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?


Yes. It is a Slater determinant.

  • $\begingroup$ So can I get the slatter determinant of only orbital wave-functions? because spin wave-function is symmetric ? $\endgroup$ – lafahi Apr 30 '17 at 16:05

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