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.


1 Answer 1


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.

  • $\begingroup$ @MomoTheSir Your welcome. Have another read, as I fixed some serious typos that made the math disagree with the text. $\endgroup$
    – JEB
    Commented Feb 3, 2018 at 21:19
  • $\begingroup$ @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. $\endgroup$
    – tparker
    Commented Feb 3, 2018 at 22:27
  • $\begingroup$ @MomoTheSir OK then, let's do 2 right handed neutrinos in a box. $\endgroup$
    – JEB
    Commented Feb 4, 2018 at 0:37
  • $\begingroup$ Does this mean the energy of the system is $2 E_0$? It sure seems like it, but I want to be sure. $\endgroup$
    – Allure
    Commented Nov 16, 2021 at 9:39
  • $\begingroup$ $H\psi_{1,2}=(H_1+H_2)E_0(x_1)E_0(x_2)\Xi=H_1E_0(x_1)E_0(x_2)\Xi+H_2E_0(x_1)E_0(x_2)\Xi=E_0E_0(x_1)E_0(x_2)\Xi+E_0E_0(x_1)E_0(x_2)\Xi=2E_0\psi_{1,2}$ $\endgroup$
    – JEB
    Commented Nov 16, 2021 at 14:26

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