Dual Quantum Instruments and Sub-observables

TL;DR

This paper introduces dual quantum instruments and sub-observables, exploring their properties, relationships, and extensions, with applications to various quantum measurement models and future research directions.

Contribution

It defines dual instruments and sub-observables, characterizes their effect algebras, and studies their convexity, sequential products, and extensions, advancing the mathematical framework of quantum measurements.

Findings

Dual instruments measure a unique observable but determine many sub-observables.

Characterization of sub-observable effect algebras and their convexity.

Discussion of sequential products and examples including L"uders, Holero, and constant state instruments.

Abstract

We introduce the concepts of dual instruments and sub-observables. We show that although a dual instruments measures a unique observable, it determines many sub-observables. We define a unique minimal extension of a sub-observable to an observable and consider sequential products and conditioning of sub-observables. Sub-observable effect algebras are characterized and studied. Moreover, the convexity of these effect algebras is considered. The sequential product of instruments is discussed. These concepts are illustrated with many examples of instruments. In particular, we discuss L\"uders, Holero and constant state instruments. Various conjectures for future research are presented.

Equations

$$tr\left[\rho A(\Delta )\right]=tr\left[\mathcal{I}(\Delta )\rho \right]$$

$$tr\left[\rho {\mathcal{I}}^{\ast }(\Delta )(a)\right]=tr\left[\mathcal{I}(\Delta )(\rho )a\right]$$

$${\Phi }_{\rho }^A(\Delta )=tr\left[\rho A(\Delta )\right]$$

$${\left[\mathcal{I}(\Delta )(\rho )\right]}^{\sim }=\frac{1}{tr\left[\mathcal{I}(\Delta )(\rho )\right]}\mathcal{I}(\Delta )(\rho )$$

$${\Phi }_{\rho }^{\mathcal{I}}(\Delta )=tr\left[\mathcal{I}(\Delta )(\rho )\right]$$

$$tr\left[\rho {\mathcal{I}}^{\ast }(\Delta )A\right]=tr\left[\mathcal{I}(\Delta )(\rho )A\right]$$

$$tr\left[\rho {\mathcal{I}}_a^{\ast }(\Delta )\right]=tr\left[\rho {\mathcal{I}}^{\ast }(\Delta )a\right]=tr\left[\mathcal{I}(\Delta )(\rho )a\right]=tr\left[\mathcal{I}(\Delta )(\rho )\right]tr\left[{\left(\mathcal{I}(\Delta )(\rho )\right)}^{\sim }a\right]$$

$${\mathcal{I}}_I^{\ast }({\Omega }_{\mathcal{I}})={\mathcal{I}}^{\ast }({\Omega }_{\mathcal{I}})(I)=I$$

$$tr\left[\rho \hat{\mathcal{I}}(\Delta )\right]=tr\left[\mathcal{I}(\Delta )(\rho )\right]$$

$${\mathcal{I}}_a^{\ast }({\Omega }_{\mathcal{I}})={\mathcal{I}}^{\ast }({\Omega }_{\mathcal{I}})(a)=I$$

$$tr\left[\mathcal{I}({\Omega }_{\mathcal{I}})(\rho )a\right]=tr\left[\rho {\mathcal{I}}^{\ast }({\Omega }_{\mathcal{I}})a\right]=tr(\rho I)=1$$

$$A(\Delta )=\sum _{x\in \Delta }{a}_x$$

$$\mathcal{L}(\Delta )(\rho )=\sum _{x\in \Delta }{\mathcal{L}}_x(\rho )=\sum _{x\in \Delta }{a}_x^{1/2}\rho a_x^{1/2}$$

$${\mathcal{L}}_b^{\ast }(\Delta )=\sum _{x\in \Delta }{a}_x^{1/2}b{a}_x^{1/2}$$

$$I={\mathcal{L}}_b^{\ast }({\Omega }_A)=\sum _{x\in {\Omega }_A}{a}_x^{1/2}b{a}_x^{1/2}$$

$$\hat{\mathcal{L}}(\Delta )={\mathcal{L}}_I^{\ast }(\Delta )=\sum _{x\in \Delta }{a}_x=A(\Delta )$$

$${\mathcal{H}}_{(\alpha ,A)}(\Delta )(\rho )=tr\left[\rho A(\Delta )\right]\alpha$$

$$({\mathcal{H}}_{(\alpha ,A)}^{\ast }{)}_a(\Delta )={\mathcal{H}}_{(\alpha ,A)}^{\ast }(\Delta )(a)=tr(\alpha a)A(\Delta )$$

$$tr(\alpha a)A({\Omega }_A)=tr(\alpha a)I=I$$

$${\mathcal{H}}_{(\alpha ,A)}(x)(\rho )=tr(\rho a_x){\alpha }_x$$

$$({\mathcal{H}}_{(\alpha ,A)}^{\ast }{)}_a(x)={\mathcal{H}}_{(\alpha ,A)}^{\ast }(x)(a)=tr({\alpha }_{x}a)a_x$$

$$({\mathcal{H}}_{(\alpha ,A)}^{\ast }{)}_a({\Omega }_A)=\sum _{x\in {\Omega }_A}tr({\alpha }_{x}a)a_x=I$$

$$tr\left[\rho {\mathcal{I}}_{\alpha }^{\ast }(\Delta )(a)\right]=tr\left[{\mathcal{I}}_{\alpha }(\Delta )(\rho )a\right]=tr\left[\mathcal{I}(\Delta )(\alpha )a\right]$$

$$({\mathcal{I}}_{\alpha }^{\ast }{)}_a(\Delta )={\mathcal{I}}_{\alpha }^{\ast }(\Delta )(a)=tr\left[\mathcal{I}(\Delta )(\rho )a\right]I$$

$$tr\left[\overline{\mathcal{I}}(\alpha )a\right]=1$$

$$tr\left[\overline{\mathcal{I}}(\alpha )a\right]=tr\left[|a{\psi }_1⟩⟨{\psi }_1|\right]=⟨{\psi }_1,a{\psi }_1⟩$$

$${\hat{\mathcal{I}}}_{\alpha }(\Delta )=tr\left[\mathcal{I}(\Delta )(\alpha )\right]I$$

$$\mathcal{L}(\Delta )(\rho )=\sum _{x\in \Delta }{a}_x\circ \rho$$

$${\mathcal{L}}_b^{\ast }(\Delta )=\sum _{x\in \Delta }{a}_x\circ b$$

$$\mathcal{J}\left(\Delta \cup \left\{y\right\}\right)(\rho )=\mathcal{I}(\Delta )(\rho )+(I-D{)}^{1/2}\rho (I-D{)}^{1/2}=\mathcal{I}(\Delta )(\rho )+D^{\prime }\circ \rho$$

Full text

1 Introduction

In this section we only present general ideas and the detailed definitions will be given in Section 2. An instrument $\mathcal{I}$ is considered to be a two-step measurement process. In the first step, an input state $\rho$ is selected and a measurement of $\mathcal{I}$ is performed. The outcome $\Delta$ of this measurement is observed and the probability of this outcome is given by the trace $tr\left[\mathcal{I}(\Delta)(\rho)\right]$. In the second step, the state $\rho$ is updated to a new state $\mathcal{I}(\Delta)(\rho)^{\sim}$ depending on the outcome $\Delta$ of the first step. When $\mathcal{I}$ produces the outcome $\Delta$ we say that the resulting effect is $A(\Delta)$. We call the map $\Delta\mapsto A(\Delta)$ an effect-valued measure or observable and say that $\mathcal{I}$ measures the observable $A$. As we shall see in Section 2, the probability distribution of $A$ in the state $\rho$ becomes:

[TABLE]

In Section 2, we introduce the concept of a sub-observable $A_{1}$ which can be considered as a deficient observable in the sense that all the possible values of $A_{1}$ need not be attainable.

