# A Note on Sequential Product of Quantum Effects

Authors: Chunyuan Deng
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

The quantum effects for a physical system can be described by the setℰ(ℋ) of positive operators on a complex Hilbert space ℋ that are bounded above by the identity operator I. For A,B∈ℰ(ℋ), let A∘B=A1/2BA1/2 be the sequential product and let A*B=(AB+BA)/2 be the Jordan product of A,B∈ℰ(ℋ). The main purpose of this note is to study some of the algebraic properties of effects. Many of our results show that algebraic conditions on A∘B and A*B imply that A and B have 3×3 diagonal operator matrix forms with Iℛ(A)¯∩ℛ(B)¯ as an orthogonal projection on closed subspace ℛ(A)¯∩ℛ(B)¯ being the common part of A and B. Moreover, some generalizations of results known in the literature and a number of new results for bounded operators are derived.


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## Body

## 1. Introduction

Letℋ, ℬ(ℋ), and 𝒫(ℋ) be complex Hilbert space, the set of all bounded linear operators on ℋ, and the set of all orthogonal projections on ℋ, respectively. For A∈ℬ(ℋ), we will denote by 𝒩(A) and ℛ(A) the null space and the range of A, respectively. An operator A∈ℬ(ℋ) is said to be injective if 𝒩(A)={0}. ℛ(A)¯ is the closure of ℛ(A). A is said to be positive if (Ax,x)≥0 for all x∈ℋ. A is said to be a contraction if ∥A∥≤1. Pℳ is the orthogonal projection on a closed subspace ℳ⊆ℋ.The elements ofℰ(ℋ)={A∈ℬ(ℋ):0≤A≤I} are called quantum effects. The elements of 𝒫(ℋ)={P∈ℰ(ℋ):P2=P} are projections corresponding to quantum events and are called sharp effects. For A,B∈ℰ(ℋ), the sequential product of A and B is A∘B=A1/2BA1/2. We interpret A∘B as the effect that occurs when A occurs first and B occurs second [1–9]. Let A*B=(AB+BA)/2 be the Jordan product of A,B∈ℰ(ℋ). If AB=BA, we say that A and B are compatible. We define the negation of A∈ℰ(ℋ) by A′=I-A.In this note, we will study some properties of the sequential product or the Jordan product. Our results show that if one tries to impose classical conditions onA∘B=A1/2BA1/2 and A*B=(AB+BA)/2, then this forces A and B to have closed relations with range relations. For example, let T=AnBn for some n∈ℤ+. Then, TT*∈𝒫(ℋ) (or A*B∈𝒫(ℋ)) if and only if A and B have 3×3 diagonal operator matrix forms as follows:
(1)A=Iℛ(A)¯∩ℛ(B)¯⊕A22⊕0,B=Iℛ(A)¯∩ℛ(B)¯⊕0⊕B33,
where Iℛ(A)¯∩ℛ(B)¯ as an orthogonal projection on closed subspace ℛ(A)¯∩ℛ(B)¯ is the common part of A and B. This results give us detailed information of matrix structures between two operators A and B. It is well known that if A or B∈𝒫(ℋ), then A∘B≤B if and only if AB=BA (see [2, Theorem 2.6(a)]  and  [10, Theorem 2.3]). We generate this result and show that, under some conditions, A∘B≤B if and only if A and B have 3×3 operator matrix forms:
(2)A=I⊕0⊕A33,B=B11⊕B22⊕0.In [11, Lemma 3.4], the authors had gotten that if A,B∈ℰ(ℋ) and dim ℋ<∞, then A∘B+A′∘B=B′ if and only if B=(1/2)I. The authors said that they did not know if the condition dim ℋ<∞ can be relaxed. By some algebraic and spectral techniques, we extend some results in [11] to ℬ(ℋ). Some generalizations of results known in the literature and a number of new results for bounded operators are derived.


## 2. Main Results

Our main interest is in sequential products of quantum effects. The next result gives some of the important properties of the sequential product.Lemma 1 (see [2]).
LetA,B∈ℰ(ℋ) and P,Q∈𝒫(ℋ).(i)
A∘B=B∘A if and only if AB=BA.(ii)
IfA∘B∈𝒫(ℋ), then AB=BA.(iii)
P∘Q∈𝒫(ℋ) if and only if PQ=QP.Lemma 2 (see [12]).
