Real-Valued Observables and Quantum Uncertainty

Abstract

We first present a generalization of the Robertson-Heisenberg uncertainty principle. This generalization applies to mixed states and contains a covariance term. For faithful states, we characterize when the uncertainty inequality is an equality. We next present an uncertainty principle version for real-valued observables. Sharp versions and conjugates of real-valued observables are considered. The theory is illustrated with examples of dichotomic observables. We close with a discussion of real-valued coarse graining.

1 Introduction

One of the basic principles of quantum theory is the Robertson-Heisenberg uncertainty inequality [4, 7]

$$\begin{aligned} \Delta _\psi (A)\Delta _\psi (B)\ge \frac{1}{4}|{\langle {\psi ,[{A,B}]\psi }\rangle }|^2 \end{aligned}$$

(1.1)

where A, B are self-adjoint operators and $\psi$ is a vector state on a Hilbert space. The inequality (1.1) is usually applied to position and momentum operators A, B in which case $|{\langle {\psi ,[{A,B}]\psi }\rangle }|^2=\hbar ^2$ where $\hbar$ is Planck’s constant. In this situation, A and B are unbounded operators, but for mathematical rigor we shall only deal with bounded operators. However, our results can be extended to the unbounded case by considering a dense subspace common to the domains of A and B. In this paper, we derive a generalization of (1.1). This generalization applies to mixed states and contains an additional covariance term that results in a stronger inequality.

The main result in Section 2 is an uncertainty principle for observable operators. This principle contains four parts: a commutator term, a covariance term, a correlation term and a product of variances term. This last term is sometimes called a product of uncertainties. In Section 2 we also characterize, for faithful states, when the uncertainty inequality is an equality. Section 3 introduces the concept of a real-valued observable. If $\rho$ is a state and A is a real-valued observable, we define the $\rho$-average, $\rho$-deviation and $\rho$-variance of A. If B is another real-valued observable, we define the $\rho$-correlation and $\rho$-covariance of A, B. An uncertainty principle for real-valued observables is given in terms of these concepts. An important role is played by the stochastic operator $\widetilde{A}$ for A. In Section 3 we also define the sharp version of a real-valued observable and characterize when two real-valued observables have the same sharp version

Section 4 illustrates the theory presented in Section 3 with two examples. The first example considers two dichotomic arbitrary real-valued observables. The second example considers the special case of two noisy spin observables. In this case, the uncertainty inequality becomes very simple. Section 5 discusses real-values coarse graining of observables.

2 Quantum Uncertainty Principle

For a complex Hilbert space H, we denote the set of bounded linear operators by $\mathcal {L}(H)$ and the set of bounded self-adjoint operators by $\mathcal {L}_S(H)$. A positive trace-class operator with trace one is a state and the set of states on H is denoted by $\mathcal {S}(H)$. A state $\rho$ is faithful if $\mathrm {tr\,} (\rho C^*C)=0$ for $C\in \mathcal {L} (H)$ implies that $C=0$. For $\rho \in \mathcal {S} (H)$ and $C,D\in \mathcal {L} (H)$ we define the sesquilinear form $\langle {C,D}\rangle _\rho =\mathrm {tr\,} (\rho C^*D)$.

Lemma 2.1

(i) If $C\in \mathcal {L} (H)$, $\rho \in \mathcal {S} (H)$, then $\mathrm {tr\,} (\rho C^*)=\overline{\mathrm {tr\,} (\rho C)}$. (ii) The form $\langle {\bullet ,\bullet }\rangle _\rho$ is a positive semi-definite inner product. (iii) A state $\rho$ is faithful if and only if $\langle {\bullet ,\bullet }\rangle _\rho$ is an inner product

Proof

(i) If D is a trace-class operator and $\left\{ \phi _i\right\}$ is an orthonormal basis for H, we have

$$\begin{aligned} \mathrm {tr\,} (D^*)=\sum \limits _i\langle {\phi _i,D^*\phi _i}\rangle =\sum \limits _i\overline{\langle {D^*\phi _i,\phi _i}\rangle }=\sum \limits _i\overline{\langle {\phi _i,D\phi _i}\rangle } =\overline{\mathrm {tr\,} (D)} \end{aligned}$$

Hence,

$$\begin{aligned} \mathrm {tr\,} (\rho C^*)=\mathrm {tr\,}\left[ {(C\rho )^*}\right] =\overline{\mathrm {tr\,} (C\rho )}=\overline{\mathrm {tr\,} (\rho C)} \end{aligned}$$

(ii) Applying (i), we have

$$\begin{aligned} \overline{\langle {C,D}\rangle _\rho }=\overline{\mathrm {tr\,} (\rho C^*D)}=\mathrm {tr\,}\left[ {\rho (C^*D)^*}\right] =\mathrm {tr\,} (\rho D^*C)=\langle {D,C}\rangle _\rho \end{aligned}$$

Moreover, since $C^*C\ge 0$ we have $\langle {C,C}\rangle _\rho =\mathrm {tr\,} (\rho C^*C)\ge 0$. Hence, $\langle {\bullet ,\bullet }\rangle _\rho$ is a positive semi-definite inner product. (iii) If $\langle {\bullet ,\bullet }\rangle _\rho$ is an inner product, then

$$\begin{aligned} \langle {C,C}\rangle _\rho =\mathrm {tr\,} (\rho C^*C)=0 \end{aligned}$$

implies $C=0$ so $\rho$ is faithful. Conversely, if $\rho$ is faithful, then

$$\begin{aligned} \mathrm {tr\,} (\rho C^*C)=\langle {C,C}\rangle _\rho =0 \end{aligned}$$

implies $C=0$ so $\langle {\bullet ,\bullet }\rangle _\rho$ is an inner product $\square$

For $A\in \mathcal {L}_S(H)$ and $\rho \in \mathcal {S} (H)$, the $\rho$-average (or $\rho$-expectation) of A is $\langle {A}\rangle _\rho =\mathrm {tr\,} (\rho A)$ and $\rho$-deviation of A is $D_\rho (A)=A-\langle {A}\rangle _\rho I$ where I is the identity map on H. If $A,B\in \mathcal {L}_S(H)$, the $\rho$-correlation of A, B is

$$\begin{aligned} \textrm{Cor}_\rho (A,B)=\mathrm {tr\,}\left[ {\rho D_\rho (A)D_\rho (B)}\right] \end{aligned}$$

Although $\textrm{Cor}_\rho (A,B)$ need not be a real number, it is easy to check that $\overline{\textrm{Cor}_\rho (A,B)}=\textrm{Cor}_\rho (B,A)$. We say that A and B are uncorrelated if $\textrm{Cor}_\rho (A,B)=0$. The $\rho$-covariance of A, B is $\Delta _\rho (A,B)=\mathrm {Re\,}\textrm{Cor}_\rho (A,B)$ and the $\rho$-variance of A is

$$\begin{aligned} \Delta _\rho (A)=\Delta _\rho (A,A)=\textrm{Cor}_\rho (A,A)=\mathrm {tr\,}\left[ {\rho D_\rho (A)^2}\right] \end{aligned}$$

It is straightforward to show that

$$\begin{aligned} \textrm{Cor}_\rho (A,B)= & {} \mathrm {tr\,} (\rho AB)-\langle {A}\rangle _\rho \langle {B}\rangle _\rho \end{aligned}$$

(2.1)

$$\begin{aligned} \Delta _\rho (A,B)= & {} \mathrm {Re\,}\mathrm {tr\,} (\rho AB)-\langle {A}\rangle _\rho \langle {B}\rangle _\rho \end{aligned}$$

(2.2)

$$\begin{aligned} \Delta _\rho (A)= & {} \langle {A^2}\rangle _\rho -\langle {A}\rangle _\rho ^2 \end{aligned}$$

(2.3)

We see from (2.1) that A and B are $\rho$-uncorrelated if and only if $\mathrm {tr\,} (\rho AB)=\langle {A}\rangle _\rho \langle {B}\rangle _\rho$. We say that A and B commute if their commutant $[{A,B}]=AB-BA=0$.

