# Can two fermions occupy the same energy level in infinite potential well?

Suppose there are two electrons in an infinite potential well, what would be the ground state for this system?

I know that two bosons can occupy the lowest energy level (n=1) because they do not have to obey the Pauli exclusion principle, but can fermions also if they have different spin?

I believe the ground state for two electrons with different spin should be that both electrons are in the lowest energy level (n=1). However, I am not sure because if both particles are in the same energy level then the ground state wave-function is not anti-symmetric.

The total wave function needs to be antisymmetric under particle interchange. Since each electron is in the same 1-particle ground state, $E_0(x)$, the spatial wave function will be symmetric under interchange; hence, the spin wave function must be antisymmetric.

The 2-particle wave function is:

$$E_{0,0}(x_1, x_2) = E_0(x_1)E_0(x_2) = E_0(x_2)E_0(x_1) = E_0(x_2, x_1).$$

The rules regarding addition of angular momentum are well documented. The antisymmetric ground state will have $S=0$, and of course $S_z=0$:

$$\Xi_{1, 2} = \frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2),$$

where the subscripts label the particle index (and the arrows indicate the z-component). Note that:

$$\Xi_{2, 1} = \frac 1{\sqrt{2}}(\uparrow_2\downarrow_1-\downarrow_2\uparrow_1) = -\frac 1{\sqrt{2}}(\uparrow_1\downarrow_2-\downarrow_1\uparrow_2)=-\Xi_{1, 2}$$

so that the spin state is indeed antisymmetric.

The total wave function is their product:

$$\psi_{1, 2} = E_0(x_1)E_0(x_2)\Xi_{1, 2}.$$

Note that the statement "the electrons have different spin" is misleading (I would even say "classical"): they have the same spin: $J = \hbar\sqrt{j(j+1)} = \sqrt{3/2}\hbar$. They even have the same projection onto the $z-$axis: $\pm\hbar/2$--it's just that their combination is antisymmetric under interchange.

Finally: nowhere did I need to refer to the quantitative solution of the square well.

• @MomoTheSir Your welcome. Have another read, as I fixed some serious typos that made the math disagree with the text.
– JEB
Feb 3, 2018 at 21:19
• @MomoTheSir I would add that this state, in which two fermions have their spins in an antisymmetric configuration and their spatial wavefunctions identical, is actually a very good description of the two electrons in a parahelium atom. In real life the parahelium electron configuration has slightly higher energy than the ground states (called "orthohelium"), because of the Coulomb repulsion between the electrons. Feb 3, 2018 at 22:27
• @MomoTheSir OK then, let's do 2 right handed neutrinos in a box.
– JEB
Feb 4, 2018 at 0:37
• Does this mean the energy of the system is $2 E_0$? It sure seems like it, but I want to be sure. Nov 16, 2021 at 9:39
• $H\psi_{1,2}=(H_1+H_2)E_0(x_1)E_0(x_2)\Xi=H_1E_0(x_1)E_0(x_2)\Xi+H_2E_0(x_1)E_0(x_2)\Xi=E_0E_0(x_1)E_0(x_2)\Xi+E_0E_0(x_1)E_0(x_2)\Xi=2E_0\psi_{1,2}$
– JEB
Nov 16, 2021 at 14:26