RAPID COMMUNICATIONS
PHYSICAL REVIEW A 69, 020301共R兲 共2004兲
Binegativity and geometry of entangled states in two qubits
Satoshi Ishizaka*
PRESTO, Japan Science and Technology Agency, 4-1-8 Honcho Kawaguchi, Saitama, Japan
and Fundamental Research Laboratories, NEC Corporation, 34 Miyukigaoka, Tsukuba, Ibaraki, Japan
共Received 10 August 2003; revised manuscript received 11 November 2003; published 17 February 2004兲
We prove that the binegativity is always positive for any two-qubit state. As a result, and as suggested by
previous work, the asymptotic relative entropy of entanglement in two qubits does not exceed the Rains bound,
and the positive partial transposed-entanglement cost for any two-qubit state is determined to be the logarithmic negativity of the state. Further, the proof reveals some geometrical characteristics of the entangled states,
and shows that partial transposition can give another separable approximation of the entangled state in two
qubits.
DOI: 10.1103/PhysRevA.69.020301
PACS number共s兲: 03.67.⫺a, 03.65.Ud
Quantum entanglement plays an essential role in many
quantum information tasks, and qualitative and quantitative
understanding of the entanglement is one of the important
topics in quantum information theory. An important mathematical operation in the theory of entanglement is partial
transposition ␴ T B 关1兴, where only the basis on one party, say,
Bob, is transposed. The states that satisfy ␴ T B ⭓0 are called
positive partial transposed 共PPT兲 states, and all separable
states are PPT states. Further, it has been shown that all PPT
states in 2 丢 2 共two qubits兲 and 2 丢 3 are separable states 关2兴.
Recently, Audenaert, De Moor, Vollbrecht, and Werner
introduced an interesting important mathematical operation,
binegativity 兩 ␴ T B 兩 T B 关3兴. They showed that, if 兩 ␴ T B 兩 T B ⭓0
holds, the asymptotic relative entropy of entanglement with
respect to PPT states does not exceed the so-called Rains
bound. Further they showed that 兩 ␴ T B 兩 T B ⭓0 holds for many
classes of states and conjectured that it holds for any twoqubit state. Subsequently, Audenaert, Plenio, and Eisert
showed that, if 兩 ␴ T B 兩 T B ⭓0 holds, the PPT-entanglement cost
for the exact preparation is given by the logarithmic negativity 关4兴. By this, they provided operational meaning to logarithmic negativity.
In this Rapid Communication, we prove that 兩 ␴ T B 兩 T B ⭓0
indeed holds for any two-qubit state. The proof is geometrical in some sense, and reveals some geometrical characteristics of the entangled states. Further, it is found that partial
transposition can give another separable approximation of
the entangled state in two qubits.
Before starting proof of 兩 ␴ T B 兩 T B ⭓0, we briefly review
several concepts necessary to the proof. The first is the entanglement witness 关2,5,6兴, which is the Hermitian operator
W such that tr W%⭓0 for all separable states %, and tr W ␴
⬍0 for some entangled states ␴ . This is expressed such that
” 0, ( 兩 ␺ 典具 ␺ 兩 ) T B is
W detects the entanglement of ␴ . For ␴ T B ⭓
the entanglement witness where 兩 ␺ 典 is the eigenstate of ␴ T B
for a negative eigenvalue 关7兴. Since the entanglement witness
cannot be a positive operator 关6兴, 兩 ␺ 典 must always be entangled.
The concept of the entanglement witness is related to the
existence of the hyperplane that separates the closed convex
set of separable states and some entangled states. W itself
*Electronic mail: isizaka@frl.cl.nec.co.jp
1050-2947/2004/69共2兲/020301共4兲/$22.50
plays the role of normal vector of the hyperplane. The concept can be applicable to another closed convex set of positive operators: if tr ␴ 兩 ␾ 典具 ␾ 兩 ⬍0, it is certain that ␴ is not
positive. In this Rapid Communication, by analogy to the
entanglement witness, we call 兩 ␾ 典具 ␾ 兩 witness of the nonpositivity, and we say that 兩 ␾ 典具 ␾ 兩 detects the nonpositivity
of ␴ .
