Single-mode squeeze statistics

Work in Schrödinger coordinates via [F1], with inner product linear in the first argument. Put $bf=(xf+f^{\mathrm{\prime }})\mathrm{/}\sqrt{2}$ and $b^{†}f=(xf-f^{\mathrm{\prime }})\mathrm{/}\sqrt{2}$ for $f\in \mathcal{S}(\mathbb{R})$; both expressions again lie in $\mathcal{S}$. For $c_n=⟨f,g_n⟩$, integration by parts gives $⟨bf,g_m⟩=⟨f,b^{†}{g}_m⟩=\sqrt{m+1}{c}_{m+1}$ and $⟨b^{†}f,g_m⟩=\sqrt{m}{c}_{m-1}$ for $m\ge 1$, with the latter coefficient zero for $m=0$. The boundary terms vanish since $f$ and each Hermite function are Schwartz. Completeness and Parseval in [F1] give ${\sum }_{n\ge 0}n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }bf{\mathrm{\parallel }}^2<\mathrm{\infty }$ and ${\sum }_{n\ge 0}(n+1)\mathrm{\mid }{c}_n{\mathrm{\mid }}^2=\mathrm{\parallel }{b}^{†}f{\mathrm{\parallel }}^2<\mathrm{\infty }$. Therefore [F5] identifies the actual closed Fock operators with $a=b$ and $a^{\ast }=b^{†}$ on every Schwartz vector, not only on finite Hermite sums. Since $bf$ is Schwartz, $f\in D(a^{\ast }a)=D(N)$, and direct differentiation gives $a^{\ast }af=b^{†}bf=(x^{2}f-f^{\mathrm{\prime }\mathrm{\prime }}-f)\mathrm{/}2=((Q^2+P^2)\mathrm{/}(2\mathrm{\hslash })-I\mathrm{/}2)f$. Thus the ladder, adjoint and number identities needed below all hold on the transported Schwartz domain.

Apply [F4] to the diffeomorphism $x\mapsto e^{-s}x$ and $\mathrm{\mid }f{\mathrm{\mid }}^2\in L^1$: ${\int }_{\mathbb{R}}\mathrm{\mid }{e}^{-s\mathrm{/}2}f(e^{-s}x){\mathrm{\mid }}^{2}dx={\int }_{\mathbb{R}}\mathrm{\mid }f(y){\mathrm{\mid }}^{2}dy,$ and $U_{-s}$ is its inverse, so $U_s$ is unitary. For Schwartz $f$, direct substitution and the chain rule give $U_s^{\ast }Q{U}_{s}f=e^{s}Qf,U_s^{\ast }P{U}_{s}f=e^{-s}Pf\mathrm{.}$ Using $a=(Q+iP)\mathrm{/}\sqrt{2\mathrm{\hslash }}$ and $a^{\ast }=(Q-iP)\mathrm{/}\sqrt{2\mathrm{\hslash }}$, this becomes $U_s^{\ast }a{U}_s=\frac{e^s+e^{-s}}{2}a+\frac{e^s-e^{-s}}{2}{a}^{\ast }=\cosh (s)a+\sinh (s)a^{\ast }\mathrm{.}$ The dilation maps Schwartz space onto itself, so all identities hold on the stated common core and the associated Weyl covariance extends everywhere by unitarity.

The ground function $g_0(x)={\pi }^{-1\mathrm{/}4}{e}^{-x^2\mathrm{/}2}$ is even, so its position mean is zero; its derivative is odd, so its momentum mean is zero. By [F3] with $A=1,B=0$, $\int e^{-x^2}dx=\sqrt{\pi }$. On $[-R,R]$, the finite-interval integration-by-parts supplier in [F3] gives ${\int }_{-R}^R{x}^2{e}^{-x^2}dx={[-\frac{x}{2}{e}^{-x^2}]}_{-R}^R+\frac{1}{2}{\int }_{-R}^R{e}^{-x^2}dx\mathrm{.}$ The boundary term tends to zero by [F3], so $\int x^2\mathrm{\mid }{g}_0(x){\mathrm{\mid }}^{2}dx={\pi }^{-1\mathrm{/}2}(\sqrt{\pi }\mathrm{/}2)=1\mathrm{/}2$. Also $g_0^{\mathrm{\prime }}=-x{g}_0$ gives both vacuum variances $\mathrm{\hslash }\mathrm{/}2$. Moreover $⟨Q\mathrm{\Omega },P\mathrm{\Omega }⟩=-i\mathrm{\hslash }\mathrm{/}2$, so the real symmetrized cross covariance is zero. For ${\psi }_s=U_s\mathrm{\Omega }$, the unitary conjugation formulas in step 2.1 give $⟨Q{⟩}_{{\psi }_s}=e^s⟨Q{⟩}_{\mathrm{\Omega }}=0,⟨Q^2{⟩}_{{\psi }_s}=e^{2s}\mathrm{\hslash }\mathrm{/}2,$ and likewise $⟨P{⟩}_{{\psi }_s}=0$ and $⟨P^2{⟩}_{{\psi }_s}=e^{-2s}\mathrm{\hslash }\mathrm{/}2$. The cross form is unchanged because the two scale factors multiply to one, so it remains zero. Therefore the covariance determinant is the product of the two variances, ${\mathrm{\hslash }}^2\mathrm{/}4$.

Since $U_s$ maps Schwartz space to itself, ${\psi }_s$ is in the domain of the number form. By step 1.1 and step 3.1, $⟨a^{\ast }a{⟩}_{{\psi }_s}=\frac{⟨Q^2+P^2{⟩}_{{\psi }_s}}{2\mathrm{\hslash }}-\frac{1}{2}=\frac{e^{2s}+e^{-2s}-2}{4}={\sinh }^2(s)\mathrm{.}$ This is the model's finite one-mode number expectation.