Identical Fermions in an infinite square well

[msumm21]

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?

 

[jrlaguna]

You've put two fermions on the ground state of the potential, with their spin making up a singlet. I think your GS is fine. Griffith's solution may be referred to spinless particles.

 

[ansgar]

In the same line as jrlaguna said, fermions in QM just refers to particles with fermi-dirac statistics. The equivalence of spin1/2 particles and fermions are only in relativistic quantum field theory.