# Physical interpretation of Quantum Mechanics Problem in Griffiths

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?

Just because $\langle (x_1 - x_2)^2 \rangle$ is unaffected by the particle statistics doesn't mean that "they can be considered as distinguishable particles." Other expectation values will be affected by the statistics.
• Thank you @tparker. And what is the physical interpretation of $I_{mn}=0$? – Clare Francis Mar 3 '17 at 18:48
• @JonathanEstévez-Fernández It just means that the net "exchange forces" due to the particle statistics happen to balance out to zero for that particular expectation value for those particular states. I don't think there's anything very fundamental there, though - I suspect that the higher separation moments, like $\langle (x_1 - x_2)^4 \rangle$ or $\langle (x_1 - x_2)^6 \rangle$, are affected by the particles' statistics. – tparker Mar 3 '17 at 23:03
• Thank you again @tparker for your answer. So, is it imposible to distinguish between two identical bosons/fermions with this potential even if $|x_1-x_2| \rightarrow \infty$ (if one particle is in Chicago and the other one in Seattle, for example)? Can it be mathematically proven ? – Clare Francis Mar 4 '17 at 9:51