# Aharonov-Bohm ring with many electrons

4 electrons are in an Aharonov-Bohm ring which the Hamiltonian is given by,

$$H=\dfrac {\hbar^2} {2mr^{2}}\sum _{n=1}^{4}\left( -i\dfrac {\partial } {\partial \theta_{n}}-4\right) ^{2}$$

How to obtain the ground state wave-function. I know that single electron states are given by,

$\phi(\theta)=\frac{1}{2\pi}$$e^{im\theta}$

Also, given that the ground state is unique. Does that means $S=0$ for ground state? so Can I written down the wave-function as a Slatter determinant?