Observable Quotients and Exact Projected Dynamics: Closure, Memory, Minimal Dynamical Completion, and Effective Field Operators

The exact observable quotient

Jeremy RodgersSection 3 of 14Original PDF ↗Publication DOI ↗

The exact observable quotient

Assumption 2.1.

$\mathcal{H}_{\mathrm{src}}$ and $\mathcal{H}_{\mathrm{obs}}$ are Hilbert spaces and $K\in\mathcal{B}(\mathcal{H}_{\mathrm{src}},\mathcal{H}_{\mathrm{obs}})$ .

Definition 2.2 (Observational equivalence).

For $\Psi,\Psi'\in\mathcal{H}_{\mathrm{src}}$ set $\Psi\sim_{K}\Psi' \quad:\Longleftrightarrow\quad K\Psi=K\Psi' \quad\Longleftrightarrow\quad \Psi-\Psi'\in\ker K .\tag{2.1}$

Since $K$ is continuous, $N:=\ker K$ is a closed subspace of $\mathcal{H}_{\mathrm{src}}$ . The quotient vector space $\mathcal{H}_{K}:=\mathcal{H}_{\mathrm{src}}/\ker K\tag{2.2}$ therefore carries the quotient norm $\bigl\lVert[\Psi]\bigr\rVert_{K}:=\inf_{n\in\ker K}\lVert\Psi+n\rVert_{\mathcal{H}_{\mathrm{src}}} =\operatorname{dist}\bigl(\Psi,\ker K\bigr),\tag{2.3}$ and $(\mathcal{H}_{K},\lVert\cdot\rVert_{K})$ is a Hilbert space: the canonical map $(\ker K)^{\perp}\to\mathcal{H}_{K}$ , $\Psi\mapsto[\Psi]$ , is a linear bijection, and it is isometric because the infimum in (2.3) is attained at the orthogonal projection of $\Psi$ onto $(\ker K)^{\perp}$ . Thus the quotient norm is induced by the inner product $\langle [\Psi] , [\Psi'] \rangle_{K}:=\langle \Pi\Psi , \Pi\Psi' \rangle_{\mathcal{H}_{\mathrm{src}}}$ , where $\Pi$ is the orthogonal projection onto $(\ker K)^{\perp}$ , and completeness is inherited from that of the closed subspace $(\ker K)^{\perp}$ .

The exact observable state associated with a source state $\Psi$ is the coset $[\Psi]_{K}=\Psi+\ker K .\tag{2.4}$ It is essential to the logic of what follows that the quotient distinguishes no representative within this coset. The orthogonal representative $\Pi\Psi$ is a choice made by the ambient inner product of $\mathcal{H}_{\mathrm{src}}$ , not by the readout.

Theorem 2.3 (Observable quotient theorem).

Under Assumption 2.1, the map $\widetilde{K}:\mathcal{H}_{K}\longrightarrow\operatorname{Ran}K, \qquad \widetilde{K}[\Psi]:=K\Psi,\tag{2.5}$ is well defined, linear, bijective, and bounded, with $\lVert\widetilde{K}\rVert_{\mathrm{op}}=\lVert K\rVert_{\mathrm{op}}$ , where $\operatorname{Ran}K$ carries the norm inherited from $\mathcal{H}_{\mathrm{obs}}$ . Moreover the following are equivalent:

$\widetilde{K}^{-1}:\operatorname{Ran}K\to\mathcal{H}_{K}$ is bounded when $\operatorname{Ran}K$ carries the inherited norm;
$\operatorname{Ran}K$ is closed in $\mathcal{H}_{\mathrm{obs}}$ ;
there exists $c>0$ with $\lVert K\Psi\rVert_{\mathcal{H}_{\mathrm{obs}}}\ge c\,\operatorname{dist}(\Psi,\ker K)$ for all $\Psi\in\mathcal{H}_{\mathrm{src}}$ .

Proof. Well-definedness and bijectivity. If $[\Psi]=[\Psi']$ then $\Psi-\Psi'\in\ker K$ , so $K\Psi=K\Psi'$ ; hence (2.5) does not depend on the representative. Linearity is immediate. Surjectivity onto $\operatorname{Ran}K$ holds by definition of the range. If $\widetilde{K}[\Psi]=0$ then $K\Psi=0$ , i.e. $\Psi\in\ker K$ , i.e. $[\Psi]=0$ ; hence $\widetilde{K}$ is injective.

