 Solution

A permutation $\pi$ of identical particles multiplies a fermionic state by $\mathrm{sgn}(\pi )$; a bosonic state is invariant. Particle exchange keeps $y_{+}$ fixed and sends $y_{-}$ to $-y_{-}$. Consequently the spatial wave function is even in $y_{-}$ for the stated bosonic model and odd for the stated fermionic model. This assumes no additional spin variables in the supplied model; exchange of full particle states is the underlying rule.

The coordinate change is orthogonal, with unit Jacobian determinant, so the kinetic energy becomeson the square $|y_{+}|,|y_{-}|<a$ with zero boundary values. Product states ${\psi }_r(y_{+}){\psi }_s(y_{-})$ have energyThe one-dimensional eigenfunction is even for odd $s$ and odd for even $s$. Bosons therefore allow odd $s$, whereas fermions require even $s$. The lowest bosonic pair is $(r,s)=(1,1)$; its first excited pair is $(2,1)$. The lowest fermionic pair is $(1,2)$. ThusThe printed bosonic ground-energy claim omits the factor ${\pi }^2/4$; the value above follows directly from the supplied one-dimensional spectrum and kinetic energy.

The separation is $|x_1-x_2|=\sqrt{2}|y_{-}|$. The normalized $y_{+}$ factor integrates out. In the fermionic ground state, the relative-coordinate density is $a^{-1}{\sin }^2(\pi y_{-}/a)$, givingFor the bosonic ground state it is $a^{-1}{\cos }^2(\pi y_{-}/(2a))$, givingBoth integrals follow by replacing squared trigonometric functions by their constant and cosine parts and integrating once by parts. Henceand the difference is $\boxed{2\sqrt{2}a/{\pi }^2}$. The odd relative state vanishes at coincidence, which increases its expected separation.