The second concept relates to the state representation
based on local filtering. According to Ref. 关8兴, all states in
two qubits can be transformed 共leaving out normalization兲 by
local filtering of full rank into either the Bell diagonal states
or the states of
␴ c⫽
1
2
冉
a⫹c
0
0
d
0
0
0
0
0
0
b⫺c
0
d
0
0
a⫺b
冊
,
共1兲
with a, b, c and d being real. Therefore, all states in two
qubits can be represented by either
␴⫽
1
N
3
兺
i⫽0
p i 共 A 丢 B 兲 兩 e i 典具 e i 兩 共 A † 丢 B † 兲
共2兲
or
␴⫽
1
共 A 丢 B 兲 ␴ c共 A † 丢 B † 兲 ,
N
共3兲
where N is the normalization, 兺 i p i ⫽1, and 兩 e i 典 is the set of
orthogonal Bell basis. Without loss of generality, we can fix
兩 e i 典 ⫽ 兵 兩 ␺ ⫺ 典 , 兩 ␺ ⫹ 典 , 兩 ␾ ⫺ 典 , 兩 ␾ ⫹ 典 其 关9,10兴, where 兩 ␺ ⫾ 典 ⫽( 兩 01典
⫾ 兩 10典 )/ 冑2 and 兩 ␾ ⫾ 典 ⫽( 兩 00典 ⫾ 兩 11典 )/ 冑2, and we can assume that p 0 is largest and p 3 is smallest among p i 关9–11兴.
Further, since A and B are full rank, we can put det A
⫽det B⫽1 without loss of generality. This leads to a convenient relation of A † Ã⫽B † B̃⫽I 关8,12兴, where the tilde is defined as Ã⬅ ␴ 2 A * ␴ 2 for local operators and 兩 ˜␺ 典 ⬅( ␴ 2
丢 ␴ 2 ) 兩 ␺ * 典 for states 关13兴. These forms of state representation 关Eq. 共2兲 or 共3兲兴 are called normal form 关8兴.
Then, let us start the proof of 兩 ␴ T B 兩 T B ⭓0. It is trivial
when ␴ is a PPT state, since 兩 ␴ T B 兩 T B ⫽ ␴ ⭓0. Therefore, we
restrict ourselves to the case where ␴ is entangled. Partial
69 020301-1
©2004 The American Physical Society
RAPID COMMUNICATIONS
PHYSICAL REVIEW A 69, 020301共R兲 共2004兲
SATOSHI ISHIZAKA
FIG. 1. Schematic of a set of positive operators 共quantum states兲
and a set of PPT operators. ␴ and 兩 ␴ T B 兩 T B are symmetrically located
with respect to P T B .
transposition of ␴ can be written as ␴ T B ⫽ P⫺␭ 兩 ␺ 典具 ␺ 兩 ,
where 兩 ␺ 典 is the 共normalized兲 eigenstate for the negative
eigenvalue of ⫺␭ 共there is only one negative eigenvalue for
an entangled state in two qubits 关14兴兲. The remainder of P is
the 共unnormalized兲 positive part ( P⭓0), which is orthogonal to 兩 ␺ 典 , and hence P 兩 ␺ 典 ⫽0. Then, ␴ and 兩 ␴ T B 兩 T B are
␴ ⫽ P T B ⫺␭ 共 兩 ␺ 典具 ␺ 兩 兲 T B ,
兩 ␴ T B 兩 T B ⫽ P T B ⫹␭ 共 兩 ␺ 典具 ␺ 兩 兲 T B .
共4兲
Here, it is worth discussing the geometrical meaning of
the problem. Let us consider the space of all Hermitian operators. The set of quantum states which are positive operators is a subset of the whole 共we do not care about normalization explicitly兲. Further, let us consider PPT operators,
which are those in which partial transposed operators are
positive 共a PPT operator is either positive or nonpositive兲.
These two sets are schematically shown in Fig. 1 共this Fig. 1
is essentially the same as Fig. 1 in Ref. 关15兴兲. Hereafter, we
call these sets the positive ball and PPT ball, although their
actual shape is not a spherical ball 关15兴. The intersection of
the two balls corresponds to the set of PPT states 共separable
states兲. Therefore, entangled state ␴ is located in the positive
ball outside the intersection. Since 兩 ␴ T B 兩 T B is a PPT operator
关 ( 兩 ␴ T B 兩 T B ) T B ⫽ 兩 ␴ T B 兩 ⭓0 兴 , it is contained in the PPT ball.