LetA∈ℬ(ℋ) be a positive operator. If A has the operator matrix representation A=(Aij)n×n with respect to the space decomposition ℋ=⨁i=1nℋi, then the following statements hold.(i)
Aii as an operator on ℋi is positive, 1≤i≤n.(ii)
Aij=Aii1/2DijAjj1/2 for some contraction Dij∈ℬ(ℋj,ℋi), 1≤i, j≤n.(iii)
IfAi0i0=0 for some 1≤i0≤n, then Ai0j=0 and Aki0=0, 1≤j, k≤n.Lemma 3 (see [13, Lemma 2.2]).
LetA∈ℬ(ℋ) be a contraction and let A as an operator from ℋ=ℋ1⊕ℋ2 into ℋ=𝒦1⊕𝒦2 have the operator matrix
(3)A=(A11A12A21A22).
If A11 is unitary from ℋ1 onto 𝒦1, then A12=0 and A21=0.In [11], Gudder had obtained that if A,B∈ℰ(ℋ) and A+B=P∈𝒫(ℋ), then A and B are compatible. Based on this result, we get the following interesting results.Theorem 4.
LetA,B∈ℰ(ℋ) and P,Q∈𝒫(ℋ).(i)
P≤A if and only if PA=AP=P; A≤P if and only if AP=PA=A.(ii)
There existP,Q∈𝒫(ℋ) such that A=P+Q if and only if A is a projection.(iii)
If there existA,B∈ℰ(ℋ) such that P=A+B, then AB=BA=A-A2. In addition, if ℛ(B)⊆ℛ(A), then P=A+B if and only if AB=BA=A-A2.Proof.
Note thatP and A, as operators on ℋ=ℛ(P)⊕𝒩(P), have the operator matrices
(4)P=(I000),A=(A1A3A3*A2),
respectively, where 0≤A1∈ℬ(ℛ(P)), 0≤A2∈ℬ(𝒩(P)), and A3∈ℬ(𝒩(P),ℛ(P)).
(i) By (4), it is clear that P≤A if PA=AP=P. On the other hand, if A-P=(A1-IA3A3*A2)≥0, then A1=I since A1⊕0=PAP≤P∈ℰ(ℋ). From
(5)A2=(I+A3A3*A3+A3A2A3*+A2A3*A22+A3*A3)≤I,
we get A3A3*=0; that is A3=0 and AP=PA=A. If AP=PA=A, then A2=0 and A3=0 in (4). We get that A≤P. On the other hand, since
(6)P-A=(I-A1-A3-A3*-A2)≥0,A2=0 and A3=0 by Lemma 2. Hence, AP=PA=A.
(ii) IfA is a projection, denote P=A and Q=0, then A=P+Q. Conversely, suppose that there exist two projections P and Q such that A=P+Q. If x∈ℛ(P) is a unit vector, then 1≥(Ax,x)=(Px,x)+(Qx,x)=1+(Qx,x), so (Qx·x)=0. That is, Qx=0 since Q is a positive operator. This shows that QP=0. Similarly, PQ=0. Hence, PQ=QP. The two projections P and Q are commutative; therefore, P+Q=A is a projection.
(iii) SinceA≤P, AP=PA=A by item (i). So, A(A+B)=AP=A=PA=(A+B)A; that is, AB=A-A2=BA. Conversely, let A,B∈ℰ(ℋ). Then there exists P∈𝒫(ℋ) such that ℛ(A)=ℛ(P). Since ℛ(B)⊆ℛ(A), A and B can be written as operator matrices A=A1⊕0, B=B1⊕0 with respect to the space decomposition ℋ=ℛ(P)⊕𝒩(P), respectively, where A1 is an injective positive operator. If AB=A-A2=BA, then A1B1=A1-A12. It follows that B1=I-A1 and A+B=P.LetPℛ(A)¯ denote the self-adjoint projection onto the closure of ℛ(A). In general, that TT* is a projection does not imply T=T*. For example, if T=(0100), then TT* and T*T are projections and T≠T*. But we have following result.Theorem 5.
LetA,B∈ℰ(ℋ) and T=AnBn for some n∈ℤ+. Then, TT*∈𝒫(ℋ) if and only if T*T∈𝒫(ℋ) if and only if A and B have 3×3 operator matrix forms as
(7)A=I⊕A22⊕0,B=I⊕0⊕B33
with respect to the space decomposition ℋ=[ℛ(A)¯∩ℛ(B)¯]⊕[ℛ(A)¯⊖(ℛ(A)¯∩ℛ(B)¯)]⊕𝒩(A); that is, AB is a range projection on ℛ(A)¯∩ℛ(B)¯.Proof.
