 Solution

Use $⟨O⟩=⟨G|O|G⟩$. The given state is a two-species uniform boson number state: only the $q=0$ mode of each species is occupied. Applying an annihilation operator to an unoccupied mode gives zero, while $A_0^{†}{A}_0$ has eigenvalue $n_A$. ThusThe two species have independent fixed occupation numbers, so mixed expectation values factorize. The inverse discrete Fourier transforms yieldIn the momentum-space Hamiltonian operator, only $p=q=r=t=0$ contributes: each annihilation operator must remove an occupied boson, and each creation operator must restore it to the same mode. HenceThere is no replacement of $n_B$ by $n_B+1$ in this energy: in the real-space calculation the relevant neighboring-site commutator is zero. Both expressions assume distinct neighbouring sites, as in ii.

To match the prescribed spin length in the mean, impose $n_A+n_B=2sL=:N$. At fixed integer $N$, the energy is minimized by maximizing $n_A{n}_B=[N^2-(n_A-n_B{)}^2]/4$. Consequently the best choices within this relaxed family areWhen $N$ is even, $n_A=n_B=sL$ and $⟨H⟩=-J{s}^{2}L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $⟨H⟩=-J(N^2-1)/(4L)$. Without fixing $N$, this auxiliary boson Hamiltonian operator has no lower bound: taking $n_A=n_B\to \infty$ sends its expectation value to $-\infty$.

Next, $S_i^x=(a_i^{†}{b}_i+b_i^{†}{a}_i)/2$. Each term changes the two species' occupation numbers, producing a state orthogonal to $|G⟩$. ThereforeFor distinct sites $i,j$ modulo $L$, expand the product of the two spin operators. The terms $S_i^{+}{S}_j^{+}$ and $S_i^{-}{S}_j^{-}$ again have zero expectation value. The remaining terms giveHere $⟨b_i{b}_j^{†}⟩=n_B/L+{\delta }_{ij}$ and the Kronecker delta vanishes at distinct sites. ConsequentlyThe same holds for the $y$ components. At the balanced choice and as $L\to \infty$, each transverse two-point correlation function tends to $s^2/2$ independently of separation, whereas the one-point transverse magnetization is zero. This is transverse spin order in a number eigenstate: the state has a definite total $S^z=(n_A-n_B)/2$ and retains the global U(1) symmetry of rotations about $z$, yet has the transverse correlations of an ordered state with no selected common azimuth. The sum of the $x$ and $y$ correlations tends to $s^2$, describing ferromagnetic long-range order within this approximation. The thermodynamic limit is understood with both the separation and $L$ growing, without identifying the two sites periodically.

Fixed-spin qualification. The proposed state is not actually in the physical fixed-$s$ Hilbert space for $s>0,L>1$. It satisfies the local spin constraint in the Schwinger boson representation only in expectation value. For example, with $L=2,s=1/2,n_A=n_B=1$,The nonzero component $a_0^{†}{b}_0^{†}|\mathrm{vac}⟩/2$ has local total occupation numbers $(2,0)$ and hence local spins $(1,0)$, instead of $(1/2,1/2)$. More generally the local total number operator $N_n$ has varianceEach species contributes a binomial distribution at a given site, with success probability $1/L$, and the two contributions are independent. In the physical subspace that variance must be zero. The proposed state even hasby the Casimir operator identity from a. Thus the calculated energy is a relaxed mean-field theory estimate, not a Rayleigh-Ritz variational principle bound for the stated fixed-$s$ quantum XY chain. Projecting onto $N_n=2s$ at every site would give a physical trial state and generally change its expectation values; in the two-site example the projected state has energy $-J$, whereas the unprojected value is $-J/2$. The calculated long-range order characterizes the supplied approximation and does not establish long-range order in the exact one-dimensional ground state.