Boundedness. For every $n\in\ker K$ , $\lVert\widetilde{K}[\Psi]\rVert_{\mathcal{H}_{\mathrm{obs}}}=\lVert K\Psi\rVert=\lVert K(\Psi+n)\rVert \le\lVert K\rVert_{\mathrm{op}}\,\lVert\Psi+n\rVert_{\mathcal{H}_{\mathrm{src}}}.$ Taking the infimum over $n\in\ker K$ gives $\lVert\widetilde{K}[\Psi]\rVert\le\lVert K\rVert_{\mathrm{op}}\lVert[\Psi]\rVert_{K}$ , so $\lVert\widetilde{K}\rVert_{\mathrm{op}}\le\lVert K\rVert_{\mathrm{op}}$ . Conversely $\lVert K\Psi\rVert=\lVert\widetilde{K}[\Psi]\rVert\le\lVert\widetilde{K}\rVert_{\mathrm{op}}\lVert[\Psi]\rVert_{K} \le\lVert\widetilde{K}\rVert_{\mathrm{op}}\lVert\Psi\rVert$ , so the two operator norms coincide.

(1) $\Rightarrow$ (2). Suppose $\lVert\widetilde{K}^{-1}\rVert_{\mathrm{op}}=:M<\infty$ . Then $\widetilde{K}$ is a linear homeomorphism from the complete space $\mathcal{H}_{K}$ onto $\operatorname{Ran}K$ with its inherited norm; hence $(\operatorname{Ran}K,\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}})$ is complete. A complete subspace of a normed space is closed, so $\operatorname{Ran}K$ is closed in $\mathcal{H}_{\mathrm{obs}}$ .

(2) $\Rightarrow$ (1). If $\operatorname{Ran}K$ is closed in the Hilbert space $\mathcal{H}_{\mathrm{obs}}$ , then $(\operatorname{Ran}K,\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}})$ is itself a Banach space. Since $\widetilde{K}$ is a bounded linear bijection between the Banach spaces $\mathcal{H}_{K}$ and $\operatorname{Ran}K$ , the bounded inverse theorem [1, Cor. 2.7] yields boundedness of $\widetilde{K}^{-1}$ .

(1) $\Leftrightarrow$ (3). Statement (1) says exactly that there is $M<\infty$ with $\lVert[\Psi]\rVert_{K}\le M\lVert K\Psi\rVert$ for all $\Psi$ ; by (2.3) this is (3) with $c=1/M$ . ◻

Remark 4 (The two norms on $\operatorname{Ran}K$ ).

Theorem 2.3 always allows us to transport the quotient norm to $\operatorname{Ran}K$ by declaring $\lVert K\Psi\rVert_{\sharp}:=\lVert[\Psi]\rVert_{K}$ ; with this transported norm $\widetilde{K}$ is by construction an isometric isomorphism, and $(\operatorname{Ran}K,\lVert\cdot\rVert_{\sharp})$ is a Hilbert space. What is not generally true is that $\lVert\cdot\rVert_{\sharp}$ coincides with, or is even equivalent to, the norm inherited from $\mathcal{H}_{\mathrm{obs}}$ . The inequality $\lVert\cdot\rVert_{\mathcal{H}_{\mathrm{obs}}}\le\lVert K\rVert_{\mathrm{op}} \lVert\cdot\rVert_{\sharp}$ always holds on $\operatorname{Ran}K$ ; the reverse inequality holds, with some constant, exactly in the closed-range case. We therefore never write $\mathcal{H}_{K}\cong\operatorname{Ran}K$ without saying which norm $\operatorname{Ran}K$ carries. Every later statement about topologies on the observable side is subject to this distinction.

Corollary 2.5 (Exact loss criterion).

The observation map $K$ loses exact source distinctions, i.e. there exist $\Psi\neq\Psi'$ with $\Psi\sim_{K}\Psi'$ , if and only if $\ker K\neq\{0\}$ . If $\ker K=\{0\}$ , then $K$ is injective and the source state is uniquely determined by its readout, even if $\widetilde{K}^{-1}=K^{-1}$ is unbounded on $\operatorname{Ran}K$ .

Proof. Immediate from (2.1): $\Psi\sim_K\Psi'$ with $\Psi\neq\Psi'$ iff $\Psi-\Psi'\in\ker K\setminus\{0\}$ . The second sentence is the content of Theorem 2.3: injectivity is the statement $\ker K=\{0\}$ and is logically independent of closedness of $\operatorname{Ran}K$ , which is what governs boundedness of the inverse. ◻

Corollary 2.5 is the formal separation of (D1) from (D2). Nonidentifiability is a statement about the kernel; instability is a statement about the range. An injective compact operator exhibits the second without the first, as we show in Example 4.5.