Self-adjoint Segal fields and explicit Weyl unitaries

Put $K=e^{\perp }$ and $G={\mathcal{F}}_s(K)$ for a unit vector $e$. Since $H$ is separable, its continuous orthogonal projection $v\mapsto v-⟨e,v{⟩}_{2}e$ sends a countable dense subset of $H$ to a countable dense subset of $K$. Finite elementary tensors from that subset have dense span in each tensor power by [F8]; applying the bounded symmetrizer gives a countable dense subset of each $K_s^{(m)}$. Rational complex finite sums over the countably many sectors give a countable dense subset of $G$ consisting of finite-particle vectors. Gram–Schmidt, omitting zero residuals, therefore gives a finite or countable orthonormal basis $(u_j{)}_{j\in J}$ of $G$ whose vectors are all finite-particle.

Map the one-mode basis $\mathrm{\mid }n⟩=e^{\otimes n}$ to $g_n$. The map $V:{\mathcal{F}}_s(\mathbb{C}e)\to L_2^2(\mathbb{R})$ is a linear unitary: the two displayed families are complete orthonormal families, and both pairings are second-variable-linear. Here $L_2^2$ has pairing $⟨f,g{⟩}_2=\int \bar{f}g$; the core and self-adjointness of $Qf=xf$ transfer from [F4] by complex conjugation. The recurrence [F3] shows that $V{2}^{-1\mathrm{/}2}(a^{†}(e)+a(e))V^{-1}$ equals $Q$ on the finite Hermite span. Together with $J$ [F2], the full field for $h=re$ acts in $L_2^2(\mathbb{R})\widehat{\otimes }G$ as multiplication by $rx$ on the dense span of $g_n\otimes u_j$.

Under the product-basis unitary of [F8], identify $L_2^2(\mathbb{R})\widehat{\otimes }G$ with ${\mathrm{\ell }}^2(J;L_2^2(\mathbb{R}))$. Define $M_r(f_j)=(rx{f}_j)$ on $D(M_r)=\{(f_j):{\sum }_j\int r^2{x}^2\mathrm{\mid }{f}_j(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty }\}$. Its domain is dense because finite coordinates supported in $[-m,m]$ are dense. It is symmetric since $rx$ is real. If $g\in D(M_r^{\ast })$ and $k=M_r^{\ast }g$, test the adjoint identity in coordinate $j$ against $f_j=(rx{g}_j-k_j){\mathbf{1}}_{[-m,m]}$ and set all other coordinates to zero. This vector lies in $D(M_r)$ because $rx$ is bounded on $[-m,m]$; the identity then gives $0={\int }_{-m}^m\mathrm{\mid }rx{g}_j-k_j{\mathrm{\mid }}^{2}dx$. Letting $m\to \mathrm{\infty }$ yields $rx{g}_j=k_j$ almost everywhere for each $j$, so ${\sum }_j\int r^2{x}^2\mathrm{\mid }{g}_j{\mathrm{\mid }}^2={\sum }_j\mathrm{\parallel }{k}_j{\mathrm{\parallel }}^2<\mathrm{\infty }$ and $g\in D(M_r)$. Conversely, for $g\in D(M_r)$, reality of $rx$ gives $⟨M_{r}f,g{⟩}_2=⟨f,M_{r}g{⟩}_2$ for every $f\in D(M_r)$, hence $M_r^{\ast }g=M_{r}g$. Thus $M_r$ is self-adjoint with the displayed domain.

The series defining $ϵ(z)$ converges in Fock norm because $\mathrm{\parallel }ϵ(z){\mathrm{\parallel }}^2={\sum }_{n\ge 0}\mathrm{\parallel }z{\mathrm{\parallel }}^{2n}\mathrm{/}n!=e^{\mathrm{\parallel }z{\mathrm{\parallel }}^2}$ [F6]. Its inner products are $⟨ϵ(z),ϵ(w){⟩}_2=e^{⟨z,w{⟩}_2}$. These vectors are total: if $\mathrm{\Psi }=({\mathrm{\Psi }}_n)$ is orthogonal to all of them, then for every $z$ and scalar $t$ the entire power series $⟨\mathrm{\Psi },ϵ(tz){⟩}_2$ vanishes, so each coefficient $⟨{\mathrm{\Psi }}_n,z^{\otimes n}{⟩}_2$ is zero. For fixed $n$, the coefficient of $t_1\cdots t_n$ in $⟨{\mathrm{\Psi }}_n,(t_1{z}_1+\cdots +t_n{z}_n{)}^{\otimes n}{⟩}_2$ is $n!⟨{\mathrm{\Psi }}_n,P_{s,n}(z_1\otimes \cdots \otimes z_n){⟩}_2$; it too vanishes. The symmetrized elementary tensors have dense span in $H_s^{(n)}$ by [F8], hence every ${\mathrm{\Psi }}_n=0$ and $\mathrm{\Psi }=0$.