Example 1

In the tensor product $H_1\otimes H_2$ let $\rho =\rho _1\otimes \rho _2\in \mathcal {S} (H_1\otimes H_2)$ be a product state and let $A_1\in \mathcal {L}_S(H_1)$, $A_2\in \mathcal {L}_S(H_2)$. Then $A=A_1\otimes I_2$, $B=I_1\otimes A_2\in \mathcal {L}_S(H_1\otimes H_2)$ are $\rho$-uncorrelated because

$$\begin{aligned} \mathrm {tr\,} (\rho AB)= & {} \mathrm {tr\,}\left[ {\rho _1\otimes \rho _2(A_1\otimes I_2)(I_2\otimes A_2)}\right] =\mathrm {tr\,}\left[ {\rho _1\otimes \rho _2(A_1\otimes A_2)}\right] \\= & {} \mathrm {tr\,} (\rho _1A_1\otimes \rho _2A_2)=\mathrm {tr\,} (\rho _1A_1)\mathrm {tr\,} (\rho _2A_2)\\= & {} \mathrm {tr\,} (\rho _1\otimes \rho _2A_1\otimes I_2)\mathrm {tr\,} (\rho _1\otimes \rho _2I_1\otimes A_2)=\langle {A}\rangle _\rho \langle {B}\rangle _\rho \end{aligned}$$

This shows that A, B are $\rho$-uncorrelated for any product state $\rho$. Of course, $[{A,B}]=0$ in this case. However, there are examples of noncommuting operators that are uncorrelated. For instance, on $H=\mathbb {C}^2$ let $\alpha =\left[ \begin{array}{c} 1\\ 0\end{array}\right]$, $\phi =\left[ \begin{array}{c} 0\\ 1\end{array}\right]$, $\psi =\frac{1}{\sqrt{2}}\left[ \begin{array}{c} 1\\ 1\end{array}\right]$. With $\rho =|{\alpha }\rangle \langle {\alpha }|$, $A=|{\phi }\rangle \langle {\phi }|$, $B=|{\psi }\rangle \langle {\psi }|$ we have

$$\begin{aligned} \mathrm {tr\,} (\rho AB)=\langle {A}\rangle _\rho \langle {B}\rangle _\rho =0 \end{aligned}$$

Hence, A, B are $\rho$-uncorrelated. However,

$$\begin{aligned} AB&=\langle {\phi ,\psi }\rangle |{\phi }\rangle \langle {\psi }|=\frac{1}{\sqrt{2}}|{\phi }\rangle \langle {\psi }|\\ BA&=\langle {\psi ,\phi }\rangle |{\psi }\rangle \langle {\phi }|=\frac{1}{\sqrt{2}}|{\psi }\rangle \langle {\phi }| \end{aligned}$$

so $[{A,B}]\ne 0$. $\square$

We now present our main result.

Theorem 2.2

If $A,B\in \mathcal {L}_S(H)$ and $\rho \in \mathcal {S} (H)$, then (i) $\frac{1}{4}|{\mathrm {tr\,}\left( {\rho [{A,B}]}\right) }|^2+\left[ {\Delta _\rho (A,B)}\right] ^2=|{\textrm{Cor}_\rho (A,B)}|^2$

(ii) $\frac{1}{4}|{\mathrm {tr\,}\left( {\rho [{A,B}]}\right) }|^2+\left[ {\Delta _\rho (A,B)}\right] ^2\le \Delta _\rho (A)\Delta _\rho (B)$

Proof

(i) Applying Lemma 2.1 we have

$$\begin{aligned} \mathrm {tr\,}\left( {[{A,B}]}\right)= & {} \mathrm {tr\,} (\rho AB)-\mathrm {tr\,} (\rho BA)=\mathrm {tr\,} (\rho AB)-\overline{\mathrm {tr\,}\left[ {\rho (BA)^*}\right] }\nonumber \\= & {} \mathrm {tr\,} (\rho AB)-\overline{\mathrm {tr\,}(\rho A^*B^*)}=\mathrm {tr\,} (\rho AB)-\overline{\mathrm {tr\,} (\rho AB)}\nonumber \\= & {} 2i\,\mathrm {Im\,}\left[ {\mathrm {tr\,} (\rho AB)}\right] \end{aligned}$$

(2.4)

From (2.2) and (2.4) we obtain

$$\begin{aligned} \frac{1}{4}|{\mathrm {tr\,}\left( {\rho [{A,B}]}\right) }|^2+\left[ {\Delta _\rho (A,B)}\right] ^2= & {} \left[ {\mathrm {Im\,} (\rho AB)}\right] ^2+\left[ {\mathrm {Re\,}\mathrm {tr\,} (\rho A B)-\langle {A}\rangle _\rho \langle {B}\rangle _\rho }\right] ^2\\= & {} |{\mathrm {Re\,}\mathrm {tr\,} (\rho AB)-\langle {A}\rangle _\rho \langle {B}\rangle _\rho +i\,\mathrm {Im\,}\mathrm {tr\,} (\rho AB)}|^2\\= & {} |{\mathrm {tr\,} (\rho AB)-\langle {A}\rangle )_\rho \langle {B}\rangle _\rho }|^2=|{\textrm{Cor}_\rho (A,B)}|^2 \end{aligned}$$

(ii) Applying Lemma 2.1(ii), the form $\langle {C,D}\rangle _\rho =\mathrm {tr\,}(\rho C^*D)$ is a positive semi-definite inner product. Hence, Schwarz’s inequality holds and we have

$$\begin{aligned} |{\textrm{Cor}_\rho (A,B)}|^2= & {} |{\mathrm {tr\,}\left[ {\rho D_\rho (A)D_\rho (B)}\right] }|^2=|{\langle {D_\rho (A),D_\rho (B)}\rangle _\rho }|^2\\\le & {} \langle {D_\rho (A),D_\rho (A)}\rangle _\rho \langle {D_\rho (B),D_\rho (B)}\rangle _\rho =\mathrm {tr\,}\left[ {\rho D_\rho (A)^2}\right] \mathrm {tr\,}\left[ {\rho D_\rho (B)^2}\right] \\= & {} \Delta _\rho (A)\Delta _\rho (B) \end{aligned}$$

$\square$

We call Theorem 2.2(i) the uncertainty equation and Theorem 2.2(ii) the uncertainty inequality. Together, they are called the uncertainty principle. Notice that Theorem 2.2(ii) is a considerable strengthening of the usual Robertson-Heisenberg inequality (1.1) since it contains the term $\left[ {\Delta _\rho (A,B)}\right] ^2$ and it applies to arbitrary states. Thus, even when $[{A,B}]=0$ we still have an uncertainty relation

$$\begin{aligned} \left[ {\Delta _\rho (A,B)}\right] ^2=|{\mathrm {tr\,}\left[ {\rho D_\rho (A)D_\rho (B)}\right] }|^2\le \Delta _\rho (A)\Delta _\rho (B) \end{aligned}$$

Lemma 2.3

A state $\rho$ is faithful if and only if the eigenvalues of $\rho$ are positive.

Proof

Suppose the eigenvalues $\lambda _i$ of $\rho$ are positive with corresponding normalized eigenvectors $\phi _i$. Then we can write $\rho =\sum \lambda _i|{\phi _i}\rangle \langle {\phi _i}|$ for the orthonormal basis $\left\{ \phi _i\right\}$. For any $A\in \mathcal {L} (H)$ we obtain

$$\begin{aligned} \mathrm {tr\,} (\rho A^*A)=\sum \lambda _i\mathrm {tr\,}\left( {|{\phi _i}\rangle \langle {\phi _i}|A^*A}\right) =\sum \lambda _i\langle {A\phi _i,A\phi _i}\rangle =\sum \lambda _i\Vert {A\phi _i}\Vert ^2 \end{aligned}$$

Hence, $\mathrm {tr\,} (\rho A^*A)=0$ implies $A\phi _i=0$ for all i. It follows that $A=0$. Conversely, if 0 is an eigenvalue of $\rho$ and $\phi$ is a corresponding unit eigenvector, then setting $P_\phi =|{\phi }\rangle \langle {\phi }|$ we have

$$\begin{aligned} \mathrm {tr\,} (\rho P_\phi ^*P_\phi )=\mathrm {tr\,} (\rho P_\phi )=\langle {\phi ,\rho \phi }\rangle =0 \end{aligned}$$