We also present the concept of a dual instrument $\mathcal{I}^{*}$ in Section 2. As we shall see, $\mathcal{I}^{*}$ satisfies the equation

[TABLE]

for states $\rho$, outcomes $\Delta$ and effects $a$. Equation (1.1) gives a duality between $\mathcal{I}$ and $\mathcal{I}^{*}$. Defining $\mathcal{I}_{a}^{*}(\Delta)=\mathcal{I}^{*}(\Delta)(a)$, we shall show in Section 2 that $\mathcal{I}_{a}^{*}$ is a sub-observable which we say is determined by $\mathcal{I}$. Thus, although $\mathcal{I}$ measures a unique observable, it determines many sub-observables.

In Section 3, we show that every sub-observable has a unique minimal extension to an observable. We also consider sequential products and conditioning of sub-observables. We next characterize and study sub-observable effect algebras. In particular, we show that the set of sub-observables $\mathcal{I}_{a}^{*}$ determined by $\mathcal{I}$ forms an effect algebra in the natural way. Moreover, the convexity of these effect algebras is considered. Section 4 studies sequential products of instruments. All of these concepts are illustrated with many examples of instruments and observables. In particular, we discuss Lüders, Holevo and constant state instruments. Various conjectures for future research are presented

2 Basic Definitions

Let $S$ be a quantum system described by a complex Hilbert space $H$ and $\mathcal{L}(H)$ be the set of bounded linear operators on $H$. For $A,B\in\mathcal{L}(H)$, we write $A\leq B$ if ${\left\langle\phi,A\phi\right\rangle}\leq{\left\langle\phi,B\phi\right\rangle}$ for all $\phi\in H$. We call $a\in\mathcal{L}(H)$ an effect if $0\leq a\leq I$ where [math], $I$ are the zero and identity operators, respectively. An effect describes $a$ two-valued true-false experiment and the set of effects is denoted by $\mathcal{E}(H)$. If an $a\in\mathcal{E}(H)$ is true, then its complement $a^{\prime}=I-a$ is false. Let $(\Omega_{A},\mathcal{F}_{A})$ be a measurable space. An observable with outcome space $(\Omega_{A},\mathcal{F}_{A})$ is a normalized effect-valued measure $A\colon\mathcal{F}_{A}\to\mathcal{E}(H)$. That is, $\Delta\mapsto A(\Delta)$ is countably additive in the strong operator topology and $A(\Omega_{A})=I$. We interpret $A(\Delta)$ as the effect that is true when a measurement of $A$ results in an outcome in $\Delta$. A state for the system $S$ is an effect $\rho$ that satisfies $tr(\rho)=1$. States describe the initial condition of the system and the set of states is denoted by $\mathcal{S}(H)$ [1, 2, 9, 11, 13]. If $\rho\in\mathcal{S}(H)$ and $A$ is an observable, the distribution of $A$ in the state $\rho$ is a probability measure given by

[TABLE]

We denote the set of observables by $Ob(H)$. If $\Delta\mapsto A(\Delta)$ is countably additive but $A(\Omega_{A})$ need not be $I$, then $A$ is called a sub-observable. We denote the set of sub-observables by $Sob(H)$.

An effect $\rho$ is called a partial state if $tr(\rho)\leq 1$. An operation is a completely positive linear map $\mathcal{O}\colon\mathcal{L}(H)\to\mathcal{L}(H)$ such that $tr\left[\mathcal{O}(\rho)\right]\leq 1$ for all $\rho\in\mathcal{S}(H)$. If $tr\left[\mathcal{O}(\rho)\right]=1$ for all $\rho\in\mathcal{S}(H)$ then $\mathcal{O}$ is called a channel [1, 9, 11, 13]. We denote the set of operations by $\mathcal{O}(H)$. Let $(\Omega_{\mathcal{I}},\mathcal{F}_{\mathcal{I}})$ be a measurable space. An instrument with outcome space $(\Omega_{\mathcal{I}},\mathcal{F}_{\mathcal{I}})$ is a normalized operation-valued measure $\mathcal{I}$. That is, for $\Delta\in\mathcal{F}_{\mathcal{I}}$, $\Delta\mapsto\mathcal{I}(\Delta)\in\mathcal{O}(H)$ is countably additive in the strong operation topology and $\overline{\mathcal{I}}=\mathcal{I}(\Omega_{\mathcal{I}})$ is a channel [9, 13]. We denote the set of instruments by $In(H)$. We interpret $\mathcal{I}\in In(H)$ as an apparatus that measures an observable $\widehat{\mathcal{I}}$ and updates states. If $\rho\in\mathcal{S}(H)$, then $\mathcal{I}(\Delta)(\rho)$ is a partial state and assuming $\mathcal{I}(\Delta)(\rho)\neq 0$, then

[TABLE]

is a state. We call $\left[\mathcal{I}(\Delta)(\rho)\right]^{\sim}$ the update of $\rho$ given that a measurement of $\mathcal{I}$ results in an outcome in $\Delta$. If $\rho\in\mathcal{S}(H)$, the distribution of $\mathcal{I}\in In(H)$ is the probability measure given by

[TABLE]

The dual instrument to $\mathcal{I}\in In(H)$ is the unique map $\mathcal{I}^{*}(\Delta)\colon\mathcal{L}(H)\to\mathcal{L}(H)$, $\Delta\in\mathcal{F}_{\mathcal{I}}$, satisfying

[TABLE]

for all $A\in\mathcal{L}(H)$ [8]. It follows that $\Delta\mapsto\mathcal{I}^{*}(\Delta)$ is countable additive and $\mathcal{I}^{*}(\Delta)$ is a completely positive linear map such that $\mathcal{I}^{*}(\Delta)\colon\mathcal{E}(H)\to\mathcal{E}(H)$ for all $\Delta\in\mathcal{F}_{\mathcal{I}}$ and $\mathcal{I}^{*}(\Omega_{\mathcal{I}})I=I$ [8]. If $a\in\mathcal{E}(H)$, we define $\mathcal{I}_{a}^{*}(\Delta)=\mathcal{I}^{*}(\Delta)(a)$. Then $\mathcal{I}_{a}^{*}\in Sob(H)$ and we say that $\mathcal{I}$ determines $\mathcal{I}_{a}^{*}$. We interpret $\mathcal{I}_{a}^{*}(\Delta)\in\mathcal{E}(H)$ as the update of the effect $a$ given that a measurement of $\mathcal{I}$ results in an outcome in $\Delta$. In this way, an instrument not only updates states, it also can be employed to update effects. For all $\rho\in\mathcal{S}(H)$ we have from (2.1) that

[TABLE]

