Identical fermions in a box - degenerate states

[dipole]

The ground state for two identical fermions in a box (in 1D) is given by:

\psi (x_{1},x_{2})_{12} = \frac{\sqrt{2}}{a}[sin(\pi x_{1}/a)sin(2\pi x_{2}/a) - sin(2\pi x_{1}/a)sin(\pi x_{2}/a)]

The book I'm reading though says that this state is non degenerate, and that the next excited state is \psi_{13}. My question is, why is the ground state not degenerate? Why can't you have either \psi_{12} or \psi_{21} which would have the same energy?

Thanks.

 

[Born2bwire]

Pauli's exclusion principle. EDIT: See below.

 

[dipole]

I don't see how that answers my question.

Pauli exclusion principle determines the form of the wave function, and it forbids the \psi_{22} state, but that doesn't explain to me why you can't have either \psi_{12} or \psi_{21} and have a doubly degenerate ground state.

 

[Born2bwire]

EDIT: Misread what you were asking exactly. Changing.

Sorry, thought you were asking about if you added another fermion.

But you have to remember that the particles are identical, so the order of your subscripts doesn't matter in this case because you cannot distinguish between the two particles. Pauli's Exlusion Principle is already in play here by the fact that one particle is in the ground state (of the one particle system) and another particle is in the first excited state (of the one particle system). You can't have the \psi_{11} and by extension neither the \psi_{22}. So the next possible superposition would be the \psi_{13} where one particle is in the ground state and the other is in the second excited state. And since this has a higher energy than the \psi_{12} state, then this represents the first excited state of the two particle system.

 

[dipole]

Ah I see, so in distinguishable particles, the \psi_{12} and \psi_{21} are degenerate, but for indistinguishable particles they really represent the same state.

That makes perfect sense, thanks.