PHYSICAL REVIEW A 79, 032330 共2009兲
Subspaces without locally distinguishable orthonormal bases Wei Jiang,1,* Xi-Jun Ren,2 Xingxiang Zhou,1 Zheng-Wei Zhou,1,† and Guang-Can Guo1
1
Laboratory of Quantum Information, University of Science and Technology of China, Hefei, Anhui 230026, People’s Republic of China 2 School of Physics and Electronics, Henan University, Kaifeng 475001, People’s Republic of China 共Received 20 October 2008; published 24 March 2009兲 In this paper, we find subspaces that have no locally distinguishable orthonormal bases. For a Cm 丢 Cn 共m , n ⱖ 3兲 bipartite tensor system, we show that the orthogonal complement of the space spanned by a Schmidt rank-3 maximally entangled state is such a subspace. We further demonstrate that, when both parties’ dimensions are greater than three, the orthogonal complement of the space spanned by any entangled state whose Schmidt rank is greater than three has no locally distinguishable orthonormal bases. Although derived using a scheme simpler than previous methods, our results are much more general than known before. DOI: 10.1103/PhysRevA.79.032330
PACS number共s兲: 03.67.Mn, 03.65.Ud
Local distinguishability of orthogonal quantum states has attracted much attention in quantum information science 关1–15兴. It is the basis for the study of many important quantum information problems. For instance, by using a set of unextendible product basis states 关1兴 which are necessarily locally indistinguishable, states with bound entanglement that cannot be distilled can be constructed 关1兴. Local distinguishability of quantum states also proves essential in the research of classical capacity of quantum channels 关2兴. Although a set of pure multipartite orthogonal quantum states can always be perfectly distinguished by global projective measurement, the distinguishability of multipartite quantum states is a difficult and nontrivial problem when allowed operations are limited to local operations and classical communications 共LOCC兲. This was recognized when Bennett et al. 关3兴 showed counterintuitively that there exist nine orthogonal product states in a 3 丢 3 system which cannot be distinguished by LOCC protocols. Nevertheless, there has been much progress in the research on this problem 关1–15兴. Lately, Watrous 关2兴 showed that, for a bipartite Cn 丢 Cn system, there are subspaces in which no orthonormal bases can be perfectly distinguished by LOCC. These are the orthogonal complement of the space spanned by a maximally rank-n 关16兴 entangled state. Watrous’ work 关2兴 is an important result because the existence of subspaces without locally distinguishable bases indicates that there are quantum channels with only suboptimal classical capacity even when the receiver may communicate classically with a third party representing the channel’s environment. However, the method used to prove the existence of such subspaces is rather complicated mathematically and is difficult to be applied in the investigation in more general settings. Also, the conclusion of 关2兴 is rather confined in the sense that the two parties must have the same dimension n 共n ⱖ 3兲 and the reference subspace must be spanned by a maximally entangled state whose Schmidt rank should be equal to the dimension of the two parties. In this paper, we are looking to find more general subspaces which have no locally distinguishable bases. For a
*
[email protected] †
[email protected]
1050-2947/2009/79共3兲/032330共4兲
