bibkey: gudder2023dual
authors: Stanley Gudder
year: 2023
title: Dual Quantum Instruments and Sub-observables
doi: null
url: https://arxiv.org/abs/2302.12243v1
claim: "Section 3 conjectures that an instrument's image of effects need not be a Sob effect algebra and asks whether an arbitrary Luders instrument has this property."
strata_touched:
  - D5/S3/Quantum/Measurement/LudersSubObservableClosure
license: citation-only
triage: anchor

Gudder, dual quantum instruments and sub-observables

Verified locator

Source: https://arxiv.org/abs/2302.12243v1

Source statements

Section 2, p. 2:

We call $a\in L(H)$ an effect if $0\le a\le I$ where $0,I$ are the zero and identity operators, respectively.

Section 2, p. 3:

If $\Delta \mapsto A(\Delta )$ is countably additive but $A({\Omega }_A)$ need not be $I$, then $A$ is called a sub-observable.

Section 3, p. 11:

Let $U\subseteq \mathrm{Sob}(H)$ and assume that all elements of $U$ have the same outcome set $(\Omega ,\mathcal{F})$. We call $U$ a Sob effect algebra if: (S1) there exists an observable $Z\in U$, (S2) if $A\in U$ then $A^{\prime }=Z-A\in U$, (S3) if $A,B\in U$ and $A+B\in \mathrm{Sob}(H)$, then $A+B\in U$.

After Theorem 3.1, p. 11:

If $\mathcal{I}\in \mathrm{In}(H)$ and $U_{\mathcal{I}}=\{{\mathcal{I}}_a^{\ast }:a\in \mathcal{E}(H)\}$, then we conjecture that $U$ need not be a Sob effect algebra.

Example 3, p. 14:

If ${\mathcal{L}}_x(\rho )=a_x\circ \rho$ is an arbitrary Lüders instrument, we do not know whether $U_{\mathcal{L}}$ is a Sob effect algebra. However, in the case where $a_x$ are projections (we then call $\mathcal{L}$ sharp) we can show that it is.

Section 2, pp. 4–5 gives the Lüders dual action ${\mathcal{L}}_x^{\ast }(b)=a_x^{1/2}b{a}_x^{1/2}$. The finite encoding uses outcome values on Fin n, matrices on Fin d → ℂ, and $K_x=a_x^{1/2}$. Event values are sums over their outcomes. Positive semidefiniteness of each outcome and of the complement of the total expresses sub-observability; equality of the total to the identity expresses observability.

Settlement and scope

The two-outcome qubit family with $K_0=\mathrm{diag}(4/5,3/5)$ and $K_1=\mathrm{diag}(3/5,4/5)$ gives a negative answer to the arbitrary Lüders question and proves the existential conjecture. For $E=\frac{1}{400}\left(\begin{array}{cc}151& 50\\ 50& 151\end{array}\right)$, both $E$ and its complement are positive. Its output $A_x=K_{x}E{K}_x$ belongs to the effect image. The doubled output is a sub-observable but has no effect preimage: invertibility of $K_0$ forces the preimage $2E$, whose complement has negative quadratic form on $(1,1)$. This does not assert failure for every unsharp instrument.

Example 3’s sharp case and Theorems 3.1–3.4 keep their stated hypotheses. In particular, Theorem 3.2 uses orthogonality of preimages, rather than ambient Sob sum closure; the counterexample distinguishes these two notions.