Can two fermions with different energylevels be at the same position?

Question

Suppose I have two fermions in a infinite square well potential, without spin or other degrees of freedom at $0 K$ temperature. Let $L$ be the width of that well. I used the two particle wave function in 1D for itentical fermions $$ \Psi_{nm}(x_1,x_2)=\frac{1}{\sqrt{2}}\left[\Psi_n(x_1)\Psi_m(x_2)-\Psi_n(x_2)\Psi_m(x_1)\right], $$ where $$ \Psi_n(x)=\sqrt{\frac{2}{L}}\sin{\frac{n\pi x}{L}} $$ is the solution of the SE for a single particle with energy level $E(n)= \hbar^2\pi^2n^2/2mL^2$ inside the well at position $x$. From this I already conclude that $n\neq m$. With that I calculated the probability density of finding one particle at $x_1$ and the other at $x_2$ with some energy levels $n$ and $m$: $$ |\Psi_{nn}(x_1,x_2)|^2=\frac{1}{2}\left[|\Psi_n(x_1)|^2|\Psi_m(x_2)|^2-2\Psi_n(x_2)\Psi_m(x_1)\Psi_n(x_1)\Psi_m(x_2)+|\Psi_n(x_2)|^2|\Psi_m(x_1)|^2\right]. $$ Since $\Psi_n(x_1)$ and $\Psi_m(x_1)$ are orthnormal the middle term is $0$.

Say one particle is found at $L/2$ what is the probability of finding the second particle at some position $x_2$, especially what happens to the probability, if we get close to the particle at $L/2$. I calculated this way: $$ \left|\Psi_{n}\left(\frac{L}{2}\right)\right|^2=\frac{2}{L}\sin^2{\frac{n\pi}{2}} =\frac{2}{L}\left\{\begin{array}{@{}lr@{}} 1 & \text{for uneven }n\\ 0 & \text{for even }n \end{array}\right\}=\frac{1+(-1)^{n+1}}{L}. $$ So inside the well, $$ \left|\Psi_{nm}\left(\frac{L}{2},x_2\right)\right|^2=\frac{1}{2}\left[\frac{1+(-1)^{n+1}}{L}\frac{2}{L}\sin^2{\frac{m\pi x_2}{L}}+\frac{1+(-1)^{m+1}}{L}\frac{2}{L}\sin^2{\frac{n\pi x_2}{L}}\right]= $$ $$ =\frac{1}{L^2}\left[(1+(-1)^{n+1})\sin^2{\frac{m\pi x_2}{L}}+(1+(-1)^{m+1})\sin^2{\frac{n\pi x_2}{L}}\right]. $$

Finally if I let $x_2\rightarrow L/2$, I get $$ \lim_{x_2\rightarrow L/2}\left|\Psi_{nm}\left(\frac{L}{2},x_2\right)\right|^2=\frac{1}{L^2}\left[(1+(-1)^{n+1})(1+(-1)^{m+1})\right]= \left\{\begin{array}{@{}lr@{}} 4L^{-2} & \text{if n and m are uneven and } n\neq m\\ 0 & \text{else} \end{array}\right\}. $$ So under these circumstances can they really be at the same spot?

EDIT2: I carried out the computation with the middle term. I now get $0$ for the probability density of $x_1=x_2=L/2$

Answer 1

No. If you calculate $$ |\Psi (x_1,x_1)| $$ directly it is zero. Your argument 'since $\Psi_n$ and $\Psi_m$ are orthogonal the middle term is zero' is wrong since the orthogonal condition requires you to integrate over the domain.

Answer 2

(This is a replacement for an answer that was wrong. "Today I Learned," as the Internet says.)

One way to answer this question is to change variables. Let's introduce

\begin{align} \Sigma x &= x_1 + x_2 & \Sigma x + \Delta x &= 2x_1 \\ \Delta x &= x_1 - x_2 & \Sigma x - \Delta x &= 2x_2 \end{align}

and try to find a probability density in terms of $\Delta x$.