But $P_\phi \ne 0$ so $\rho$ is not faithful. $\square$

Theorem 2.4

If $\rho$ is faithful. then the following statements are equivalent. (i) The uncertainty inequality of Theorem 2.2(ii) is an equality. (ii) $D_\rho (B)=\alpha D_\rho (A)$ for $\alpha \in \mathbb {R}$. (iii) $B=\alpha A+\beta I$ for $\alpha ,\beta \in \mathbb {R}$. If one of the conditions holds, then

$$\begin{aligned} \left[ {\Delta _\rho (A,B)}\right] ^2=|{\textrm{Cor}_\rho (A,B)}|^2=\Delta _\rho (A)\Delta _\rho (B) \end{aligned}$$

(2.5)

Proof

(i)$\Rightarrow$(ii) If the uncertainty inequality is an equality, then

$$\begin{aligned} |{\mathrm {tr\,}\left[ {\rho D_\rho (A)D_\rho (B)}\right] }|^2=\Delta _\rho (A)\Delta _\rho (B) \end{aligned}$$

(2.6)

We can rewrite (2.6) as

$$\begin{aligned} |{\langle {D_\rho (A),D_\rho (B)}\rangle _\rho }|^2=\langle {D_\rho (A),D_\rho (A)}\rangle _\rho \langle {D_\rho (B),D_\rho (B)}\rangle _\rho \end{aligned}$$

Since we have equality in Schwarz’s inequality and $\langle {\bullet ,\bullet }\rangle _\rho$ is an inner product, it follows that $D_\rho (B)=\alpha D_\rho (A)$ for some $\alpha \in \mathbb {C}$. Since $D_\rho (B)^*=D_\rho (B)$ and $D_\rho (A)^*=D_\rho (A)$ we conclude that $\alpha \in \mathbb {R}$. (ii)$\Rightarrow$(iii) If $D_\rho (B)=\alpha D_\rho (A)$ for $\alpha \in \mathbb {R}$, we have

$$\begin{aligned} B-\langle {B}\rangle _\rho I=\alpha \left( {A-\langle {A}\rangle _\rho I)}\right) \end{aligned}$$

Hence, letting $\beta =\langle {B}\rangle _\rho -\alpha \langle {A}\rangle _\rho$ we have $B=\alpha A+\beta I$. Since $A,B\in \mathcal {L}_S(H)$ and $\alpha \in \mathbb {R}$, we have that $\beta \in \mathbb {R}$. (iii)$\Rightarrow$(i) If (iii) holds, then

$$\begin{aligned} \langle {B}\rangle _\rho =\mathrm {tr\,} (\rho B)=\alpha \mathrm {tr\,} (\rho A)+\beta =\alpha \langle {A}\rangle _\rho +\beta \end{aligned}$$

Hence, $\beta =\langle {B}\rangle _\rho -\alpha \langle {A}\rangle _\rho$ so that

$$\begin{aligned} D_\rho (B)= & {} B-\langle {B}\rangle _\rho I=\alpha A+\beta I-\langle {B}\rangle _\rho I\\= & {} \alpha A+\langle {B}\rangle _\rho I-\alpha \langle {A}\rangle _\rho I-\langle {B}\rangle _\rho I=\alpha D_\rho (A) \end{aligned}$$

Thus, (ii) holds and it follows that (2.6) holds and this implies (i). Equation (2.5) holds because (2.6) holds. $\square$

Example 2

The simplest faithful state when $\dim H=n<\infty$ is $\rho =I/n$. Then $\langle {A,B}\rangle _\rho =\frac{1}{n}\,\mathrm {tr\,} (A^*B)$ which is essentially the Hilbert-Schmidt inner product $\langle {A,B}\rangle _{HS}=\mathrm {tr\,} (A^*B)$. In this case for $A,B\in \mathcal {L}_S(H)$ we have $\langle {A}\rangle _\rho =\frac{1}{n}\,\mathrm {tr\,} (A)$, $D_\rho (A)=A-\frac{1}{n}\,\mathrm {tr\,} (A)I$. The other statistical concepts become:

$$\begin{aligned} \textrm{Cor}_\rho (A,B)= & {} \mathrm {tr\,}\left[ {\rho D_\rho (A)D_\rho (B)}\right] =\frac{1}{n}\,\mathrm {tr\,} (AB)-\frac{1}{ n^2}\,\mathrm {tr\,} (A)\mathrm {tr\,} (B)\\ \Delta _\rho (A,B)= & {} \frac{1}{n}\,\mathrm {Re\,}\mathrm {tr\,} (AB)-\frac{1}{n^2}\,\mathrm {tr\,} (A)\mathrm {tr\,} (B)\\ \Delta _\rho (A)= & {} \frac{1}{n}\,\mathrm {tr\,} (A^2)-\left[ {\frac{1}{n}\,\mathrm {tr\,} (A)}\right] ^2\\ \mathrm {tr\,}\left( {\rho \left[ {A,B}\right] }\right)= & {} \frac{2i}{n}\,\mathrm {Im\,}\mathrm {tr\,} (AB) \end{aligned}$$

The uncertainty principle is given by:

$$\begin{aligned} \left[ {\mathrm {Im\,}\mathrm {tr\,} (AB)}\right] ^2+ & {} \left[ {\mathrm {Re\,}\mathrm {tr\,}(AB)-\frac{1}{n}\,\mathrm {tr\,} (A)\mathrm {tr\,} (B)}\right] ^2 =|{\mathrm {tr\,} (AB)-\frac{1}{n}\,\mathrm {tr\,} (A)\mathrm {tr\,} (B)}|^2\\\le & {} \left[ {\mathrm {tr\,} (A^2)-\frac{1}{n}\,\mathrm {tr\,} (A)^2}\right] \left[ {\mathrm {tr\,} (B^2)-\frac{1}{n}\,\mathrm {tr\,} (B)^2}\right] \square \end{aligned}$$

3 Real-Valued Observables

An effect is an operator $C\in \mathcal {L}_S(H)$ that satisfies $0\le C\le I$ [1, 4,5,6]. Effects are thought of as two outcomes yes-no measurements. When the result of measuring C is yes , we say that C occurs and when the result is no , then C does not occur. A real-valued observable is a finite set of effects $A=\left\{ A_x: x\in \Omega _A\right\}$ where $\sum \limits _{x\in \Omega _A}A_x=I$ and $\Omega _A\subseteq \mathbb {R}$ is the outcome space for A. The effect $A_x$ occurs when the result of measuring A is the outcome x. The condition $\sum \limits _{x\in \Omega _A}A_x=I$ specifies that one of the possible outcomes of A must occur. An observable is also called a positive operator-valued measure (POVM). We say A is sharp if $A_x$ is a projection for all $x\in \Omega _A$ and in this case, A is a projection-valued measure [4, 7]. Corresponding to A we have the stochastic operator $\widetilde{A}\in \mathcal {L} (H)$ given by $\widetilde{A} =\sum \limits _{x\in \Omega _A}xA_x$. Notice that we need A to be real-valued in order for $\widetilde{A}$ to exist.

