Exact onsite Bose and Fermi interacting atoms

The one-mode case of [F1] identifies the number basis with the standard basis of ${\mathrm{\ell }}^2({\mathbb{N}}_0)$, whose finite-support span is dense. For any real diagonal multiplier with entries ${\lambda }_n$, testing the adjoint identity on each basis vector forces the adjoint image to have coordinates ${\lambda }_n{c}_n$, so its adjoint domain is contained in $\{c:{\sum }_n{\lambda }_n^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}$. Conversely, on that weighted domain [F11] makes the pairing bounded and identifies the adjoint action with the same multiplier. Thus its maximal diagonal domain is self-adjoint. Applying this calculation to $E_B(n)$, $n$, and $K_B(n)$ establishes the three domains in the Statement.

Consecutive Bose grand energies on the number basis [F1] satisfy $K_B(n+1)-K_B(n)=ϵ-\mu +Un$, a strictly increasing affine sequence in $n$. It is positive at every $n\ge 0$ when $\mu <ϵ$, so $0$ uniquely minimizes. For $r\ge 1$, it is negative through $n=r-1$ and positive from $n=r$ exactly when $ϵ+U(r-1)<\mu <ϵ+Ur$, so $r$ uniquely minimizes. At $\mu =ϵ+Ur$, the increment at $n=r$ is zero while all earlier increments are negative and all later increments positive; exactly $r,r+1$ minimize.

The four Fermi basis states have grand energies $0$, $ϵ-\mu -h$, $ϵ-\mu +h$, and $2(ϵ-\mu )+U$, respectively. Summing their four positive weights gives the displayed ${\mathrm{\Xi }}_F$, and summing $n_{\uparrow }+n_{\downarrow }$ or $n_{\uparrow }{n}_{\downarrow }$ against those weights gives the stated mean and double-occupation probability. The state is well-defined because this is a finite positive sum, as required by [A1] and [F3].

For fixed real $a$ and $n\ge 0$, completing the square gives $\mathrm{\mid }a\mathrm{\mid }n\le U{n}^2\mathrm{/}4+a^2\mathrm{/}U$. Hence, for $P_a(n)=U{n}^2\mathrm{/}2+an$, $U{n}^2\mathrm{/}4\le \mathrm{\mid }{P}_a(n)\mathrm{\mid }+a^2\mathrm{/}U$, so $n^2\le (4\mathrm{/}U)\mathrm{\mid }{P}_a(n)\mathrm{\mid }+4a^2\mathrm{/}{U}^2$. Fix an energy reference $E_{\mathrm{r}\mathrm{e}\mathrm{f}}>0$ in joules. Since $\mathrm{\mid }{P}_a(n)\mathrm{\mid }\le E_{\mathrm{r}\mathrm{e}\mathrm{f}}+P_a(n{)}^2\mathrm{/}{E}_{\mathrm{r}\mathrm{e}\mathrm{f}}$, this bound shows that the maximal diagonal domain associated with either quadratic sequence $E_B(n)$ or $K_B(n)$ is contained in $D(N_B)$. On $D(H_B)$, $K_{B}c=H_{B}c-\mu N_{B}c$ is square-summable, so $D(H_B)\subseteq D(K_B)$; on $D(K_B)$, $H_{B}c=K_{B}c+\mu N_{B}c$ is square-summable, so $D(K_B)\subseteq D(H_B)$. Thus $D(H_B)=D(K_B)\subseteq D(N_B)$, and the operator sum $H_B-\mu N_B$, whose domain is $D(H_B)\cap D(N_B)$, equals $K_B$ on its maximal domain.

In a strict interval with unique minimizer $r$, let $\delta$ be the smaller positive adjacent energy gap from $r$ (use only the upward gap when $r=0$). For $n>r$, the gap is a sum of increments from step 1.2, each at least the upward gap at $r$; for $n<r$, it is a sum of reversed increments, each at least the downward gap from $r-1$. Hence $K_B(n)-K_B(r)\ge \delta \mathrm{\mid }n-r\mathrm{\mid }$. The total relative excited weight is therefore at most $2{\sum }_{\mathrm{\ell }\ge 1}{e}^{-\beta \delta \mathrm{\ell }}=2e^{-\beta \delta }\mathrm{/}(1-e^{-\beta \delta })$, which tends to zero. At a boundary $\mu =ϵ+Ur$, the increments are $K_B(n+1)-K_B(n)=U(n-r)$, so exactly $r,r+1$ minimize. To the right, $K_B(r+1+\mathrm{\ell })-K_B(r+1)=U(1+\cdots +\mathrm{\ell })\ge U\mathrm{\ell }$; to the left, $K_B(r-\mathrm{\ell })-K_B(r)=U(1+\cdots +\mathrm{\ell })\ge U\mathrm{\ell }$ whenever $r\ge \mathrm{\ell }$. Thus the total relative excited weight is at most $2{\sum }_{\mathrm{\ell }\ge 1}{e}^{-\beta U\mathrm{\ell }}=2e^{-\beta U}\mathrm{/}(1-e^{-\beta U})$. The normalized Gibbs probabilities from [F3] therefore concentrate on the unique number state in a strict interval, or tend to $1\mathrm{/}2,1\mathrm{/}2$ on the adjacent pair at a boundary.