Then, the geometrical meaning to prove 兩 ␴ T B 兩 T B ⭓0 is to
prove that 兩 ␴ T B 兩 T B is always located in the intersection.
Further, let us pay attention to the geometrical location of
P T B : it is located on the middle of the line connecting ␴ and
兩 ␴ T B 兩 T B since P T B ⫽ ␴ /2⫹ 兩 ␴ T B 兩 T B /2. In addition, P T B must
be located on the edge of the PPT ball, since P T B itself is a
PPT operator and its partial transposition is rank deficient
( P 兩 ␺ 典 ⫽0). However, so that 兩 ␴ T B 兩 T B is located at the intersection, it is geometrically obvious that P T B must be located
on the edge of the intersection 共thick solid curve in Fig. 1兲.
This corresponds to P T B ⬎0, which is indeed necessary for
兩 ␴ T B 兩 T B ⭓0 because 兩 ␴ T B 兩 T B ⫽2 P T B ⫺ ␴ .
Then, the proof is as follows: We first prove a lemma
which simplifies proof of 兩 ␴ T B 兩 T B ⭓0. Second, we prove that
P T B ⬎0 共positive definite兲 whenever a given state ␴ is entangled. Finally, we search for an operator X at the intersection such that 兩 ␴ T B 兩 T B is located on the line connecting P T B
and X 共see Fig. 1兲. As a result, it is found that 兩 ␴ T B 兩 T B can be
always written as a convex sum of two positive operators
( P T B and X), which can complete the proof.
The key point of the proof for 兩 ␴ T B 兩 T B ⭓0 is to represent
P 共not ␴ ) in the normal form mentioned before. A lemma we
first prove is concerned with the state representation of P.
Lemma 1. Let ␴ be any entangled state in two qubits and
write it as ␴ T B ⫽ P⫺␭ 兩 ␺ 典具 ␺ 兩 where P⭓0 and P 兩 ␺ 典 ⫽0. If P
is rank 3, P is always represented in the normal form of Eq.
共2兲. If the rank of P is less than 3, there always exist ␴ ⬘ in
the vicinity of ␴ such that ␴ ⬘ T B ⫽ P ⬘ ⫺␭ 兩 ␺ 典具 ␺ 兩 where
P ⬘ 兩 ␺ 典 ⫽0, P ⬘ is rank 3, and P ⬘ is represented in the normal
form of Eq. 共2兲.
Proof. In the case where P is rank 3, let us assume that P
is represented in the normal form of Eq. 共3兲 as P⫽1/N(A
丢 B) ␴ c (A † 丢 B † ) where ␴ c is given by Eq. 共1兲. Since P is
assumed to be rank 3 and A 丢 B is full rank, ␴ c must be rank
3. By using A † Ã⫽B † B̃⫽I, it can be easily checked that only
the state of 兩 ␺ 典 ⫽1/ 冑M (Ã 丢 B̃) 兩 01典 satisfies P 兩 ␺ 典 ⫽0 where
M is the normalization. However, this state is a product state
which contradicts that 兩 ␺ 典 must be an entangled state in order that ( 兩 ␺ 典具 ␺ 兩 ) T B is an entanglement witness that detects
the entanglement of ␴ 共an entanglement witness cannot be a
positive operator as mentioned before兲. Therefore, P of rank
3 must be represented in the normal form of Eq. 共2兲 as
1
P⫽
N
3
兺
i⫽0
p i 共 A 丢 B 兲 兩 e i 典具 e i 兩 共 A † 丢 B † 兲 ,
共5兲
where p 3 ⫽0 in order that P is rank 3 since p 3 is smallest
among the p i . Further, by fixing the Bell basis as 兩 e i 典
⫽ 兵 兩 ␺ ⫺ 典 , 兩 ␺ ⫹ 典 , 兩 ␾ ⫺ 典 , 兩 ␾ ⫹ 典 其 , it is found that only the state of
兩 ␺ 典 ⫽ 共 1/ 冑M 兲 共 Ã 丢 B̃ 兲 兩 ␾ ⫹ 典
共6兲
satisfies P 兩 ␺ 典 ⫽0.