Cm 丢 Cn system 共m , n ⱖ 3, m and n not necessarily equal兲, these are the orthogonal complement of the space spanned by Schmidt rank-3 maximally entangled states. Further, when the dimensions of both parties are greater than 3 共m , n ⬎ 3兲, we demonstrate that the orthogonal complement of the space spanned by any entangled state with a Schmidt rank greater than three is a subspace without locally distinguishable orthonormal bases. Our results are much more general than known before 关2兴 and are thus a significant progress in understanding local distinguishability of quantum states. However, although our conclusions incorporate previous results 关2兴 as a special case, we did not derive them by generalizing earlier results using previously known methods. Instead, we followed an approach that is based on an easily understandable idea and much simpler mathematics. Therefore, both our results and methods are valuable for the study of local distinguishability of quantum states. Our idea is as follows. As the set of LOCC operators is a subset of separable operators, if we can prove that all protocols discriminating a set of states must involve some inseparable operators, then we can conclude that this set of states cannot be perfectly distinguished locally. To exploit this idea, we need the following key Lemma: Lemma 1. For a Cm 丢 Cn 共m , n ⱖ 3兲 system, if a Schmidt rank-3 maximally entangled state 兩⌿典 is orthogonal to a Schmidt rank-2 entangled state 兩⌽典, the superposition of these two states cannot be a product state. We prove this by, as a start, assuming that Alice and Bob own the m-dimensional and n-dimensional parts of a Cm 丢 Cn system. We denote the quantum state of a Cm 丢 Cn system 兩典AB. We define span关A共兩典具兩兲兴 as the subspace spanned by the eigenstates of A共兩典具兩兲 with nonzero eigenvalues, where A共兩典具兩兲 is the reduced density matrix of Alice. We consider three different situations. When span关A共兩⌽典具⌽兩兲兴 ⬜ span关A共兩⌿典具⌿兩兲兴, without loss of generality, 兩⌿典 and 兩⌽典 can be denoted as 兩⌿典 = 兩0典兩0典 + 兩1典兩1典 + 兩2典兩2典,
共1兲
兩⌽典 = 兩3典兩0典 + 兩4典兩1典,
共2兲
where span兵兩0典 , 兩1典 , 兩2典其 = span关A共兩⌿典具⌿兩兲兴 and span兵兩3典 , 兩4典其 = span关A共兩⌽典具⌽兩兲兴, and Eqs. 共1兲 and 共2兲 are 032330-1
©2009 The American Physical Society
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Schmidt decompositions of 兩⌿典 and 兩⌽典. It can be directly seen that the superposition of 兩⌽典 and 兩⌿典 is not a product state. When span关A共兩⌽典具⌽兩兲兴 傺 span关A共兩⌿典具⌿兩兲兴, without loss of generality, 兩⌿典 and 兩⌽典 can be denoted as 兩⌿典 = 兩0典兩0典 + 兩1典兩1典 + 兩2典兩2典,
共3兲
兩⌽典 = 兩0典兩0典 + 兩1典兩1典,
共4兲
where span兵兩0典 , 兩1典 , 兩2典其 = span关A共兩⌿典具⌿兩兲兴 and span兵兩0典 , 兩1典其 = span关A共兩⌽典具⌽兩兲兴. The proof is by contradiction. Suppose a linear combination x兩⌿典 + y兩⌽典 共xy ⫽ 0兲 is a product state, we can write it as x兩⌿典 + y兩⌽典 = 兩0典共x兩0典 + y兩0典兲 + 兩1典共x兩1典 + y兩1典兲 + 兩2典共x兩2典兲.
共5兲
Since x兩0典 + y兩0典, x兩1典 + y兩1典, and 兩2典 must be proportional, 兩0典 can be expressed as the linear superposition of 兩0典 and 兩2典, and 兩1典 can be expressed as the linear superposition of 兩1典 and 兩2典: 兩0典 = a0兩0典 + b0兩2典,
共6兲
兩1典 = a1兩1典 + b1兩2典.
共7兲
x兩0典 + y兩0典 = 共x + a0y兲兩0典 + b0y兩2典 ⬀ 兩2典,
共8兲
x兩1典 + y兩1典 = 共x + a1y兲兩1典 + b1y兩2典 ⬀ 兩2典,
共9兲
Thus,
which leads to
where span兵兩␣典 , 兩典 , 兩␥典其 ⬜ span兵兩0典 , 兩1典其. Then, the superposition of 兩⌽典 and 兩⌿典 could be written as 共xy ⫽ 0兲: x兩⌽典 + y兩⌿典 = x关兩0⬘典兩0典 + 兩1⬘典兩1典 + 兩2⬘典兩2典兴 + y关兩0典兩0典 + 兩1典兩1典兴 = 兩0典关xa0兩0典 + xa1兩1典 + xa2兩2典 + y兩0典兴 + 兩1典关xb0兩0典 + xb1兩1典 + xb2兩2典 + y兩1典兴 + 兩␣典共xc0兲兩0典 + 兩典共xc1兲兩1典 + 兩␥典 ⫻共xc2兲兩2典.