We now apply the theory presented in Section 2 to real-valued observables. For $\rho \in \mathcal {S} (H)$, the $\rho$-average (or $\rho$-expectation) of A is defined by

$$\begin{aligned} \langle {A}\rangle _\rho =\langle {\widetilde{A}\,}\rangle _\rho =\mathrm {tr\,} (\rho \widetilde{A}\,)=\sum \limits _{x\in \Omega _A}x\mathrm {tr\,} (\rho A_x) \end{aligned}$$

(3.1)

We interpret $\mathrm {tr\,} (\rho A_x)$ as the probability that a measurement of A results in the outcome x when the system is in state $\rho$. Thus, (3.1) says that the $\rho$-average of A is the sum of its outcomes times the probabilities these outcomes occur. We define the $\rho$-deviation of A by

$$\begin{aligned} D_\rho (A)= & {} D_\rho (\widetilde{A}\,)=\widetilde{A} -\langle {A}\rangle _\rho I=\sum \limits _{x\in \Omega _A}xA_x-\sum \limits _{x\in \Omega _A}x\mathrm {tr\,} (\rho A_x)I\\= & {} \sum \limits _{x\in \Omega _A}x\left[ {A_x-\mathrm {tr\,} (\rho A_x)I}\right] \end{aligned}$$

If A, B are real-valued observables, the $\rho$-correlation of A, B is $\textrm{Cor}_\rho (A,B)=\textrm{Cor}_\rho (\widetilde{A} ,\widetilde{B}\,)$, $\rho$-covariance of A, B is $\Delta _\rho (A,B)=\Delta _\rho (\widetilde{A} ,\widetilde{B}\,)$ and the $\rho$-variance of A is $\Delta _\rho (A)=\Delta _\rho (\widetilde{A}\,)$. Applying (2.1) we obtain

$$\begin{aligned} \textrm{Cor}_\rho (A,B)= & {} \mathrm {tr\,} (\rho \widetilde{A}\widetilde{B}\,)-\langle {\widetilde{A}\,}\rangle _\rho \langle {\widetilde{B}\,}\rangle _\rho =\mathrm {tr\,}\left( {\rho \sum \limits _{x,y}xyA_xB_y}\right) -\langle {\widetilde{A}\,}\rangle _\rho \langle {\widetilde{B}\,}\rangle _\rho \nonumber \\= & {} \sum \limits _{x,y}xy\left[ {\mathrm {tr\,} (\rho A_xB_y)-\mathrm {tr\,} (\rho A_x)\mathrm {tr\,} (\rho B_y)}\right] \end{aligned}$$

(3.2)

It follows that

$$\begin{aligned} \Delta _\rho (A,B)=\sum \limits _{x,y}xy\left[ {\mathrm {Re\,}\mathrm {tr\,} (\rho A_xB_y)-\mathrm {tr\,} (\rho A_x)\mathrm {tr\,} (\rho B_y)}\right] \end{aligned}$$

(3.3)

and

$$\begin{aligned} \Delta _\rho (A)=\sum \limits _{x,y}xy\left[ {\mathrm {tr\,} (\rho A_xA_y)-\mathrm {tr\,} (\rho A_x)\mathrm {tr\,} (\rho A_y)}\right] \end{aligned}$$

(3.4)

We also have by (2.4) that

$$\begin{aligned} \mathrm {tr\,}\left( {\rho \left[ {\widetilde{A} ,\widetilde{B}\,}\right] }\right)= & {} 2i\,\mathrm {Im\,}\mathrm {tr\,} (\rho \widetilde{A}\widetilde{B}\,)=2i\,\mathrm {Im\,}\mathrm {tr\,}\left( {\rho \sum \limits _{x,y}xyA_xB_y)}\right) \nonumber \\= & {} 2i\sum \limits _{x,y}xy\,\mathrm {Im\,}\mathrm {tr\,} (\rho A_xB_y) \end{aligned}$$

(3.5)

Substituting $\widetilde{A},\widetilde{B}$ for A, B in Theorem 2.2 gives an uncertainty principle for real-valued observables.

Two observables A, B are compatible (or jointly measurable) if there exists a joint observable $C_{(x,y)}$, $(x,y)\in \Omega _A\times \Omega _B$, such that $A_x=\sum \limits _yC_{(x,y)}$, $B_y=\sum \limits _xC_{(x,y)}$ for all $x\in \Omega _A$, $y\in \Omega _B$. If $\left[ {A_x,B_y}\right] =0$ for all x, y, then A, B are compatible with $C_{(x,y)}=A_xB_y$ for all $(x,y)\in \Omega _A\times \Omega _B$. However, if A, B are compatible, they need not commute [4]. If A, B are compatible real-valued observables, then

$$\begin{aligned} \widetilde{A}= & {} \sum \limits _xxA_x=\sum \limits _{x,y}xC_{(x,y)}\\ \widetilde{B}= & {} \sum \limits _yyB_y=\sum \limits _{x,y}yC_{(x,y)} \end{aligned}$$

Using (3.2), (3.3), (3.4), (3.5) we can write $\textrm{Cor}_\rho (A,B),\Delta _\rho (A,B),\Delta _\rho (A),\Delta _\rho (B)$ and $\mathrm {tr\,}\left( {\rho \left[ {\widetilde{A} ,\widetilde{B}\,}\right] }\right)$ in terms of $C_{(x,y)}$. Hence, we can express the uncertainty principle in terms of $C_{(x,y)}$.

If $A=\left\{ A_x: x\in \Omega _A\right\}$ is a real-valued observable, then $\widetilde{A}$ has spectral decomposition $\widetilde{A} =\sum \limits _{i=1}^n\lambda _iP_i$ where $\lambda _i\in \mathbb {R}$ are the distinct eigenvalues of $\widetilde{A}$ and $P_i$ are projections with $\sum P_i=I$. We call $\widehat{A} =\left\{ P_i: i=1,2,\ldots ,n\right\}$ the sharp version of A. Then $\widehat{A}$ is a real-valued observable with outcome space $\Omega _{\widehat{A}}=\left\{ \lambda _i: i=1,2,\ldots ,n\right\}$ and $P _{\lambda _i}=P_i$. Since $(\widehat{A}\,)^\sim =\widetilde{A}$, A and $\widehat{A}$ have the same stochastic operator. It follows that $\langle {A}\rangle _\rho =\langle {\widehat{A}\,}\rangle _\rho$, $\Delta _\rho (A)=\Delta _\rho (\widehat{A}\,)$ and if B is another real-valued observable, then $\textrm{Cor}_\rho (A,B)=\textrm{Cor}_\rho (\widehat{A} ,\widehat{B}\,)$ and $\Delta _\rho (A,B)=\Delta _\rho (\widehat{A} ,\widehat{B}\,)$.

Lemma 3.1

The following statements are equivalent. (i) $\widehat{A} =\widehat{B}$. (ii) $\widetilde{A} =\widetilde{B}$. (iii) $\langle {A}\rangle _\rho =\langle {B}\rangle _\rho$ for all $\rho \in \mathcal {S} (H)$.

Proof

(i)$\Rightarrow$(ii) If $\widehat{A} =\widehat{B}$ then

$$\begin{aligned} \widetilde{A} =(\widehat{A}\,)^\sim =(\widehat{B}\,)^\sim =\widetilde{B} \end{aligned}$$

(ii)$\Rightarrow$(iii) If $\widetilde{A} =\widetilde{B}$ then

$$\begin{aligned} \langle {A}\rangle _\rho =\langle {\widetilde{A}\,}\rangle _\rho =\langle {\widetilde{B}\,}\rangle _\rho =\langle {B}\rangle _\rho \end{aligned}$$

(iii)$\Rightarrow$(i) If $\langle {A}\rangle _\rho =\langle {B}\rangle _\rho$ for all $\rho \in \mathcal {S} (H)$, then $\langle {\widetilde{A}\,}\rangle _\rho =\langle {\widetilde{B}\,}\rangle _\rho$ for all $\rho \in \mathcal {S} (H)$. It follows that $\widehat{A} =\widehat{B}$. $\square$

Let $\widetilde{A} =\sum xA_x=\sum \lambda _iP_i$ so $\widehat{A} =\left\{ P_i: i=1,2,\ldots ,n\right\}$ is a sharp version of A. Let $B=\left\{ B_x: x\in \Omega _A\right\}$ be the real-valued observable given by $B_x=\sum \limits _{i=1}^nP_iA_xP_i$. We conclude that A and B have the same sharp version because

$$\begin{aligned} \widetilde{B}= & {} \sum \limits _xxB_x=\sum \limits _iP_i\sum \limits _xxA_xP_i=\sum \limits _iP_i\widetilde{A} P_i=\sum \limits _iP_i\sum \limits _j\lambda _jP_jP_i\\= & {} \sum \limits _{i,j}\lambda _iP_iP_jP_i=\sum \limits _i\lambda _iP_i=\widetilde{A} \end{aligned}$$

so by Lemma 3.1, $\widehat{A} =\widehat{B}$. We say that B is a conjugate of A. Letting $C_{ix}=P_iA_xP_i$, we have that

$$\begin{aligned} \left\{ C_{ix}: i=1,2,\ldots ,n, x\in \Omega _A\right\} \end{aligned}$$

is an observable and $\sum \limits _iC_{ix}=B_x$, $\sum \limits _xC_{ix}=P_i$. It follows that B and $\widehat{A}$ are compatible with joint observable $\left\{ C_{ix}\right\}$. We say that an observable $A=\left\{ A_x: x\in \Omega _A\right\}$ is commutative if $\left[ {A_x,A_y}\right] =0$ for all $x,y\in \Omega _A$. Notice that if A is sharp, then A is commutative. However, there are many unsharp observables that are commutative.

