Boson occupation ladder and fermion one-mode exclusion

Let $S_2={\sum }_{k=1}^{\mathrm{\infty }}{k}^{-2}$. For each $j\ge 0$, the block $2^j\le k<2^{j+1}$ contains $2^j$ terms, each at most $2^{-2j}$, so its sum is at most $2^{-j}$. Summing these geometric bounds proves $1\le S_2<\mathrm{\infty }$. Therefore $C=S_2^{-1\mathrm{/}2}$ is a positive finite constant, and the orthonormal bosonic mode states from [F1] give $\mathrm{\Psi }={\sum }_{n=0}^{\mathrm{\infty }}C(n+1{)}^{-1}\mathrm{\mid }n{⟩}_h$ with $\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2=C^2{S}_2=1$.

For ${\xi }_n$ in the $n$th antisymmetric sector and ${\eta }_{n+1}$ in the next sector, the insertion formula, $P_{a,n+1}{\eta }_{n+1}={\eta }_{n+1}$, and self-adjointness of this orthogonal projection give $⟨c_n^{†}(u){\xi }_n,{\eta }_{n+1}{⟩}_2=\sqrt{n+1}⟨u\otimes {\xi }_n,{\eta }_{n+1}{⟩}_2=⟨{\xi }_n,\sqrt{n+1}(C_u\otimes I^{\otimes n}){\eta }_{n+1}{⟩}_2=⟨{\xi }_n,c_{n+1}(u){\eta }_{n+1}{⟩}_2$ by the tensor product pairing and contraction formula in [F3]. Summing over finite-sector vectors proves $⟨c^{†}(u)\xi ,\eta {⟩}_2=⟨\xi ,c(u)\eta {⟩}_2$ there. Boundedness in [F2] and density in [F3] extend this identity to the full Fock space, so $c^{†}(u)=c(u{)}^{\ast }$. The CAR, vacuum relation, and nilpotence in [F2]–[F3] then give $\mathrm{\parallel }{c}^{†}(u){\mathrm{\Omega }}_D{\mathrm{\parallel }}^2=⟨{\mathrm{\Omega }}_D,c(u)c^{†}(u){\mathrm{\Omega }}_D{⟩}_2=⟨{\mathrm{\Omega }}_D,(I-c^{†}(u)c(u)){\mathrm{\Omega }}_D{⟩}_2=1$. Thus the one-mode state is normalized and a second creation in that mode is zero.

By the ladder in [F1], the squared norm of the annihilation-sector sequence for this vector is $C^2{\sum }_{n=1}^{\mathrm{\infty }}n\mathrm{/}(n+1{)}^2$. Since $n+1\le 2n$ for $n\ge 1$, $n\mathrm{/}(n+1{)}^2\ge 1\mathrm{/}(4n)$. Each dyadic block $2^j\le n<2^{j+1}$ contributes at least $1\mathrm{/}2$ to the harmonic series, so ${\sum }_{n\ge 1}1\mathrm{/}n$ diverges and therefore ${\sum }_{n\ge 1}n\mathrm{/}(n+1{)}^2=\mathrm{\infty }$. By [F1], the annihilation images of the different particle sectors lie in orthogonal sectors, so this is exactly ${\sum }_{n\ge 1}\mathrm{\parallel }{a}_n(h){\mathrm{\Psi }}_n{\mathrm{\parallel }}^2$. The exact domain in [F4] now gives $\mathrm{\Psi }\mathrm{\notin }D(a(h))$, despite $\mathrm{\Psi }$ being a normalized Fock vector.

The adjoint identity and CAR in [F2] give $N_u^{\ast }=N_u$ and $N_u^2=c^{†}(u)(I-c^{†}(u)c(u))c(u)=N_u$, so $N_u$ is a projection. The vacuum relation gives $N_u{\mathrm{\Omega }}_D=0$, while $N_u{c}^{†}(u){\mathrm{\Omega }}_D=c^{†}(u){\mathrm{\Omega }}_D$. Thus its eigenvalues on the one-mode subspace are $0$ and $1$. These are conditional model-state calculations, not measured outcomes.