共17兲
If x兩⌽典 + y兩⌿典 was a product state, c0兩␣典 , c1兩典 and c2兩␥典 共at least one of c0, c1, and c2 is nonzero; should be proportional. This suppose c0 ⫽ 0兲 implies that span兵兩0⬘典 , 兩1⬘典 , 兩2⬘典 兩 其 = span兵兩0典 , 兩1典 , 兩␣典其, thus span兵兩0典 , 兩1典其 = span关A共兩⌽典具⌽兩兲兴 傺 span关A共兩⌿典具⌿兩兲兴 = span兵兩0典 , 兩1典 , 兩␣典其. This contradicts the fact that span关A共兩⌽典具⌽兩兲兴 is not the subspace of span关A共兩⌿典具⌿兩兲兴. This completes the proof of Lemma 1. We need to introduce the following two more lemmas. Lemma 2. For a Cm 丢 Cn共m , n ⱖ 4兲 system, the superposition of any entangled state 兩⌿典 with a Schmidt rank ⱖ4 and any Schmidt rank-2 entangled state 兩⌽典 is not a product state. We prove this Lemma by examining the situation where the Schmidt rank of 兩⌿典 is four. The same argument applies to situations where the Schmidt rank of 兩⌿典 is greater than four. When span关A共兩⌽典具⌽兩兲兴 ⬜ span关A共兩⌿典具⌿兩兲兴, the superposition of 兩⌽典 and 兩⌿典 is not a product state. When span关A共兩⌽典具⌽兩兲兴 傺 span关A共兩⌿典具⌿兩兲兴, without loss of generality and by using the Schmidt decomposition, 兩⌿典 and 兩⌽典 can be denoted as
x + a0y = 0,
共10兲
兩⌿典 = 兩0典兩0典 + 兩1典兩1典 + 兩2典兩2典 + 兩3典兩3典,
共18兲
x + a1y = 0.
共11兲
兩⌽典 = 兩0⬘典兩0典 + 兩1⬘典兩1典,
共19兲
共12兲
where span兵兩0典 , 兩1典 , 兩2典 , 兩3典其 = span关A共兩⌿典具⌿兩兲兴; span兵兩0⬘典 , 兩1⬘典其 = span关A共兩⌽典具⌽兩兲兴; 兩0⬘典 = a0兩0典 + b0兩1典 + c0兩2典 + d0兩3典 and 兩1⬘典 = a1兩0典 + b1兩1典 + c1兩2典 + d1兩3典. Then, the superposition of 兩⌽典 and 兩⌿典 could be written as 共xy ⫽ 0兲:
共13兲
x兩⌿典 + y兩⌽典 = x关兩0典兩0典 + 兩1典兩1典 + 兩2典兩2典 + 兩3典兩3典兴 + y关兩0⬘典
Therefore 2x + 共a0 + a1兲y = 0. From the orthogonality of 兩⌿典 and 兩⌽典, we get 具0兩0典 + 具1兩1典 = 0.
⫻兩0典 + 兩1⬘典兩1典兴 = 兩0典关x兩0典 + ya0兩0典
This implies that 1 a0具0兩0典 + a1具1兩1典 = 共a0 + a1兲 = 0, 3
+ ya1兩1典兴 + 兩1典关x兩1典 + yb0兩0典 + yb1兩1典兴
共14兲
which means a0 + a1 = 0. Thus, we get x = 0 from Eq. 共12兲. This contradicts the condition xy ⫽ 0. When span关A共兩⌽典具⌽兩兲兴 is neither orthogonal to span关A共兩⌿典具⌿兩兲兴 nor the subspace of span关A共兩⌿典具⌿兩兲兴, without loss of generality, 兩⌿典 and 兩⌽典 can be denoted as 兩⌿典 = 兩0⬘典兩0典 + 兩1⬘典兩1典 + 兩2⬘典兩2典,
共15兲
兩⌽典 = 兩0典兩0典 + 兩1典兩1典,
共16兲
+ 兩2典关x兩2典 + yc0兩0典 + yc1兩1典兴 + 兩3典关x兩3典 + yd0兩0典 + yd1兩1典兴.