Given your wavefunctions, $$ \psi_m(x_1) = \sqrt\frac2L\sin\frac{m\pi x_1}L, $$ some wrangling with trig identities shows that

\begin{align} \sqrt2\psi_{mn}(x_1,x_2) &= \psi_m(x_1) \psi_n(x_2) - \psi_n(x_1) \psi_m(x_2) \\ &= \frac2L\left[ \sin\left(\tfrac{m+n}2\tfrac{\pi\Sigma x}L\right) \sin\left(\tfrac{m-n}2\tfrac{\pi\Delta x}L\right) - \sin\left(\tfrac{m-n}2\tfrac{\pi\Sigma x}L\right) \sin\left(\tfrac{m+n}2\tfrac{\pi\Delta x}L\right) \right] \end{align}

If we integrate this distribution over all the allowed values of $\Sigma x$, we're left with

\begin{align} {\Large\vert}\psi(\Delta x){\Large\vert}^2 &= \int_{\Sigma x = |\Delta x|}^{2L-|\Delta x|}\mathrm d(\Sigma x) {\Large\vert}\psi(\Sigma x,\Delta x){\Large\vert}^2 \\ &= 2\frac {L-|\Delta x|}{L^2} \left[ \sin^2\left(\frac{m-n}2\frac{\pi\Delta x}L\right) + \sin^2\left(\frac{m+n}2\frac{\pi\Delta x}L\right) \right] \end{align}

which is the probability distribution for finding your two particles separated by a distance $\Delta x$. As $\Delta x$ becomes small, the term in the square brackets becomes proportional to $(m^2+n^2)\Delta x^2$: you're more likely to find highly-excited particles near each other than particles in the lower states, but the probability density for finding two particles with $x_1=x_2$ vanishes.

Here are some numerical results for a particular $m,n$ (click to embiggen). The null line in the joint probability density at $x_1=x_2$ (horizontally centered) is pretty easy to pick out. The local minima in probability density at $|\Delta x|/L = 0.42, 0.84$ are not actually zeros, though it's hard to tell from this particular presentation of the graph.

Answer 3

This has nothing to do with dynamics; it is true without any mention of "energy levels" and follows purely as a consequence of general principles of quantizing fermionic systems.

The correct statement for a continuous observable $x$ is a bit technical and not any more enlightening (specifically, the joint probability density function $f(x_1,x_2)$ goes to $0$ as $x_2 \rightarrow x_1$) so instead let's consider a finite-dimensional single-particle Hilbert space of dimension $n$ with a self-adjoint operator $\hat A$ with a nondegenerate spectrum of eigenvalues $\alpha_1, \ldots, \alpha_n$ and corresponding eigenvectors $|\alpha_1 \rangle, \ldots, |\alpha_n \rangle$.

If we want to describe a system with 2 noninteracting identical fermions of this type, we have $n(n-1)/2$ basis vectors which can be chosen as $|\beta_{ij}\rangle = \frac{1}{\sqrt2}(|\alpha_i\rangle \otimes | \alpha_j \rangle - |\alpha_j\rangle \otimes | \alpha_i \rangle)$, where $1 \le i < j \le n$. Fermions are only allowed to occupy these states, not the full $n^2$ dimensional tensor product space, because the states must be antisymmetric under exchange of particles. However, a state where both particles are measured to have $A = \alpha_i$ would necessarily be $|\alpha_i \rangle \otimes | \alpha_i \rangle$, and no such state exists as a linear combination of the $|\beta_{ij}\rangle$. Of course, such a state exists in the full tensor product space, but is orthogonal to the fermionic subspace because it is symmetric rather than antisymmetric under particle exchange.

There is a way to avoid this though. If you let $\hat A$ have a degenerate spectrum with $\alpha_i = \alpha_j$, then you have 2 distinct vectors $|\alpha_i \rangle$ and $|\alpha_j\rangle$ with the same observed value of $A$, and the antisymmetric state $|\beta_{ij}\rangle$ is a state in which we would measure both particles with $A = \alpha_i = \alpha_j$. The analogous thing in the continuous case would be to allow additional quantum numbers, which is to say that the states $\{ |x\rangle \}$ are no longer sufficient as a basis. This is crucially important e.g. if we want to have fermions with spin, but the question specifically rules it out of consideration, and without that it is true that the two fermions will always have different positions.