Thus, the probability of the effect $\mathcal{I}_{a}^{*}(\Delta)$ when $S$ is in the state $\rho$ is the probability that a measurement of $\mathcal{I}$ results in an outcome in $\Delta$ times the probability of $a$ in the updated state $\left(\mathcal{I}(\Delta)(\rho)\right)^{\sim}$. Notice that

[TABLE]

so $\mathcal{I}_{I}^{*}$ is an observable. We call $\mathcal{I}_{I}^{*}$ the observable measured by $\mathcal{I}$ and we write $\widehat{\mathcal{I}}=\mathcal{I}_{I}^{*}$. Notice that $\widehat{\mathcal{I}}$ is the unique observable satisfying

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$, $\rho\in\mathcal{S}(H)$. We have that $\mathcal{I}_{a}^{*}$ is an observable if and only if

[TABLE]

which is equivalent to

[TABLE]

We now present some examples that illustrate the previous definitions. An observable $A$ is finite if $\Omega_{A}$ is a finite set. In this case we assume that $\mathcal{F}_{A}=2^{\Omega_{A}}$ so we need not specify the $\sigma$-algebra $\mathcal{F}_{A}$. We then write $A=\left\{a_{x}\colon x\in\Omega_{A}\right\}$ and we have that

[TABLE]

for all $\Delta\subseteq\Omega_{A}$. Corresponding to a finite observable $A$ we have the Lüders instrument $\mathcal{L}_{x}(\rho)=a_{x}^{1/2}\rho a_{x}^{1/2}$ for all $\rho\in\mathcal{S}(H)$, $x\in\Omega_{A}$ [12]. It follows that

[TABLE]

for all $\rho\in\mathcal{S}(H)$, $\Delta\subseteq\Omega_{A}=\Omega_{\mathcal{L}}$. The dual instrument satisfies $\mathcal{L}_{x}^{*}(b)=a^{1/2}ba^{1/2}$ for all $x\in\Omega_{A}$, $b\in\mathcal{E}(H)$ [8]. The sub-observables determined by $\mathcal{L}$ have the form $\mathcal{L}_{b}^{*}$, $b\in\mathcal{E}(H)$ where

[TABLE]

If $\mathcal{L}_{b}^{*}\in Ob(h)$, we have that

[TABLE]

It follows that $b=I$. Hence,

[TABLE]

for all $\Delta\subseteq\Omega_{A}$ so $\widehat{\mathcal{L}}=A$ is the only observable determined by $\mathcal{L}$.

A Holevo instrument with state $\alpha$ and observable $A$ has the form

[TABLE]

for all $\rho\in\mathcal{S}(H)$, $\Delta\in\mathcal{F}_{A}$ [10]. The sub-observables determined by $\mathcal{H}_{(\alpha,A)}$ become

[TABLE]

Then $(\mathcal{H}{(\alpha,A)}^{*})_{a}$ is an observable if and only if

[TABLE]

This is equivalent to $tr(\alpha a)=1$ which is equivalent to $(\mathcal{H}_{(\alpha,A)}^{*})_{a}(\Delta)=A(\Delta)$. Thus, $A=(\mathcal{H}_{(\alpha,A)}^{*})_{I}$ is the only observable determined by $\mathcal{H}_{(\alpha,A)}$.

Let $A=\left\{a_{x}\colon x\in\Omega_{A}\right\}$ be a finite observable and $\left\{\alpha_{x}\colon x\in\Omega_{A}\right\}\subseteq\mathcal{S}(H)$. A finite Holevo instrument with states $\left\{\alpha_{x}\colon x\in\Omega_{A}\right\}$ and observable $A$ has the form

[TABLE]

The sub-observables determined by $\mathcal{H}_{(\alpha,A)}$ becomes

[TABLE]

We see that $(\mathcal{H}_{(\alpha,A)}^{*})_{a}$ is an observable if and only if

[TABLE]

which is equivalent to $tr(\alpha_{x}a)=1$ for all $x\in\Omega_{A}$. Again, $A$ is the only observable determined by $\mathcal{H}_{(\alpha,A)}$. $\mathcal{H}_{(\alpha,A)}$ is also called a conditional state preparator [9].

A constant-state instrument has the form $\mathcal{I}_{\alpha}(\Delta)(\rho)=\mathcal{I}(\Delta)(\alpha)$ where $\alpha\in\mathcal{S}(H)$, $\mathcal{I}\in In(H)$ [8]. Since

[TABLE]

for all $\rho\in\mathcal{S}(H)$, we conclude that $\mathcal{I}_{\alpha}^{*}(\Delta)(a)=tr\left[\mathcal{I}(\Delta)(\alpha)a\right]I$ for all $\Delta\in\mathcal{F}_{\mathcal{I}}$. Hence, for all $a\in\mathcal{E}(H)$ we obtain

[TABLE]

It follows that $(\mathcal{I}_{\alpha}^{*})_{a}$ is an observable if and only if

[TABLE]

There can be many $a\in\mathcal{E}(H)$ that satisfy (2.3). For example, suppose $\overline{\mathcal{I}}(\rho)=\sum P_{i}\rho P_{i}$ where $P_{i}={\left|\psi_{i}\right>}{\left<\psi_{i}\right|}$ and $\left\{\psi_{i}\right\}$ is an orthonormal basis for $H$. Also, suppose $\alpha={\left|\psi_{1}\right>}{\left<\psi_{1}\right|}$. Then $\overline{\mathcal{I}}(\alpha)=P_{1}$ and we have

[TABLE]

Then ${\left\langle\psi_{1},a\psi_{1}\right\rangle}=1$ if and only if $a\psi_{1}=\psi_{1}$ and there are many $a\in\mathcal{E}(H)$ that satisfy this. We conclude that $\mathcal{I}_{\alpha}$ can determine many sub-observables. The unique observable measured by $\mathcal{I}_{\alpha}$ is $\widehat{\mathcal{I}}_{\alpha}=(\mathcal{I}_{\alpha}^{*})_{I}$ where

[TABLE]

We now discuss the algebraic structure of $\mathcal{E}(H)$. An effect algebra is a four-tuple $(E,0,1,\oplus)$ where $E$ is a set, $0,1$ are elements of $E$ and $\oplus$ is a partial binary operation on $E$ [3, 5, 6, 9]. When $a\oplus b$ is defined, we say that $a\oplus b$ exists and write $a\perp b$. An effect algebra satisfies the following axioms:

• (E1)

If $a\perp b$, then $b\perp a$ and $a\oplus b=b\oplus a$.

• (E2)

If $a\perp b$, $c\perp(a\oplus b)$, then $b\perp c$, $a\perp(b\oplus c)$ and $a\oplus(b\oplus c)=(a\oplus b)\oplus c$.

• (E3)

If $a\in E$, there exists a unique $a^{\prime}\in E$ such that $a^{\prime}\perp a$ and $a\oplus a^{\prime}=1$.

• (E4)

If $a\perp 1$, then $a=0$.

It is easy to check that $\left(\mathcal{E}(H),0,I,\oplus\right)$ is an effect algebra where $a\perp b$ when $a+b\leq I$ and in this case we define $a\oplus b=a+b$. We interpret the effect $a\oplus b$ to be a parallel stochastic sum of $a$ and $b$.