As we know,σ(TT*)∖{0}=σ(T*T)∖{0} (see [14, Section 1.2.1]). So, for arbitrary T∈ℬ(ℋ), TT* is a projection if and only if T*T is a projection. If A and B have the forms (7), then T=T*=Pℛ(A)¯∩ℛ(B)¯ and TT*=T*T∈𝒫(ℋ).
Necessity. LetS=TT*=AnB2nAn∈𝒫(ℋ). Then, S≤A2n≤I, and hence S≤SA2nS≤SAS≤S. It follows that S=SAS. If we consider S as 2×2 matrix form S=I⊕0 with respective space decomposition ℋ=ℛ(S)⊕𝒩(S), then A has the corresponding matrix form A=(IA3A3*A2). By Lemma 3, we that get A3=0. Hence, S=AS=SA and ℛ(S)⊆ℛ(A). From
(8)S=SAnB2nAn=SB2nAn=AnB2nAnB2nAn=AnB2nS,
we get S=SAnB2nS=SB2nS≤SBS≤S. By similar proof that S=SBS implies that S=SB=BS and ℛ(S)⊆ℛ(B). Now, from AnB2nAn=AnB2nAnAnB2nAn we derive that AnBn(I-BnA2nBn)BnAn=0; that is, AnBn=AnB2nA2nBn=SAnBn=S. We get that S=AnBn=BnAn. Hence, AB=BA. If we denote
(9)ℋ=[ℛ(A)¯∩ℛ(B)¯]⊕[ℛ(A)¯⊖(ℛ(A)¯∩ℛ(B)¯)]⊕𝒩(A),
then A and B can be rewritten as A=A11⊕A22⊕0 and B=B11⊕0⊕B33, where A11,A22, and B11 are injective, densely defined operators and A11B11=B11A11. Since S=AnB2nAn=(AB)2n=(A11B11)2n⊕0⊕0 is projection, this implies that A11B11=B11A11=I. So, A11-1=B11∈ℰ(ℋ). Hence, A11=B11=I; A and B have the matrix forms as in (7).In Theorem5, T=AnBn=BnAn=AB=BA=I⊕0⊕0=Pℛ(A)¯∩ℛ(B)¯.Theorem 6.
LetA,B∈ℰ(ℋ). Then, A*B∈𝒫(ℋ) if and only if A and B have 3×3 operator matrix forms as
(10)A=I⊕A22⊕0,B=I⊕0⊕B33
with respect to the space decomposition ℋ=[ℛ(A)¯∩ℛ(B)¯]⊕[ℛ(A)¯⊖(ℛ(A)¯∩ℛ(B)¯)]⊕𝒩(A). In particular, A*B=0 if and only if AB=0.Proof.
By (10), if AB=BA=Pℛ(A)¯∩ℛ(B)¯, then clearly A*B∈𝒫(ℋ).
Necessity. Observing thatA and B as operators on ℋ=R(A)¯⊕𝒩(A) have the forms as A=(A1000) and B=(B1B3B3*B33), where A1 is injective, densely defined. Then
(11)A*B=((A1B1+B1A1)2A1B32B3*A120)
is a projection implies that A1B3=0 by Lemma 2. So, B3=0 because A1 is injective, densely defined. B1 can be further written as B1=B11⊕0 with respect to space decomposition ℛ(A)¯=(ℛ(A)¯∩ℛ(B)¯)⊕(ℛ(A)¯⊖(ℛ(A)¯∩ℛ(B)¯)), where B11 is injective, densely defined. Similarly, A1 has corresponding form as A1=A11⊕A22 with A11 and A22 being injective, densely defined. So
(12)A*B=(A11B11+B11A11)2⊕0⊕0.
We say that(A11B11+B11A11)/2 is injective. In fact, if 𝒩(A11B11+B11A11)≠{0}, then A11B11=-B11A11 on 𝒩(A11B11+B11A11) and hence A112B112=B112A112 on 𝒩(A11B11+B11A11). Therefore, A11B11=B11A11 on 𝒩(A11B11+B11A11). Hence, for every 0≠x∈𝒩(A11B11+B11A11),
(13)A11B11+B11A112x=A11B11x=0.
