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

The exact projected-dynamics identity

Jeremy RodgersSection 7 of 14Original PDF ↗Publication DOI ↗

The exact projected-dynamics identity

Assumption 6.1.

For the remainder of Sections 6–7: $\mathcal{H}$ is a Hilbert space; $P\in\mathcal{B}(\mathcal{H})$ is an orthogonal projection onto a closed resolved subspace $\mathcal{H}_{P}:=P\mathcal{H}$ ; $Q:=I-P$ is the orthogonal projection onto the unresolved subspace $\mathcal{H}_{Q}:=Q\mathcal{H}$ ; and $L\in\mathcal{B}(\mathcal{H})$ generates the source evolution $\dot\Psi(t)=L\Psi(t), \qquad \Psi(0)=\Psi_{0}\in\mathcal{H},\tag{6.1}$ whose unique solution is $\Psi(t)=e^{tL}\Psi_{0}$ .

This is the case $K=P$ of the previous section: the readout literally is the resolved component, $\mathcal{H}_{\mathrm{obs}}=\mathcal{H}_{P}$ , $\operatorname{Ran}P=\mathcal{H}_{P}$ is closed, and $\ker P=\mathcal{H}_{Q}$ . It is the setting in which the cost of failed descent can be computed in closed form. Write $x(t):=P\Psi(t)\in\mathcal{H}_{P}, \qquad z(t):=Q\Psi(t)\in\mathcal{H}_{Q},\tag{6.2}$ so that $\Psi(t)=x(t)+z(t)$ and $x(0)=P\Psi_{0}$ , $z(0)=:z_{0}=Q\Psi_{0}$ . Applying $P$ and $Q$ to (6.1) and inserting $I=P+Q$ gives the exact block system $\begin{aligned} \dot x(t)&=PLP\,x(t)+PLQ\,z(t),\qquad\text{(6.3)}\\ \dot z(t)&=QLP\,x(t)+QLQ\,z(t).\qquad\text{(6.4)} \end{aligned}$ Note that $PLQ:\mathcal{H}_{Q}\to\mathcal{H}_{P}$ measures precisely how the unresolved sector drives the resolved one, and $QLP:\mathcal{H}_{P}\to\mathcal{H}_{Q}$ how the resolved sector feeds back into the unresolved one.

Theorem 6.2 (Exact projected-dynamics equation).

Under Assumption 6.1, for every source initial state $\Psi_{0}=x(0)+z_{0}$ the resolved component $x(t)=Pe^{tL}\Psi_{0}$ satisfies, for all $t\ge0$ , $\boxed{\ \ \dot x(t) =PLP\,x(t) \;+\;PLQ\,e^{tQLQ}z_{0} \;+\;\int_{0}^{t}PLQ\,e^{(t-s)QLQ}\,QLP\,x(s)\,ds . \ \ }\tag{6.5}$ Defining the unresolved-initial-state term $\eta_{z_{0}}(t):=PLQ\,e^{tQLQ}z_{0}\ \in\mathcal{H}_{P}\tag{6.6}$ and the memory kernel $\mathcal{K}(\tau):=PLQ\,e^{\tau QLQ}\,QLP\ \in\mathcal{B}(\mathcal{H}_{P}),\tag{6.7}$ equation (6.5) reads $\dot x(t)=PLP\,x(t)+\eta_{z_{0}}(t)+\int_{0}^{t}\mathcal{K}(t-s)\,x(s)\,ds .\tag{6.8}$ Both $\eta_{z_{0}}$ and $\mathcal{K}$ are continuous in $t$ , with the bounds $\lVert\eta_{z_{0}}(t)\rVert\le\lVert PLQ\rVert\,e^{t\lVert QLQ\rVert}\lVert z_{0}\rVert$ and $\lVert\mathcal{K}(\tau)\rVert_{\mathrm{op}}\le\lVert PLQ\rVert\,\lVert QLP\rVert\,e^{\tau\lVert QLQ\rVert}$ .