For $a,b\in\mathcal{E}(H)$ we define their standard sequential product by $a\circ b=a^{1/2}ba^{1/2}$ [3, 4]. We interpret $a\circ b$ as the effect that results from first measuring $a$ and then measuring $b$. In this way, the measurement of $a$ can interfere with the measurement of $b$ but not vice-versa. It is shown in [4] that $a\circ b=b\circ a$ if and only if $ab=ba$. Most of the following properties of the standard sequential product are straightforward to show [3].

Lemma 2.1**.**

(1)* $a\circ(b\oplus c)=a\circ b\oplus a\circ c$. (2) $I\circ a=a\circ I=a$. (3) If $a\circ b=0$, then $ab=ba$. (4) If $ab=ba$, then $a\circ(b\circ c)=(a\circ b)\circ c$ for all $c\in\mathcal{E}(H)$. (5) If $ac=ca$ and $bc=cb$, then $c(a\circ b)=(a\circ b)c$ and $c(a\oplus b)=(a\oplus b)c$. (6) $a\circ b\leq a$ for all $a,b\in\mathcal{E}(H)$. (7) If $a\leq b$, then $c\circ a\leq c\circ b$ for all $c\in\mathcal{E}(H)$.*

Notice that if $A=\left\{a_{x}\colon x\in\Omega_{A}\right\}$ is a finite observable, then the corresponding Lüders instrument has the form

[TABLE]

and its determined sub-observables are given by

[TABLE]

3 Sub-Observables

Let $A$ be a sub-observable that is not an observable and let $a=A(\Omega_{A})\neq I$. For the outcome space $(\Omega_{A},\mathcal{F}_{A})$, let $y\notin\Omega_{A}$ and define $\Omega_{B}=\Omega_{A}\cup\left\{y\right\}$, $\mathcal{F}_{B}=\mathcal{F}_{A}\cup\left\{\Delta\cup\left\{y\right\}\colon\Delta\in\mathcal{F}_{A}\right\}$. Then $\mathcal{F}_{B}$ is a $\sigma$-algebra of subsets of $\Omega_{B}$ and we call $(\Omega_{B},\mathcal{F}_{B})$ the one-point extension of $(\Omega_{A},\mathcal{F}_{A})$. For $\Gamma\in\mathcal{F}_{B}$ define $B(\Gamma)=A(\Gamma)$ if $\Gamma\in\mathcal{F}_{A}$ and $B(\Gamma)=A(\Delta)+a^{\prime}$ if $\Gamma=\Delta\cup\left\{y\right\}$, $\Delta\in\mathcal{F}_{A}$. Then $B$ is an observable with outcome space $(\Omega_{B},\mathcal{F}_{B})$ that we call the minimal extension of $A$. Notice that if $A=Ob(H)$, then $B$ is essentially the same as $A$. For a simple example, if $a\in\mathcal{E}(H)$ with $a\neq 0,I$, then $A_{x}=a$ is a sub-observable with outcome space $\left\{\left\{x\right\},\left\{\emptyset,\left\{x\right\}\right\}\right\}$. The minimal extension of $A$ is $B=\left\{B_{x},B_{y}\right\}$ where $B_{x}=a$, $B_{y}=a^{\prime}$ and the outcome space is $\left\{\left\{x,y\right\},\left\{\emptyset,\left\{x\right\},\left\{y\right\},\left\{x,y\right\}\right\}\right\}$.

A sub-instrument $\mathcal{I}$ satisfies the conditions for an instrument except $\mathcal{I}(\Omega_{\mathcal{I}})$ need not be a channel. Let the Kraus decomposition of $\mathcal{I}(\Omega_{\mathcal{I}})$ be $\mathcal{I}(\Omega)(\rho)=\sum C_{i}\rho C_{i}^{*}$ [9, 11, 13] where $C_{i}\in\mathcal{L}(H)$ and $\sum C_{i}^{*}C_{i}=D<I$. Notice that $D\in\mathcal{E}(H)$. Let $(\Omega_{\mathcal{J}},\mathcal{F}_{\mathcal{J}})$ be the one-point extension of $(\Omega_{\mathcal{J}},\mathcal{F}_{\mathcal{J}})$ and define $\mathcal{J}(\Delta)=\mathcal{I}(\Delta)$ if $\Delta\in\Omega_{\mathcal{J}}$ and

[TABLE]

Then $\mathcal{J}$ is an instrument called the minimal extension of $\mathcal{I}$.

Let $A\in Sob(H)$ with outcome space $(\Omega_{A},\mathcal{F}_{A})$ and let $A_{1}$ be the minimal extension of $A$ with outcome space $(\Omega_{A_{1}},\mathcal{F}_{A_{1}})$. Let $\mathcal{J}_{1}$ be an instrument with outcome space $(\Omega_{A_{1}},\mathcal{F}_{A_{1}})$ such that $\widehat{\mathcal{J}}_{1}=A_{1}$. (Such an instrument exists although it need not be unique.) Then for all $\Delta\in\mathcal{F}_{A_{1}}$ we have

[TABLE]

for all $\rho\in\mathcal{S}(H)$. Hence, for all $\Delta\in\mathcal{F}_{A}$ we have $tr\left[\mathcal{J}_{1}(\Delta)(\rho)\right]=tr\left[\rho A(\Delta)\right]$ for all $\rho\in\mathcal{S}(H)$. For example, let $\mathcal{I}_{a}^{*}$ be a sub-observable determined by $\mathcal{I}\in In(H)$. Let $\mathcal{I}_{a,1}^{*}$ be the minimal extension of $\mathcal{I}_{a}^{*}$ with outcome space $(\Omega_{1},\mathcal{F}_{1})$ and let $\mathcal{J}_{1}$ be an instrument with outcome space $(\Omega_{1},\mathcal{F}_{1})$ such that $(\Omega_{1},\mathcal{F}_{1})$ and $\widehat{\mathcal{J}}_{1}=\mathcal{I}_{a,1}^{*}$. Then for every $\Delta\in\mathcal{F}_{\mathcal{I}}$ we have

[TABLE]

Let $A,B\in Sob(H)$ and let $A_{1}$ be the minimal extension of $A$. If $\mathcal{I}\in In(H)$ satisfies $\widehat{\mathcal{I}}=A_{1}$, then define the $\mathcal{I}$-sequential product of $A$ then $B$ with outcome space $(\Omega_{A}\times\Omega_{B},\mathcal{F}_{A}\times\mathcal{F}_{B})$ to be the sub-observable $A\left[\mathcal{I}\right]B=\mathcal{I}^{*}(B)$. This is shorthand notation for

[TABLE]

for all $\Delta\in\mathcal{F}_{A}$, $\Gamma\in\mathcal{F}_{B}$ [5, 6, 7]. We also define $B$ $\mathcal{I}$-conditioned by $A$ to be the sub-observable with outcome space $(\Omega_{B},\mathcal{F}_{B})$ given by

[TABLE]

for all $\Gamma\in\mathcal{F}_{B}$ [5, 6, 7].