Theorem 3.2

If A is commutative, then B is conjugate to A if and only if $B=A$.

Proof

If A is commutative, we show that A is conjugate to A. Since

$$\begin{aligned} \widehat{A} =\sum xA_x=\sum \lambda _iP_i \end{aligned}$$

we have that $\left[ {\widehat{A} ,A_x}\right] =0$ for all $x\in \Omega _A$. By the spectral theorem, $\left[ {A_x,P_i}\right] =0$ for all x, i so $A_x=\sum P_iA_xP_i$. Therefore, A is conjugate to A. Conversely, suppose A is commutative and B is conjugate to A. Then $B_x=\sum \limits _iP_iA_xP_i$ for all $x\in \Omega _A$. As before, we have that $\left[ {\widehat{A} _x,A_x}\right] =0$ for all $x\in \Omega _A$ so $\left[ {A_x,P_i}\right] =0$ for all x, i. Hence,

$$\begin{aligned} B_x=\sum \limits _iP_iA_xP_i=A_x\sum \limits _iP_i=A_x \end{aligned}$$

for all $x\in \Omega _B=\Omega _A$ so $B=A$. $\square$

Thus, nontrivial conjugates only occur in the nonclassical case where A is noncommutative.

4 More Examples

This section illustrates the theory in Sections 2 and 3 with two examples.

Example 3

A two outcome observable is called a dichotomic observable. Of course, a dichotomic observable is commutative but it need not be sharp. Let $A=\left\{ A_1,I-A_1\right\}$ be a dichotomic observable with $\Omega _A=\left\{ 1,-1\right\}$. Then

$$\begin{aligned} \widetilde{A}= & {} A_1-(I-A_1)=2A_1-I\\ \langle {A}\rangle _\rho= & {} \mathrm {tr\,} (\rho \widetilde{A}\,)=\mathrm {tr\,}\left[ {\rho (2A_1-I)}\right] =2\,\mathrm {tr\,} (\rho A_1)-1\\ D_\rho (A)= & {} \widetilde{A} -\langle {A}\rangle _\rho I=2A_1-I-2\,\mathrm {tr\,} (\rho A_1)I+I=2\left[ {A_1-\mathrm {tr\,} (\rho A_1)I}\right] \end{aligned}$$

If $B=\left\{ B_1,I-B_1\right\}$ is another dichotomic observable with $\Omega _B=\left\{ 1,-1\right\}$, then

$$\begin{aligned} \textrm{Cor}_\rho (A,B)= & {} \mathrm {tr\,} (\rho \widetilde{A}\widetilde{B}\,)-\langle {A}\rangle _\rho \langle {B}\rangle _\rho \nonumber \\= & {} \mathrm {tr\,}\left[ {\rho (2A_1-I)(2B_1-I)}\right] -\left[ {2\,\mathrm {tr\,} (\rho A_1-1)}\right] \left[ {2\,\mathrm {tr\,} (\rho B_1-1)}\right] \nonumber \\= & {} \mathrm {tr\,}\left[ {\rho (4A_1B_1-2A_1-2B_1+I)}\right] -4\,\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)\nonumber \\{} & {} \quad +2\,\mathrm {tr\,} (\rho A_1)+2\mathrm {tr\,} (\rho B_1)-1\nonumber \\= & {} 4\left[ {\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}\right] \end{aligned}$$

(4.1)

Hence,

$$\begin{aligned} \Delta _\rho (A,B)= & {} 4\left[ {\mathrm {Re\,}\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}\right] \end{aligned}$$

and

$$\begin{aligned} \Delta _\rho (A)= & {} \Delta _\rho (A,A)=4\left[ {\mathrm {tr\,} (\rho A_1^2)-\left( {\mathrm {tr\,} (\rho A_1)}\right) ^2}\right] \end{aligned}$$

We also have

$$\begin{aligned} \left[ {\widetilde{A} ,\widetilde{B}\,}\right]= & {} \left[ {2A_1-I,2B_1-I}\right] =(2A_1-I)(2B_1-I)-(2B_1-I)(2A_1-I)\\= & {} 4\left[ {A_1,B_1}\right] \end{aligned}$$

We conclude that $\left[ {\widetilde{A} ,\widetilde{B}\,}\right] =0$ if and only if $\left[ {A_1,B_1}\right] =0$ and this does not hold in general so $\widetilde{A},\widetilde{B}$ need not commute. The uncertainty principle becomes

$$\begin{aligned}{} & {} \left[ {\mathrm {Im\,}\mathrm {tr\,} (\rho A_1B_1)}\right] ^2+\left[ {\mathrm {Re\,}\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho A_2)}\right] ^2\nonumber \\{} & {} =|{\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}|^2\nonumber \\{} & {} \le \left[ {\mathrm {tr\,} (\rho A_1^2)-\left( {\mathrm {tr\,} (\rho A_1)}\right) ^2}\right] \left[ {\mathrm {tr\,} (\rho B_1^2)-\left( {\mathrm {tr\,} (\rho B_1)}\right) ^2}\right] \end{aligned}$$

(4.2)

$\square$

Example 4

We now consider a special case of Example 3. For $H\in \mathbb {C}^2$ we define the Pauli matrices

$$\begin{aligned} \sigma _x=\left[ \begin{array}{cc} 0&{}1\\ 1&{}0 \end{array}\right] , \quad \sigma _y=\left[ \begin{array}{cc} 0&{}i\\ -i&{}0 \end{array}\right] ,\quad \sigma _z=\left[ \begin{array}{cc} 1&{}0\\ 0&{}-1 \end{array}\right] \end{aligned}$$

Let $\mu \in \left[ {0,1}\right]$ and define the dichotomic observable $A=\left\{ A_1,I-A_1\right\}$, where

$$\begin{aligned} A_1=\frac{1}{2}(I+\mu \sigma _x)=\frac{1}{2}\left[ \begin{array}{cc} 1&{}\mu \\ \mu &{}1 \end{array}\right] \end{aligned}$$

and $\Omega _A=\left\{ 1,-1\right\}$. Similarly, let $B=\left\{ B_1,I-B_1\right\}$, where

$$\begin{aligned} B_1=\frac{1}{2}(I+\mu \sigma _y)=\frac{1}{2}\left[ \begin{array}{cc} 1&{}i\mu \\ -i\mu &{}1 \end{array}\right] \end{aligned}$$

and $\Omega _B=\left\{ 1,-1\right\}$. We call A and B noisy spin observables along the x and y directions, respectively, with noise parameter $1-\mu$ [7].

Any state $\rho \in \mathcal {S} (H)$ has the form $\rho =\frac{I}{2}(I+\overrightarrow{r}\bullet \overrightarrow{\sigma })$ where $\overrightarrow{r}\in \mathbb {R} ^3$ with $\Vert {\overrightarrow{r}}\Vert \le 1$ [1, 2]. This is called the Block sphere representation of $\rho$ [4, 7]. The eigenvalues of $\rho$ are $\lambda _{\pm } =\frac{1}{2}\left( {1\pm \Vert {\overrightarrow{r}}\Vert }\right)$. Then $\lambda _+=1$, $\lambda _-=0$ if and only if $\Vert {\overrightarrow{r}}\Vert =1$ and these are precisely the pure states. Letting $\sigma _1=\sigma _x$, $\sigma _2=\sigma _y$, $\sigma _3=\sigma _z$ we obtain

$$\begin{aligned} \rho= & {} \frac{1}{2}\left[ \begin{array}{cc} 1+r_3&{}r_1-ir_2\\ r_1+ir_2&{}1-r_3\end{array}\right] \end{aligned}$$

and

$$\begin{aligned} \rho A_1= & {} \frac{1}{4}\left[ \begin{array}{cc} 1+r_3&{}r_1-ir_2\\ r_1+ir_2&{}1-r_3\end{array}\right] \left[ \begin{array}{cc} 1&{}\mu \\ \mu &{}1\end{array}\right] \\= & {} \left[ \begin{array}{cc} 1+r_3+(r_1-ir_2)\mu &{}(1+r_3)\mu +r_1-ir_2\\ (1-r_3)\mu +r _1+ir_2&{}1-r_3+(r_1+ir_2)\mu \end{array}\right] \end{aligned}$$