In the case where the rank of P is less than 3, there always
exist P ⬘ of rank 3 in the vicinity of P such that P ⬘ 兩 ␺ 典 ⫽0
共for example, if P is rank 2, using 兩 ␺⬜ 典 orthogonal to 兩 ␺ 典
and satisfying P 兩 ␺⬜ 典 ⫽0, let P ⬘ ⫽ P⫹ ⑀ 兩 ␺⬜ 典具 ␺⬜ 兩 with ⑀ being an infinitesimally small positive value兲. This P ⬘ must be
represented in the normal form of Eq. 共2兲 for the same reason
discussed above ( 兩 ␺ 典 satisfying P ⬘ 兩 ␺ 典 ⫽0 is entangled兲.
Since P ⬘ is in the vicinity of P, ␴ ⬘ ⫽ P ⬘ T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B is
䊏
also in the vicinity of ␴ .
It should be noted that the rank of P will be shown to be
3, and the possibility of the second case in Lemma 1 will be
denied 共see the corollary 1 below兲.
The next task for proof of 兩 ␴ T B 兩 T B ⭓0 is to prove P T B
⬎0 whenever ␴ is entangled. To this end, it suffices to show
” 0, ␴ ⫽ P T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B cannot be
that, if we assume P T B ⬎
any entangled state for ␭⬎0 and for 兩 ␺ 典 satisfying P 兩 ␺ 典
⫽0. According to Lemma 1, P 共or P ⬘ in the close vicinity of
P) must be written like Eq. 共5兲 at least so that ␴ is an entangled state. For those P 共or P ⬘ ) of rank 3, Eq. 共6兲 is only
the state that satisfies P 兩 ␺ 典 ⫽0 共or P ⬘ 兩 ␺ 典 ⫽0). Therefore,
” 0, ␴
in the following, we shall show that, if P T B ⬎
⫽ P T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B cannot be positive 共and hence cannot
be an entangled state兲 for every P of Eq. 共5兲 and for 兩 ␺ 典 of
Eq. 共6兲. According to this, when the rank of P is less than 3,
since ␴ ⬘ ⫽ P ⬘ T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B cannot be positive as well, ␴
in the close vicinity of ␴ ⬘ cannot be positive.
020301-2
RAPID COMMUNICATIONS
PHYSICAL REVIEW A 69, 020301共R兲 共2004兲
RAPID COMMUNICATION
Partial transposition of Eq. 共5兲 is calculated as
1
P ⫽
N
TB
3
兺
i⫽0
1
⫽
2N
p i 共 A 丢 B * 兲共 兩 e i 典具 e i 兩 兲 T B 共 A † 丢 B T 兲
3
兺 共 1⫺2 p 3⫺i 兲共 A 丢 B * 兲 兩 e i 典具 e i兩 共 A † 丢 B T 兲 ,
i⫽0
共7兲
where we fixed 兩 e i 典 as 兵 兩 ␺ ⫺ 典 , 兩 ␺ ⫹ 典 , 兩 ␾ ⫺ 典 , 兩 ␾ ⫹ 典 其 and used
共 兩 ␺ ⫾ 典具 ␺ ⫾ 兩 兲 T B ⫽ 21 共 兩 ␺ ⫺ 典具 ␺ ⫺ 兩 ⫹ 兩 ␺ ⫹ 典具 ␺ ⫹ 兩
⫿ 兩 ␾ ⫺ 典具 ␾ ⫺ 兩 ⫾ 兩 ␾ ⫹ 典具 ␾ ⫹ 兩 兲 ,
共 兩 ␾ ⫾ 典具 ␾ ⫾ 兩 兲 T B ⫽ 21 共 ⫿ 兩 ␺ ⫺ 典具 ␺ ⫺ 兩 ⫾ 兩 ␺ ⫹ 典具 ␺ ⫹ 兩
⫺
⫺
⫹
共8兲
⫹
⫹ 兩 ␾ 典具 ␾ 兩 ⫹ 兩 ␾ 典具 ␾ 兩 兲 .
