Closed-form diagonalization in Landau gauge

On the dense tensor core $a(x)b(y)$ with $a,b\in \mathcal{S}(\mathbb{R})$, ${\mathcal{F}}_y$ is the identity on the first factor and the unitary Fourier transform on the second, by [F2] and the Euclidean $L^2$ product identification in [F2]. It extends uniquely to a unitary on $L^2({\mathbb{R}}^2)$. For each $k$, translation $x=\xi +x_0(k)$ preserves the $x$-integral by [F4]; Fubini then shows that $Wg(\xi ,k)=g(\xi +x_0(k),k)$ is an isometry with inverse $W^{-1}h(x,k)=h(x-x_0(k),k)$. Hence $U=W{\mathcal{F}}_y$ is unitary.

Let $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ and write $F(\xi ,k)=Uu(\xi ,k)$. For each fixed $k$, $F(\cdot ,k)$ is smooth and compactly supported in $\xi$. Plancherel in $y$, the derivative identity in [F2], and the substitution $x=\xi +x_0(k)$ give $q_B[u]={\int }_{{\mathbb{R}}^2}(\frac{{\mathrm{\hslash }}^2}{2m}\mathrm{\mid }{\mathrm{\partial }}_{\xi }F{\mathrm{\mid }}^2+\frac{(qB{)}^2}{2m}{\xi }^2\mathrm{\mid }F{\mathrm{\mid }}^2)d\xi dk\mathrm{.}$ For a compactly supported smooth $g(\xi )$, put $Lg=-{\mathrm{\hslash }}^2{g}^{\mathrm{\prime }\mathrm{\prime }}\mathrm{/}(2m)+m{\omega }_c^2{\xi }^{2}g\mathrm{/}2\in L^2$. Integrating by parts twice on a finite interval containing the support of $g$ gives $⟨Lg,{\psi }_n⟩=E_n⟨g,{\psi }_n⟩$, since $L{\psi }_n=E_n{\psi }_n$ and the boundary terms vanish. Completeness and Parseval therefore give ${\sum }_{n=0}^{\mathrm{\infty }}{E}_n^2\mathrm{\mid }⟨g,{\psi }_n⟩{\mathrm{\mid }}^2=\mathrm{\parallel }Lg{\mathrm{\parallel }}_2^2<\mathrm{\infty }\mathrm{.}$ The exact domain characterization in [F3] now implies $g\in D(H_{\mathrm{o}\mathrm{s}\mathrm{c}})$, and its diagonal action gives $H_{\mathrm{o}\mathrm{s}\mathrm{c}}g=Lg$. Hence $⟨H_{\mathrm{o}\mathrm{s}\mathrm{c}}g,g⟩={\sum }_{n=0}^{\mathrm{\infty }}{E}_n\mathrm{\mid }⟨g,{\psi }_n⟩{\mathrm{\mid }}^2\mathrm{.}$ Integration by parts also identifies the left side with $\frac{1}{2m}({\mathrm{\hslash }}^2\mathrm{\parallel }{g}^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2+(qB{)}^2\mathrm{\parallel }\xi g{\mathrm{\parallel }}_2^2)\mathrm{.}$ Apply this identity to $F(\cdot ,k)$ for almost every $k$. Tonelli [F6] permits integrating the nonnegative series, so the displayed core form is the coefficient form on every transformed core vector.

