Identical Fermions in an infinite square well

Main Question or Discussion Point

If you have 2 identical, noninteracting Fermions in an infinite 1 dimensional square well of width a, I was thinking the state would be:
$$\frac{1}{\sqrt{2}}\psi_1(x_1)\psi_1(x_2)(\uparrow\downarrow - \downarrow\uparrow )$$
where $$\psi_1$$ is the ground state of the single particle well problem.

However, I just looked in Griffths "Introduction to QM, 2nd Ed" and it says there is no state with the energy of that state and that in fact the ground state is $$\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)$$.

The state I gave seems to be (1) antisymmetric, (2) an eigenstate of the Hamiltonian, and (3) have lower energy than the ground state given by Griffths.

Do you see a problem with ground state I gave?

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