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Identical Fermions in an infinite square well

  1. Apr 8, 2010 #1
    If you have 2 identical, noninteracting Fermions in an infinite 1 dimensional square well of width a, I was thinking the state would be:
    [tex]\frac{1}{\sqrt{2}}\psi_1(x_1)\psi_1(x_2)(\uparrow\downarrow - \downarrow\uparrow )[/tex]
    where [tex]\psi_1[/tex] is the ground state of the single particle well problem.

    However, I just looked in Griffths "Introduction to QM, 2nd Ed" and it says there is no state with the energy of that state and that in fact the ground state is [tex]\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)[/tex].

    The state I gave seems to be (1) antisymmetric, (2) an eigenstate of the Hamiltonian, and (3) have lower energy than the ground state given by Griffths.

    Do you see a problem with ground state I gave?
     
    Last edited: Apr 8, 2010
  2. jcsd
  3. Apr 9, 2010 #2
    You've put two fermions on the ground state of the potential, with their spin making up a singlet. I think your GS is fine. Griffith's solution may be referred to spinless particles.
     
  4. Apr 9, 2010 #3
    In the same line as jrlaguna said, fermions in QM just refers to particles with fermi-dirac statistics. The equivalence of spin1/2 particles and fermions are only in relativistic quantum field theory.
     
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