- #1

- 155

- 3

## 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:

[tex]\frac{1}{\sqrt{2}}\psi_1(x_1)\psi_1(x_2)(\uparrow\downarrow - \downarrow\uparrow )[/tex]

where [tex]\psi_1[/tex] 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 [tex]\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)[/tex].

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?

[tex]\frac{1}{\sqrt{2}}\psi_1(x_1)\psi_1(x_2)(\uparrow\downarrow - \downarrow\uparrow )[/tex]

where [tex]\psi_1[/tex] 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 [tex]\psi_1(x_1)\psi_2(x_2)-\psi_2(x_1)\psi_1(x_2)[/tex].

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?

Last edited: