#### Abstract

Let be the Hilbert space effect algebra on a Hilbert space with , two positive numbers with and a bijective map. We show that if holds for all , then there exists a unitary or an antiunitary operator on such that for every .

#### 1. Introduction and Notations

Let be a Hilbert space. Denote by the algebra of all bounded linear operators acting on . The operator interval in : , where is the identity, is called Hilbert space effect algebra on , and the elements in are called effects. The concept of effect algebras plays a fundamental role in mathematical description of quantum measurement, and the range of general quantum observables (POV-measurement) consists of effects [1].

The Hilbert space effect algebra can be equipped with some algebraic operations. Denote the space of all projections on by . Clearly . Let . is a partial ordered set with the order “” if . Obviously, the sum may not be in . In the case that , we can define an operation of conditional addition in . Particularly, since , this gives an orthogonal addition operation in . Generally speaking, the product may not be in either. However, the Jordan semitriple product is a well-defined operation in since for all . More generally, for any fixed positive real numbers , the generalized Jordan triple product induced by is well defined, too, in and here .

Recently many mathematicians pay their attention to the problem of characterizing certain maps on Hilbert space effect algebras or other quantum structures [2–8]. Let be a Hilbert space with and a bijective map. It seems that [5] is the first paper discussing the problem of characterizing the Jordan semitriple maps on Hilbert effect algebras. In [5] Molnár showed that if for all , then there exists a unitary or an antiunitary operator on such that for every ; that is, is implemented by a unitary or an antiunitary operator on . Molnár and Šemrl proved in [7] that if satisfies where denotes and , then there is a positive number and a unitary or an antiunitary operator such that for every (we will show that in fact). The generalized Jordan semitriple product is called sequence product. Kim showed in [4] that if is separable and satisfies for , then is also implemented by a unitary or an antiunitary operator on . Our purpose in this paper is to generalize the results of [5, 7] to generalized Jordan semitriple maps, that is, the maps satisfying Let be a Hilbert space with , two positive real numbers with , and a bijective map satisfying for all , . We first show that is orthoadditive on , that is, for any orthogonal projections , , . As an application, we prove that all bijective generalized Jordan semitriple maps on Hilbert space effect algebras, that is, the maps satisfying (4), are implemented by a unitary or an antiunitary operator on .

#### 2. Orthoadditivity of Generalized Jordan Semitriple Maps

Additivity of multiplicative maps is studied by many authors. In this section, we show that every generalized Jordan semitriple map on the effect algebra is orthoadditive on . To do this we need a description which gives a description of the zero product preservers on . Recall that a map preserves zero products (in both directions) if , whenever (if and only if) .

Lemma 1. *Let be a real or complex Hilbert space with . Assume that is a bijective map preserving zero products of operators in both directions. *(1)*If is real, then there is a unitary operator on such that for all rank one .*(2)*If is complex, then there is a unitary operator or an antiunitary operator on such that for all rank one .*

*Proof. *Denote by the set of all projections of rank one. For any , write .*Claim **1*. For any , if and only if ; is singleton; that is, it contains exactly one element, if and only if is of rank one.

If is obvious, assume that and then and .

If contains only one element, then and . If is not of rank one, for any unit vector , one has but , a contradiction. So, . The converse is obvious.*Claim **2*. preserves rank one projections in both directions.

For any rank one projection , let ; then . By the property of that it preserves zero products in both directions, one has . Thus is singleton, which forces that is of rank one by Claim 1, and vice versa.*Claim **3*. There exists a unitary or antiunitary operator on such that
holds for all . is unitary in real case.

By Claim 3, for every unit vector , there is a unit vector such that .

Let be the map on the projective space defined by , where stands for the 1-dimensional linear subspace spanned by . Then it is clear that is bijective since preserves rank one projection in both directions. For any linearly independent unit vectors and any scalars , let . For any unit , there is a unit vector such that . If , then . This implies that and consequently . Hence . It follows from the arbitrariness of that ; that is, . Now we can use the fundamental theorem of projective geometry to get a semilinear bijection such that . Therefore, for any unit , . It is obvious that preserves orthogonality in both directions; that is, . Hence for some unitary or antiunitary operator on . In the case that is real, is unitary. Clearly, , so for all rank one projection .