Further, P T B ⬎
” 0 corresponds to p 0 ⭓1/2 in Eq. 共7兲, since 1
⫺2p 0 is smallest among the 1⫺2p i ( P T B is positive
semidefinite for p 0 ⫽1/2, and has a negative eigenvalue for
p 0 ⬎1/2). By introducing the state of
兩 ␾ 典 ⫽ 共 1/ 冑L 兲 共 Ã 丢 B̃ * 兲 兩 ␾ ⫹ 典 ,
共9兲
where L is the normalization, it is found that
具 ␾ 兩 ␴ 兩 ␾ 典 ⫽ 具 ␾ 兩 P T B 兩 ␾ 典 ⫺␭ 具 ␾ 兩 共 兩 ␺ 典具 ␺ 兩 兲 T B 兩 ␾ 典
⫽
1⫺2p 0
⫺␭ tr兩 ␾ 典具 ␾ 兩 共 兩 ␺ 典具 ␺ 兩 兲 T B
2NL
⫽
1⫺2p 0
␭
⫺
tr共 C 丢 I 兲 V 共 C † 丢 I 兲 P ⫹
2NL
2M L
⫽
␭
1⫺2 p 0
⫺
tr CC * ,
2NL
4M L
共10兲
where V⫽2( 兩 ␾ ⫹ 典具 ␾ ⫹ 兩 ) T B is the flip operator, and C
⬅H 1 H 2 is the product of two positive definite operators,
H 1 ⬅Ã † Ã and H 2 ⬅(B̃ † B̃) * ⫽B̃ T B̃ * . In the third equality, we
Using
used
(Y 丢 Z)V(Y † 丢 Z † )⫽(Y Z † 丢 I)V(ZY † 丢 I).
det H 1 ⫽det H 2 ⫽det C⫽1, it can be shown that tr CC * ⭓2
关16兴. As a result, it is found that 具 ␾ 兩 ␴ 兩 ␾ 典 ⬍0 for ␭⬎0, and
␴ cannot be positive where it is assumed that P T B ⬎
” 0. In this
way, 兩 ␾ 典具 ␾ 兩 works as a witness operator that detects the
nonpositivity of ␴ . Then, the following theorem was proven.
Theorem 1. For any two-qubit state ␴ , the positive part
共P兲 of ␴ T B is a PPT state. Further, if ␴ is entangled, the
partial transposition of the positive part ( P T B ) is full rank.
It has been shown that, if the rank of a separable state is
less than 3, the rank of its partial transposition agrees with
the rank of the original separable state 关17,18兴. Therefore, the
fact that P T B is separable and full rank implies that P, which
is rank deficient, must be rank 3, we obtain the following
corollary.
Corollary 1. For any entangled state ␴ in two qubits, the
positive part 共P兲 of ␴ T B is rank 3 关19兴.
Theorem 1 states that, in some sense, partial transposition
in two qubits can also give separable approximations of the
FIG. 2. P T B is located at the point at which two edges cross.
Two hyperplanes of entanglement witness 共EW兲 and nonpositivity
witness 共NPW兲 are indicated by gray lines.
entangled states as well as the best separable approximation
共BSA兲 关18,20兴, the closest disentangled state in the relative
entropy measure 关21兴, and so on. Every entangled state in
two qubits can be decomposed into the separable approximation 1/(1⫹␭) P T B 共normalized兲 and the deviation from it
关 ( 兩 ␺ 典具 ␺ 兩 ) T B 兴 .
Further, it is important to discuss the geometrical meaning
of 兩 ␾ 典 in Eq. 共9兲. In the case of p 0 ⫽1/2, P T B in Eq. 共7兲
becomes positive semidefinite 共rank 3兲 and it can be seen
that 兩 ␾ 典 satisfies P T B 兩 ␾ 典 ⫽0. Geometrically, P T B of rank 3 is
just located at the crossing point of two edges as shown in
Fig. 2. The hyperplane that corresponds to the entanglement
witness of ( 兩 ␺ 典具 ␺ 兩 ) T B is in contact with the PPT ball at the
point of crossing, and P T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B is located in the
direction perpendicular to this hyperplane 共since the entanglement witness plays the role of normal vector兲. In addition, the nonpositivity witness 兩 ␾ 典具 ␾ 兩 , which is the kernel of
P T B , also specifies a hyperplane which is in contact with the
positive ball at the crossing point. What we showed in the
proof of theorem 1 is that these two hyperplanes always
cross with shallow angles so that the nonpositivity of P T B
⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B is always detected by the hyperplane specified by 兩 ␾ 典具 ␾ 兩 . The inner product of two normal vectors of
the hyperplanes corresponds to tr( 兩 ␺ 典具 ␺ 兩 ) T B 兩 ␾ 典具 ␾ 兩 , which
was shown to always be positive. It is important to note that,
T
in the case of the BSA (% s ), two kernels of % s and % s B also
play a crucial role 关18兴.