Example 1.  Let $A=\left\{a_{x}\colon x\in\Omega\right\}$ be a finite observable and let $\mathcal{L}$ be the Lüders instrument given by $\mathcal{L}_{x}(\rho)=a_{x}\circ\rho$. Then for $b\in\mathcal{E}(H)$ we have $\mathcal{L}_{b}^{*}(\Delta)=\sum\limits_{x\in\Delta}a_{x}\circ b$ and we define $b_{1}\in\mathcal{E}$ by

[TABLE]

Let $\Omega_{1}=\Omega\cup\left\{y\right\}$ be the one-point extension of $\Omega$ and the minimal extension of $\mathcal{L}_{b}^{*}$ satisfies $\mathcal{L}_{b,1}^{*}(x)=a_{x}\circ b$ for $x\in\Omega$, $\mathcal{L}_{b,1}^{*}(y)=b^{\prime}$ and

[TABLE]

for $\Delta\subseteq\Omega$. Let $\mathcal{J}$ be the instrument on $\Omega_{1}$ given by

[TABLE]

for $x\in\Omega$ and $\mathcal{J}(y)(\rho)=(b^{\prime}_{1})\circ\rho$. Then $\widehat{\mathcal{J}}=\mathcal{L}_{b,1}^{*}$ so $\mathcal{J}$ measures $\mathcal{L}_{b,1}^{*}$. The $\mathcal{J}$-sequential product of $\mathcal{L}_{b}^{*}$ then $\mathcal{L}_{c}^{*}$ becomes [7]

[TABLE]

We then obtain

[TABLE]

Moreover [5, 6],

[TABLE]

Another instrument that measures $\mathcal{L}_{b,1}^{*}$ is the finite Holevo instrument $\mathcal{H}_{(\alpha,\mathcal{L}_{b,1}^{*})}$ with states $\alpha_{x}$ and observable $A$. We then have

[TABLE]

Moreover,

[TABLE]

Example 2.  Let $\mathcal{H}_{(\alpha,A)}(\Delta)(\rho)=tr\left[\rho A(\Delta)\right]\alpha$ be a Holevo instrument and let

[TABLE]

be a sub-observable determined by $\mathcal{H}_{(\alpha,A)}$. The minimal extension of $(\mathcal{H}_{(\alpha,A)}^{*})_{a}$ satisfies

[TABLE]

Notice that $D$ is additive because if $\Delta\cap\Gamma=\emptyset$ and $y\notin\Delta\cup\Gamma$, then

[TABLE]

Since $\mathcal{H}_{(\beta,D)}$ measures $D$ we have

[TABLE]

Moreover,

[TABLE]

If $A\in Sob(H)$ and $\Delta_{1},\Delta_{2}\in\mathcal{F}_{A}$ with $\Delta_{1}\subseteq\Delta_{2}$, then $A(\Delta_{1})\leq A(\Delta_{2})$ and in particular, $A(\Delta)\leq A(\Omega_{A})$ for all $\Delta\in\mathcal{F}_{A}$. For $A,B\in Sob(H)$ we write $A\leq B$ if $(\Omega_{A},\mathcal{F}_{A})=(\Omega_{B},\mathcal{F}_{B})$ and $A(\Delta)\leq B(\Delta)$ for all $\Delta\in\mathcal{F}_{A}$. Let $\mathcal{U}\subseteq Sob(H)$ and assume that all elements of $\mathcal{U}$ have the same outcome set $(\Omega,\mathcal{F})$. We call $\mathcal{U}$ a $Sob$ effect algebra if:

• (S1)

there exists an observable $Z\in\mathcal{U}$,

• (S2)

if $A\in\mathcal{U}$ then $A^{\prime}=Z-A\in\mathcal{U}$,

• (S3)

if $A,B\in\mathcal{U}$ and $A+B\in Sob(H)$, then $A+B\in\mathcal{U}$.

Notice that $0\in\mathcal{U}$ because $0=Z-Z\in\mathcal{U}$. Also, $Z=A$ is the only observable in $\mathcal{U}$. Indeed, suppose $A$ is an observable in $\mathcal{U}$, then $A^{\prime}=Z-A\in\mathcal{U}$. Since

[TABLE]

we conclude that $A^{\prime}(\Delta)\leq A^{\prime}(\Omega)=0$ for all $\Delta\in\mathcal{F}$. Hence, $A^{\prime}=0$ so $A=Z$. If $A,B\in Sob(H)$ with $A+B\in Sob(H)$, we write $A\perp B$. When $A\perp B$ we define $A\oplus B=A+B$ and say that $A\oplus B$ exists.

Theorem 3.1**.**

If $\mathcal{U}$ is a $Sob$ effect algebra, then $(\mathcal{U},0,Z,\oplus)$ is an effect algebra.

Proof.

There are four conditions to be satisfied which we now check.

(E1) If $A,B\in\mathcal{U}$ with $A\perp B$, then $B\perp A$ and $A\oplus B=B\oplus A=A+B\in\mathcal{U}$.

(E2) If $A,B,C\in\mathcal{U}$ with $A\perp B$ and $C\perp(A\oplus B)$, then $(A\oplus B)\oplus C=A+B+C\in\mathcal{U}$, Hence, $B\perp C$ and $A\perp(B\oplus C)$ so

[TABLE]

(E3) If $A\in\mathcal{U}$, then $A^{\prime}=Z-A$ is the unique element of $\mathcal{U}$ satisfying $A\oplus A^{\prime}=Z$.

(E4) If $A\in\mathcal{U}$ and $A\perp Z$, then

[TABLE]

Hence, $-A\in\mathcal{U}$ and it follows that $A=0$. ∎

If $\mathcal{I}\in In(H)$ and $\mathcal{U}_{\mathcal{I}}=\left\{\mathcal{I}_{a}^{*}\colon a\in\mathcal{E}(H)\right\}$, then we conjecture that $\mathcal{U}$ need not be a $Sob$ effect algebra. However, we can change the definition of $\perp$ so that $\mathcal{U}_{\mathcal{I}}$ becomes an effect algebra. We write $\mathcal{I}_{a}^{*}\perp\mathcal{I}_{b}^{*}$ is $a\perp b$ and if $a\perp b$ we define $\mathcal{I}_{a}^{*}\oplus\mathcal{I}_{b}^{*}=\mathcal{I}_{(a+b)}^{*}$.

Theorem 3.2**.**

If $\mathcal{I}\in In(H)$, then $(\mathcal{U}_{\mathcal{I}},0,\mathcal{I}_{I}^{*},\oplus)$ is an effect algebra and $F(a)=\mathcal{I}_{a}^{*}$ is a morphism from $\mathcal{E}(H)$ onto $\mathcal{U}_{\mathcal{I}}$. Moreover, $(\mathcal{I}_{a}^{*})^{\prime}=\mathcal{I}_{a^{\prime}}^{*}$.

Proof.

If $a\perp b$, then

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$. Hence, $\mathcal{I}_{a}^{*}\oplus\mathcal{I}_{b}^{*}=\mathcal{I}_{a+b}^{*}=\mathcal{I}_{b}^{*}+\mathcal{I}_{b}^{*}$. We now check the four conditions for an effect algebra.