共20兲
If x兩⌿典 + y兩⌽典 was a product state, then x兩0典 + ya0兩0典 + ya1兩1典, x兩1典 + yb0兩0典 + yb1兩1典, x兩2典 + yc0兩0典 + yc1兩1典, and x兩3典 + yd0兩0典 + yd1兩1典 should be proportional. Thus, we obtain
where span兵兩0⬘典 , 兩1⬘典 , 兩2⬘典其 = span关A共兩⌿典具⌿兩兲兴; span兵兩0典 , 兩1典其 = span关A共兩⌽典具⌽兩兲兴. Generally, 兩0⬘典 = a0兩0典 + b0兩1典 + c0兩␣典, 兩1⬘典 = a1兩0典 + b1兩1典 + c1兩典, and 兩2⬘典 = a2兩0典 + b2兩1典 + c2兩␥典, 032330-2
兩0典 = 0兩0典 + 共0兩0典 + 0兩1典兲,
共21兲
兩1典 = 1兩0典 + 共1兩0典 + 1兩1典兲,
共22兲
兩2典 = 2兩0典 + 共2兩0典 + 2兩1典兲,
共23兲
PHYSICAL REVIEW A 79, 032330 共2009兲
SUBSPACES WITHOUT LOCALLY DISTINGUISHABLE…
兩3典 = 3兩0典 + 共3兩0典 + 3兩1典兲.
共24兲
It is impossible that span兵兩0典 , 兩1典 , 兩0典其 = span兵兩0典 , 兩1典 , 兩2典 , 兩3典其. When span关A共兩⌽典具⌽兩兲兴 is neither orthogonal to span关A共兩⌿典具⌿兩兲兴 nor the subspace of span关A共兩⌿典具⌿兩兲兴, we can use the same method as in the proof of Lemma 1 to prove that the superposition of 兩⌽典 and 兩⌿典 is not a product state. This completes the proof. Lemma 3. If a set of orthogonal states 兵兩i典其 can be distinguished by some LOCC protocol, every state in 兵兩i典其 共say 兩k典兲 can be written as 兩 k典 =
兩1k 典
+
兩2k 典
+
兩3k 典
+ ... ,
共25兲
where 兩lk典’s are all product states, and every 兩lk典 is orthogonal to the states in 兵兩i典其i⫽k. Lemma 3 is actually the main conclusion of Ref. 关4兴 and will be used in the following proof. Before we proceed to prove our main results, we review how a LOCC protocol works in local distinguishability 关2,15兴. In order to distinguish states in the set = 兵兩i典其, Alice and Bob perform successive rounds of local measurement and classical communication. Suppose Alice is the first to perform a positive operator-valued measurement 共POVM兲 兵A†j A j1其 j1. If the outcome is j1, 兵兩i典其 becomes 兵A j1兩i典其i, up 1 to normalization. Following this, Alice communicates her result to Bob who performs his POVM. Notice that Bob’s choice of measurement is dependent on Alice’s result 关15兴. His POVM operators must be nontrivial, meaning they are not proportional to the identity operator, and they must preserve the orthogonality between the states already measured by Alice. Designate Bob’s first-round POVM operators satisfying these constraints 兵Bk† 共j1兲Bk1共j1兲其k1. Assume Bob’s 1 outcome is k1, states 兵A j1兩i典其i become 兵A j1 丢 Bk1共j1兲兩i典其i .
共26兲
Alice and Bob then repeat this process. After N0 rounds of measurement and communication, there are many possible outcomes corresponding to measurement operators acting on Alice and Bob’s Hilbert space, each of which is a product of the N0 operators, that is, 关A jN 共j1 , k1 , . . . , jN0−1 , kN0−1兲 0 丢 Bk 共j 1 , k1 , . . . , j N 兲兴 . . . 关A j 丢 Bk 共j 1兲兴. We denote these opN0 0 1 1 erators 兵Ai 丢 Bi其. They satisfy
兺i A†i 丢 B†i Ai 丢 Bi = 兺i M i = I,
共27兲
where M i = A†i Ai 丢 B†i Bi. We now introduce our main results summarized in the following theorems. Theorem 1. For a Cm 丢 Cn 共m , n ⱖ 3兲 system, in the orthogonal complement of the space spanned by any Schmidt rank-3 maximally entangled state 兩⌿典, there do not exist orthonormal bases that can be perfectly distinguished by LOCC protocols. mn−1 be a set of orthonormal To prove this, let = 兵兩i典其i=1 basis of the subspace. Suppose elements in can be distinguished by some LOCC protocol 兵M i其, then for every 兩i典
there exists at least one measurement operator M i that satisfies the conditions: M i兩 j典 = 0,
all j ⫽ i,
M i兩i典 ⫽ 0,
共28兲 共29兲
and we say M i identifies 兩i典. We sum all the operators identifying 兩i典 and define the sum as M i共s兲. Thus, 兵M i共s兲其i is a set of Hermitian positive separable operators satisfying 兺iM i共s兲 = I. As M i共s兲 is separable, and its support is orthogonal to 兵兩1典 , . . . , 兩i−1典 , 兩i+1典 , . . . , 兩mn−1典其, M i共s兲 can be written as M i共s兲 = 兺 y ki 兩ki 典具兩ki 兩 丢 兩ki 典具ki 兩,
共30兲
k
