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?

closed as unclear what you're asking by ACuriousMind, user36790, Ryan Unger, Kyle Kanos, Prahar Oct 28 '15 at 3:59

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  • 1
    How do you mean the second electron should get trapped? In what type of process? You can't say "there was one electron, and on the next moment of time a second one appeared". – Ruslan Oct 25 '15 at 12:13
  • Well you couldn't get another electron of the same energy trapped in the well because the PEP forbids it, isn't that kinda the point of the PEP? – Kyle Kanos Oct 26 '15 at 10:18
  • @KyleKanos what is PEP? – Muhsin Ibn Al Azeez Oct 26 '15 at 10:57
  • PEP = Pauli Exclusion Principle. – Kyle Kanos Oct 26 '15 at 11:44
  • If you try to add a 2nd electron, the potential energy function inside the well is no longer constant but is a function of the positions of the electrons. It's no longer a square well. – Bill N Oct 26 '15 at 20:36
up vote 5 down vote accepted

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: $\Psi=\frac{1}{\sqrt{2}}[\psi_{n1}(1)\psi_{n2}(2)+\psi_{n1}(2)\psi_{n2}(1)]\frac{1}{\sqrt{2}}[|+,-\rangle-|-,+\rangle]$

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