Put $a=ϵ-\mu -U\mathrm{/}2$. Completing the square gives $K_B(n)=\frac{U}{4}{n}^2+\frac{U}{4}(n+2a\mathrm{/}U{)}^2-a^2\mathrm{/}U\ge \frac{U}{4}{n}^2-a^2\mathrm{/}U$. For $n\ge 1$, $n^2\ge n$, so $e^{-\beta K_B(n)}\le e^{\beta a^2\mathrm{/}U}{q}^n$ with $q=e^{-\beta U\mathrm{/}4}$. Since $\beta U\mathrm{/}4>0$, [F5] gives $0<q<1$, and [F6] makes the geometric upper bound summable. The $n=0$ term is $1$, so $0<{\mathrm{\Xi }}_B<\mathrm{\infty }$ and [A1], [F3] give a well-defined normalized grand state for the exact operator $H_B-\mu N_B=K_B$ from step 2.1. The free-mode product trace formula in [F2] is not used.

Let $g=1$ in a strict interval and $g=2$ at a boundary, and shift all energies by their minimum so the ground energy is zero. The diagonal entropy from [F3] is $S_B\mathrm{/}{k}_B=\log D_{\beta }+\beta {\sum }_{n\ge 0}{p}_n{K}_B(n)$, where $D_{\beta }=g+R_{\beta }$ and $R_{\beta }$ is the total relative excited weight from step 2.2. Thus $D_{\beta }\to g$ by step 2.2 and $\log D_{\beta }\to \log g$ by continuity [F10]. In either case, for each distance $\mathrm{\ell }\ge 1$ there are at most two excited states; their lower gaps are $\delta \mathrm{\ell }$ by step 2.2 and their upper gaps are $C(1+{\mathrm{\ell }}^2)$ because $K_B(n)$ is quadratic in $n$. Here $C>0$ is a fixed energy in joules and $\delta =U$ at a boundary. Fix ${\beta }_0>0$ in inverse joules. For $\beta \ge {\beta }_0$, [F9] makes $(1+{\mathrm{\ell }}^2)e^{-{\beta }_0\delta \mathrm{\ell }\mathrm{/}2}$ bounded; splitting $e^{-\beta \delta \mathrm{\ell }}=e^{-\beta \delta \mathrm{\ell }\mathrm{/}2}{e}^{-\beta \delta \mathrm{\ell }\mathrm{/}2}$ therefore bounds $\beta {\sum }_{\mathrm{\ell }\ge 1}(1+{\mathrm{\ell }}^2)e^{-\beta \delta \mathrm{\ell }}$ by a constant times $\beta {\sum }_{\mathrm{\ell }\ge 1}{e}^{-\beta \delta \mathrm{\ell }\mathrm{/}2}=\beta q_{\beta }\mathrm{/}(1-q_{\beta })$, where $q_{\beta }=e^{-\beta \delta \mathrm{/}2}$. This tends to zero by [F5], [F6], and [F9]. Consequently the excited contribution to the entropy vanishes, while the normalized ground weights contribute $0$ for $g=1$ and $\log 2$ for $g=2$.

