# Identical Fermions in an infinite square well

1. Apr 8, 2010

### 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?

Last edited: Apr 8, 2010
2. Apr 9, 2010

### 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.

3. Apr 9, 2010

### 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.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook