Creation contraction bounds

[The contraction map] Fix $f\in H$, $n\ge 0$ and $\chi \in H^{\otimes (n+1)}$. The map $\phi \mapsto ⟨f\otimes \phi ,\chi ⟩$ is linear, and by [F4] and [F3] $\mathrm{\mid }⟨f\otimes \phi ,\chi ⟩\mathrm{\mid }\le \mathrm{\parallel }f\otimes \phi \mathrm{\parallel }\mathrm{\parallel }\chi \mathrm{\parallel }=\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }\phi \mathrm{\parallel }\mathrm{\parallel }\chi \mathrm{\parallel }$, so it is bounded with norm at most $\mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }\chi \mathrm{\parallel }$; by [F5] there is a unique ${\mathrm{\Psi }}_f^{(n)}\chi \in H^{\otimes n}$ representing it. The assignment $\chi \mapsto {\mathrm{\Psi }}_f^{(n)}\chi$ is linear (the representing vectors of $\chi \mapsto ⟨f\otimes \phi ,\chi ⟩$ depend linearly on $\chi$), and $\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}\chi \mathrm{\parallel }={\sup }_{\mathrm{\parallel }\phi \mathrm{\parallel }\le 1}\mathrm{\mid }⟨f\otimes \phi ,\chi ⟩\mathrm{\mid }\le \mathrm{\parallel }f\mathrm{\parallel }\mathrm{\parallel }\chi \mathrm{\parallel }$ gives $\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }$. For an elementary tensor, the tensor-product inner product and the first-linear convention give [F3, F4, F5, algebra] $⟨f\otimes \phi ,u_1\otimes \cdots \otimes u_{n+1}⟩=⟨f,u_1⟩⟨\phi ,u_2\otimes \cdots \otimes u_{n+1}⟩=⟨\phi ,⟨u_1,f⟩u_2\otimes \cdots \otimes u_{n+1}⟩\mathrm{.}$ This equality for every $\phi$ and uniqueness of the representing vector imply ${\mathrm{\Psi }}_f^{(n)}(u_1\otimes \cdots \otimes u_{n+1})=⟨u_1,f⟩u_2\otimes \cdots \otimes u_{n+1}$. Finally ${\mathrm{\Psi }}_{cf}^{(n)}=\bar{c}{\mathrm{\Psi }}_f^{(n)}$ for $c\in \mathbb{C}$ since $⟨u_1,cf⟩=\bar{c}⟨u_1,f⟩$ on elementary tensors, so $f\mapsto {\mathrm{\Psi }}_f^{(n)}$ is conjugate-linear.

[Symmetry preservation] Let $\chi \in \mathrm{ran}{P}_{+}^{(n+1)}$ and let $\sigma \in S_n$ act on the last $n$ tensor factors by $U_{1\otimes \sigma }$; by [F3] applied to the permutation $(1,\sigma )\in S_{n+1}$ one has $U_{1\otimes \sigma }\chi =\chi$. Then for every $\phi \in H^{\otimes n}$, $⟨\phi ,{\mathrm{\Psi }}_f^{(n)}\chi ⟩=⟨f\otimes \phi ,\chi ⟩=⟨U_{1\otimes \sigma }^{-1}(f\otimes \phi ),\chi ⟩=⟨f\otimes U_{{\sigma }^{-1}}\phi ,\chi ⟩=⟨U_{{\sigma }^{-1}}\phi ,{\mathrm{\Psi }}_f^{(n)}\chi ⟩=⟨\phi ,U_{\sigma }{\mathrm{\Psi }}_f^{(n)}\chi ⟩$, the third equality because $U_{1\otimes \sigma }^{-1}=U_{1\otimes {\sigma }^{-1}}$ permutes only the last factors and fixes $f$. Hence $U_{\sigma }{\mathrm{\Psi }}_f^{(n)}\chi ={\mathrm{\Psi }}_f^{(n)}\chi$ for every $\sigma \in S_n$, so ${\mathrm{\Psi }}_f^{(n)}\chi \in \mathrm{ran}{P}_{+}^{(n)}$ by the converse direction of [F3].

