Anisotropic Hermite oscillator and quadratic form basis

In [F2] choose $\mathrm{\hslash }=1$, $m_j=(2a_j{)}^{-1}$ and ${\omega }_j=2\sqrt{a_{j}c}$. Then ${\mathrm{\hslash }}^2\mathrm{/}(2m_j)=a_j$ and $m_j{\omega }_j^2\mathrm{/}2=c$. The eigenvalue from [F2] in coordinate $j$ is ${\omega }_j(n_j+1\mathrm{/}2)=(2n_j+1)\sqrt{a_{j}c}$. Summing over the $N$ coordinates gives the displayed differential eigenvalue ${ϵ}_{\mathbf{n}}$. In the same scaling, $b_j=m_j{\omega }_j=\sqrt{c\mathrm{/}{a}_j}$ and the normalized $n_j=0$ Gaussian becomes $(b_j\mathrm{/}\pi {)}^{1\mathrm{/}4}{e}^{-b_j{q}_j^2\mathrm{/}2}$; taking the product gives the stated ground representative.

Each product Hermite function is a polynomial times a Gaussian, so it and all first derivatives have finite squared integrals after multiplication by any fixed polynomial. Choose a smooth radial cutoff ${\chi }_R(q)=\chi (q\mathrm{/}R)$ with $\chi =1$ on $\mathrm{\mid }q\mathrm{\mid }\le 1$, $\chi =0$ on $\mathrm{\mid }q\mathrm{\mid }\ge 2$, and $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. Such a $\chi$ can be formed from $\eta (s)=e^{-1\mathrm{/}s}$ for $s>0$ and $\eta (s)=0$ for $s\le 0$, by taking $\chi (q)=\eta (4-\mathrm{\mid }q{\mathrm{\mid }}^2)\mathrm{/}(\eta (4-\mathrm{\mid }q{\mathrm{\mid }}^2)+\eta (\mathrm{\mid }q{\mathrm{\mid }}^2-1))$. Then $0\le {\chi }_R\le 1$ and ${\chi }_R{\psi }_{\mathbf{n}}\in C_c^{\mathrm{\infty }}$. The weighted-function error is supported on $\mathrm{\mid }q\mathrm{\mid }>R$ and bounded in squared integral by the integrable tail $(c\mathrm{\mid }q{\mathrm{\mid }}^2+2)\mathrm{\mid }{\psi }_{\mathbf{n}}{\mathrm{\mid }}^2$, so its $L^2$ norm tends to zero. For each derivative, ${\mathrm{\partial }}_j((1-{\chi }_R){\psi }_{\mathbf{n}})=(1-{\chi }_R){\mathrm{\partial }}_j{\psi }_{\mathbf{n}}-({\mathrm{\partial }}_j{\chi }_R){\psi }_{\mathbf{n}};$ the first term tends to zero by its Gaussian tail and the second has norm at most $C\mathrm{\parallel }{\psi }_{\mathbf{n}}{\mathrm{\parallel }}_2\mathrm{/}R$. Thus ${\chi }_R{\psi }_{\mathbf{n}}$ converges in the form norm, proving ${\psi }_{\mathbf{n}}\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$.

For $v\in C_c^{\mathrm{\infty }}$, integration by parts and the eigenvalue identity in step 1.1 give $a_{\mathrm{o}\mathrm{s}\mathrm{c}}({\psi }_{\mathbf{n}},v)=({ϵ}_{\mathbf{n}}+2)⟨{\psi }_{\mathbf{n}},v⟩\mathrm{.}$ The left side is continuous in the form norm of $v$ by Cauchy--Schwarz [F4] in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$, and the right side is continuous since $\mathrm{\parallel }v{\mathrm{\parallel }}_2\le \mathrm{\parallel }v{\mathrm{\parallel }}_{Q_{\mathrm{o}\mathrm{s}\mathrm{c}}}$. Density of $C_c^{\mathrm{\infty }}$ in its defining completion extends the identity to every $v\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. Taking $v={\psi }_{\mathbf{m}}$ proves that ${\xi }_{\mathbf{n}}:=({ϵ}_{\mathbf{n}}+2{)}^{-1\mathrm{/}2}{\psi }_{\mathbf{n}}$ is orthonormal in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. If $u\in Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$ is orthogonal in this form inner product to every ${\xi }_{\mathbf{n}}$, conjugating the displayed identity gives $⟨u,{\psi }_{\mathbf{n}}⟩=0$ for every $\mathbf{n}$. Completeness in $L^2$ from [F2] makes the image of $u$ zero, and the injective inclusion in [F1] makes $u=0$ in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$. Thus $({\xi }_{\mathbf{n}})$ is complete there.

Parseval in $Q_{\mathrm{o}\mathrm{s}\mathrm{c}}$ with the complete orthonormal family $({\xi }_{\mathbf{n}})$ gives $a_{\mathrm{o}\mathrm{s}\mathrm{c}}(u,u)={\sum }_{\mathbf{n}}({ϵ}_{\mathbf{n}}+2)\mathrm{\mid }⟨u,{\psi }_{\mathbf{n}}⟩{\mathrm{\mid }}^2\mathrm{.}$ Parseval in $L^2$ gives $\mathrm{\parallel }u{\mathrm{\parallel }}_2^2={\sum }_{\mathbf{n}}\mathrm{\mid }⟨u,{\psi }_{\mathbf{n}}⟩{\mathrm{\mid }}^2$. Subtracting $2\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ from the first identity yields the asserted expansion of the unshifted oscillator form.