In the second edition of "Introduction to Quantum Mechanics" of David J. Griffiths (2016) there is an example that says (p.209, example 5.1)
Suppose we have two noninteracting --they pass right through one another ... -- particles, both of mass $m$, in the infinite square well. The one-particle states are $$ \psi_n (x) = \sqrt{\dfrac{2}{a}} \sin \left(\dfrac{n\pi }{a} x \right), \qquad E_n = n^2 K$$ (where $K\equiv \pi^2 \hbar^2 /2ma^2$, for convenience). If the particles are $\textit{distinguishable}$, with #1 in state $n_1$ and #2 in state $n_2$, the composite wave function is a simple product: $$ \psi_{n_1 n_2} (x_1, x_2) = \psi_{n_1} (x_1) \psi (x_2), \qquad E_{n_1 n_2} = (n_1^2+n_2^2) K.$$ For example, the ground state is $$ \psi_{11} = \frac{2}{a} \sin (\pi x_1/a) \sin (\pi x_2/a), \qquad E_{11} = 2K$$ [...]
and a problem 5.6. (p-214)
Imagine two noninteracting particles, each of mass $m$, in the infinite square well. If one is in the state $\psi_n$, and the other in state $\psi_m (m \neq n)$, calculate $<(x_1-x_2)^2>$, assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions.
If they are distinguishable particles: $$ <(x_1-x_2)^2>\equiv g_{nm}(a).$$ If they are identical bosons: $$ <(x_1-x_2)^2> = g_{nm}(a) -|I_{nm}|^2.$$ If they are identical fermions: $$ <(x_1-x_2)^2> = g_{nm}(a) + |I_{mn}|^2$$ where $$ g_{nm}(a) = a^2 \left[ \dfrac{1}{6} -\dfrac{1}{2\pi^2} \left( \dfrac{1}{n^2} +\dfrac{1}{m^2}\right)\right], \qquad n,m \in \mathbb{N}$$ and $$ I_{mn} =\dfrac{4a}{\pi^2} \dfrac{mn}{(n^2-m^2)^2} \left\{ (-1)^{m+n} -1 \right\}, \qquad n,m \in \mathbb{N}.$$ So, for bosons and fermions if $I_{nm}=0$ they can be considered as distinguishable particles. This occurs when the wave functions have the same parity on $n,m$. What is the physical interpretation of the last sentence?
If the sentence "they can be considered as distinguishable particles" is false, what is the physical interpretation of $|I_{mn}|=0$? What expectation values are needed to prove that the bosons or the fermions can be considered as distinguished particles?