[Contraction bounds] For $\psi \in F_0$, using $\mathrm{\parallel }{P}_{+}^{(n)}\mathrm{\parallel }\le 1$ ([F3]) and the tensor norm identity of [F3], $\mathrm{\parallel }{S}_{+}(f)\psi {\mathrm{\parallel }}^2={\sum }_{n\ge 1}n\mathrm{\parallel }{P}_{+}^{(n)}(f\otimes {\psi }_{n-1}){\mathrm{\parallel }}^2\le {\sum }_{n\ge 1}n\mathrm{\parallel }f\otimes {\psi }_{n-1}{\mathrm{\parallel }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}^2{\sum }_{n\ge 0}(n+1)\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}_1^2$. Similarly $\mathrm{\parallel }{S}_{-}(f)\psi {\mathrm{\parallel }}^2={\sum }_{n\ge 0}(n+1)\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}{\psi }_{n+1}{\mathrm{\parallel }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}^2{\sum }_{n\ge 0}(n+1)\mathrm{\parallel }{\psi }_{n+1}{\mathrm{\parallel }}^2\le \mathrm{\parallel }f{\mathrm{\parallel }}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}_1^2$, where step 1.1 supplied $\mathrm{\parallel }{\mathrm{\Psi }}_f^{(n)}\mathrm{\parallel }\le \mathrm{\parallel }f\mathrm{\parallel }$.

[The operators on the core] For $\psi \in F_0$ only finitely many coordinates are nonzero, each in the appropriate range ([F1]); by [F2] and step 2.1 all displayed expressions $(S_{\pm }(f)\psi {)}_n$ lie in $\mathrm{ran}{P}_{+}^{(n)}$ and are nonzero only for finitely many $n$, so $S_{\pm }(f)\psi \in F_0$. Linearity of $S_{+}(f)$ in $\psi$ and of $S_{+}(\cdot )\psi$ in $f$, and conjugate-linearity of $S_{-}(f)$ in $f$ from step 1.1, are immediate from the formulas.

[Per-sector commutator computation] For $n\ge 0$ define $A_g^{(n)}\chi :=P_{+}^{(n+1)}(g\otimes \chi )$ from $\mathrm{ran}{P}_{+}^{(n)}$ to $\mathrm{ran}{P}_{+}^{(n+1)}$, and for $n\ge 1$ define $B_f^{(n)}\chi :={\mathrm{\Psi }}_f^{(n-1)}\chi$ from $\mathrm{ran}{P}_{+}^{(n)}$ to $\mathrm{ran}{P}_{+}^{(n-1)}$. Let $\chi =P_{+}^{(n)}(v_1\otimes \cdots \otimes v_n)$. Since $P_{+}^{(n+1)}(g\otimes \chi )=P_{+}^{(n+1)}(g\otimes v_1\otimes \cdots \otimes v_n)$, averaging over the $(n+1)!$ permutations and grouping by the first tensor factor gives [F3, step 1.1, step 2.1, algebra] $B_f^{(n+1)}{A}_g^{(n)}\chi =\frac{1}{n+1}(⟨g,f⟩\chi +{\sum }_{j=1}^n⟨v_j,f⟩P_{+}^{(n)}(g\otimes v_1\otimes \cdots \otimes \widehat{v_j}\otimes \cdots \otimes v_n))\mathrm{.}$ The same grouping over the $n!$ permutations of $\chi$ gives $B_f^{(n)}\chi =\frac{1}{n}{\sum }_{j=1}^n⟨v_j,f⟩P_{+}^{(n-1)}(v_1\otimes \cdots \otimes \widehat{v_j}\otimes \cdots \otimes v_n),$ so applying $A_g^{(n-1)}$ and multiplying by $n$ yields exactly the sum over $j$ in the preceding display. Therefore $(n+1)B_f^{(n+1)}{A}_g^{(n)}\chi -n{A}_g^{(n-1)}{B}_f^{(n)}\chi =⟨g,f⟩\chi$. The maps in this identity are bounded by step 1.1 and contractivity of $P_{+}^{(m)}$ ([F3]); the displayed $\chi$ have dense span in $\mathrm{ran}{P}_{+}^{(n)}$ ([F3]), so the identity extends to every vector in that sector. For the vacuum scalar $c\in \mathbb{C}$, ${\mathrm{\Psi }}_f^{(0)}{P}_{+}^{(1)}(g\otimes c)=⟨g,f⟩c$ and the annihilation-then-creation term vanishes, so the commutator identity also holds at $n=0$.

[Adjoint relation] Let $\phi ,\psi \in F_0$. Using the definition of $S_{+}$ and the self-adjointness of $P_{+}^{(n+1)}$, and the fact that ${\psi }_{n+1}\in \mathrm{ran}{P}_{+}^{(n+1)}$, $⟨S_{+}(f)\phi ,\psi ⟩={\sum }_{n\ge 0}\sqrt{n+1}⟨P_{+}^{(n+1)}(f\otimes {\phi }_n),{\psi }_{n+1}⟩={\sum }_{n\ge 0}\sqrt{n+1}⟨f\otimes {\phi }_n,{\psi }_{n+1}⟩$. By the defining property of ${\mathrm{\Psi }}_f^{(n)}$ in step 1.1 this equals ${\sum }_{n\ge 0}\sqrt{n+1}⟨{\phi }_n,{\mathrm{\Psi }}_f^{(n)}{\psi }_{n+1}⟩=⟨\phi ,S_{-}(f)\psi ⟩$, both sums being finite.

[Canonical commutation relation] Let $f,g\in H$ and $\psi \in F_0$. For $n\ge 1$, the definitions of $S_{\pm }$ and step 3.2 give $(S_{-}(f)S_{+}(g)\psi {)}_n=(n+1)B_f^{(n+1)}{A}_g^{(n)}{\psi }_n$ and $(S_{+}(g)S_{-}(f)\psi {)}_n=n{A}_g^{(n-1)}{B}_f^{(n)}{\psi }_n$. For $n=0$, the first composition is ${\mathrm{\Psi }}_f^{(0)}{P}_{+}^{(1)}(g\otimes {\psi }_0)=⟨g,f⟩{\psi }_0$ and the second is zero. Thus each coordinate of the difference is $⟨g,f⟩{\psi }_n$; all coordinates outside a finite set vanish, so the two sides agree.

[The norm identity] Let $g\in H$ and $\psi \in F_0$. Taking the inner product of the commutation relation of step 4.2 (with $f=g$) with $\psi$ gives $⟨S_{-}(g)S_{+}(g)\psi ,\psi ⟩-⟨S_{+}(g)S_{-}(g)\psi ,\psi ⟩=\mathrm{\parallel }g{\mathrm{\parallel }}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$. By the adjoint relation of step 4.1, $⟨S_{-}(g)S_{+}(g)\psi ,\psi ⟩=⟨S_{+}(g)\psi ,S_{+}(g)\psi ⟩=\mathrm{\parallel }{S}_{+}(g)\psi {\mathrm{\parallel }}^2$ and $⟨S_{+}(g)S_{-}(g)\psi ,\psi ⟩=⟨S_{-}(g)\psi ,S_{-}(g)\psi ⟩=\mathrm{\parallel }{S}_{-}(g)\psi {\mathrm{\parallel }}^2$; hence $\mathrm{\parallel }{S}_{+}(g)\psi {\mathrm{\parallel }}^2=\mathrm{\parallel }{S}_{-}(g)\psi {\mathrm{\parallel }}^2+\mathrm{\parallel }g{\mathrm{\parallel }}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$. This completes the proof of (a)-(d).