Separable State Decompositions for a Class of Mixed States
Ting-Gui Zhang1 , Xiaofen Huang∗,2 ,Xianqing Li-Jost1 , Naihuan Jing3,4 and Shao-Ming
arXiv:1307.6182v1 [quant-ph] 23 Jul 2013
Fei1,5 1
Max-Planck-Institute for Mathematics in the Sciences, Leipzig, 04103, Germany
2
School of Mathematics and Statistics, Hainan Normal University, Haikou, 571158, China
3
School of Sciences, South China University of Technology, Guangzhou, 510640, China
4
Department of Mathematics, North Carolina State University, Raleigh, NC27695, USA
5
School of Mathematical Sciences, Capital Normal University, Beijing 100048, China
∗ Corresponding author, e-mail address:
[email protected]
Abstract We study certain quantum states for which the PPT criterion is both sufficient and necessary for separability. A class of n × n bipartite mixed states is presented and the conditions of PPT for these states are derived. The separable pure state decompositions of these states are explicitly constructed when they are PPT.
Quantum entangled states have become one of the key resources in quantum information processing. The study of quantum teleportation, quantum cryptography, quantum dense coding, quantum error correction and parallel computation [1–3] has spurred a flurry of activities in the investigation of quantum entanglement. Despite the potential applications of quantum entangled states, there are many open questions in the theory of quantum entanglement. The separability of quantum mixed states is one of the important problems in the theory of quantum entanglement. Let H be an n-dimensional complex Hilbert space, with |ii, i = 1, ..., n the orthonormal basis. A bipartite mixed state in H ⊗ H is said to be separable if the density matrix can be 1
written as ρ=
X i
where 0 < pi ≤ 1,
P
i
pi = 1,
ρ1i
and
ρ2i
pi ρ1i ⊗ ρ2i ,
(1)
are density matrices associated with the first and
the second Hilbert spaces respectively. It is a challenge to find a decomposition like (1) or to prove that such a decomposition does not exist for a given state ρ. With considerable effort in analyzing the separability, there have been some (necessary) criterias for separability in recent years, for instance, Bell inequalities [4], PPT (positive partial transposition) [5], reduction criterion [6, 7], majorization criterion [8], entanglement witnesses [9, 10], realignment [11, 12] and generalized realignment [13], range criterion [14], criteria based on the local uncertainty relations [15], correlation matrix approach [16], as well as some necessary and sufficient criterias for low rank density matrices [17–19]. The PPT criterion is generally a necessary condition for separability. It becomes also sufficient for the cases of 2 ⊗ 2 and 2 ⊗ 3 bipartite states [21]. In [22], it has been shown that a state ρ supported on m × n Hilbert space (m ≤ n) with rank(ρ) ≤ m is separable if and only if ρ is PPT. However, it is generally a difficult problem to find the concrete PPT conditions for a given such state within this class. Moreover, even if the PPT conditions are satisfied and hence the state is separable, it is still a challenging problem to find the detailed separable pure state decompositions (1). For separable two-qubit states, an elegant separable pure state decompositions has been given in [20]. In [23] a class of 3 ⊗ 3 mixed states ρ with rank(ρ) = 3 has been investigated. The PPT conditions are derived. And the explicit separable pure state decompositions are constructed. In this paper we generalize the results in [23] to a class of n ⊗ n quantum mixed states. We derive the PPT conditions and construct explicitly the separable pure state decompositions for states satisfying the PPT conditions. We consider a set of mixed states defined in H ⊗ H space which has the following form of spectral decomposition: ρ=
n X l=1
λl |Vl ihVl |,
2
(2)
with
Pn
l=1
λl = 1, 0 < λl < 1, and n X
|Vl i = where 0 6= vlj ∈ C and
j=1
vlj |ji ⊗ |j + l − 1i, l = 1, 2, · · · , n,
(3)
vl j vlj = 1. (z denoting the complex conjugation of z). When √ n = 3, the state ρ becomes the object of study in [23]. For simplicity we denote xjl = λl vlj , √ |Xl i = λl |Vl i. Then ρ has the form, ρ=
n X l=1
|Xl ihXl | =
P
j
n X
l,j,k=1
xjl xkl |ji⊗|j+l−1ihk|⊗hk+l−1|
=
n X
l,j,k=1
xjl xkl |jihk|⊗|j+l−1ihk+l−1|.
We first deduce the PPT conditions of ρ. The partial transposed matrix of ρ is given by T1
ρ
n X
=
l,j,k=1
xjl xkl |kihj| ⊗ |j + l − 1ihk + l − 1|,
(4)
where T1 stands for partial transpose with respect to the first Hilbert space. That ρ is PPT P means that ρT1 ≥ 0. Namely, for any vector |Y i = nr,s=1 yrs+r−1|ri ⊗ |si in H ⊗ H, we
obtain hY |ρT1 |Y i ≥ 0. Here and later, we use s + r − 1 to represent s + r − 1 mod n, mod denoted modulo arithmetic. We have n X
T1
hY |ρ |Y i = hY | = hY | = = =
l,j,k=1 n X
l,j,k=1
n X
′
xjl
xkl |kihj|
⊗ |j + l − 1ihk + l − 1|
r,s=1
yrs+r−1|ri ⊗ |si
xjl xkl yrs+r−1δjr δs(k+l−1) |ki ⊗ |j + l − 1i ′
yrs′ +r −1 |r ′ i ⊗ |s′ i
r ′ ,s′ =1 n X
n X
n X
l,j,k=1
xjl xkl yjk+j+l−2|ki ⊗ |j + l − 1i
xjl xkl yjk+j+l−2 y rr′ +s −1 δrk′ δsj+l−1 ′
l,j,k,r ′,s′ =1 n X xjl xkl l,j,k=1
′
′
yjk+j+l−2 y kk+j+l−2 ≥ 0.