Theorem 2. *Let be a Hilbert space with , positive numbers with , and a bijective map. If satisfies
**
for all , then for any projections , implies that
*

*Proof. *For , we have
where . Note . For , the spectral solution theorem implies that . By considering , we see that preserves projections in both directions.

For any ,
Similarly, we have
So
holds for all . By the surjectivity of and , we must have . Since is bijective, there is such that . Thus
that is, .

Now we show that preserves zero product in both directions on , that it preserves orthogonality in both directions. For any in ,
Then
Therefore, by Lemma 1, there exists a unitary or an antiunitary operator on such that holds for all rank one projections . Without loss of generality we may assume that for every rank one projection .

For any , if , let , , and . We will show that . To do this, it is enough to show that . Note that for any in with norm 1,
Thus we have
This forces . So whenever .

#### 3. Characterization of Generalized Jordan Semitriple Maps

The following is our main result.

Theorem 3. *Let be a Hilbert space with . Assume that are positive numbers with and a bijective map. If
**
for all , then there exists a unitary or an antiunitary operator on such that for all .*

The following corollary gives a better form of sequence product multiplicative maps, comparing with [7].

Corollary 4. *Let be a Hilbert space with and a bijective map. If
**
for all , then there exists an unitary or an antiunitary operator on such that for every .*

*Proof of Theorem 3. *By Theorem 2 we know that preserves projections in both directions, , , and is orthoadditive on .

Furthermore, preserves the order of projections in both directions. To see this, let . It is clear that
So, implies that
Consequently, . Similarly, using , one can check that implies that .

So, by Theorem 2 and [9], the restriction of on can be extended to a bounded linear map . Note that preserves projections in both directions; it must be a Jordan star-isomorphism. Since every Jordan star-automorphism of is either a star-automorphism or star-antiautomorphism, there exists a unitary operator or an antiunitary operator such that
or
It follows that there is a unitary or an antiunitary operator such that for all .

Without loss of generality, we can assume that holds for every projection .

Next we show that holds for every and .

By the definition of that we have for any and any rank one projection , we have
So there is a such that . Denote by ; that is, . is a function such that . It follows that for every
On the other hand,
Thus and . Hence, we have holds for all ; that is, is multiplicative.

Obviously, by the surjectivity of , . So is continuous on . It is well known that every multiplicative continuous bijection is of the form (). Hence there is a such that for every . It follows that
for every rank one projection .

We will show that is independent of . For any rank one projections and , we have
If , we get
Letting and comparing two sides of the above equation yield . So . If , pick a rank one projection so that and . Then . By the fact we just checked, we get . It follows that there is a common such that
holds for all rank one projection and .

We claim that . To see this, it is enough to check that whenever , where . Pick unit vectors with and let , , and . Then
Similarly, we have
Since is orthoadditive on , we have
Note that is a rank one projection, and for any , we have
Thus and
This forces that whenever , and hence, .

Now, let us show that holds for all and all . By the orthoadditivity of , holds for every and every finite rank projection . Assume have infinite rank. For any with , we have
It follows that and are linearly dependent; that is, there is a scalar , dependent on such that with and . Similarly, for every , we have . These entail that there exists a function having values in and satisfying for all and all . Hence, and . As before, for a fixed , we have that for some positive . Next we claim that . Taking the rank one projection for , so , we have
which entails that . In sum, now we have that for and for . Let , since ,

Finally, we show that for every . In fact, letting , for every unit vector , we have
Thus there exists a scalar with such that . This entails that for some scalar with . Since both and are positive, we see that . Hence and the uniqueness of the positive root of a positive operator implies that , as desired.

#### Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work was supported partially by National Science Foundation of China (11201329, 11171249, and 11271217).