SinceA11 and B11 are injective, we get x=0, which contradicts the assumption. Now, A*B∈𝒫(ℋ) implies that A11B11+B11A11=2I. For every unit vector x∈ℛ(A)¯∩ℛ(B)¯,
(14)1=〈x,x〉=12〈A11B11x,x〉+12〈B11A11x,x〉.
Since A11B11 is contraction, we derive that 〈A11B11x,x〉=1 and 〈B11A11x,x〉=1 for every unit vector x∈ℛ(A)¯∩ℛ(B)¯. This concludes that A11B11=B11A11=I. So, A11-1=B11∈ℰ(ℋ). Hence, A11=B11=I, A and B have the matrix forms as in (7).
In particular, ifAB=0, then BA=0 and A*B=(AB+BA)/2=0. On the other hand, if A*B=0, then B3=0 and A1B1=-B1A1 in (11). We have A12B1=-A1B1A1=B1A12. Therefore, A1B1=B1A1; that is, AB=0.Next, we are now interested in the question of whenA∘B≥B or A∘B≤B. In Theorem 2.6 of [2] it is proved that, if ℋ is finite dimensional and A∘B≥B, then AB=BA=B, and it is asked whether this holds for infinite-dimensional spaces ℋ. In [5, Theorem 2.6], the authors answer this question positively. Here, we include a different proof because it is very short.Theorem 7.
LetA,B∈ℰ(ℋ) such that A∘B≥B if and only if
(15)A=I⊕A1,B=B1⊕0,whereA1∈ℬ(𝒩(I-A)⊥),B1∈ℬ(𝒩(I-A)).Proof.
IfAB=BA=B, then clearly A∘B≥B. On the other hand, for arbitrary 0<δ<1, let Δ1=[1-δ,1]∩σ(A) and Δ2=[0,1-δ)∩σ(A). Let A=∫0∥A∥‍λdEλ be the spectral representation of A. Thus, A has the operator matrix form A=A1⊕A2 with respect to the space decomposition ℋ=ℋ1⊕ℋ2, where ℋ1=E(Δ1)ℋ and ℋ2=E(Δ2)ℋ. It is clear that A2≤(1-δ)Iℋ2. Let B have corresponding matrix form. Since B≤A1/2BA1/2≤ABA≤A3/2BA3/2≤⋯≤AnBAn≤⋯, n∈ℕ. Hence
(16)(A1n00A2n)(B1B3B3*B2)(A1n00A2n)mm=(A1nB1A1nA1nB3A2nA2nB3*A1nA2nB2A2n)≥(B1B3B3*B2).
It follows A2nB2A2n≥B2≥0 for all n∈ℕ. Since A2n is convergence by strong operator topology to zero, we get that B2=0. By Lemma 2, we know that B3=0. Hence, A1nB1A1n≥B1 for arbitrary 0<δ<1. Note that ⋂0<δ<1‍[[1-δ,1]∩σ(A)]⊆{1}. Hence. A1=Iℋ1 and A, B have the form (15).Note that ifA,B∈ℰ(ℋ), then (i) AB=BA=B⇔B(I-A)=0⇔ℛ(I-A)⊆𝒩(B); (ii) B∘A=B⇔B1/2(I-A)B1/2=0⇔B(I-A)=0; (iii) B∘A≥B⇔-B1/2(I-A)B1/2≥0 ⇔B(I-A)=0. By Theorem 7, it is easy to get the following results.Corollary 8.
ConsiderA,B∈ℰ(ℋ). 							(17)B=P𝒩(I-A)B⟺A∘B≥B⟺A∘B=B⟺B∘A≥B⟺B∘A=B⟺A=I⊕A1,B=B1⊕0,A1,B1aredefinedin(15)⟺AB=BA=B.From Corollary10, we know that APℛ(B)¯=Pℛ(B)¯A=Pℛ(A)¯∩ℛ(B)¯. However, A∘B≤B does not imply AB=BA. One can check this fact by choices A=(1/41/41/41/4) and B=(3/4001/4) in ℂ2 (see [2]). However, we obtain the following result.Theorem 9.
LetA,B∈ℰ(ℋ) and P∈𝒫(ℋ) such that ℛ(P)=ℛ(A)¯∩ℛ(B)¯.(i)
IfAP=P, then A∘B≤B if and only if A and B have 3×3 operator matrix forms
(18)A=I⊕0⊕A33,B=B11⊕B22⊕0
with respect to the space decomposition ℋ=ℛ(P)⊕[ℛ(B)¯⊖ℛ(P)]⊕𝒩(B).(ii)
IfBP=P, then A∘B≤B if and only if A and B have 3×3 operator matrix forms
(19)A=A11⊕A22⊕0,B=I⊕0⊕B33
with respect to the space decomposition ℋ=ℛ(P)⊕[ℛ(A)¯⊖ℛ(P)]⊕𝒩(A).Proof.