Hence, $\mathrm {tr\,} (\rho A_1)=\frac{1}{2}(1+r_1\mu )$ and as in Example 3, $\langle {A}\rangle _\rho =r_1\mu$. Similarly, $\mathrm {tr\,} (\rho B_1)=\frac{1}{2}(1+r_2\mu )$ and $\langle {B}\rangle _\rho =r_2\mu$. We also obtain

$$\begin{aligned} \mathrm {tr\,} (\rho A_1B_1)=\frac{1}{4}\left[ {1+(r_1+r_2)\mu +ir_2\mu ^2}\right] \end{aligned}$$

and it follows from (4.1) that

$$\begin{aligned} \textrm{Cor}_\rho (A,B)= & {} 4\left[ {\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}\right] \\= & {} 1+(r_1+r_2)\mu +ir_3\mu ^2-(1+r_1\mu )(1+r_2\mu )=-r_1r_2\mu ^2+ir_3\mu ^2 \end{aligned}$$

Therefore, $\Delta _\rho (A,B)=-r_1r_2\mu ^2$. A straightforward calculation shows that

$$\begin{aligned} \mathrm {tr\,} (\rho A_1^2)= & {} \frac{1}{4}(1+\mu ^2)+\frac{1}{2}\mu r_1\\ \mathrm {tr\,} (\rho B_1^2)= & {} \frac{1}{4}(1+\mu ^2)+\frac{1}{2}\mu r_2 \end{aligned}$$

It follows that

$$\begin{aligned} \Delta _\rho (A)=4\left[ {\mathrm {tr\,} (\rho A_1^2)-\left( {\mathrm {tr\,} (\rho A_1)}\right) ^2}\right] =\mu ^2(1-r_1^2) \end{aligned}$$

and similarly, $\Delta _\rho (B)=\mu ^2(1-r_2^2)$.

The commutator term in (4.2) becomes

$$\begin{aligned} \left[ {\mathrm {Im\,}\mathrm {tr\,} (\rho A_1B_1)}\right] ^2=\frac{1}{16}\,r_3^2\mu ^4 \end{aligned}$$

The covariance term in (4.2) is

$$\begin{aligned} \left[ {\mathrm {Re\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}\right] ^2=\frac{1}{16}\,r_1^2r_2^2\mu ^4 \end{aligned}$$

and the correlation term in (4.2) is

$$\begin{aligned} |{\mathrm {tr\,} (\rho A_1B_1)-\mathrm {tr\,} (\rho A_1)\mathrm {tr\,} (\rho B_1)}|^2=\frac{1}{16}\,(r_3^2+r_1^2r_2^2)\mu ^4 \end{aligned}$$

Finally, the variance term in (4.2) is given by

$$\begin{aligned} \Delta _\rho (A_1)\Delta _\rho (B_1)=\frac{1}{16}\,(1-r_1^2)(1-r_2^2)\mu ^4 \end{aligned}$$

The inequality in (4.2) reduces to

$$\begin{aligned} \frac{1}{16}(r_3^2+r_1^2+r_2^2)\mu ^4\le \frac{1}{16}(1-r_1^2)(1-r_2^2)\mu ^4 \end{aligned}$$

(4.3)

If $\mu \ne 0$, (4.3) is equivalent to the inequality

$$\begin{aligned} \Vert {\overrightarrow{r}}\Vert ^2=r_1^2+r_2^2+r_3^2\le 1 \end{aligned}$$

If the commutator term vanishes and $\mu \ne 0$, the uncertainty inequality becomes

$$\begin{aligned} r_1^2r_2^2\le (1-r_1^2)(1-r_2^2) \end{aligned}$$

(4.4)

which is equivalent to $r_1^2+r_2^2\le 1$. If A and B are $\rho$-uncorrelated and $\mu \ne 0$, the uncertainty inequality becomes $r_3^2\le (1-r_1^2)(1-r_2^2)$ which is equivalent to $\Vert {\overrightarrow{r}}\Vert ^2\le 1+r_1^2r_2^2$. This inequality and (4.4) are weaker than (4.3). $\square$

5 Real-Valued Coarse Graining

Let $A=\left\{ A_x: x\in \Omega _A\right\}$ be an arbitrary observable. We assume that A is not necessarily real-valued so the outcome space $\Omega _A$ is an arbitrary finite set. For $f:\Omega _A\rightarrow \mathbb {R}$ with range $\mathcal {R} (f)$ we define the real-valued observable f(A) by $\Omega _{f(A)}=\mathcal {R} (f)$ and for all $z\in \Omega _{f(A)}$

$$\begin{aligned} f(A)_z=A_{f^{-1}(z)}=\sum \left\{ A_x: f(x)=z\right\} \end{aligned}$$

We call f(A) a real-valued coarse graining of A [2,3,4]. Then f(A) has stochastic operator

$$\begin{aligned} f(A)^\sim =\sum \limits _zzf(A)_z=\sum \limits _zzA_{f^{-1}(z)}=\sum \limits _z\sum \limits _{x\in f^{-1}(z)}zA_x=\sum \limits _xf(x)A_x \end{aligned}$$

It follows that $\langle {f(A)}\rangle _\rho =\sum \limits _xf(x)\mathrm {tr\,} (\rho A_x)$ for all $\rho \in \mathcal {S} (H)$. If B is another observable and $g:\Omega _B\rightarrow \mathbb {R}$ we have

$$\begin{aligned} \textrm{Cor}_\rho \left[ {f(A),g(B)}\right]= & {} \sum \limits _{x,y}f(x)g(y)\mathrm {tr\,} (\rho A_xB_y)-\langle {f(A)}\rangle _\rho \langle {g(B)}\rangle _\rho \\ \Delta _\rho \left[ {f(A),g(B)}\right]= & {} \sum \limits _{x,y}f(x)g(y)\mathrm {Re\,}\mathrm {tr\,} (\rho A_xB_y)-\langle {f(A)}\rangle _\rho \langle {g(B)}\rangle _\rho \\ \Delta _\rho \left[ {f(A)}\right]= & {} \sum \limits _{x,y}f(x)f(y)\mathrm {tr\,} (\rho A_xA_y)-\langle {f(A)}\rangle _\rho ^2 \end{aligned}$$

Moreover, we have the uncertainty inequality

$$\begin{aligned} |{\textrm{Cor}_\rho \left[ {f(A),g(B)}\right] }|^2\le \Delta _\rho \left[ {f(A)}\right] \Delta _\rho \left[ {g(B)}\right] \end{aligned}$$

We denote the set of trace-class operators on H by $\mathcal {T} (H)$. An operation on H is a completely positive, trace reducing, linear map $\mathcal {O}:\mathcal {T} (H)\rightarrow \mathcal {T} (H)$ [1,2,3,4]. If $\mathcal {O}$ preserves the trace, then $\mathcal {O}$ is called a channel. A (finite) instrument is a finite set of operators $\mathcal {I} =\left\{ \mathcal {I}_x: x\in \Omega _\mathcal {I}\right\}$ such that $\overline{\mathcal {I}} =\sum \left\{ \mathcal {I}_x: x\in \Omega _\mathcal {I}\right\}$ is a channel [1,2,3,4]. We say that $\mathcal {I}$ measures an observable A if $\Omega _\mathcal {I} =\Omega _A$ and $\mathrm {tr\,}\left[ {\mathcal {I}_x(\rho )}\right] =\mathrm {tr\,} (\rho A_x)$ for all $x\in \Omega _\mathcal {I}$. It can be shown that $\mathcal {I}$ measures a unique observable which we denote by $J(\mathcal {I})$ [2, 3]. Conversely, any observable is measured by many instruments [1,2,3,4]. Corresponding to an operation $\mathcal {O}$ we have its dual-operation $\mathcal {O} ^*:\mathcal {L} (H)\rightarrow \mathcal {L} (H)$ defined by $\mathrm {tr\,}\left[ {\rho \mathcal {O} ^*(C)}\right] =\mathrm {tr\,}\left[ {\mathcal {O} (\rho )C}\right]$ for all $\rho \in \mathcal {S} (H)$ [2, 3]. It can be shown that $J(\mathcal {I})_x=\mathcal {I}_x^*(I)$ for all $x\in \Omega _\mathcal {I}$ where I is the identity operator [2, 3].

