Finite bosonic and fermionic mode partition and occupation formulas

Assume AC and Countable Choice and adopt the quantum Gibbs preparation of The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature at assigned temperature $T>0$ kelvins, with $\beta =(k_{\mathrm{B}}T{)}^{-1}$ in ${\mathrm{J}}^{-1}$. A finite independent mode list means an integer $m\ge 1$ and, for each $r\in \{1,\dots ,m\}$, a declared type B or F and an excitation energy ${\epsilon }_r>0$ joules. For type B take the one-mode Fock Hilbert space with complete number basis $(e_n{)}_{n\in {\mathbb{N}}_0}$ and self-adjoint occupation operator $N_r{e}_n=n{e}_n$ on its exact diagonal domain as in Single-mode thermal trace. For type F take ${\mathcal{H}}_r={\mathbb{C}}^2$ with orthonormal basis $f_0,f_1$ and $N_r{f}_n=n{f}_n$. In either case put $H_r={\epsilon }_r{N}_r$, and write $u_{r,n}=e_n$ or $f_n$ according to the factor type. Define ${\mathcal{H}}^{(1)}={\mathcal{H}}_1$ and recursively ${\mathcal{H}}^{(r+1)}={\mathcal{H}}^{(r)}\widehat{\otimes }{\mathcal{H}}_{r+1}$ by the completed Hilbert tensor construction of Hilbert tensor construction: inner product, completion and product bases; let $\mathcal{H}={\mathcal{H}}^{(m)}$. The canonical orthonormal product basis is $u_{\mathbf{n}}={⨂}_r{u}_{r,n_r}$, indexed by $\mathbf{n}=(n_1,\dots ,n_m)\in \mathcal{C}:={\prod }_r{\mathcal{C}}_r$, where ${\mathcal{C}}_r={\mathbb{N}}_0$ for type B and ${\mathcal{C}}_r=\{0,1\}$ for type F. If all factors are F then $\mathcal{C}$ is finite; otherwise it is countably infinite and is enumerated by increasing ${\sum }_r{n}_r$ and lexicographic order within each finite level, which is finite because every $n_r\le {\sum }_r{n}_r$. For each factor put ${\widehat{\epsilon }}_r={\epsilon }_r\mathrm{/}(1\mathrm{J})$. Define the numerical energy map $\widehat{E}:\mathcal{C}\to \mathbb{R}$ by $\widehat{E}(\mathbf{n})={\sum }_{r=1}^m{\widehat{\epsilon }}_r{n}_r$ and the physical energy $E(\mathbf{n})=(1\mathrm{J})\widehat{E}(\mathbf{n})$. Define the numerical self-adjoint diagonal operator $\widehat{H}$ by $\widehat{H}{u}_{\mathbf{n}}=\widehat{E}(\mathbf{n})u_{\mathbf{n}}$ on the exact domain $D(\widehat{H})=\{\psi \in \mathcal{H}:{\sum }_{\mathbf{n}\in \mathcal{C}}\widehat{E}(\mathbf{n}{)}^2\mathrm{\mid }⟨\psi ,u_{\mathbf{n}}⟩{\mathrm{\mid }}^2<\mathrm{\infty }\};$ when $\mathcal{C}$ is finite, $\widehat{H}$ is a finite real diagonal endomorphism and is self-adjoint; when $\mathcal{C}$ is countably infinite, the diagonal-operator lemma Diagonal adjoint and graph-core lemma gives $\widehat{H}$ self-adjoint for the displayed enumeration. The physical total energy operator is $H=(1\mathrm{J})\widehat{H}$, with $H{u}_{\mathbf{n}}=E(\mathbf{n})u_{\mathbf{n}}$ and $D(H)=D(\widehat{H})$; multiplication by a nonzero real scalar preserves self-adjointness. This sum-energy Hamiltonian with no interaction term is the definition of independence for this quantum model.