For $N<\mathrm{\infty }$ and $f_0,\dots ,f_N\in C_c^{\mathrm{\infty }}({\mathbb{R}}_k)$, $F_N(\xi ,k)={\sum }_{n=0}^N{f}_n(k){\psi }_n(\xi )$ has inverse $U^{-1}{F}_N(x,y)=\frac{1}{\sqrt{2\pi }}{\sum }_{n=0}^N{\int }_{\mathbb{R}}{e}^{iky}{f}_n(k){\psi }_n\text{\hspace{0.17em}⁣}(x-\frac{\mathrm{\hslash }k}{qB})dk\mathrm{.}$ Write $v:=U^{-1}{F}_N$. The integrands have compact $k$-support and are Schwartz in $x$ uniformly there by the Hermite form in [F3]; differentiating under the integral and integrating by parts in $k$ show rapid decay in $y$, so the inverse is Schwartz. Define ${\chi }_R(x,y)={\prod }_{s\in \{x,y\}}\sigma (2-s\mathrm{/}R)\sigma (2+s\mathrm{/}R),$ using [F5]. These cutoffs are one on $[-R,R{]}^2$, vanish outside $[-2R,2R{]}^2$, and have $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. For each covariant derivative, ${\mathrm{\Pi }}_j^A({\chi }_{R}v)={\chi }_R{\mathrm{\Pi }}_j^{A}v-i\mathrm{\hslash }({\mathrm{\partial }}_j{\chi }_R)v\mathrm{.}$ Take integer $R\ge 1$. The step-function bounds in [F5] give $0\le {\chi }_R\le 1$, and for every fixed $(x,y)$, ${\chi }_R(x,y)=1$ once $R\ge \max \{\mathrm{\mid }x\mathrm{\mid },\mathrm{\mid }y\mathrm{\mid }\}$, so ${\chi }_R\to 1$ pointwise. Since $v$ is Schwartz and $A=(0,Bx)$ is linear, ${\mathrm{\Pi }}_1^{A}v=-i\mathrm{\hslash }{\mathrm{\partial }}_{x}v$ and ${\mathrm{\Pi }}_2^{A}v=-i\mathrm{\hslash }{\mathrm{\partial }}_{y}v-qBxv$ are Schwartz and belong to $L^2({\mathbb{R}}^2)$. For each $w\in \{v,{\mathrm{\Pi }}_1^{A}v,{\mathrm{\Pi }}_2^{A}v\}$, the functions $\mathrm{\mid }({\chi }_R-1)w{\mathrm{\mid }}^2$ are measurable, tend pointwise to zero, and are bounded by the single measurable $L^1$ majorant $\mathrm{\mid }w{\mathrm{\mid }}^2$. Applying [F9] on the Lebesgue measure space ${\mathbb{R}}^2$ gives $\mathrm{\parallel }({\chi }_R-1)w{\mathrm{\parallel }}_2\to 0$. The derivative-cutoff term separately satisfies $\mathrm{\parallel }\mathrm{\hslash }({\mathrm{\partial }}_j{\chi }_R)v{\mathrm{\parallel }}_2\le C\mathrm{\hslash }\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{/}R\to 0$. Thus the displayed product identity and the triangle inequality give $\mathrm{\parallel }{\mathrm{\Pi }}_j^A({\chi }_{R}v-v){\mathrm{\parallel }}_2\to 0$ for both $j$, as well as $\mathrm{\parallel }{\chi }_{R}v-v{\mathrm{\parallel }}_2\to 0$. The form identity of step 2.1 also holds for Schwartz functions: the Fourier and shear maps preserve Schwartz regularity, and for each Schwartz $g(\xi )$ one has $Lg\in L^2$; its integration-by-parts pairings with the polynomial-Gaussian ${\psi }_n$ have vanishing boundary terms at both infinities. The same Parseval calculation and exact diagonal operator-domain criterion in [F3] therefore give $g\in D(H_{\mathrm{o}\mathrm{s}\mathrm{c}})$ and the same oscillator energy identity. All resulting nonnegative energies can again be integrated by [F6]. Apply this extended identity to the Schwartz difference ${\chi }_{R}v-v$. Its weighted coefficient norm squared is exactly $\frac{1}{2m}{\sum }_{j=1}^2\mathrm{\parallel }{\mathrm{\Pi }}_j^A({\chi }_{R}v-v){\mathrm{\parallel }}_2^2$, which tends to zero by the preceding bounds. Since ${\chi }_{R}v\in C_c^{\mathrm{\infty }}$, each $F_N$ therefore lies in the weighted-norm closure of the transformed original core, without assuming closability in advance. Finite coefficient families are dense in the weighted target: truncate its convergent nonnegative series, then approximate the remaining finitely many $L^2(k)$ coefficients by $C_c^{\mathrm{\infty }}(k)$ using [F7].

Let ${\mathcal{K}}_E$ be the coefficient space $\{(f_n):{\sum }_n{E}_n\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2^2<\mathrm{\infty }\}$ with squared norm ${\sum }_n{E}_n\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2^2$, and let $J:{\mathcal{K}}_E\hookrightarrow L^2({\mathbb{N}}_0\times \mathbb{R},d\mathrm{\#}dk)$ be the identity inclusion. Step 2.1 maps the transformed core isometrically into ${\mathcal{K}}_E$. Since $E_n\ge E_0=\mathrm{\hslash }{\omega }_c\mathrm{/}2>0$, $J$ is continuous and injective, with $\mathrm{\parallel }Jf{\mathrm{\parallel }}_2^2\le E_0^{-1}{\sum }_n{E}_n\mathrm{\parallel }{f}_n{\mathrm{\parallel }}_2^2$. The coefficient map is unitary on the underlying $L^2$ space by [F2] and [F3]. Thus $q_B[u]\ge E_0\mathrm{\parallel }u{\mathrm{\parallel }}_2^2$ on the core. For every $\epsilon >0$, $q_B[u]\le q_B[u]+\epsilon \mathrm{\parallel }u{\mathrm{\parallel }}_2^2\le (1+\epsilon \mathrm{/}{E}_0)q_B[u],$ so each shifted form norm is equivalent to $q_B[u{]}^{1\mathrm{/}2}$. Suppose a core sequence is Cauchy in $q_B[u{]}^{1\mathrm{/}2}$ and tends to zero in $L^2$. Its weighted coefficient image converges in ${\mathcal{K}}_E$ to some $f$; unweighted $L^2$ convergence and continuity of $J$ give $Jf=0$. Injectivity of $J$ gives $f=0$, so $q_B[u]\to 0$ along that sequence. The equivalence above then gives convergence to zero in the defining shifted form norm, proving closability of the core form. By [F8] the target is complete, so the isometric image of the form completion is closed there. Step 3.1 places a dense subspace of that target in the image; hence the image is the whole weighted space and the form closure is exactly the stated one.