if electrons were bosons, they would tend to congregate between the two nuclei, which would account for the covalent bonding. Then he explains that since electrons are fermions they would actually do the opposite (again, disregarding spin). Then he says that the entire state of the particle is what needs to be antisymmetrical (since electrons are fermions), which includes spin and the spatial wave function. He concludes by saying that in order for the two-particle system to be conducive to covalent bonding, the spatial wave function must be symmetric and the spin states must be antisymmetric.

    My question is how you make the leap from saying that a state reverses its sign when two identical fermions are exchanged, to saying that two identical fermions with antisymmetric spin, but a symmetric wave function, group together.
     
  2. jcsd
  3. Apr 26, 2012 #2
    The total wave function above for the two electrons must be antisymmetric, space x spin, but for the hydrogen atoms to bond the space part of the total wave function must be symmetric? Is that correct?

    zmitchel, when you use symmetric above does that refer to the space part of the WF or to the total WF?
     
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