Assign each factor readout $(N_r,1)$ and the composite readout $({\mathcal{N}}_r,1)$ as sharp observables in the sense of clause (c) of The canonical equilibrium-ensemble assumption: the Gibbs preparation of a declared model at assigned temperature, with dimensionless particle-count unit. On $\mathcal{H}$ define ${\mathcal{N}}_r\psi ={\sum }_{\mathbf{n}\in \mathcal{C}}{n}_r⟨\psi ,u_{\mathbf{n}}⟩u_{\mathbf{n}},D({\mathcal{N}}_r)=\{\psi :{\sum }_{\mathbf{n}\in \mathcal{C}}{n}_r^2\mathrm{\mid }⟨\psi ,u_{\mathbf{n}}⟩{\mathrm{\mid }}^2<\mathrm{\infty }\}\mathrm{.}$ To specify the joint occupation readout using a single sharp observable, let $j(\mathbf{n})$ be the zero-based position of $\mathbf{n}$ in the enumeration above (the same ordering on the finite set in the all-F case). Define $J\psi ={\sum }_{\mathbf{n}}j(\mathbf{n})⟨\psi ,u_{\mathbf{n}}⟩u_{\mathbf{n}}$ on $D(J)=\{\psi :{\sum }_{\mathbf{n}}j(\mathbf{n}{)}^2\mathrm{\mid }⟨\psi ,u_{\mathbf{n}}⟩{\mathrm{\mid }}^2<\mathrm{\infty }\}$. Assign $(J,1)$ as the dimensionless label readout, decoding $j(\mathbf{n})$ as $\mathbf{n}$. These diagonal operators are self-adjoint by the same diagonal lemma (or finite diagonal matrices). Their spectral projections act by retaining the coordinates with $n_r\in B$ or $j(\mathbf{n})\in B$, respectively, for a Borel set $B\subseteq \mathbb{R}$. Thus the decoded joint projections are $Q(A)\psi ={\sum }_{\mathbf{n}\in A}⟨\psi ,u_{\mathbf{n}}⟩u_{\mathbf{n}}$ for $A\subseteq \mathcal{C}$; these sums converge in $\mathcal{H}$, and countable additivity holds in the strong operator topology by square-summable coordinate tails. The adopted outcome law is $\mathrm{tr}({\rho }_{\beta }Q(A))$, whose $r$th marginal is the outcome law of ${\mathcal{N}}_r$.

For one bosonic mode with excitation energy $\epsilon >0$ joules, let $q=e^{-\beta \epsilon }$ and let $N$ be its dimensionless occupation number. Its partition trace and mean occupation are $Z_B=(1-q{)}^{-1},⟨N{⟩}_B=\frac{q}{1-q}=\frac{1}{e^{\beta \epsilon }-1}\mathrm{.}$ For a fermionic two-level occupation system with energies $0,\epsilon$, the partition sum and excited-state occupation are $Z_F=1+q,⟨N{⟩}_F=\frac{q}{1+q}=\frac{1}{e^{\beta \epsilon }+1}\mathrm{.}$ For a finite independent collection as defined above, write $K_{\beta }=e^{-\beta H}$ and $Z=\mathrm{Tr}{K}_{\beta }$. Then $0<Z<\mathrm{\infty }$ and the joint occupation-basis law is $P_{\beta }(\mathbf{n}):=e^{-\beta E(\mathbf{n})}\mathrm{/}Z$. For each factor $r$, define its marginal law by $P_{\beta ,r}(k):={\sum }_{\mathbf{n}\in \mathcal{C}:n_r=k}{P}_{\beta }(\mathbf{n})$. The trace satisfies $Z={\prod }_{r=1}^m{Z}_r$, the joint law factorizes as $P_{\beta }(\mathbf{n})={\prod }_{r=1}^m(q_r^{n_r}\mathrm{/}{Z}_r)$, and each marginal mean is the corresponding one-mode mean displayed above. These are exact model identities under the adopted preparation.