Ground state of 3 identical fermions laying over a circumference We are asked to show that the ground state in a system of 3 free identical fermions that are on a circle of length, 2$\pi$, is equivalent to $\sin(q_1-q_2)+\sin(q_2-q_3)+\sin(q_3-q_1)$ up to some scalar multiplication. Furthermore we are asked to find the dimension of the eigenspace corresponding to the first excited energy level.
Any help is appreciated!
Edit: For a circle the eigenfunctions are;
$$\psi_n = e^{(2\pi inq/L)}$$
 A: The eigenfunctions are
$$\psi_n={1\over\sqrt L}e^{2i\pi nq/L},\quad n\in\mathbb Z$$
with the associated energies $E_n={\hbar^2\over 2m}\left({2\pi\over L}\right)^2n^2$. Due to Pauli principle, the ground state of 3 fermions corresponds to one fermion in the state $n=0$, the second one in the state $n=+1$ and the third one in the state $n=-1$. The total energy is therefore $E_0+E_{-1}+E_1={4\hbar^2\pi^2\over mL^2}$. The wavefunction has to be anti-symmetric under the exchange of two fermions and can be written as a Slater determinant
$$\psi(q_1,q_2,q_3)=\left|\matrix{
     \psi_{0}(q_1) & \psi_{1}(q_1) & \psi_{-1}(q_1) \cr
     \psi_{0}(q_2) & \psi_{1}(q_2) & \psi_{-1}(q_2) \cr             \psi_{0}(q_3) & \psi_{1}(q_3) & \psi_{-1}(q_3) \cr
}\right|$$
which leads to
$$\eqalign{
 \psi(q_1,q_2,q_3)&={1\over L^{3/2}}\left|\matrix{
   1& e^{2i\pi q_1/L} & e^{-2i\pi q_1/L} \cr
   1& e^{2i\pi q_2/L} & e^{-2i\pi q_2/L} \cr
   1& e^{2i\pi q_3/L} & e^{-2i\pi q_3/L} \cr
  }\right|\cr
 &={2i\over L^{3/2}}\Big(\sin {2\pi\over L}(q_2-q_3)
+\sin {2\pi\over L}(q_1-q_2)+\sin {2\pi\over L}(q_3-q_1)\Big)
}$$