(E1) If $\mathcal{I}_{a}^{*},\mathcal{I}_{b}^{*}\in\mathcal{U}_{\mathcal{I}}$ with $\mathcal{I}_{a}^{*}\perp\mathcal{I}_{b}^{*}$ we have $\mathcal{I}_{b}^{*}\perp\mathcal{I}_{a}^{*}$ and

[TABLE]

(E2) If $\mathcal{I}_{a}^{*},\mathcal{I}_{b}^{*},\mathcal{I}_{c}^{*}\in\mathcal{U}_{\mathcal{I}}$ with $\mathcal{I}_{a}^{*}\perp\mathcal{I}_{b}^{*}$ and $\mathcal{I}_{c}^{*}\perp(\mathcal{I}_{a}^{*}\oplus\mathcal{I}_{b}^{*})$ then $a\perp b$ and $c\perp(a\oplus b)$. Hence, $a+b+c\in\mathcal{E}(H)$ so $b\perp c$ and $a\perp(b\oplus c)$. Hence, $\mathcal{I}_{b}^{*}\perp\mathcal{I}_{c}^{*}$ and $\mathcal{I}_{a}^{*}\perp(\mathcal{I}_{b}^{*}\oplus\mathcal{I}_{c}^{*})$ and we have

[TABLE]

(E3) If $\mathcal{I}_{a}^{*}\in\mathcal{U}_{\mathcal{I}}$, then $\mathcal{I}_{a^{\prime}}^{*}\perp\mathcal{I}_{a}^{*}$ and $\mathcal{I}_{a}^{*}\oplus\mathcal{I}_{a^{\prime}}^{*}=\mathcal{I}_{a+a^{\prime}}^{*}=\mathcal{I}_{I}^{*}$. If $\mathcal{I}_{a}^{*}\oplus\mathcal{I}_{b}^{*}=\mathcal{I}_{I}^{*}$, then $\mathcal{I}_{b}^{*}=\mathcal{I}_{I}^{*}-\mathcal{I}_{a}^{*}=\mathcal{I}_{a^{\prime}}^{*}$ so $(\mathcal{I}_{a}^{*})^{\prime}=\mathcal{I}_{a^{\prime}}^{*}$ is unique.

(E4) If $\mathcal{I}_{a}^{*}\perp\mathcal{I}_{I}^{*}$, then $a\perp I$ so $a=0$ and hence, $\mathcal{I}_{a}^{*}=0$.

To show that $F$ is a morphism, if $a\perp b$ we have

[TABLE]

Moreover, $F(I)=\mathcal{I}_{I}^{*}$ so $F$ is a morphism. ∎

We now show that certain subsets $\mathcal{U}_{\mathcal{I}}\subseteq Sob(H)$ are $Sob$ effect algebras.

Theorem 3.3**.**

If $\mathcal{I}=\mathcal{H}_{(\alpha,A)}$ is a Holevo instrument, then $\mathcal{U}_{\mathcal{I}}$ is a $Sob$ effect algebra.

Proof.

We have that $\mathcal{U}_{\mathcal{I}}=\left\{\mathcal{I}_{a}^{*}\colon a\in\mathcal{E}(H)\right\}$. We now check the three conditions for a $Sob$ effect algebra.

(1) $\mathcal{I}_{I}^{*}\in\mathcal{U}_{\mathcal{I}}$ and $\mathcal{I}_{I}^{*}\in Ob(H)$.

(2) If $\mathcal{I}_{a}^{*}\in\mathcal{U}_{\mathcal{I}}$, then for all $\Delta\in\mathcal{F}_{\mathcal{I}}$ we obtain

[TABLE]

Hence, $\mathcal{I}_{I}^{*}-\mathcal{I}_{a}^{*}=\mathcal{I}_{a^{\prime}}^{*}\in\mathcal{U}_{\mathcal{I}}$.

(3) Let $\mathcal{I}_{a}^{*},\mathcal{I}_{b}^{*}\in\mathcal{U}_{\mathcal{I}}$ and suppose $\mathcal{I}_{a}^{*}+\mathcal{I}_{b}^{*}\in Sob(H)$. Then for all $\Delta\in\mathcal{F}_{\mathcal{I}}$ we have

[TABLE]

It follows that $0\leq tr(\alpha a)+tr(\alpha b)\leq 1$. If $C=\left[tr(\alpha a)+tr(\alpha b)\right]I$ we have that $C\in\mathcal{E}(H)$ and

[TABLE]

Therefore, $\mathcal{I}_{a}^{*}+\mathcal{I}_{b}^{*}=\mathcal{I}_{c}^{*}\in\mathcal{U}_{\mathcal{I}}$.

We conclude that $\mathcal{U}_{\mathcal{I}}$ is a $Sob$ effect algebra. ∎

Theorem 3.4**.**

If $\mathcal{I}_{\alpha}(\Delta)(\rho)=\mathcal{I}(\Delta)(\alpha)$ is a constant state instrument, then $\mathcal{U}_{\mathcal{I}_{\alpha}}$ is a $Sob$ effect algebra if and only if for every $a,b\in\mathcal{E}(H)$ satisfying $tr\left[\overline{\mathcal{I}}(\alpha)(a+b)\right]\leq 1$ there exists a $c\in\mathcal{E}(H)$ such that

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$.

Proof.

We have that $(\mathcal{I}_{\alpha}^{*})_{a}(\Delta)=tr\left[\mathcal{I}(\Delta)(\alpha)a\right]I$ for all $a\in\mathcal{E}(H)$, $\Delta\in\mathcal{F}_{\mathcal{I}}$. The three conditions for $\mathcal{U}_{\mathcal{I}}$ to be a $Sob$ effect algebra are the following:

(1) $(\mathcal{I}_{\alpha}^{*})_{I}=tr\left[\mathcal{I}(\Delta)(\alpha)\right]I\in Ob(H)$.

(2) $(\mathcal{I}_{\alpha}^{*})_{I}-(\mathcal{I}_{\alpha}^{*})_{a}=(\mathcal{I}_{\alpha}^{*})_{a^{\prime}}\in\mathcal{U}_{\mathcal{I}_{\alpha}}$.

(3) Let $(\mathcal{I}_{\alpha}^{*})_{a},(\mathcal{I}_{\alpha}^{*})_{b}\in\mathcal{U}_{\alpha}$ and suppose that $(\mathcal{I}_{\alpha}^{*})_{a}+(\mathcal{I}_{\alpha}^{*})_{b}\in Sob(H)$. We then have

[TABLE]

and hence, $tr\left[\mathcal{I}(\Delta)(\alpha)(a+b)\right]\leq 1$ for all $\Delta\in\mathcal{F}_{\mathcal{I}}$.

It follows that $\left[\overline{\mathcal{I}}(\alpha)(a+b)\right]\leq 1$. If the given condition holds, there exists a $c\in\mathcal{E}(H)$ such that

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$ so that $(\mathcal{I}_{\alpha}^{*})_{c}=(\mathcal{I}_{\alpha}^{*})_{a}+(\mathcal{I}_{\alpha}^{*})_{b}$.