The finite Hermite span is graph dense in $D(Q)$ by [F4]. The vectors $u_j$ from step 1.1 are finite-particle, so every $g_n\otimes u_j$ belongs to the image of ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ under $J^{-1}$ and $V\otimes I$. For any $(f_j)\in D(M_r)$, truncating the countable coordinate sum converges in the graph norm of $M_r$ because both ${\sum }_j\mathrm{\parallel }{f}_j{\mathrm{\parallel }}^2$ and ${\sum }_j\mathrm{\parallel }rx{f}_j{\mathrm{\parallel }}^2$ converge. For each retained coordinate, if $r>0$ then $f_j\in D(Q)$ and we approximate it in the graph norm of $Q$ by finite Hermite sums using [F4]; multiplying the graph component by the fixed scalar $r$ preserves convergence. If $r=0$, finite Hermite sums are dense in $L^2$ by [F3], and the operator graph norm is just the $L^2$ norm. Hence the image of ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is graph dense in $D(M_r)$. Step 1.2 identifies the restriction with ${\varphi }_0(e)$, and scaling gives the same for $h=re$. Therefore ${\varphi }_0(h)$ is essentially self-adjoint and its closure is the self-adjoint $M_r$ transported through these unitaries, with exactly the domain stated. When $h=0$, it is the zero operator on the dense ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ and its closure is zero on all Fock space.

Write ${\lambda }_h(z)=\exp (-\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{/}4+i⟨h,z{⟩}_2\mathrm{/}\sqrt{2})$. Using the exponential-vector inner product from step 1.4 and $⟨z+ih\mathrm{/}\sqrt{2},w+ih\mathrm{/}\sqrt{2}{⟩}_2$ $=⟨z,w{⟩}_2+i⟨z,h{⟩}_2\mathrm{/}\sqrt{2}-i⟨h,w{⟩}_2\mathrm{/}\sqrt{2}+\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{/}2$, direct cancellation gives $⟨{\lambda }_h(z)ϵ(z+ih\mathrm{/}\sqrt{2}),{\lambda }_h(w)ϵ(w+ih\mathrm{/}\sqrt{2}){⟩}_2=e^{⟨z,w{⟩}_2}$. Thus the formula in the Statement preserves inner products on the span of total exponential vectors and extends uniquely to an isometry $W(h)$. Substituting the formula twice, using $⟨h,ik\mathrm{/}\sqrt{2}{⟩}_2=i⟨h,k{⟩}_2\mathrm{/}\sqrt{2}$, gives the left-hand scalar $\exp \text{\hspace{0.17em}⁣}(-\frac{\mathrm{\parallel }h{\mathrm{\parallel }}^2+\mathrm{\parallel }k{\mathrm{\parallel }}^2}{4}+\frac{i⟨h+k,z{⟩}_2}{\sqrt{2}}-\frac{1}{2}⟨h,k{⟩}_2)\mathrm{.}$ Since $\mathrm{\parallel }h+k{\mathrm{\parallel }}^2=\mathrm{\parallel }h{\mathrm{\parallel }}^2+\mathrm{\parallel }k{\mathrm{\parallel }}^2+2\mathrm{Re}⟨h,k{⟩}_2$, this scalar is exactly the right-hand scalar below, including its Weyl phase: $W(h)W(k)ϵ(z)=e^{-i\mathrm{Im}⟨h,k{⟩}_2\mathrm{/}2}W(h+k)ϵ(z)\mathrm{.}$ Taking $k=-h$ shows $W(-h)W(h)=W(h)W(-h)=I$, so each isometry is unitary and the Weyl identity holds on all Fock space. The vacuum case $z=0$ gives $⟨\mathrm{\Omega },W(h)\mathrm{\Omega }{⟩}_2=e^{-\mathrm{\parallel }h{\mathrm{\parallel }}^2\mathrm{/}4}$.

