Finite-mode implementability does not supply an infinite-mode limit

Apply [F5] to $x\mapsto e^{-s}x$ and $\mathrm{\mid }f{\mathrm{\mid }}^2$ to get ${\int }_{\mathbb{R}}\mathrm{\mid }{U}_{s}f(x){\mathrm{\mid }}^{2}dx={\int }_{\mathbb{R}}\mathrm{\mid }f(y){\mathrm{\mid }}^{2}dy,$ and $U_{-s}$ is its inverse, so $U_s$ is unitary. The chain rule for $Q=\sqrt{\mathrm{\hslash }}x$ and $P=-i\sqrt{\mathrm{\hslash }}{\mathrm{\partial }}_x$ gives $U_s^{\ast }a{U}_s=\cosh (s)a+\sinh (s)a^{\ast }$. The scalar pair obeys $\mathrm{\mid }u{\mathrm{\mid }}^2-\mathrm{\mid }v{\mathrm{\mid }}^2=1$ and $uv=vu$. For the $N$-mode product take $U_N=u{I}_N$ and $V_N=v{I}_N$; the finite Bogoliubov result [F1] supplies an implementer for these coefficients. Write ${\psi }_s(x)=U_s{g}_0(x)=e^{-s\mathrm{/}2}{\pi }^{-1\mathrm{/}4}{e}^{-e^{-2s}{x}^2\mathrm{/}2}$. Then $a{\psi }_s=\frac{1-e^{-2s}}{\sqrt{2}}x{\psi }_s\mathrm{.}$ By [F4] with $A=1,B=0$, $\int e^{-y^2}dy=\sqrt{\pi }$. On $[-R,R]$, integration by parts gives ${\int }_{-R}^R{y}^2{e}^{-y^2}dy={[-\frac{y}{2}{e}^{-y^2}]}_{-R}^R+\frac{1}{2}{\int }_{-R}^R{e}^{-y^2}dy\mathrm{.}$ The boundary term tends to zero by [F4], so $\int y^2\mathrm{\mid }{g}_0(y){\mathrm{\mid }}^{2}dy=1\mathrm{/}2$. The substitution $y=e^{-s}x$ in [F5] gives ${\int }_{\mathbb{R}}{x}^2\mathrm{\mid }{\psi }_s(x){\mathrm{\mid }}^{2}dx=e^{2s}{\int }_{\mathbb{R}}{y}^2\mathrm{\mid }{g}_0(y){\mathrm{\mid }}^{2}dy=e^{2s}\mathrm{/}2.$ Consequently $\mathrm{\parallel }a{\psi }_s{\mathrm{\parallel }}^2=\frac{(1-e^{-2s}{)}^2}{2}\frac{e^{2s}}{2}={\sinh }^2(s)\mathrm{.}$ By [F3], the $N$-mode Fock–Schrödinger unitary sends its vacuum ${\mathrm{\Omega }}_N$ to $g_0^{\otimes N}$. On $L^2({\mathbb{R}}^N)$ define $[U_s^{(N)}f](x_1,\dots ,x_N)=e^{-Ns\mathrm{/}2}f(e^{-s}{x}_1,\dots ,e^{-s}{x}_N)\mathrm{.}$ The $C^1$ change-of-variables formula [F5] in ${\mathbb{R}}^N$ gives $\mathrm{\parallel }{U}_s^{(N)}f{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2$, so this is unitary. It is the product of the one-mode dilations, and the coordinatewise chain rule gives $U_s^{(N)\ast }{a}_j{U}_s^{(N)}=u{a}_j+v{a}_j^{\ast }$ for each $j$. The finite Bogoliubov statement [F1] also supplies an implementer for this coefficient pair. Put ${\mathrm{\Psi }}_N=(U_s\mathrm{\Omega }{)}^{\otimes N}$ under the same identification; its $j$th number expectation is the one-mode value $\mathrm{\parallel }a{\psi }_s{\mathrm{\parallel }}^2={\sinh }^2(s)$ already computed above. Each factor is a Schwartz Gaussian, so the product is in every $D(a_j)$. By the explicitly defined total-number form, the finite sum therefore gives $n_N({\mathrm{\Psi }}_N)={\sum }_{j=1}^N\mathrm{\parallel }{a}_j{\mathrm{\Psi }}_N{\mathrm{\parallel }}^2=N{\sinh }^2(s)\mathrm{.}$ The product unitary exists because it is the tensor product of finitely many copies of this unitary. The same coefficient pair is covered by the finite Bogoliubov statement [F2]; no infinite product is taken.

In Schrödinger coordinates, $\mathrm{\Omega }=g_0$ and $U_s{g}_0(x)=e^{-s\mathrm{/}2}{\pi }^{-1\mathrm{/}4}{e}^{-e^{-2s}{x}^2\mathrm{/}2}$. Set $A=(1+e^{-2s})\mathrm{/}2>0$. By [F4] with $B=0$, ${\int }_{\mathbb{R}}{e}^{-A{x}^2}dx=\sqrt{\pi \mathrm{/}A}=\sqrt{\frac{2\pi }{1+e^{-2s}}}\mathrm{.}$ Hence $⟨g_0,U_s{g}_0⟩={\pi }^{-1\mathrm{/}2}{e}^{-s\mathrm{/}2}\sqrt{\frac{2\pi }{1+e^{-2s}}}=\sqrt{\frac{2}{e^s+e^{-s}}}=\frac{1}{\sqrt{\cosh s}}\mathrm{.}$ Since $U_s$ and $g_0$ are real and positive here, this is also the overlap magnitude; the formula holds for every real $s$. Product inner products factor over the $N$ modes, yielding $\mathrm{\mid }⟨{\mathrm{\Omega }}_N,{\mathrm{\Psi }}_N⟩\mathrm{\mid }=(\cosh s{)}^{-N\mathrm{/}2}\mathrm{.}$ For $s\mathrm{\ne }0$, $\cosh s>1$, so this magnitude tends to zero as $N\to \mathrm{\infty }$. Together with step 1.1 this proves only the stated finite-$N$ divergence and overlap limit. Finite-dimensional implementation does not itself provide an infinite-mode implementer, and this example does not prove that none exists.