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Eigenfunction of a system of three fermions

  1. Aug 23, 2014 #1
    I have to find the eigenfunction of the ground state [itex]\Psi_0 [/itex] of a three independent s=1/2 particle system.

    The eigenfunctions [itex]\phi_{n,s}(x) = \varphi_n(x) \ \chi_s [/itex] and eigenvalues [itex] E_n [/itex] of the single particle Hamiltonian are known.

    Becuse of the Pauli exclusion principle, there must be two particles with opposite z component of the spin in the lowest energy single particle level and one particle in the first excited single particle level.
    I have attempted to solve the problem saying that the eigenfunction [itex]\Psi_0 [/itex] of the three particle system must be either the Slater determinant of ##\phi_{0,+}(x), \phi_{0,-}(x)## and ##\phi_{1,+}(x)## or the Slater determinant of ##\phi_{0,+}(x), \phi_{0,-}(x)## and ##\phi_{1,-}(x)## (two possible excited levels).

    Now, my question is: since the particle with higher energy can be both in the state ##\phi_{1,+}(x)## and in the state ##\phi_{1,-}(x)## without any preference, is it fine to consider these two single particle states in order to calculate the two Slater determinants, or should I consider two linear combinations of them (i.e. ##\varphi_1(x) \ (\chi_+ + \chi_-)## and ##\varphi_1(x) \ (\chi_+ - \chi_-)##)?
     
  2. jcsd
  3. Aug 23, 2014 #2

    Orodruin

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    No, it does not matter. As long as the states are degenerate you may use either combination as a basis.
     
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