The formula for $W(h)ϵ(z)$ is norm-continuous in $h$ for every $z$. Since the exponential vectors have dense span and $\mathrm{\parallel }W(h)\mathrm{\parallel }=1$, this implies strong continuity: approximate an arbitrary vector by a vector in that span and bound the two remaining errors by twice the approximation error. To identify its generator, write $h=re$, $z=\alpha e+w$ with $w\in e^{\perp }$. Expanding the permutation average in the one-mode splitting [F2] gives $ϵ(z)=J(ϵ(\alpha e)\otimes ϵ(w))$. The absolutely convergent exponential series and their Cauchy product [F6] give $e^{2xs-s^2}={\sum }_{n\ge 0}{c}_n(x)s^n\mathrm{/}n!$, where $c_n(x)=n!{\sum }_{j=0}^{\lfloor n\mathrm{/}2\rfloor }(-1{)}^j(2x{)}^{n-2j}\mathrm{/}((n-2j)!j!)$. Here $c_0=1$ and $c_1=2x$. In $2x{c}_n-2n{c}_{n-1}$, reindex the second finite sum by $j\mapsto j+1$; for each common power $(2x{)}^{n+1-2j}$, the combined numerator is $(n+1-2j)+2j=n+1$, giving $c_{n+1}=2x{c}_n-2n{c}_{n-1}$. The recurrence and initial values in [F3] therefore give $c_n=H_n$ by induction, so ${\sum }_n{H}_n(x)s^n\mathrm{/}n!=e^{2xs-s^2}$. Since $g_n=H_n{g}_0\mathrm{/}\sqrt{2^{n}n!}$, substituting $s=\alpha \mathrm{/}\sqrt{2}$ shows that the Hermite series for $Vϵ(\alpha e)$ converges pointwise to $g_0(x)e^{\sqrt{2}\alpha x-{\alpha }^2\mathrm{/}2}$. Let $S_N(x)={\sum }_{n=0}^N{\alpha }^n{g}_n(x)\mathrm{/}\sqrt{n!}$. The $S_N$ also converge in $L^2$ by the Fock norm series. To identify these limits, choose a subsequence $S_{N_j}$ whose squared $L^2$ errors from the Hilbert-space limit are below $2^{-3j}$. The sets where $\mathrm{\mid }{S}_{N_j}-Vϵ(\alpha e)\mathrm{\mid }>2^{-j}$ have measure at most $2^{-j}$, by integrating the squared error; their tail unions have measure at most ${\sum }_{j\ge J}{2}^{-j}\to 0$. Thus this subsequence converges almost everywhere to the $L^2$ limit, which must equal the pointwise generating-function limit. Multiplication by $e^{itrx}$ sends this function to $e^{itr\alpha \mathrm{/}\sqrt{2}-t^2{r}^2\mathrm{/}4}{g}_0(x)e^{\sqrt{2}(\alpha +itr\mathrm{/}\sqrt{2})x-(\alpha +itr\mathrm{/}\sqrt{2}{)}^2\mathrm{/}2},$ which is exactly the image of $W(th)ϵ(z)$ under $J^{-1}$ and $V\otimes I$. By totality, $W(th)$ is the coordinate multiplier $e^{itrx}$ for every real $t$. Its difference quotient converges on the displayed domain to $i{M}_r$: if $rx{f}_j\in L^2$, [F6] gives $\mathrm{\mid }(e^{itrx}-1)\mathrm{/}t\mathrm{\mid }\le 2r\mathrm{\mid }x\mathrm{\mid }$, so the squared error from $irx{f}_j$ is bounded by $9r^2{x}^2\mathrm{\mid }{f}_j{\mathrm{\mid }}^2$ and dominated convergence gives the norm limit; conversely, if a difference quotient converges, Fatou's lemma along $t_n=1\mathrm{/}n$ gives ${\sum }_j\int r^2{x}^2\mathrm{\mid }{f}_j{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus its exact generator domain is $D(M_r)$ and the derivative is $i{M}_r=i\mathrm{\Phi }(h)$; $h=0$ is immediate.