By (18) and (19), it is clear that AB=BA and A∘B=A1/2BA1/2=B1/2AB1/2≤B.
Necessity. (i) If AP=P, by Lemma 3, A and B as operators on ℛ(P)⊕[ℛ(B)¯⊖ℛ(P)]⊕𝒩(B) have the operator matrix forms
(20)A=(I0000000A33),B=(B11B120B12*B220000).
IfA∘B≤B, then 0≤ABA≤A1/2BA1/2≤B. So
(21)B-ABA=(0B120B12*B220000)≥0.
By Lemma2, we have B12=0. So, (18) holds.
(ii) IfBP=P, then A and B as operators on ℛ(P)⊕[ℛ(A)¯⊖ℛ(P)]⊕𝒩(A) can be denoted as
(22)A=(A11A120A12*A220000),B=(I0000000B33).
We have(23)B-ABA=(I-A112-A11A120-A12*A11-A12*A12000B33)≥0.
By Lemma2, we have A12*A12=0; that is, A12=0 and (18) holds.LetA,B∈ℰ(ℋ) and P=Pℛ(A)¯∩ℛ(B)¯. Theorem 9 implies that if AP=P or BP=P, then A∘B≤B⇔AB=BA. In particular, if A or B∈𝒫(ℋ), then APℛ(A)¯∩ℛ(B)¯=Pℛ(A)¯∩ℛ(B)¯ or BPℛ(A)¯∩ℛ(B)¯=Pℛ(A)¯∩ℛ(B)¯ hold automatically. We get the following corollary.Corollary 10 (see [2, Theorem 2.6(a)] and [10, Theorem 2.3]).
LetA,B∈ℰ(ℋ). If A or B∈𝒫(ℋ), then A∘B≤B if and only if AB=BA.In [11, Lemma 3.4], the authors had gotten that if A,B∈ℰ(ℋ) and dim ℋ<∞, then A∘B+A′∘B=B′ if and only if B=(1/2)I. The authors said they did not know if the condition dim ℋ<∞ can be relaxed. In the following, we show that the condition dim ℋ<∞ in [11, Lemma 3.4] can be relaxed.Theorem 11.
ConsiderA,B∈ℰ(ℋ). A∘B+A′∘B=B′ if and only if B=(1/2)I.Proof.
IfA∘B+A′∘B=A1/2BA1/2+(I-A)1/2B(I-A)1/2=B′, then
(24)A1/2BA1/2=I-B-(I-A)1/2B(I-A)1/2.
So
(25)ABA=A1/2(I-B)A1/2m-(I-A)1/2A1/2BA1/2(I-A)1/2=A-A1/2BA1/2m-(I-A)1/2A1/2BA1/2(I-A)1/2=A-[I-B-(I-A)1/2B(I-A)1/2]m-(I-A)1/2[I-B-(I-A)1/2B(I-A)1/2]m×(I-A)1/2=2A-2I+2B-AB-BA+ABAm+2(I-A)1/2B(I-A)1/2=2A-2I+2B-AB-BA+ABAm+2[I-B-A1/2BA1/2]=2A-AB-BA+ABA-2A1/2BA1/2.
We get(26)2A=AB+BA+2A1/2BA1/2,
which is equal to A-A1/2BA1/2-AB=-[A-BA-A1/2BA1/2]. Put T=A1/2-BA1/2-A1/2B. Then, T=T* and A1/2T=-TA1/2. Product T from right, we get
(27)A1/2T2=-TA1/2T=T2A1/2.
