 Solution

Add the one-particle energy term $H_0={ϵ}_1{n}_1+{ϵ}_2{n}_2$. It is diagonal in the same Fock state basis as the contact interaction between two ring modes. For $N=2$, the ordered basis $(|2,0⟩,|1,1⟩,|0,2⟩)$ givesFor $g>0$, the middle eigenvalue exceeds the mean of the two endpoint eigenvalues by $g/L$, so it cannot be the smallest. The ground state is therefore $|2,0⟩$ if ${ϵ}_1<{ϵ}_2$, and $|0,2⟩$ if ${ϵ}_2<{ϵ}_1$, withIf ${ϵ}_1={ϵ}_2$, every normalized linear combination of the two endpoint Fock states is a ground state. The same endpoint argument works for general fixed $N$, since the interaction contribution is a concave quadratic function of the mode occupation number.