For the case of $N$ identical fermions in a three-dimensional box, the Pauli Exclusion Principle necessitates that the overall wavefunction of the system is antisymmetric. No two fermions can occupy the same state in momentum space, and so form a Fermi sphere where lower energy states are typically occupied first.

However, two electrons in isolated infinite wells can surely be treated independently, and both occupy an identical lowest momentum state. What about two electrons in finite wells whose wavefunctions overlap to a small degree? What about two electrons in an infinite well large enough that a causal connection may be interrupted? In general, what is the precise distinguishing condition necessary to treat a set of identical fermions as interacting versus non-interacting?

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    $\begingroup$ You are describing superposition, not interactions. The difference is that superpositions "add" wavefunctions, interactions are observed when energy momentum is exchanged and seen in the complex conjugate squared of the scattering wavefunctions. $\endgroup$
    – anna v
    Commented Oct 24, 2018 at 6:14

1 Answer 1


Let's put this in detail: consider the one-dimensional potential

$$ V_\ell(x) = \begin{cases} \infty \quad x \in \left(-\infty,-\ell-\frac{1}{2}\right) \cup \left(-\ell + \frac{1}{2}, \ell - \frac{1}{2}\right) \cup \left( \ell + \frac{1}{2},\infty \right) \\ 0 \quad \ \ \text{else} \end{cases} $$

Let $\psi_{L,R}(x)$ be wave-functions for a particle centered in the left respectively right well. Then the wave-functions of two particles is

$$ \Psi(x,y) = \frac{1}{\sqrt{2}} \psi_L(x) \psi_R(y) - \frac{1}{\sqrt{2}} \psi_R(x) \psi_L(y) $$

regardless of the value of $\ell$.

However,the important question is which observables we want to consider. For example, let's say you want to compute the probability of finding one particle in the an interval $I$ which is supposed to be contained in, say, the left well. This probability is

$$ p(I) = 2 \int_I d x \int_{\mathbb{R}\setminus I} d y \ |\Psi(x,y)|^2 \ . $$

The factor $2$ in front arises because we have to integrate over the regions $x\in I, y \in \mathbb{R}\setminus I$ and $ x \in \mathbb{R}\setminus I, y \in I$, but their contribution is the same. Because of the support properties of the $\psi_{L,R}$ wave functions:

$$ p(I) = \left[\int_I |\psi_L(x) | dx \right] \left[ \int_{\mathbb{R}\setminus I} |\psi_R(x)|^2 dx \right] = \int_I |\psi_L(x) | dx $$

where in the second equality the normalization was used. So when we want to calculate properties only concerning one well, we may use wavefunction describing one well only. Note that in reality there are no infinite wells, so that the wavefunctions will have some exponential tails outside the well. Then the error we make calculating properties with one-well wave-functions decays as $e^{- C \ell}$, with $C$ the size of the barrier.


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