Fix $({\beta }_0,{\mu }_0)$ with ${\beta }_0>0$, put $u=\beta -{\beta }_0$, $v=\mu -{\mu }_0$, $a_0=ϵ-{\mu }_0-U\mathrm{/}2$, and $K_n^0=U{n}^2\mathrm{/}2+a_{0}n$. Then $-\beta K_B(n;\mu )=-{\beta }_0{K}_n^0-u{K}_n^0+{\beta }_{0}vn+uvn$, so [F5] expands each summand as $e^{-{\beta }_0{K}_n^0}{e}^{-u{K}_n^0}{e}^{{\beta }_{0}vn}{e}^{uvn}$. On radii $r_{\beta }={\beta }_0\mathrm{/}2$ and any fixed $r_{\mu }>0$, the sum of the absolute values of all terms of these three exponential power series, over all $n$, is bounded by ${\sum }_{n\ge 0}\exp (-{\beta }_0{K}_n^0+r_{\beta }\mathrm{\mid }{K}_n^0\mathrm{\mid }+{\beta }_0{r}_{\mu }n+r_{\beta }{r}_{\mu }n)$. For all sufficiently large $n$, $K_n^0\ge 0$ and $\mathrm{\mid }{K}_n^0\mathrm{\mid }\le U{n}^2\mathrm{/}2+\mathrm{\mid }{a}_0\mathrm{\mid }n$, so the exponent is at most $-{\beta }_{0}U{n}^2\mathrm{/}4+Cn$ for a finite constant $C$. Increasing the cutoff so $Cn\le {\beta }_{0}U{n}^2\mathrm{/}8$ leaves a tail bounded by $e^{-{\beta }_{0}Un\mathrm{/}8}$, whose series converges by [F5]–[F6]; the finitely many initial terms are finite. Thus the total absolute coefficient sum is finite. Grouping $r+t=p$ and $s+t=q$ gives $c_{pq}={\sum }_{n\ge 0}{e}^{-{\beta }_0{K}_n^0}{\sum }_{t=0}^{\min (p,q)}\frac{(-K_n^0{)}^{p-t}({\beta }_{0}n{)}^{q-t}{n}^t}{(p-t)!(q-t)!t!}$. The absolute bound and [F7] justify these coefficient sums and rearrangement, and give ${\sum }_{p,q}\mathrm{\mid }{c}_{pq}\mathrm{\mid }{r}_{\beta }^p{r}_{\mu }^q<\mathrm{\infty }$, hence $\mathrm{\mid }{c}_{pq}\mathrm{\mid }\le M{r}_{\beta }^{-p}{r}_{\mu }^{-q}$ for some finite $M$. Applying the multivariable power-series theorem [F7] gives a holomorphic sum on the complex polydisc; its real restriction is the absolutely convergent germ in [F4]. Since $({\beta }_0,{\mu }_0)$ was arbitrary, ${\mathrm{\Xi }}_B$ is jointly real analytic on $(0,\mathrm{\infty })\times \mathbb{R}$.

Fix $\beta >0$ and let $\mu$ range over a compact interval. The quantity $a=ϵ-\mu -U\mathrm{/}2$ is then bounded, so the square bound in step 3.1 gives $e^{-\beta K_B(n;\mu )}\le C{e}^{-\beta U{n}^2\mathrm{/}4}$ for a constant $C$ uniform on that interval. Put $q=e^{-\beta U\mathrm{/}8}\in (0,1)$. Since $n^2\ge n$, $(1+n^2)e^{-\beta K_B(n;\mu )}\le C(1+n^2)q^{2n}$. By [F9], $(1+n^2)q^n$ is bounded, so the right side is at most $C^{\mathrm{\prime }}{q}^n$, a summable geometric envelope. Apply [F8] on ${\mathbb{N}}_0$ with the counting measure [A2], weight $w(n)=e^{-\beta E_B(n)}$, observable $X(n)=n$, and natural parameter $\theta =\beta \mu$. Its hypotheses hold by this compact envelope and step 3.1; it gives ${\mathrm{\partial }}_{\theta }\mathbb{E}(N_B)=\mathrm{Var}(N_B)$. The chain rule [F12] yields the displayed response, while the normalized weights give the expectation formula.

The normalized means are $⟨n_{\uparrow }⟩=(w_{\uparrow }+w_2)\mathrm{/}{\mathrm{\Xi }}_F$ and $⟨n_{\downarrow }⟩=(w_{\downarrow }+w_2)\mathrm{/}{\mathrm{\Xi }}_F$, while $⟨n_{\uparrow }{n}_{\downarrow }⟩=w_2\mathrm{/}{\mathrm{\Xi }}_F$. Subtracting the product of the means yields $\mathrm{Cov}(n_{\uparrow },n_{\downarrow })=(w_2-w_{\uparrow }{w}_{\downarrow })\mathrm{/}{\mathrm{\Xi }}_F^2$. Since $w_{\uparrow }{w}_{\downarrow }=e^{-2\beta (ϵ-\mu )}$ and $w_2=e^{-2\beta (ϵ-\mu )}{e}^{-\beta U}$, this is the displayed formula; $U\ge 0$ makes it nonpositive, with equality exactly at $U=0$.