 Solution

For distinct positive integers $p,q$, use normalized box modes ${\phi }_1=\sqrt{2/L}\sin (p\pi x/L)$ and ${\phi }_2=\sqrt{2/L}\sin (q\pi x/L)$. Their contact interaction overlaps obeyUnlike traveling ring modes, these standing modes also allow pair conversion. For the first two box modes, or any distinct pair without a third-harmonic resonance, the contact interaction between two box modes isThe last two terms change an individual mode occupation number by two while conserving total particle number. Thus the Hamiltonian operator is not diagonal in the Fock state basis, and minimizing diagonal entries alone does not find its ground state. For example, at $N=2$ with no dispersion, $\left(|2,0⟩-|0,2⟩\right)/\sqrt{2}$ has the lowest energy $g/(2L)$, below either unmixed endpoint's $3g/(2L)$.

The wording permits modes beyond the first two. For completeness, the identity ${\sin }^{3}u=(3\sin u-\sin 3u)/4$ givesWhen $q=3p$, add $-g(a_1^{†2}{a}_1{a}_2+a_1^{†}{a}_2^{†}{a}_1^2)/(2L)$; when $p=3q$, add $-g(a_1^{†}{a}_2^{†}{a}_2^2+a_2^{†2}{a}_1{a}_2)/(2L)$. These extra terms change an individual occupation number by one and require diagonalizing the corresponding fixed-$N$ matrix as well.