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 $x_1=x_2=L/2$