Hence, $\mathcal{U}_{\mathcal{I}_{\alpha}}$ is a $Sob$ effect algebra. Conversely, suppose $\mathcal{U}_{\mathcal{I}_{\alpha}}$ is a $Sob$ effect algebra and $a,b\in\mathcal{E}(H)$ satisfy $tr\left[\overline{\mathcal{I}}(\alpha)(a+b)\right]\leq 1$. Then by (3.1)

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$ so $(\mathcal{I}_{\alpha}^{*})_{a}+(\mathcal{I}_{\alpha}^{*})_{b}\in Sob(H)$. Therefore, there exists a $c\in\mathcal{E}(H)$ such that $(\mathcal{I}_{\alpha}^{*})_{a}+(\mathcal{I}_{\alpha}^{*})_{b}=(\mathcal{I}_{\alpha}^{*})_{c}$. Again by (3.1)

[TABLE]

Example 3.  If $\mathcal{L}_{x}(\rho)=a_{x}\circ\rho$ is an arbitrary Lüders instrument, we do not know whether $\mathcal{U}_{\mathcal{L}}$ 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. Indeed, if $\mathcal{L}_{a}^{*}+\mathcal{L}_{b}^{*}\in Sob(H)$ then

[TABLE]

We conclude that $c\in\mathcal{E}(H)$ and since $a_{x}a_{y}=\delta_{xy}a_{x}$ for every $x,y\in\Omega_{\mathcal{L}}$ we have

[TABLE]

Hence, $\mathcal{L}_{c}^{*}=\mathcal{L}_{a}^{*}+\mathcal{L}_{b}^{*}$ so $\mathcal{U}_{\mathcal{L}}$ is a $Sob$ effect algebra. ∎

A subset $\mathcal{V}\subseteq Sob(H)$ is convex if $A_{i}\in\mathcal{V}$, $0\leq\lambda_{i}\leq 1$, $i=1,2,\ldots,n$, $\sum_{i=1}^{n}\lambda_{i}=1$, implies $\sum_{i=1}^{n}\lambda_{i}A_{i}\in\mathcal{V}$. A $Sob$ effect algebra need not be convex. For example $\mathcal{V}=\left\{\left\{0,0\right\},\left\{0,I\right\}\right\}$ is a $Sob$ effect algebra that is not convex because

[TABLE]

Theorem 3.5**.**

A $Sob$ effect algebra $\mathcal{U}$ is convex if and only if $A\in\mathcal{U}$, $0\leq\lambda\leq 1$ imply $\lambda A\in\mathcal{U}$.

Proof.

Suppose $\mathcal{U}$ is convex, $A\in\mathcal{U}$ and $0\leq\lambda\leq 1$. We then have

[TABLE]

Conversely, suppose $\lambda A\in\mathcal{U}$ whenever $A\in\mathcal{U}$ and $0\leq\lambda\leq 1$. We need to show that if $A_{i}\in\mathcal{U}$, $0\leq\lambda_{i}\leq 1$, $\sum_{i=1}^{n}\lambda_{i}=1$, then $\sum_{i=1}^{n}\lambda_{i}A_{i}\in\mathcal{U}$. We employ induction on $n$. If $n=2$, suppose that $A_{1},A_{2}\in\mathcal{U}$, $0\leq\lambda_{1},\lambda_{2}\leq 1$ and $\lambda_{1}+\lambda_{2}=1$. By assumption $\lambda_{1}A_{1},\lambda_{2}A_{2}\in\mathcal{U}$ and since

[TABLE]

we have that $\lambda_{a}A_{1}+\lambda_{2}A_{2}\in Sob(H)$. Since $\mathcal{U}$ is a $Sob$ effect algebra $\lambda_{1}A_{1}+\lambda_{2}A_{2}\in\mathcal{U}$ so the result holds for $n=2$. Proceeding by induction, suppose the result holds for $n\geq 2$, $A_{i}\in\mathcal{U}$, $i=1,2,\ldots,n+1$, $0\leq\lambda_{i}\leq 1$ and $\sum_{i=1}^{n+1}\lambda_{i}=1$. Letting $\mu=\sum_{i=1}^{n}\lambda_{i}$ we can assume that $\mu\neq 0$. We have that $\sum_{i=1}^{n}\lambda_{i}/\mu=1$ so by hypothesis $\tfrac{1}{\mu}\sum_{n=1}^{n}\lambda_{i}A_{i}\in\mathcal{U}$. Since $0\leq\mu\leq 1$ we obtain $\sum_{i=1}^{n}\lambda_{i}A_{i}\in\mathcal{U}$. Since $A_{n+1}\in\mathcal{U}$ we have that $\lambda_{n+1}A_{n+1}\in\mathcal{U}$. Moreover, $\sum_{i=1}^{n}\lambda_{i}A_{i}+\lambda_{n+1}A_{n+1}\in Sob(H)$ so $\sum_{i=1}^{n+1}\lambda_{i}A_{i}\in\mathcal{U}$. ∎

Notice that $\mathcal{U}_{\mathcal{I}}=\left\{\mathcal{I}_{a}^{*}\colon a\in\mathcal{E}(H)\right\}$ is convex because if $\mathcal{I}_{a_{i}}^{*}\in\mathcal{U}_{\mathcal{I}}$ and $0\leq\lambda_{i}\leq 1$ with $\sum\lambda_{i}=1$, then $b=\sum\lambda_{i}a_{i}\in\mathcal{E}(H)$. We conclude that

[TABLE]

If $a,b\in\mathcal{E}(H)$ we define the sequential product of $\mathcal{I}_{a}^{*}$ and $\mathcal{I}_{b}^{*}$ to be the sub-observable $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}(\Delta)=\mathcal{I}_{a\circ b}^{*}(\Delta)$ for all $\Delta\in\mathcal{F}_{\mathcal{I}}$. If $ab=ba$, it follows that $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}=\mathcal{I}_{b}^{*}\circ\mathcal{I}_{a}^{*}$. The next theorem follows from Lemma 2.1.

Theorem 3.6**.**

*The sequential product $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}$ satisfies the following conditions: (1) If $b\perp c$, then $\mathcal{I}_{a}^{*}\circ(\mathcal{I}_{b}^{*}+\mathcal{I}_{c}^{*})=\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}+\mathcal{I}_{a}^{*}\circ\mathcal{I}_{c}^{*}$.

(2) If $0\leq\lambda_{i}\leq 1$, $\sum_{i=1}^{n}\lambda_{i}=1$, then for any $b_{1},b_{2},\ldots,b_{n}\in\in\mathcal{E}(H)$ we have*

[TABLE]

(3)* $\mathcal{I}_{I}^{*}\circ\mathcal{I}_{a}^{*}=\mathcal{I}_{a}^{*}\circ\mathcal{I}_{I}^{*}=\mathcal{I}_{a}^{*}$ for all $a\in\mathcal{E}(H)$.

(4) If $a\circ b=0$, then $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}=\mathcal{I}_{b}^{*}\circ\mathcal{I}_{a}^{*}$.

(5) If $ab=ba$, then $\mathcal{I}_{a}^{*}\circ(\mathcal{I}_{b}^{*}\circ\mathcal{I}_{c}^{*})=(\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*})\circ\mathcal{I}_{c}^{*}$.

(6) If $ac=ca$ and $bc=cb$ then $\mathcal{I}_{c}^{*}\circ(\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*})=(\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*})\circ\mathcal{I}_{c}^{*}$ and

$\mathcal{I}_{c}^{*}\circ(\mathcal{I}_{a}^{*}+\mathcal{I}_{b}^{*})=(\mathcal{I}_{a}^{*}+\mathcal{I}_{b}^{*})\circ\mathcal{I}_{c}^{*}$ when $a\perp b$.

