= Solution
Use $\langle O\rangle=\langle G|O|G\rangle$. 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 $A_0^\dagger A_0$ has <eigenvalue> $n_A$. Thus
$$
\boxed{\langle A_q^\dagger A_k\rangle=n_A\delta_{q0}\delta_{k0},\qquad
\langle B_q^\dagger B_k\rangle=n_B\delta_{q0}\delta_{k0}.}
$$
The two species have independent fixed <occupation numbers>, so mixed <expectation values> factorize. The inverse <discrete Fourier transforms> yield
$$
\langle a_i^\dagger a_j\rangle=\frac{n_A}{L},\qquad
\langle b_i^\dagger b_j\rangle=\frac{n_B}{L},\qquad
\boxed{\langle a_n^\dagger a_n+b_n^\dagger b_n\rangle
=\frac{n_A+n_B}{L}.}
$$
In 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. Hence
$$
\langle A_p^\dagger B_q^\dagger B_r A_t\rangle
=n_An_B\delta_{p0}\delta_{q0}\delta_{r0}\delta_{t0},
\qquad
\boxed{\langle H\rangle=-\frac JLn_An_B.}
$$
There 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 <2023/2/b/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 $n_An_B=[N^2-(n_A-n_B)^2]/4$. Consequently the best choices within this relaxed family are
$$
\boxed{\{n_A,n_B\}=\{\lfloor sL\rfloor,\lceil sL\rceil\}.}
$$
When $N$ is even, $n_A=n_B=sL$ and $\langle H\rangle=-Js^2L$. When $N$ is odd, either assignment of $(N-1)/2,(N+1)/2$ gives $\langle H\rangle=-J(N^2-1)/(4L)$. 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, $S_i^x=(a_i^\dagger b_i+b_i^\dagger a_i)/2$. Each term changes the two species' <occupation numbers>, producing a state <orthogonal> to $|G\rangle$. Therefore
$$
\boxed{\langle S_i^x\rangle=0,\qquad \langle S_i^y\rangle=0,\qquad
\langle S_i^z\rangle=\frac{n_A-n_B}{2L}.}
$$
For 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 give
$$
\begin{aligned}
\langle S_i^+S_j^-\rangle
&=\langle a_i^\dagger a_j\rangle\langle b_i b_j^\dagger\rangle
=\frac{n_A}{L}\frac{n_B}{L},\\
\langle S_i^-S_j^+\rangle&=\frac{n_An_B}{L^2}.
\end{aligned}
$$
Here $\langle b_i b_j^\dagger\rangle=n_B/L+\delta_{ij}$ and the <Kronecker delta> vanishes at distinct sites. Consequently
$$
\boxed{\langle S_m^xS_{m+n}^x\rangle=\frac{n_An_B}{2L^2}
\quad(n\not\equiv0\pmod L).}
$$
The 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.
\b[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 $L=2,s=1/2,n_A=n_B=1$,
$$
|G\rangle=\tfrac12(a_0^\dagger+a_1^\dagger)(b_0^\dagger+b_1^\dagger)|\mathrm{vac}\rangle.
$$
The nonzero component $a_0^\dagger b_0^\dagger|\mathrm{vac}\rangle/2$ 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 <variance>
$$
\boxed{\operatorname{Var}(N_n)
=\frac{n_A+n_B}{L}\left(1-\frac1L\right)
=2s\left(1-\frac1L\right)>0.}
$$
Each 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 has
$$
\langle S_n^2\rangle=s(s+1)+\tfrac14\operatorname{Var}(N_n),
$$
by the <Casimir operator> identity from <2023/2/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>.
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