SinceA≥0 and T=T*, we derive that T2 is positive, and hence A1/2T2=-TA1/2T=T2A1/2≥0. Note that -TA1/2T≤0. We get that TA1/2T=TA1/4[TA1/4]*=0; that is, TA1/4=0. Therefore, A1/2T=TA1/2=0. Since A1/2T=A-A1/2BA1/2-AB and TA1/2=A-A1/2BA1/2-BA, we obtain that AB=BA. In this case, A,B, as operators on ℛ(A)¯⊕𝒩(A), have 2×2 operator matrix form
(28)A=(A1000),B=(B100B2),whereA1,B1∈ℬ(ℛ(A)¯),B2∈ℬ(𝒩(A))
and A1 is injective, densely defined. By (26), we get that A=2AB. By (28), we get, B1=(1/2)Iℛ(A). By (24), we get that B2=(1/2)I𝒩(A). Hence, B=(1/2)I. Conversely, by (28), it is clear that B=(1/2)I implies that A∘B+A′∘B=B′.For𝒜,ℬ⊆ℰ(ℋ) with 𝒜={Ai} and ℬ={Bj}, the sequential product of 𝒜 and ℬ is defined by 𝒜∘ℬ={Ai∘Bj}. We interpret 𝒜∘ℬ to be the measurement obtained when 𝒜 is performed first and ℬ is performed second. The sequential product is noncommutative and nonassociative in general. We write 𝒜≈ℬ if the nonzero elements of 𝒜 are a permutation of the nonzero elements of ℬ. “≈” is an equivalence relation, and when 𝒜≈ℬ we say that 𝒜 and ℬ are equivalent. In this case, the two submeasurements are identical up to an ordering of their outcomes [11].The results in [11, Theorem 3.1] could be modified as the following. Note that, in [2, Theorem 4.4], it had proved that A∘B+A′∘B=B if and only if AB=BA.Theorem 12.
Suppose,A,B∈ℰ(ℋ), 𝒜={A,A′}, and ℬ={B,B′}. If 𝒜∘ℬ≈ℬ∘𝒜, then AB=BA.Proof.
Denote(29)T=:(A∘B,A∘B′,A′∘B,A′∘B′)=:(X00,X01,X10,X11),(B∘A,B′∘A,B∘A′,B2∘A′)m=:(X00T,X01T,X10T,X11T),
respectively. If there exists one corresponding term Xij=XijT, 0≤i,j≤1, then AB=BA by Lemma 1. Next, we consider equality for noncorresponding terms.
Case
I. If T=(X01T,X00T,X11T,X10T), then by comparing the third and the fourth components in two sides, we get that X10+X11=X10T+X11T; that is, B∘A′+B′∘A′=A′. So, AB=BA.
Case
II. If T=(X01T,X10T,X11T,X00T) or T=(X11T,X00T,X01T,X10T), then by comparing the first and the third components in two sides, we get that X00+X10=X01T+X11T, that is, A∘B+A′∘B=B′. By Theorem 11, we get AB=BA.
Case
III. If T=(X01T,X11T,X00T,X10T), then by comparing the first and the second components in two sides, we get that X00+X01=X01T+X11T; that is, A=B′, and hence AB=BA.
Case
IV. If T=(X10T,X00T,X11T,X01T), then by comparing the first and the second components in two sides, we get that X00+X01=X00T+X10T; that is, A=B. So, AB=BA.
Case
V. If T=(X10T,X11T,X00T,X01T), then by comparing the first and the third components in two sides, we get that X00+X10=X00T+X10T; that is, A∘B+A′∘B=B. So AB=BA.
Case
VI. If T=(X10T,X11T,X01T,X00T), then by comparing the third and the fourth components in two sides we get X10+X11=X00T+X01T, that is, B∘A+B′∘A=A′. By Theorem 11, we get that A=(1/2)I and AB=BA.
Case
VII. If T=(X11T,X10T,X00T,X01T) or T=(X11T,X10T,X01T,X00T), then by comparing the first and the second components in two sides, we get X00+X01=X10T+X11T; that is, B∘A′+B′∘A′=A. By Theorem 11, we get that A=(1/2)I and AB=BA.The converse does not hold. Indeed,𝒜∘𝒜≈𝒜∘𝒜 and yet the elements in 𝒜 need not be commutative. In the following, we give a characterization of the two submeasurements that are identical up to an arbitrary ordering of their outcomes.Corollary 13.
Suppose thatA,B∈ℰ(ℋ), 𝒜={A,A′}, and ℬ={B,B′}. An arbitrary permutation of the elements in ℬ∘𝒜 is equivalent to 𝒜∘ℬ if and only if A=B=(1/2)I.Proof.
IfA=B=(1/2)I, then A=A′=B=B′, and clearly an arbitrary permutation of the elements in ℬ∘𝒜 is equivalent to 𝒜∘ℬ.
Conversely, by Cases IV and VII in the proof of Theorem12, we have A=B=(1/2)I.



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*Source: 520436-2013-09-09.xml*