1
$\begingroup$

I am studying Vassili N. Kolokoltsov's paper "On the Mathematical Theory of Quantum Stochastic Filtering Equations for Mixed States" and need to understand the role of the control operator $C$ from the Mora-Rebolledo (MR) Hypothesis.

The MR Hypothesis (Section 2.1) introduces $C$ as a strictly positive self-adjoint operator with discrete spectrum $0 \leq \lambda_1 \leq \lambda_2 \leq \cdots$ and corresponding eigenbasis $\{e_m\}$. It requires:

  1. Domain Control: $Dom(C) \subset Dom(H) \cap Dom(L) \cap Dom(L^*L)$
  2. Operator Bounds: $\|Hx\|^2 \leq K\|x\|_C^2$, $\|L^*Lx\|^2 \leq K\|x\|_C^2$ for all $x \in Dom(C)$
  3. Generalized Dissipativity: $-2Re(Cx, iCP_mHx) - Re(Cx, CP_mL^*Lx) + \|CP_mLx\|^2 \leq \alpha(\|x\|_C^2 + \beta)$

Additionally, Hypothesis A provides more structure: there exists $l \in \mathbb{N}$ such that $L: \mathcal{H}_m \to \mathcal{H}_{m+l}$ and $H: \mathcal{H}_m \to \mathcal{H}_{m+l}$ for all $m$.

!!!!! What specific mathematical pathology in quantum filtering equations necessitates introducing $C$? The paper proves well-posedness for the linear Belavkin equation (Theorem 2.1): $d\chi(t) = -[iH\chi(t) + \frac{1}{2}L^*L\chi(t)]dt + L\chi(t)dY(t)$ Does $C$ primarily prevent solutions from escaping to "too singular" states in $Dom(C)^\perp$, or is it essential for controlling the stochastic integral $\int_0^t L\chi(s)dY(s)$ when $L$ is unbounded?

I'm trying to understand if $C$ addresses a physical obstruction in quantum stochastic evolution with unbounded operators, or if alternative approaches could circumvent its necessity.

$\endgroup$
3
  • $\begingroup$ Perhaps this question would be better suited in Math SE. $\endgroup$ Commented Nov 16 at 4:23
  • $\begingroup$ @flippiefanus There they deleted my questions and sent them back for revision. $\endgroup$ Commented Nov 16 at 10:56
  • $\begingroup$ Did they give you a reason? $\endgroup$ Commented Nov 17 at 3:04

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.