Eigenfunction of a system of three fermions

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I have to find the eigenfunction of the ground state \Psi_0 of a three independent s=1/2 particle system.

The eigenfunctions \phi_{n,s}(x) = \varphi_n(x) \ \chi_s and eigenvalues E_n 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 \Psi_0 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_-)##)?
 
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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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