Pauli's exclusion principle and square well potential [closed]

Question

Consider an infinite square potential well with potential
$$V(x)=\begin{cases} 0 & \text{if} -a<x<a\\ \infty & \text{otherwise.}\\ \end{cases}$$
The stationary state energies for a quantum particle trapped in such a potential are given by
$$E_n=\frac{n^2\pi^2\hbar^2}{8ma^2},n\in\mathbb{N}$$ where $m$ is the mass of the particle. Now Pauli's exclusion principle states that no two particles having the same quantum state exist in a quantum system.

Suppose an electron trapped in this well and now it is in the quantum state with $n=1$ then its energy (if we set $a=1$) is $E\approx {1.51}\cdot{10^{-38}}\ \mathrm{J}$ .

What would happen if another electron of the same energy as above was trapped in this well? If it was trapped inside the well, then it must violate Pauli's exclusion principle because above electron must have state $n\geq 2$ in this well. And also electron cannot occupy any other positions outside the well because potential is infinite outside the well. So what happend to the electron?

Answer 1

The process of having an electron inside the infinite well and then trapping another one is not possible, since the well is infinite.

If you have from the beginning two electrons inside the infinite well, then ignoring the spin of the electrons, it is not possible for them to have the same n. So as you say the one of them will be in the state n=1 and the other in the state n=2, with the coordinate wavefunction $\Psi=\frac{1}{\sqrt{2}}[\psi_{n1}(1)\psi_{n2}(2)-\psi_{n1}(2)\psi_{n2}(1) ]$ being antisymmetric.

On the other hand if you take into account the spin of the electrons, it is possible for them to have the same n. The coordinate wavefunction will now be symmetric (and therefore not zero when n is the same) and the spin state antisymmetric.The overall wavefunction will be: $\Psi=\frac{1}{\sqrt{2}}[\psi_{n1}(1)\psi_{n2}(2)+\psi_{n1}(2)\psi_{n2}(1)]\frac{1}{\sqrt{2}}[|+,-\rangle-|-,+\rangle]$