The remaining task for proof of 兩 ␴ T B 兩 T B ⭓0 is to search
for a positive X operator. According to lemma 1 and corol-
FIG. 3. An upper bound, ␭ 0 , is obtained so the nonpositivity of
␴ is not detected by 兩 ␾ 典具 ␾ 兩 . X is located apart from P T B by ␭ 0 on
a side opposite that of ␴ .
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SATOSHI ISHIZAKA
lary 1, P and P T B are represented by Eqs. 共5兲 and 共7兲, respectively, when ␴ is an entangled state. Further, according to
P T B ⬎0 共theorem 1兲, p 0 ⬍1/2 共and p 3 ⫽0 since P is rank 3兲.
The range of ␭ is limited so that ␴ ⫽ P T B ⫺␭( 兩 ␺ 典具 ␺ 兩 ) T B
⭓0, and an upper bound of ␭ must be found. It is slightly
surprising that the hyperplane of 兩 ␾ 典具 ␾ 兩 also plays a crucial
role in this. Using 兩 ␾ 典 in Eq. 共9兲 and 具 ␾ 兩 ␴ 兩 ␾ 典 in Eq. 共10兲
共but p 0 ⬍1/2 here兲, the condition 具 ␾ 兩 ␴ 兩 ␾ 典 ⭓0 leads to
␭⭐(1⫺2 p 0 )M /N⬅␭ 0 共we again used tr CC * ⭓2). Then,
we define operator X as X⬅ P T B ⫹␭ 0 ( 兩 ␺ 典具 ␺ 兩 ) T B , whose
geometrical location is shown in Fig. 3. This X is always
positive as shown below. Let us introduce
X ⬘ ⫽2N 共 Ã † 丢 B̃ T 兲 X 共 Ã 丢 B̃ * 兲
2
⫽2
兺 共 p 0 ⫺p 3⫺i 兲 兩 e i 典具 e i兩
i⫽0
⫹ 共 1⫺2 p 0 兲关 I 丢 I⫹ 共 H 1 丢 H 2 兲 V 共 H 1 丢 H 2 兲兴
2
⫽2
兺 共 p 0 ⫺p 3⫺i 兲 兩 e i 典具 e i兩 ⫹ 共 1⫺2 p 0 兲
i⫽0
⫻ 共 C 丢 I 兲关 C̃ † C̃ 丢 I⫹V 兴共 C † 丢 I 兲 .
共11兲
Since A and B are full rank, X⭓0 if and only if X ⬘ ⭓0. The
first term of X ⬘ is positive since p 0 is largest. According to
Ref. 关20兴, for a given R⭓0, if 兩 ␰ 典 belongs to the range of R
and ␬ ⭐1/具 ␰ 兩 R ⫺1 兩 ␰ 典 , then R⫺ ␬ 兩 ␰ 典具 ␰ 兩 ⭓0. Since det(C̃ † C̃)
⫽1, the eigenvalues of C̃ † C̃ are written as 兵 t,1/t 其 , and we
obtain
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共2002兲.
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010101 共2001兲.
关9兴 First convert the given Bell diagonal state by a suitable local
unitary into canonical Hilbert–Schmidt form where the basis
set is 兩 e i 典 ⫽ 兵 兩 ␺ ⫺ 典 , 兩 ␺ ⫹ 典 , 兩 ␾ ⫺ 典 , 兩 ␾ ⫹ 典 其 共Ref. 关10兴兲. After that, if
p k is largest, convert 兩 e k 典 to 兩 ␺ ⫺ 典 by unilateral ␲ rotation
共Ref. 关11兴兲. Finally, if p l is smallest, convert 兩 e l 典 to 兩 ␾ ⫹ 典 by
bilateral ␲ /2 rotation 共Ref. 关11兴兲, where 兩 ␺ ⫺ 典 remains unchanged. These transformations can be absorbed into A and B.
关10兴 R. Horodecki and M. Horodecki, Phys. Rev. A 54, 1838
共1996兲.
关11兴 C.H. Bennett et al., Phys. Rev. A 54, 3824 共1996兲.