As with observables, if $\mathcal {I}$ is an instrument, and $f:\Omega _\mathcal {I}\rightarrow \mathbb {R}$ we define the real-valued instrument $f(\mathcal {I})$ such that $\Omega _{f(\mathcal {I})}=\mathcal {R} (f)$ and

$$\begin{aligned} f(\mathcal {I})_z=\sum \left\{ \mathcal {I}_x: f(x)=z\right\} \end{aligned}$$

If $J(\mathcal {I})=A$, then $J\left[ {f(\mathcal {I})}\right] =f(A)$ because

$$\begin{aligned} \mathrm {tr\,}\left[ {f(\mathcal {I})_z(\rho )}\right]= & {} \mathrm {tr\,}\left[ {\sum \left\{ \mathcal {I}_x(\rho ): f(x)=z\right\} }\right] =\sum \left\{ \mathrm {tr\,}\left[ {\mathcal {I}_x(\rho )}\right] : f(x)=z\right\} \\= & {} \sum \left\{ \mathrm {tr\,} (\rho A_x): f(x)=z\right\} =\mathrm {tr\,}\left[ {\rho \sum \left\{ A_x: f(x)=z\right\} }\right] \\= & {} \mathrm {tr\,}\left[ {\rho f(A)_z}\right] \end{aligned}$$

for all $z\in \Omega _{f(A)}=\Omega _{f(\mathcal {I})}$. If $\mathcal {I}$ is real-valued, we define $\widetilde{\mathcal {I}}$ on $\mathcal {L} (H)$ by $\widetilde{\mathcal {I}}(C)=\sum x\mathcal {I}_x(C)$ and $\langle {\mathcal {I}}\rangle _\rho =\mathrm {tr\,}\left[ {\widetilde{\mathcal {I}} (\rho )}\right]$. If $J(\mathcal {I})=A$, then

$$\begin{aligned} \langle {\mathcal {I}}\rangle _\rho =\mathrm {tr\,}\left[ {\sum x\mathcal {I}_x(\rho )}\right] =\sum x\mathrm {tr\,}\!\left[ {\mathcal {I}_x(\rho )}\right] =\sum x\mathrm {tr\,} (\rho A_x) =\langle {A}\rangle _\rho \end{aligned}$$

for all $\rho \in \mathcal {S} (H)$. We also define $\Delta _\rho (\mathcal {I})=\Delta _\rho (A)$. It follows that $\langle {f(\mathcal {I})}\rangle _\rho =\langle {f(A)}\rangle _\rho$, $\Delta _\rho \left[ {f(\mathcal {I})}\right] =\Delta _\rho \left[ {f(A)}\right]$ and $f(\mathcal {I})^\sim =\sum f(x)\mathcal {I}_x$.

Let $A=\left\{ A_x: x\in \Omega _A\right\}$, $B=\left\{ B_y: y\in \Omega _B\right\}$ be arbitrary observables and suppose $\mathcal {I}$ is an instrument with $J(\mathcal {I})=A$. Define the $\mathcal {I}$-product observable $A\circ B$ with $\Omega _{A\circ B}=\Omega _A\times \Omega _B$ given by $(A\circ B)_{(x,y)}=\mathcal {I}_x(B_y)$ [2, 3]. Then $A\circ B$ is indeed an observable because

$$\begin{aligned} \sum \limits _{x,y}(A\circ B)_{(x,y)}=\sum \limits _{x,y}\mathcal {I}_x^*(B_y)=\sum \limits _x\mathcal {I}_x^*\left( {\sum \limits _yB_y}\right) =\sum \limits _x\mathcal {I}_x^*(I)=\sum \limits _xA_x=I \end{aligned}$$

Although $A\circ B$ depends on $\mathcal {I}$, we shall not indicate this for simplicity. We interpret $A\circ B$ as the observable obtained by first measuring A using $\mathcal {I}$ and then measuring B. If $f:\Omega _A\times \Omega _B\rightarrow \mathbb {R}$ we obtain the real-valued observable $f(A,B)=f(A\circ B)$. We then have

$$\begin{aligned} f(A,B)_z= & {} (A\circ B)_{f^{-1}(z)}=\sum \left\{ (A\circ B)_{(x,y)}: f(x,y)=z\right\} \\= & {} \sum \left\{ \mathcal {I}_x^*(B_y): f(x,y)=z\right\} \\ f(A,B)^\sim= & {} \sum \limits _{x,y}f(x,y)(A\circ B)_{(x,y)}=\sum \limits _{x,y}f(x,y)\mathcal {I}_x^*(B_y)\\ \langle {f(A,B)}\rangle _\rho= & {} \sum \limits _{x,y}f(x,y)\mathrm {tr\,}\left[ {\rho (A\circ B)_{(x,y)}}\right] =\sum \limits _{x,y}f(x,y)\mathrm {tr\,}\left[ {\rho \mathcal {I}_x^*(B_y)}\right] \\ \Delta _\rho \left[ {f(A,B)}\right]= & {} \sum \limits _{x,y,x',y'}f(x,y)f(x',y')\mathrm {tr\,}\left[ {\rho (A\circ B)_{(x,y)}(A\circ B)_{(x',y')}}\right] -\langle {f(A,B)}\rangle _\rho ^2\\= & {} \mathrm {tr\,}\left\{ \rho \left[ {\sum \limits _{x,y}f(x,y)\mathcal {I}_x^*(B_y)}\right] ^2\right\} -\langle {f(A,B)}\rangle _\rho ^2 \end{aligned}$$

If f is a product function $f(x,y)=g(x)h(y)$ we obtain

$$\begin{aligned} f(A,B)_z=\sum \limits _z\left\{ \mathcal {I}_x^*(B_y): g(x)h(y)=z\right\} \end{aligned}$$

We then have the simplification

$$\begin{aligned} f(A,B)^\sim= & {} \sum \limits _{x,y}g(x)h(y)\mathcal {I}_x^*(B_y)=\sum \limits _xg_x\mathcal {I}_x^*\left( {\sum \limits _yh(y)B_y}\right) \\= & {} \sum \limits _xg(x)\mathcal {I}_x^*\left[ {h(B)^\sim }\right] \\ \end{aligned}$$

Hence,

$$\begin{aligned} \langle {f(A,B)}\rangle _\rho= & {} \mathrm {tr\,}\left[ {\rho f(A,B)^\sim }\right] =\mathrm {tr\,}\left\{ \rho \sum \limits _xg(x)\mathcal {I}_x^*\left[ {h(B)^\sim }\right] \right\} \\= & {} \sum \limits _xg(x)\mathrm {tr\,}\left\{ \rho \mathcal {I}_x^*\left[ {h(B)^\sim }\right] \right\} =\sum \limits _xg(x)\mathrm {tr\,}\left\{ \mathcal {I}_x(\rho )\left[ {h(B)^\sim }\right] \right\} \\= & {} \mathrm {tr\,}\left\{ \sum \limits _xg(x)\mathcal {I}_x(\rho )\left[ {h(B)^\sim }\right] \right\} =\mathrm {tr\,}\left\{ g(\mathcal {I})^\sim (\rho )\left[ {h(B)^\sim }\right] \right\} \end{aligned}$$

In a similar way we obtain

$$\begin{aligned} \Delta _\rho \left[ {f(A,B)}\right] =\mathrm {tr\,}\left\{ \left( {g(\mathcal {I})^\sim (\rho )\left[ {h(B)^\sim }\right] }\right) ^2\right\} -\langle {f(A,B)}\rangle _\rho ^2 \end{aligned}$$