Proof. Equation (6.4) is an inhomogeneous linear ODE in the Banach space $\mathcal{H}_{Q}$ with bounded generator $QLQ\in\mathcal{B}(\mathcal{H}_{Q})$ and continuous inhomogeneity $t\mapsto QLP\,x(t)$ (continuous because $t\mapsto e^{tL}\Psi_{0}$ is norm-continuous and $QLP$ is bounded). Variation of constants gives the unique solution $z(t)=e^{tQLQ}z_{0}+\int_{0}^{t}e^{(t-s)QLQ}\,QLP\,x(s)\,ds ,\tag{6.9}$ the integral being a Riemann (equivalently Bochner) integral of a continuous $\mathcal{H}_{Q}$ -valued integrand. Indeed, put $w(t):=e^{-tQLQ}z(t)$ ; then $w$ is differentiable with $\dot w(t)=e^{-tQLQ}\bigl(\dot z(t)-QLQ\,z(t)\bigr)=e^{-tQLQ}QLP\,x(t)$ by (6.4), and integrating from $0$ to $t$ and multiplying by $e^{tQLQ}$ yields (6.9). Uniqueness holds because the difference of two solutions solves the homogeneous equation with zero data.

Substituting (6.9) into (6.3) and applying the bounded operator $PLQ$ under the integral sign gives $\dot x(t) =PLP\,x(t) +PLQ\,e^{tQLQ}z_{0} +\int_{0}^{t}PLQ\,e^{(t-s)QLQ}\,QLP\,x(s)\,ds ,$ which is (6.5); (6.8) is a rewriting using (6.6) and (6.7), noting $\mathcal{K}(t-s)$ has argument $t-s\ge0$ throughout the integration range. The stated bounds follow from $\lVert e^{\tau QLQ}\rVert_{\mathrm{op}}\le e^{\tau\lVert QLQ\rVert}$ and submultiplicativity; continuity in $t$ follows from norm-continuity of $\tau\mapsto e^{\tau QLQ}$ . ◻

Remark 24 (What (6.5) is, and what it is not).

Equation (6.5) is an identity, valid for every source solution, obtained by nothing more than variation of constants. It is the projection-operator identity of Mori and Zwanzig [20, 10, 21] in the special case of a bounded generator and an orthogonal projection. In that literature the second term is often called the “fluctuating force” or “noise”, and $\mathcal{K}$ the “friction kernel”. We use neither word without qualification, and we record explicitly what has not been done here:

$\eta_{z_{0}}$ is not stochastic. It is a deterministic, linear, continuous function of the single vector $z_{0}\in\mathcal{H}_{Q}$ , given by the closed formula (6.6). It becomes a random process only if one additionally posits a probability measure on the unresolved initial data, an equilibrium ensemble, say, and that is an extra modelling hypothesis with its own physical content. We impose no such measure.
$\mathcal{K}$ is not a friction coefficient. It is the exact kernel (6.7). No fluctuation–dissipation relation has been derived, because no ensemble has been introduced with respect to which one could be stated.
No Markov approximation has been made. The convolution in (6.8) is over the entire history $[0,t]$ . Replacing $\mathcal{K}$ by $\delta$ -correlated kernel, or taking a timescale-separation limit, would produce a different (approximate) equation; we take no such limit.
No ensemble average or coarse-graining functional appears. $P$ is an orthogonal projection on a Hilbert space, chosen by the modeller, and nothing else.

The physical content of (6.8) is therefore modest and exact: if the resolved sector does not close, then the resolved trajectory is determined by resolved initial data plus an inhomogeneity carrying unresolved initial data plus a convolution over its own past. These are three structurally distinct dependences, and (6.8) says they are the only ones.

Theorem 6.4 (Autonomous-closure criterion).

Under Assumption 6.1, the following are equivalent.

There exists a single operator $Y\in\mathcal{B}(\mathcal{H}_{P})$ such that, for every source initial state $\Psi_{0}\in\mathcal{H}$ , that is, for every $x(0)\in\mathcal{H}_{P}$ and every $z_{0}\in\mathcal{H}_{Q}$ , the resolved component $x(t)=Pe^{tL}\Psi_{0}$ satisfies the autonomous first-order equation $\dot x(t)=Y\,x(t),\qquad t\ge0 .\tag{6.10}$
$PLQ=0$ .
$L(\ker P)\subseteq\ker P$ , i.e. $L\,\mathcal{H}_{Q}\subseteq\mathcal{H}_{Q}$ .

When they hold, $Y=PLP|_{\mathcal{H}_{P}}$ is unique, and both $\eta_{z_{0}}\equiv0$ (for every $z_{0}$ ) and $\mathcal{K}\equiv0$ .

