Bosonic and fermionic Fock operators with exact domains

Each sector $H_s^{(n)}$ or $H_a^{(n)}$ is a closed subspace of its Hilbert tensor power, being the range of the corresponding orthogonal projection [F2]. The square-summable direct sum of these sectors is complete: for a Cauchy sequence, each coordinate converges in its complete sector; finite partial sums of squared coordinate norms are bounded by the common sequence bound, so the coordinate limits are square-summable; taking coordinate limits in the Cauchy estimate over each finite set of coordinates and then the supremum proves convergence in the direct-sum norm. Truncating a square-summable sequence after sector $N$ converges in norm, so ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is dense.

On sector $n$, creation and contraction have norms at most $\sqrt{n+1}\mathrm{\parallel }h\mathrm{\parallel }$ and $\sqrt{n}\mathrm{\parallel }h\mathrm{\parallel }$, respectively: the contraction has norm at most $\mathrm{\parallel }h\mathrm{\parallel }$ on its first factor by Cauchy–Schwarz and extends to the tensor product with the same bound [F4]; each symmetrizer or antisymmetrizer has norm at most one [F2]. For $\mathrm{\Phi }\in H_s^{(n)}$ and $\mathrm{\Psi }\in H_s^{(n+1)}$, use $P_{s,n+1}\mathrm{\Psi }=\mathrm{\Psi }$ and self-adjointness of $P_{s,n+1}$ to compute $⟨a_n^{†}(h)\mathrm{\Phi },\mathrm{\Psi }{⟩}_2=\sqrt{n+1}⟨h\otimes \mathrm{\Phi },\mathrm{\Psi }{⟩}_2=\sqrt{n+1}⟨\mathrm{\Phi },(C_h\otimes I^{\otimes n})\mathrm{\Psi }{⟩}_2=⟨\mathrm{\Phi },a_{n+1}(h)\mathrm{\Psi }{⟩}_2$; the tensor product pairing gives the middle equality because $C_h(v)=⟨h,v{⟩}_2$. The same calculation with $P_{a,n+1}$ proves the fermionic relation.

The maximal bosonic direct-sum maps are closed. For example, if ${\mathrm{\Phi }}^{(j)}\to \mathrm{\Phi }$ and $a(h){\mathrm{\Phi }}^{(j)}\to \mathrm{\Theta }$ in Fock norm, boundedness of each sector map gives coordinatewise ${\mathrm{\Theta }}_{n-1}=a_n(h){\mathrm{\Phi }}_n$; since $\mathrm{\Theta }$ is square-summable, the defining sum for $D(a(h))$ is finite and $a(h)\mathrm{\Phi }=\mathrm{\Theta }$. The same coordinate argument applies to $a^{†}(h)$. Truncating a vector and its image after sector $N$ shows ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is a graph core for each. For the adjoint, testing against vectors supported in sector $n+1$ forces the adjoint output coordinate in sector $n+1$ to be $a_n^{†}(h){\mathrm{\Phi }}_n$; square-summability of these coordinates is exactly the displayed domain of $a^{†}(h)$. Conversely, if that sum is finite, the sector pairing of step 1.2 and Cauchy–Schwarz in the direct sum give the adjoint identity for every vector in $D(a(h))$. Thus $a(h{)}^{\ast }=a^{†}(h)$. Testing $a^{†}(h{)}^{\ast }$ against input sector $n$ similarly forces its output coordinate to be $a_{n+1}(h){\mathrm{\Phi }}_{n+1}$; square-summability and the same pairing give exactly $D(a(h))$ and the converse inclusion. Hence the maximal maps are mutually adjoint.

For $n\ge 1$ and a symmetric elementary tensor $v_1\vee \cdots \vee v_n:=P_{s,n}(v_1\otimes \cdots \otimes v_n)$, contraction and creation give $a(h)(v_1\vee \cdots \vee v_n)=n^{-1\mathrm{/}2}{\sum }_{j=1}^n⟨h,v_j{⟩}_2{v}_1\vee \cdots \widehat{v_j}\cdots \vee v_n$ and $a^{†}(k)(v_1\vee \cdots \vee v_n)=\sqrt{n+1}{P}_{s,n+1}(k\otimes v_1\otimes \cdots \otimes v_n)$, by expanding the permutation averages [F2]. In the composition $a(h)a^{†}(k)$, the contraction term for the newly inserted $k$ is $⟨h,k{⟩}_2(v_1\vee \cdots \vee v_n)$; each of the other $n$ contraction terms is exactly the corresponding term of $a^{†}(k)a(h)$. Hence their difference is $⟨h,k{⟩}_{2}I$. On sector zero, $a(h)\mathrm{\Omega }=0$, $a^{†}(k)\mathrm{\Omega }=k$, and $a_1(h)k=⟨h,k{⟩}_2$ give the same commutator directly. Symmetry of $P_{s,n}$ makes two creations commute; two contractions commute because contracting distinct tensor slots in either order gives the same sum. Finite elementary symmetric tensors are dense in each sector and the sector maps are bounded there, so the identities hold on all of ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$.

For a unit vector $h$, put $K=h^{\perp }$. In the $n$th antisymmetric sector, every $v\in H$ decomposes as $v=⟨h,v{⟩}_{2}h+(v-⟨h,v{⟩}_{2}h)$ with the second term in $K$, so $H=\mathbb{C}h\oplus K$. Tensors split orthogonally into those with no $h$ factor and those with exactly one: expand each tensor factor in $\mathbb{C}h\oplus K$; terms with two $h$ factors vanish under antisymmetrization, and different placements of the single $h$ are orthogonal. The map $\sqrt{n+1}{P}_{a,n+1}(h\otimes \cdot )$ is an isometry from the no-$h$ part of sector $n$ onto the one-$h$ part of sector $n+1$, since the $n+1$ placements are orthogonal and each has coefficient $1\mathrm{/}(n+1)$ in the antisymmetrizer. It vanishes on the one-$h$ part. Its adjoint $c(h)$ reverses this isometry and vanishes on the no-$h$ part. Thus $c(h)c^{†}(h)+c^{†}(h)c(h)=I$ and both maps have norm one. Scaling gives $\{c(h),c^{†}(h)\}=\mathrm{\parallel }h{\mathrm{\parallel }}^{2}I$ and norms $\mathrm{\parallel }h\mathrm{\parallel }$ for $h\mathrm{\ne }0$; for $h=0$ these statements hold because both operators vanish. Antisymmetry gives $c^{†}(h{)}^2=0$; taking adjoints gives $c(h{)}^2=0$. Polarizing the diagonal anticommutator at $h+k$ and $h+ik$, using that $c(\cdot )$ is conjugate-linear and $c^{†}(\cdot )$ is linear, yields $\{c(h),c^{†}(k)\}=⟨h,k{⟩}_{2}I$: if $q(u)=\{c(u),c^{†}(u)\}$, then the sesquilinear polarization identity is $B(h,k)=\frac{1}{4}(q(h+k)-q(h-k)-iq(h+ik)+iq(h-ik))$, where $B(h,k)=\{c(h),c^{†}(k)\}$; substituting $q(u)=\mathrm{\parallel }u{\mathrm{\parallel }}^{2}I$ gives the claimed scalar. Polarizing $c(u{)}^2=0$ and $c^{†}(u{)}^2=0$ gives the two same-type anticommutators. All fermionic maps are bounded, so these identities hold on the whole Fock space.