Closed-form diagonalization with free longitudinal motion

First apply the unitary planar map from [F1] at each fixed $z$, then the partial Fourier transform in $z$ from [F2]. On the dense product core both maps are unitary; the Euclidean $L^2$ product identifications in [F2] extend their composition uniquely to a unitary on $L^2({\mathbb{R}}^3)$. The inverse uses the inverse Fourier transforms and the opposite translation, which is measure preserving by [F3]. The coefficient map in the transverse variable is unitary because $({\psi }_n)$ is a complete orthonormal basis. Thus $U_3$ is unitary onto the stated coefficient space.

Let $f=(f_n)$ have finite weighted norm and fix $\epsilon >0$. Since the defining nonnegative series converges, choose $N$ so the norm of its tail over $n>N$ is below $\epsilon \mathrm{/}3$. For the finitely many indices $n\le N$, the weighted integrals outside $K_R=[-R,R{]}_{k_y}\times [-R,R{]}_{k_z}$ tend to zero as $R\to \mathrm{\infty }$ by dominated convergence, so choose one $R$ making their combined tail norm below $\epsilon \mathrm{/}3$. Define ${\eta }_R(k_y,k_z)={\prod }_{s\in \{k_y,k_z\}}\sigma (R+1-s)\sigma (R+1+s)\mathrm{.}$ It equals one on $K_R$, is between zero and one, and has support in $K_{R+1}$ by [F4]. For each $n\le N$, truncate $f_n$ to $K_R$ and approximate that truncation in unweighted $L^2$ by ${\phi }_n\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ using [F4]. The function $g_n={\eta }_R{\phi }_n$ is smooth and supported in $K_{R+1}$, and on that box $E_n+c{k}_z^2\le E_n+c(R+1{)}^2$. Since ${\eta }_R=1$ on $K_R$ and $0\le {\eta }_R\le 1$, choose each of the finitely many $L^2$ errors small enough that the weighted norm difference between $g_n$ and the truncation has combined norm below $\epsilon \mathrm{/}3$. Set $g_n=0$ for $n>N$. The triangle inequality then gives a finite family $g_0,\dots ,g_N\in C_c^{\mathrm{\infty }}({\mathbb{R}}^2)$ within $\epsilon$ of $f$ in the weighted coefficient norm.

For $u\in C_c^{\mathrm{\infty }}({\mathbb{R}}^3)$, every $z$-slice $u(\cdot ,\cdot ,z)$ is a planar test function. Apply the form identity in [F1] on those slices and integrate in $z$ by Fubini. The $z$ derivative commutes with the planar transform; after Fourier transformation in $z$, Plancherel and the derivative identity in [F2] turn $\mathrm{\parallel }-i\mathrm{\hslash }{\mathrm{\partial }}_{z}u{\mathrm{\parallel }}_2^2\mathrm{/}(2m)$ into the multiplier $c{k}_z^2$. Tonelli [F5] permits summing the nonnegative transverse energies and integrating in $(k_y,k_z)$. Therefore $q_B^{(3)}[u]={\sum }_{n=0}^{\mathrm{\infty }}{\int }_{{\mathbb{R}}^2}(E_n+c{k}_z^2)\mathrm{\mid }{f}_n(k_y,k_z){\mathrm{\mid }}^{2}d{k}_{y}d{k}_z\mathrm{.}$ This makes the transformed core an isometric subspace of the stated weighted coefficient space.

For the finite family from step 1.2 put $G(\xi ,k_y,k_z)={\sum }_{n=0}^N{g}_n(k_y,k_z){\psi }_n(\xi )$. Its inverse is $v(x,y,z)=\frac{1}{2\pi }{\sum }_{n=0}^N{\int }_{{\mathbb{R}}^2}{e}^{i(k_{y}y+k_{z}z)}{g}_n(k_y,k_z){\psi }_n\text{\hspace{0.17em}⁣}(x-\frac{\mathrm{\hslash }{k}_y}{qB})d{k}_{y}d{k}_z\mathrm{.}$ Each coefficient is smooth and compactly supported in $(k_y,k_z)$, and each ${\psi }_n$ is a polynomial times a Gaussian by [F7]. Every derivative of the integrand is smooth with compact wave-number support and has uniformly rapid decay in $x$ because $k_y$ remains in a bounded set. Repeated integration by parts in $k_y,k_z$ gives arbitrary polynomial decay in $y,z$; the same estimates hold after any fixed $x,y,z$ derivatives. Thus $v$ is Schwartz. Define ${\chi }_R(x,y,z)={\prod }_{s\in \{x,y,z\}}\sigma (2-s\mathrm{/}R)\sigma (2+s\mathrm{/}R),$ using the smooth step from [F4]. These cutoffs equal one on $[-R,R{]}^3$, vanish outside $[-2R,2R{]}^3$, and satisfy $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$. For each covariant derivative, ${\mathrm{\Pi }}_j({\chi }_{R}v)={\chi }_R{\mathrm{\Pi }}_{j}v-i\mathrm{\hslash }({\mathrm{\partial }}_j{\chi }_R)v,$ where ${\mathrm{\Pi }}_x=-i\mathrm{\hslash }{\mathrm{\partial }}_x$, ${\mathrm{\Pi }}_y=-i\mathrm{\hslash }{\mathrm{\partial }}_y-qBx$, and ${\mathrm{\Pi }}_z=-i\mathrm{\hslash }{\mathrm{\partial }}_z$. Dominated convergence [F8] sends $({\chi }_R-1){\mathrm{\Pi }}_{j}v$ and $({\chi }_R-1)v$ to zero in $L^2$, while the added term has norm at most $C\mathrm{\hslash }\mathrm{\parallel }v{\mathrm{\parallel }}_2\mathrm{/}R$. Hence ${\chi }_{R}v$ approximates $v$ in the form norm and belongs to the original compactly supported core. Thus each finite smooth coefficient family from step 1.2 lies in the transformed form closure.

Step 2.1 embeds the transformed core isometrically into the weighted coefficient space. Because the form completion is complete and the weighted target is complete by [F6], the isometric image of the completion is closed in that target. Steps 1.2 and 2.2 show that a dense subspace of the entire weighted space lies in the image, so the image is the whole target. This identifies the completion and its quadratic form as stated.