Unitary identifications of the component-valued wave-function space

[Well-definedness of ${\mathrm{\Psi }}_0$] Let $(f_l)\in {⨁}_{l}H$ and let $({\tilde{f}}_l)$ be another choice of representatives with ${\tilde{f}}_l=f_l$ almost everywhere for each $l$; then $({\tilde{f}}_1,\dots ,{\tilde{f}}_r)=(f_1,\dots ,f_r)$ almost everywhere, so ${\mathrm{\Psi }}_0(f)$ is the same class of $H^{(r)}$ and ${\mathrm{\Psi }}_0$ is well defined [F3]. It is complex-linear because addition and scalar multiplication in $H^{(r)}$ are componentwise.

[Isometry] For $f=(f_l)$ and $g=(g_l)$ in ${⨁}_{l}H$, the function $x\mapsto {\sum }_l\mathrm{\mid }{f}_l(x){\mathrm{\mid }}^2$ is nonnegative and measurable, so [F4] gives ${\int }_{{\mathbb{R}}^d}{\sum }_l\mathrm{\mid }{f}_l{\mathrm{\mid }}^2={\sum }_l{\int }_{{\mathbb{R}}^d}\mathrm{\mid }{f}_l{\mathrm{\mid }}^2={\sum }_l\mathrm{\parallel }{f}_l{\mathrm{\parallel }}_H^2$, that is $\mathrm{\parallel }{\mathrm{\Psi }}_0(f){\mathrm{\parallel }}^2=\mathrm{\parallel }f{\mathrm{\parallel }}_{\oplus }^2$; the map ${\mathrm{\Psi }}_0$ is thus an isometry. For the inner product, Cauchy-Schwarz gives ${\int }_{{\mathbb{R}}^d}{\sum }_l\mathrm{\mid }{f}_l\mathrm{\mid }\mathrm{\mid }{g}_l\mathrm{\mid }\le {\sum }_l\mathrm{\parallel }{f}_l{\mathrm{\parallel }}_H\mathrm{\parallel }{g}_l{\mathrm{\parallel }}_H\le \left({\sum }_l\mathrm{\parallel }{f}_l{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}\right({\sum }_l\mathrm{\parallel }{g}_l{\mathrm{\parallel }}^2{)}^{1\mathrm{/}2}<\mathrm{\infty }$. Each product $f_l\overline{g_l}$ is integrable by Cauchy-Schwarz, and finite linearity of the Lebesgue integral therefore gives $⟨{\mathrm{\Psi }}_0(f),{\mathrm{\Psi }}_0(g)⟩=\int {\sum }_l{f}_l\overline{g_l}={\sum }_l⟨f_l,g_l{⟩}_H$.

[The tensor identification on the algebraic tensor product] On elementary tensors put ${\mathrm{\Theta }}_0(\kappa (u\otimes z)):={\sum }_l{z}_l{\iota }_l(u)$ for $z={\sum }_l{z}_l{e}_l$. The map $(u,z)\mapsto {\sum }_l{z}_l{\iota }_l(u)$ $H\times {\mathbb{C}}^r\to H^{(r)}$ is $\mathbb{C}$-bilinear, so ${\mathrm{\Theta }}_0$ factors through the defining bilinearity relations of the algebraic tensor product $H\odot {\mathbb{C}}^r$ and is a well-defined complex-linear map (Hilbert tensor construction: inner product, completion and product bases clause (a)). Every element of $H\odot {\mathbb{C}}^r$ can be written as ${\sum }_l{w}_l\otimes e_l$: a finite sum ${\sum }_i{u}_i\otimes z_i$ equals ${\sum }_l\left({\sum }_i{z}_{i,l}{u}_i\right)\otimes e_l$ by bilinearity. For such an element, [F1] gives ${\mathrm{\Theta }}_0({\sum }_l{w}_l\otimes e_l)={\sum }_l{\iota }_l(w_l)=(w_1,\dots ,w_r)$ and $\mathrm{\parallel }{\sum }_l{w}_l\otimes e_l{\mathrm{\parallel }}^2={\sum }_l\mathrm{\parallel }{w}_l{\mathrm{\parallel }}^2=\mathrm{\parallel }{\mathrm{\Theta }}_0({\sum }_l{w}_l\otimes e_l){\mathrm{\parallel }}_{H^{(r)}}^2$. By [F3], every element of $H^{(r)}$ is an $r$-tuple of classes in $H$, so ${\mathrm{\Theta }}_0$ is onto $H^{(r)}$. Hence ${\mathrm{\Theta }}_0$ is an isometry of $H\odot {\mathbb{C}}^r$ onto $H^{(r)}$.