Proof. (2) $\Rightarrow$ (1) and the value of $Y$ . If $PLQ=0$ then (6.3) reduces immediately to $\dot x(t)=PLP\,x(t)$ for every source solution, which is (6.10) with $Y=PLP|_{\mathcal{H}_{P}}$ . Moreover $\eta_{z_{0}}(t)=PLQ\,e^{tQLQ}z_{0}=0$ for all $t$ and all $z_{0}$ , and $\mathcal{K}(\tau)=PLQ\,e^{\tau QLQ}QLP=0$ , so the last two terms of (6.8) vanish identically, consistently.

(1) $\Rightarrow$ (2). This is where the quantifier “for every source initial state” does the work. Fix any $x_{0}\in\mathcal{H}_{P}$ and any two unresolved data $z_{0},z_{0}'\in\mathcal{H}_{Q}$ . The two source initial states $\Psi_{0}=x_{0}+z_{0}$ and $\Psi_{0}'=x_{0}+z_{0}'$ have the same resolved component $x(0)=x'(0)=x_{0}$ . By hypothesis both resolved trajectories obey (6.10) with the same $Y$ , so evaluating at $t=0$ , $\dot x(0)=Y x_{0}=\dot x'(0).$ On the other hand (6.3) at $t=0$ gives the true initial velocities $\dot x(0)=PLP\,x_{0}+PLQ\,z_{0}, \qquad \dot x'(0)=PLP\,x_{0}+PLQ\,z_{0}' .$ Equating and cancelling $PLP\,x_{0}$ yields $PLQ\,(z_{0}-z_{0}')=0$ . Since $z_{0},z_{0}'$ range over all of $\mathcal{H}_{Q}$ , the difference $z_{0}-z_{0}'$ ranges over all of $\mathcal{H}_{Q}$ , whence $PLQ=0$ as an operator $\mathcal{H}_{Q}\to\mathcal{H}_{P}$ . Taking $z_{0}=0$ in (6.10) and (6.3) then forces $Yx_{0}=PLP\,x_{0}$ for all $x_{0}\in\mathcal{H}_{P}$ , i.e. $Y=PLP|_{\mathcal{H}_{P}}$ , which also gives uniqueness of $Y$ .

(2) $\Leftrightarrow$ (3). $\ker P=\mathcal{H}_{Q}=\operatorname{Ran}Q$ . Now $L\mathcal{H}_{Q}\subseteq\mathcal{H}_{Q}$ means $PL\zeta=0$ for every $\zeta\in\mathcal{H}_{Q}$ , i.e. $PLQ\Psi=0$ for every $\Psi\in\mathcal{H}$ , i.e. $PLQ=0$ . Conversely $PLQ=0$ gives $PL\zeta=PLQ\zeta=0$ for $\zeta\in\mathcal{H}_{Q}$ , so $L\zeta=(P+Q)L\zeta=QL\zeta\in\mathcal{H}_{Q}$ . ◻

Remark 26 (The quantifier is not decorative).

Theorem 6.4 is a statement about all source states. Three distinct weaker phenomena must not be mistaken for autonomous closure:

Closure on a prepared state. If $z_{0}\in N_{Q}$ , see (7.2), then $\eta_{z_{0}}\equiv0$ and the initial-state term drops out for that state. The memory kernel $\mathcal{K}$ is unaffected: by (6.7) it is determined by $L$ and $P$ alone and does not depend on $z_{0}$ at all, so no preparation of the unresolved initial data can remove it. The resolved motion of such a state is therefore still history-dependent, and other initial states fail even the weaker condition. Vanishing of the initial-state term in one specially prepared state is not closure.
Static quotient. That $\mathcal{H}_{P}\cong\mathcal{H}/\mathcal{H}_{Q}$ is a perfectly good Hilbert space says nothing whatever about $PLQ$ . This is (D1)/(D4) again.
Small singular values. That $PLQ$ has small norm makes the coupling weak and may justify a perturbative approximation, with a controllable error. It does not make the projected equation autonomous, and no truncation of (6.8) is exact unless $PLQ=0$ identically. This is (D3)/(D4).

Finally, observe that Theorem 6.4(3) is precisely the invariance condition (5.8) of Proposition 5.5 for the observation operator $K=P$ , whose kernel is $\mathcal{H}_{Q}$ . Theorems 5.2 and 6.4 therefore agree, as they must: the semigroup descends exactly when the projected equation closes.

Example 6.6 (A completely calculated block).