where 兵y ki 其’s are positive constant coefficients, and every element in 兵兩ki 典兩ki 典其 is a linear superposition of 兩⌿典 and 兩i典. When there exists in an entangled state whose Schmidt rank is greater than or equal to three 共say 兩i典兲, if elements in can be distinguished by some LOCC protocol, then according to Lemma 3, 兩i典 can be written as the superposition of at least three linearly independent product states, each of which is orthogonal to other elements in . This implies that there exist at least mn + 1 linearly independent states in a m 丢 n system. When there exists a Schmidt rank-2 entangled state in 共say 兩i典兲, then M i共s兲, the sum of all operators identifying 兩i典, is inseparable, as the superposition of 兩⌿典 and 兩i典 is inseparable according to Lemma 1. When all 兩i典’s are product states, then every M i共s兲 should be expressed as 兩i典具i兩. In order to keep the completeness of the LOCC protocol, we need to add measurement operators, the sum of which can be denoted as M 0 = I − 兺 兩i典具i兩 = 兩⌿典具⌿兩.
共31兲
i
M 0 is inseparable and cannot be realized locally. Therefore, elements in 兵兩i典其 cannot be distinguished by LOCC. This completes the proof of Theorem 1. Theorem 2. For a Cm 丢 Cn 共m , n ⬎ 3兲 system, in the orthogonal compliment of the space spanned by any entangled state whose Schmidt rank is greater than or equal to four, there do not exist orthonormal bases that can be perfectly distinguished by LOCC protocols. mn−1 be a set of orthonormal To prove this, let = 兵兩i典其i=1 basis of the subspace. When there exists in an entangled state with a Schmidt rank greater than or equal to three, states in cannot be distinguished by LOCC protocols according to Lemma 3. When there exists a Schmidt rank-2 entangled state in 共say 兩i典兲, then M i共s兲, the sum of all operators identifying 兩i典, is inseparable, as the superposition of 兩⌿典 and 兩i典 is inseparable according to Lemma 2. When all 兩i典’s are product states, the same corresponding method as that used in the proof of Theorem 1 can be used to prove the local indistinguishability of states in . This completes the proof. In comparison with Watrous’ results in 关2兴, the subspaces we find that have no locally distinguishable orthonormal bases are more general. In Watrous’ subspaces, the dimen-
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sions of Alice and Bob’s systems must be the same. The one-dimensional space is spanned by a maximally entangled state whose Schmidt rank must be the same with the dimensions of the systems of Alice and Bob. In our case, as proved in Theorem 1, the one dimensional space is spanned by a maximally entangled rank-3 state 兩⌿典 in a Cm 丢 Cn 共m , n ⱖ 3兲 system, and the Schmidt rank of the state is irrelevant to m and n. In Theorem 2, the one-dimensional space is spanned by any entangled state whose Schmidt rank is greater than three. Obviously, Watrous’ subspaces are a special case of our results. The advantage of our method is, by making use of the separability of LOCC operators and Lemmas 1–3, our conclusions can be derived without complicated mathematics and the idea can be comprehended easily.
In conclusion, we find subspaces without locally distinguishable orthonormal bases that are more general than known before. Our results are useful in studying important quantum information problems such as the classical capacity of quantum channels where local distinguishability of quantum states plays an essential role.
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We thank Xiang-Fa Zhou for helpful discussions and suggestions. This work was funded by National Fundamental Research Program No. 2006CB921900, the Innovation funds from Chinese Academy of Sciences, and National Natural Science Foundation of China 共Grants No. 60621064, No. 10574126, No. 10875110, and No. 60836001兲.
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