If A and B are arbitrary observables, we define the observable B conditioned by A to be

$$\begin{aligned} (B\mid A)_y=\mathcal {I}_{\Omega _A}^*(B_y)=\sum \limits _{x\in \Omega _A}\mathcal {I}_x^*(B_y) \end{aligned}$$

where $\Omega _{B\mid A}=\Omega _B$ [2, 3]. We interpret $(B\mid A)$ as the observable obtained by first measuring A without taking the outcome into account and then measuring B. If B is real-valued we have

$$\begin{aligned} (B\mid A)^\sim= & {} \sum \limits _yy(B\mid A)_y=\sum \limits _{x,y}y\mathcal {I}_x^*(B_y)=\mathcal {I}_{\Omega (A)}^*(\widetilde{B} )\\ \langle {(B\mid A)}\rangle _\rho= & {} \sum \limits _yy\mathrm {tr\,}\left[ {\rho \mathcal {I}_{\Omega (A)}^*(B_y)}\right] =\sum y\mathrm {tr\,}\left[ {\overline{\mathcal {I}} (\rho )B_y}\right] =\mathrm {tr\,}\left[ {\overline{\mathcal {I}} (\rho )\widetilde{B}\,}\right] =\langle {B}\rangle _{\overline{\mathcal {I}}_(\rho )}\\ \Delta _\rho \left[ {(B\mid A)}\right]= & {} \Delta _\rho \left[ {(B\mid A)^\sim }\right] =\Delta _\rho \left[ {\mathcal {I}_{\Omega (A)}^*(\widetilde{B} )}\right] =\mathrm {tr\,}\left\{ \left[ {\mathcal {I}_{\Omega (A)}^*(\widetilde{B} )}\right] ^2\right\} -\left[ {\langle {B}\rangle _{\overline{\mathcal {I}} (\rho )}}\right] ^2 \end{aligned}$$

We now illustrate the theory of this section with some examples.

Example 5

The simplest example of an instrument is a trivial instrument $\mathcal {I}_x(\rho )=\omega (x)\rho$ where $\omega$ is a probability measure on the finite set $\Omega _\mathcal {I}$. It is clear that $\mathcal {I}$ measures the trivial observable $A_x=\omega (x)I$. Let B be an arbitrary observable and let $f:\Omega _A\times \Omega _B\rightarrow \mathbb {R}$. We then have

$$\begin{aligned} (A\circ B)_{(x,y)}= & {} \mathcal {I}_x^*(B_y)=\omega (x)B_y\\ f(A,B)_z= & {} f(A\circ B)_z=\sum \left\{ \omega (x)B_y: f(x,y)=z\right\} \end{aligned}$$

We conclude that

$$\begin{aligned} f(A,B)^\sim= & {} \sum \limits _{x,y}f(x,y)\omega (x)B_y\\ \langle {f(A,B)}\rangle _\rho= & {} \sum \limits _{x,y}f(x,y)\omega (x)\mathrm {tr\,} (\rho B_y)\\ \Delta _\rho \left[ {f(A,B))}\right]= & {} \mathrm {tr\,}\left\{ \rho \left[ {\sum \limits _{x,y}f(x,y)\omega (x)B_y}\right] ^2\right\} -\langle {f(A,B)}\rangle _\rho ^2 \end{aligned}$$

Moreover, since

$$\begin{aligned} (B\mid A)_y=\sum \limits _x\mathcal {I}_x^*(B_y)=\sum \limits _x\omega (x)(B_y)=B_y \end{aligned}$$

we have that $(B\mid A)=B$. $\square$

Example 6

Let $A=\left\{ A_x: x\in \Omega _A\right\}$ and $B=\left\{ B_y: y\in \Omega _B\right\}$ be arbitrary observables and let $\mathcal {H}_x(\rho )=\mathrm {tr\,} (\rho A_x)\alpha _x$, $\alpha _x\in \mathcal {S} (H)$ be a Holevo instrument [2, 3]. Then $\mathcal {H}$ measure A because

$$\begin{aligned} \mathrm {tr\,}\left[ {\mathcal {H}_x(\rho )}\right] =\mathrm {tr\,}\left[ {\mathrm {tr\,} (\rho A_x)\alpha _x}\right] =\mathrm {tr\,} (\rho A_x) \end{aligned}$$

Since $\mathcal {H}_x^*(a)=\mathrm {tr\,} (\alpha _xa)A_x$ for all $x\in \Omega _A$ [2, 3], we have

$$\begin{aligned} (A\circ B)_{(x,y)}=\mathcal {H}_x^*(B_y)=\mathrm {tr\,} (\alpha _xB_y)A_x \end{aligned}$$

If $f:\Omega _A\times \Omega _B\rightarrow \mathbb {R}$, we obtain the real-valued observable

$$\begin{aligned} f(A,B)_z=\sum \left\{ \mathrm {tr\,} (\alpha _xB_y)A_x: f(x,y)=z\right\} \end{aligned}$$

We conclude that

$$\begin{aligned} f(A,B)_z= & {} \sum \limits _{x,y}f(x,y)\mathcal {H}_x^*(B_y)=\sum \limits _{x,y}f(x,y)\mathrm {tr\,} (\alpha _xB_y)A_x\\ \langle {f(A,B)}\rangle _\rho= & {} \sum \limits _{x,y}f(x,y)\mathrm {tr\,} (\alpha _xB_y)\mathrm {tr\,} (\rho A_x)\\ \Delta _\rho \left[ {f(A,B)}\right]= & {} \sum \limits _{x,y,x',y'}f(x,y)f(x',y')\mathrm {tr\,}\left[ {\rho \mathrm {tr\,} (\alpha _xB_y)A_x\mathrm {tr\,} (\alpha _{x'}B_{y'})A_{x'}}\right] \\{} & {} \quad -\langle {f(A,B)}\rangle _\rho ^2\\= & {} \mathrm {tr\,}\left\{ \rho \left[ {\sum \limits _{x,y}f(x,y)\mathrm {tr\,} (\alpha _xB_y)A_x}\right] ^2\right\} -\langle {f(A,B)}\rangle _\rho ^2 \end{aligned}$$

Moreover, we have

$$\begin{aligned} (B\mid A)_y=\sum \limits _x\mathcal {H}_x^*(B_y)=\sum \limits _x\mathrm {tr\,} (\alpha _xB_y)A_x \square \end{aligned}$$

Example 7

Let A, B be arbitrary observables and let $\mathcal {L}$ be the Lüders instrument given by $\mathcal {L}_x(\rho )=A_x^{1/2}\rho A_x^{1/2}$ [2, 3, 6]. Then

$$\begin{aligned} \mathrm {tr\,}\left[ {\mathcal {L}_x(\rho )}\right] =\mathrm {tr\,} (A_x^{1/2}\rho A_x^{1/2})=\mathrm {tr\,} (\rho A_x) \end{aligned}$$

so $\mathcal {L}$ measures A. Since $\mathcal {L}_x^*(a)=A_x^{1/2}aA_x^{1/2}$ [2, 3] we have

$$\begin{aligned} (A\circ B)_{(x,y)}=A_x^{1/2}B_yA_x^{1/2} \end{aligned}$$

If $f:\Omega _A\times \Omega _B\rightarrow \mathbb {R}$, we obtain the real-valued observable

$$\begin{aligned} f(A,B)_z=\sum \left\{ A_x^{1/2}B_yA_x^{1/2}: f(x,y)=z\right\} \end{aligned}$$

We conclude that

$$\begin{aligned} f(A,B)^\sim= & {} \sum \limits _{x,y}f(x,y)A_x^{1/2}B_yA_x^{1/2}\\ \langle {f(A,B)}\rangle _\rho= & {} \sum \limits _{x,y}f(x,y)\mathrm {tr\,} (\rho A_x^{1/2}B_yA_x^{1/2})=\sum \limits _{x,y}f(x,y)\mathrm {tr\,} (A_x^{1/2}\rho A_x^{1/2}B_y)\\ \Delta _\rho \left[ {f(A,B)}\right]= & {} \mathrm {tr\,}\left\{ \rho \left[ {\sum \limits _{x,y}f(x,y)A_x^{1/2}B_yA_x^{1/2}}\right] ^2\right\} -\langle {f(A,B)}\rangle _\rho ^2 \end{aligned}$$

Moreover, we have

$$\begin{aligned} (B\mid A)_y=\sum \limits _x\mathcal {L}_x^*(B_y)=\sum \limits _xA_x^{1/2}B_yA_x^{1/2}\square \end{aligned}$$