[Surjectivity] Let $g\in H^{(r)}$. By [F3] each component $g_l$ is an $L^2$ class and $g=(g_1,\dots ,g_r)={\mathrm{\Psi }}_0((g_l))$; hence ${\mathrm{\Psi }}_0$ is surjective, and (a) is proved.

[Extension and surjectivity] First $H^{(r)}$ is complete: if $(g^{(n)}{)}_n$ is Cauchy, each component sequence $g_l^{(n)}$ is Cauchy in the complete scalar space $H$ by [F6] because $\mathrm{\parallel }{g}_l^{(n)}-g_l^{(m)}{\mathrm{\parallel }}_H\le \mathrm{\parallel }{g}^{(n)}-g^{(m)}{\mathrm{\parallel }}_{H^{(r)}}$. Let $g_l\in H$ be its limit. By [F3] the finite tuple $g=(g_l)$ lies in $H^{(r)}$, and $\mathrm{\parallel }{g}^{(n)}-g{\mathrm{\parallel }}_{H^{(r)}}^2={\sum }_{l=1}^r\mathrm{\parallel }{g}_l^{(n)}-g_l{\mathrm{\parallel }}_H^2\to 0$, so $H^{(r)}$ is complete. By [F5] the isometry ${\mathrm{\Theta }}_0$ extends uniquely to a linear isometry $\mathrm{\Theta }:H\widehat{\otimes }{\mathbb{C}}^r\to H^{(r)}$, because $\kappa (H\odot {\mathbb{C}}^r)$ is dense in the completion (Hilbert tensor construction: inner product, completion and product bases clause (c)); since ${\mathrm{\Theta }}_0$ is onto by step 1.3, so is $\mathrm{\Theta }$. Thus $\mathrm{\Theta }$ is unitary. This proves (b) except for the displayed formula, which holds on elementary tensors by construction and extends to all $w$ by continuity of the bounded operators involved.

[Intertwining identities] For $u\in H$ and $z={\sum }_l{z}_l{e}_l$, ${\mathrm{\Theta }}^{-1}(a\mathrm{\Theta }(\kappa (u\otimes z)))={\mathrm{\Theta }}^{-1}(a{\sum }_l{z}_l{\iota }_l(u))={\mathrm{\Theta }}^{-1}({\sum }_l{z}_l{\iota }_l(au))=\kappa (au\otimes z)=(M_a\otimes I)\kappa (u\otimes z)$, where the last identity is [F2]; similarly ${\mathrm{\Theta }}^{-1}(T_v^{(r)}\mathrm{\Theta }(\kappa (u\otimes z)))=\kappa (T_{v}u\otimes z)=(T_v\otimes I)\kappa (u\otimes z)$, since $T_v^{(r)}$ acts componentwise and $T_v^{(r)}{\iota }_l(u)={\iota }_l(T_{v}u)$ (Self-adjointness, Fourier representation and translation identities of position and momentum clause (c)). Both sides of each identity are bounded operators (the amplifications are bounded by [F2], and $\mathrm{\Theta }$ is unitary by step 2.2), and the elementary tensors have dense span in $H\widehat{\otimes }{\mathbb{C}}^r$ [F1]; hence the identities hold on all of $H\widehat{\otimes }{\mathbb{C}}^r$. This proves (c).