Take $\mathcal{H}=\mathbb{C}^{3}$ with the standard basis $e_{1},e_{2},e_{3}$ , let $P$ project onto $\mathcal{H}_{P}=\mathop{\mathrm{span}}\{e_{1}\}$ and $Q$ onto $\mathcal{H}_{Q}=\mathop{\mathrm{span}}\{e_{2},e_{3}\}$ , and let $L=\begin{pmatrix} a & b_{1} & b_{2}\\ c_{1} & d & 0\\ c_{2} & 0 & \mu\end{pmatrix},$ so that $PLP=(a),\quad PLQ=\begin{pmatrix}b_{1}&b_{2}\end{pmatrix},\quad QLP=\begin{pmatrix}c_{1}\\ c_{2}\end{pmatrix},\quad QLQ=\begin{pmatrix}d&0\\0&\mu\end{pmatrix},$ with $d\neq\mu$ . Write $x(t)\in\mathbb{C}$ for the resolved component and $z_{0}=(z_{0}^{(2)},z_{0}^{(3)})^{\mathsf T}$ .

(i) Same readout, different projected velocity. Suppose $b_{1}\neq0$ . The two source states $\Psi_{0}=x_{0}e_{1}$ and $\Psi_{0}'=x_{0}e_{1}+e_{2}$ have identical resolved component $x_{0}$ , hence identical readout at $t=0$ . By (6.3), $\dot x(0)=a x_{0}, \qquad \dot x'(0)=a x_{0}+b_{1}\neq\dot x(0).$ No autonomous law $\dot x=Yx$ can produce both, confirming Theorem 6.4: here $PLQ=(b_{1}\ b_{2})\neq0$ .

(ii) The initial-state term and memory kernel. Since $QLQ$ is diagonal, $e^{\tau QLQ}=\operatorname{diag}(e^{\tau d},e^{\tau\mu})$ , and (6.6), (6.7) give in closed form $\eta_{z_{0}}(t)=b_{1}e^{td}\,z_{0}^{(2)}+b_{2}e^{t\mu}\,z_{0}^{(3)}, \qquad \mathcal{K}(\tau)=b_{1}c_{1}e^{\tau d}+b_{2}c_{2}e^{\tau\mu},$ so the exact projected equation (6.8) is the scalar Volterra equation $\dot x(t)=a\,x(t) +b_{1}e^{td}z_{0}^{(2)}+b_{2}e^{t\mu}z_{0}^{(3)} +\int_{0}^{t}\Bigl(b_{1}c_{1}e^{(t-s)d}+b_{2}c_{2}e^{(t-s)\mu}\Bigr)x(s)\,ds .$ Everything here is deterministic: $\eta_{z_{0}}$ is a sum of two exponentials with coefficients read off from $z_{0}$ ; there is no noise in sight.

(iii) $N_{Q}$ and $E_{\mathrm{dyn}}$ . Here $m=\dim\mathcal{H}_{Q}=2$ , so Proposition 7.7 gives $N_{Q}=\ker(PLQ)\cap\ker(PLQ\cdot QLQ)$ , i.e. the solutions of $\begin{pmatrix} b_{1} & b_{2}\\ b_{1}d & b_{2}\mu\end{pmatrix} \begin{pmatrix}z^{(2)}\\ z^{(3)}\end{pmatrix}=0 .$ The observability matrix has determinant $b_{1}b_{2}(\mu-d)$ . Case A: $b_{1},b_{2}\neq0$ and $d\neq\mu$ . Then the determinant is nonzero, so $N_{Q}=\{0\}$ and $E_{\mathrm{dyn}}=\mathcal{H}_{Q}\cong\mathbb{C}^{2}$ : both unresolved directions are dynamically distinguishable and must be retained. Case B: $b_{2}=0$ , $b_{1}\neq0$ . Then both rows are multiples of $(1\ 0)$ , so $N_{Q}=\mathop{\mathrm{span}}\{e_{3}\}$ and $E_{\mathrm{dyn}}\cong\mathbb{C}$ : the direction $e_{3}$ never influences the resolved trajectory through the initial-state term, and the minimal completion retains only $z^{(2)}$ (indeed $\mathcal{K}(\tau)=b_{1}c_{1}e^{\tau d}$ also loses its $e_3$ contribution). Note that $e_{3}$ is not dynamically trivial, it evolves as $e^{t\mu}$ , it is merely unobservable in the resolved response, which is exactly the relativization Theorem 7.4 makes precise.