# Eigenfunction of a system of three fermions

1. Aug 23, 2014

### vabite

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 $\varphi_1(x) \ (\chi_+ - \chi_-)$)?

2. Aug 23, 2014

### Orodruin

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