1
2
具 ␺ ⫺ 兩 关共 C̃ † C̃⫹I 兲 丢 I 兴 ⫺1 兩 ␺ ⫺ 典 ⫽ tr共 C̃ † C̃⫹I 兲 ⫺1
⫽
冉
冊
1 1
1
1
⫽ ,
⫹
2 t⫹1 1/t⫹1
2
共12兲
and C̃ † C̃ 丢 I⫹V⫽(C̃ † C̃⫹I) 丢 I⫺2 兩 ␺ ⫺ 典具 ␺ ⫺ 兩 ⭓0. As a result, since p 0 ⬍1/2, the second term of X ⬘ is also positive
and X ⬘ is found to be positive. Since 0⬍␭⭐␭ 0 , 兩 ␴ T B 兩 T B can
be always written as a convex sum of two positive operators
(X and P T B ), the following theorem was proven.
Theorem 2. 兩 ␴ T B 兩 T B ⭓0 for any two-qubit state ␴ .
Finally, we briefly discuss the case in higher dimensional
systems. It has been already mentioned that 兩 ␴ T B 兩 T B ⭓0 does
not hold in general, and that the states violating 兩 ␴ T B 兩 T B ⭓0
are called binegative states 关3兴. In order to obtain some insights into how the binegative states emerge, we numerically
generated the random binegative states of full rank in two
qutrits, and confirmed that P T B ⬅ ␴ /2⫹ 兩 ␴ T B 兩 T B /2 is not positive in general. This implies that the necessary condition corresponding to theorem 1 is already violated in the higher
dimensional systems 共binegative states that satisfy theorem 1
also exist兲. This seems to imply that the two hyperplanes at
the crossing point 共like those shown in Fig. 2兲 sometimes
cross with steep angles 共it was shown to be always shallow
in two qubits兲.
It will be important to further clarify the geometry of state
space, and it might lead to geometrical understanding of
quantum information tasks.
关12兴
关13兴
关14兴
关15兴
L.-X. Cen et al., Phys. Rev. A 65, 052318 共2002兲.
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F. Verstraete et al., J. Phys. A 34, 10327 共2001兲.
F. Verstraete, J. Dehaene, and B. De Moor, J. Mod. Opt. 49,
1277 共2002兲.
关16兴 For C⫽H 1 H 2 ⫽( ␥␣ ␦␤ ) and det C⫽ ␣ ␦ ⫺ ␤␥ ⫽1, tr CC *
⫽ 兩 ␣ 兩 2 ⫹ 兩 ␦ 兩 2 ⫹ ␤␥ * ⫹ ␤ * ␥ ⫽ 兩 ␣ ⫺ ␦ * 兩 2 ⫹ ␣ ␦ ⫹ ␣ * ␦ * ⫹ ␤␥ *
⫹ ␤ * ␥ ⫽ 兩 ␣ ⫺ ␦ * 兩 2 ⫹ ( ␤ ⫹ ␤ * )( ␥ ⫹ ␥ * ) ⫹ 2. Then, putting
a
b ⫺ic
H k ⫽( b ⫹k ic k d k ), where a k ,d k ⬎0, and b k and c k are real,
k
关17兴
关18兴
关19兴
关20兴
关21兴
020301-4
k
k
tr CC * ⫽ (a 1 a 2 ⫺d 1 d 2 ) 2 ⫹ 4(a 1 b 2 ⫹ d 2 b 1 )(a 2 b 1 ⫹ d 1 b 2 ) ⫹ 2
2
2
⫽(a 2 ⫺d 2 ) 2 ⫹(a⫹d) 2 b ⫹
⫺(a⫺d) 2 b ⫺
⫹2, where a⫽ 冑a 1 a 2 ,
1/4
d⫽ 冑d 1 d 2 and b ⫾ ⫽(a 2 d 2 /a 1 d 1 ) b 1 ⫾(a 1 d 1 /a 2 d 2 ) 1/4b 2 .
Since det H k ⫽1 and c k must be real, b 2k ⭐a k d k ⫺1. Then,
2
⭐4ad and tr CC * ⭓(a⫺d) 4 ⫹ (a ⫹ d) 2 b 2⫹ ⫹ 2⭓2.
b⫺
A. Sanpera, R. Tarrach, and G. Vidal, Phys. Rev. A 58, 826
共1998兲.
T. Wellens and M. Kuś, Phys. Rev. A 64, 052302 共2001兲.
This is consistent with the fact, discussed in Ref. 关14兴, that
partial transpose of an entangled state is always full rank.
M. Lewenstein and A. Sanpera, Phys. Rev. Lett. 80, 2261
共1998兲.
V. Vedral et al., Phys. Rev. Lett. 78, 2275 共1997兲.