(7) $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}\leq\mathcal{I}_{a}^{*}$

(8) If $a\leq b$, then $\mathcal{I}_{c}^{*}\circ\mathcal{I}_{a}^{*}\leq\mathcal{I}_{c}^{*}\circ\mathcal{I}_{b}^{*}$ for every $c\in\mathcal{E}(H)$.*

The distribution of $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}$ in the state $\rho$ becomes

[TABLE]

Example 4.  For the Holevo instrument $\mathcal{H}_{(\alpha,A)}$ we have

[TABLE]

Not only is $(\mathcal{H}_{(\alpha,A)}^{*})_{a}\circ(\mathcal{H}_{(\alpha,A)}^{*})_{b}\leq(\mathcal{H}_{(\alpha,A)}^{*})_{a}$ as in Theorem 3.6(7) but

[TABLE]

for a constant $\lambda$. Writing $\mathcal{I}=\mathcal{H}_{(\alpha,A)}$, since $\mathcal{I}(\Delta)(\rho)=tr\left[\rho A(\Delta)\right]$ it follows from (3) that the distribution of $\mathcal{I}_{a}^{*}\circ\mathcal{I}_{b}^{*}$ is

[TABLE]

which is a constant times the distribution of $\mathcal{I}$. If $\mathcal{I}_{\alpha}(\Delta)=\mathcal{I}(\Delta)\alpha$ is a constant state instrument, we obtain

[TABLE]

In particular, if $\mathcal{I}=\mathcal{H}_{(\beta,A)}$ is a Holevo instrument, then

[TABLE]

4 Sequential Products of Instruments

Let $\mathcal{I},\mathcal{J}\in In(H)$ with outcome spaces $(\Omega_{\mathcal{I}},\mathcal{F}_{\mathcal{I}})$, $(\Omega_{\mathcal{J}},\mathcal{F}_{\mathcal{J}})$, respectively. We define the sequential product of $\mathcal{I}$ then $\mathcal{J}$ to be the instrument [7] with outcome space $(\Omega_{\mathcal{I}}\times\Omega_{\mathcal{J}},\mathcal{F}_{\mathcal{I}}\times\mathcal{F}_{\mathcal{J}})$ that satisfies

[TABLE]

for all $\Delta\in\mathcal{F}_{\mathcal{I}}$, $\Gamma\in\mathcal{F}_{\mathcal{J}}$, $\rho\in\mathcal{S}(H)$. We also define $\mathcal{J}$ conditioned by $\mathcal{I}$ to be the instrument give by [7].

[TABLE]

Theorem 4.1**.**

(1)* For all $a\in\mathcal{E}(H)$ we have*

[TABLE]

(2)* The observable measured by $\mathcal{I}\circ\mathcal{J}$ satisfies*

[TABLE]

(3)* For all $a\in\mathcal{E}(H)$ we have*

[TABLE]

(4)* The observable measured by $(\mathcal{J}\mid\mathcal{I})$ satisfies*

[TABLE]

Proof.

(1) For all $\rho\in\mathcal{S}(H)$, $a\in\mathcal{E}(H)$ we have

[TABLE]

Hence,

[TABLE]

It follows that

[TABLE]

(2) It follows from (1) that

[TABLE]

(3) Since

[TABLE]

We conclude that

[TABLE]

(4) It follows from (3) that

[TABLE]

Example 5.  Let $\mathcal{I}=\mathcal{H}_{(\alpha,A)}$, $\mathcal{J}=\mathcal{H}_{(\beta,B)}$ be Holevo instruments so that $\mathcal{I}_{a}^{*}(\Delta)=tr(\alpha a)A(\Delta)$ and $\mathcal{J}_{a}^{*}(\Gamma)=tr(\beta a)B(\Gamma)$. We then obtain

[TABLE]

For all $a\in\mathcal{E}(H)$ we have

[TABLE]

The observable measured by $\mathcal{I}\circ\mathcal{J}$ satisfies

[TABLE]

The instrument $\mathcal{J}$ conditioned by $\mathcal{I}$ becomes

[TABLE]

We then obtain

[TABLE]

The observable measured by $(\mathcal{J}\mid\mathcal{I})$ is $(\mathcal{J}\mid\mathcal{I})_{I}^{*}(\Gamma)=tr\left[\alpha B(\Gamma)\right]I\hfill\square$

Example 6.  Let $\mathcal{I}_{x}(\rho)=a_{x}\circ\rho$, $\mathcal{J}_{y}(\rho)=b_{y}\circ\rho$ be Lüders instruments. We then have

[TABLE]

The dual instruments satisfying $\mathcal{I}_{x}^{*}(a)=a_{x}\circ a$, $\mathcal{J}_{y}^{*}(a)=b_{y}\circ a$ and we obtain

[TABLE]

The observable measured by $\mathcal{I}\circ\mathcal{J}$ becomes

[TABLE]

which is the standard sequential product of the observable $A=\left\{a_{x}\colon x\in\Omega_{A}\right\}$ and $B=\left\{b_{y}\colon y\in\Omega_{B}\right\}$ [6, 7]. The instrument $\mathcal{J}$ conditioned by $\mathcal{I}$ becomes

[TABLE]

We then obtain

[TABLE]

The observable measured by $(\mathcal{J}\mid\mathcal{I}$ is

[TABLE]

which is again of the standard form [6, 7].                                            $\square$

Example 7.  Let $\mathcal{I}_{\alpha}$, $\mathcal{J}_{\beta}$ be constant-state instruments so that

[TABLE]

The sequential product becomes

[TABLE]

The sub-observables determined by $\mathcal{I}_{\alpha}\circ\mathcal{J}_{\beta}$ are given by

[TABLE]

The observable measured by $\mathcal{I}_{\alpha}\circ\mathcal{J}_{\beta}$ is

[TABLE]

The instrument $\mathcal{J}_{\beta}$ conditioned by $\mathcal{I}_{\alpha}$ satisfies

[TABLE]

so we conclude that $(\mathcal{J}_{\beta}\mid\mathcal{I}_{\alpha})=\mathcal{J}_{\beta}$. We have that

[TABLE]

and the observable measured by $(\mathcal{J}_{\beta}\mid\mathcal{I}_{\alpha})$ becomes

[TABLE]

So far we have considered examples of sequential products for two instruments of the same types. We now discuss sequential products of instruments of different types.

Example 8.  Let $\mathcal{I}$ be the Lüders instrument $\mathcal{I}_{x}(\rho)=a_{x}\circ\rho$ and $\mathcal{J}$ the finite Holevo instrument $\mathcal{J}_{y}(\rho)=tr\left[\rho B_{y}\right]\beta_{y}$. The sequential product becomes

[TABLE]

The sub-observables determined by $\mathcal{I}\circ\mathcal{J}$ are given by

[TABLE]

As in Example 6, we obtain

[TABLE]

which is the standard sequential product of the observables $A=\left\{a_{x}\colon x\in\Omega_{A}\right\}$ and $B$. Letting

[TABLE]

we obtain

[TABLE]

We have that

[TABLE]

The observable measured by $(\mathcal{J}\mid\mathcal{I})$ becomes

[TABLE]

We now consider the other order. These are given by the following equations.

[TABLE]