Because of the independence of the variables yrα , the above inequality is equivalent to the
3
following inequities: < y n−1|A1 |y n−1 > ≡ n
xjl xkl yjn−1 y kn−1 ≥ 0,
n X
xjl xkl yjn y nk ≥ 0,
l,j,k=1,l+j+k=1 mod n
n
< y |A2 |y > ≡
n X
l,j,k=1,l+j+k=2 mod n
··· < y n−3|An−3 |y n−1 > ≡ < y n−2 |An |y n−2 > ≡
n X
l,j,k=1,l+j+k=n−1 mod n n X xjl l,j,k=1,l+j+k=0 mod n
xjl xkl yjn−3 y kn−3 ≥ 0, xkl yjn−2 y kn−2 ≥ 0,
where |y i >= (y1i , y2i , ..., yni )t (t stands for transpose), i = 1, 2, ..., n. A1 , A1 , · · · , An are non-
negative, hermitian matrices, with the entries of Am given by xjl xkl for l + j + k = m mod (n). For fixed m, l = [m − (j + k)] mod n only depends on j, k. For example, when n = 5, one has x14 x14 x13 x23 x2 x1 x2 x2 3 3 2 2 3 2 3 1 A1 = x2 x2 x1 x1 x4 x1 x4 x2 1 1 5 5
x12 x32 x11 x41 x15 x55
x21 x31 x25 x45 x24 x54 x35 x35 x34 x44 x33 x53 ,··· . x44 x34 x43 x43 x42 x52 5 1 5 2 5 3 5 4 5 5 x5 x5 x4 x4 x3 x3 x2 x2 x1 x1
(5)
Due to the non-negativity of the matrices A1 , ..., An , all the principal minors of Am , m = 1, 2, ..., n, are non-negative. We have Theorem 1: The entries of the matrices Am , ∀ m = 1, 2, · · · , n satisfy the following quadratic relations, xip xjq = xiq xjp ,
(6)
where xip = xil(i,p) xpl(i,p) , l(i, p) = (m mod n) − (i + p) ≡ m − (i + p), the other marks have the same meaning. Proof: First, we consider order two principal minors {(i, j), (i, j)} of the matrix Am . From the non-negativity of Am , we get that xii xjj ≥ xij xji . The inequality is in fact an equality. Q Q Because if for some m, xii xjj > xij xji , then nm=1 xii xjj > nm=1 xij xji . On the other Q Q hand, from straightforward calculation, we have nm=1 xii xjj = nm=1 xij xji for fixed i, j. 4
Therefore, for any m we have xii xjj = xij xji .
(7)
Second, from the non-negativity of the order three principal minors {(i, j, k), (i, j, k)} of the matrix Am , we have 0 ≤ xij xjk xki − xii xjk xkj + xik xji xkj − xik xjj xki = xij xjk xki − 2xii xjj xkk + xik xji xkj = 2Re(xij xjk xki ) − 2xii xjj xkk ≤ 2(|xij ||xjk ||xki | − xii xjj xkk ) = √ √ √ 2( xii xjj xjj xkk xkk xii − xii xjj xkk ) = 0, where we have used the condition (7) in the first and the third equations. Therefore, we get the following relations: xij xjk xki = xii xjj xkk .
(8)
For a nonzero 3 × 3 hermitian matrix, if its order two and three principal minors are all 0, then it has only one eigenvalue, and all of its order two minors are 0. Therefore, we have xij xjk = xik xjj . Third,
combining
the
non-negativity
of
the
(9) order
four
principal
minors
{(i, j, k, l), (i, j, k, l)} of the matrix Am with (7) and (8), we have 6xii xjj xkk xll − 2Re(xil xlk xkj xji + xjl xlk xki xij + xjl xli xik xkj ) ≥ 0. Using relations (9), we have xii xjj xkk xll −Re(xil xlk xkj xji ) = xii xjj xkk xll −Re(xii xjj xkk xll ) = xii xjj xkk xll − xii xjj xkk xll = 0.
Hence all order four principal minors are all 0 and
xil xlk xkj xji = xii xjj xkk xll . Since order two, three and four principal minors are equivalent to 0, therefore, the nonzero 4 × 4 hermitian matrix (order four principal minors) has only one eigenvalue, then all of its order two minors are 0. Furthermore, all order two minors are included in one order four principal minors. Therefore the entries of Am satisfy the relations (6). From the condition (7), xjm−2j xjm−2j xkm−2k xkm−2k = xjm−(j+k) xkm−(j+k) xkm−(j+k) xjm−(j+k) , we have the following relations: jk
xjm−(j+k) xkm−(j+k) eiθm = xjm−2j xkm−2k
(10)
or jk
xjm xkm eiθm = xjm−(j−k) xkm+(j−k) , 5
(11)
jk where 0 ≤ θm ≤ 2π. jk Theorem 2: The number of independent θm is at most n − 1.
proof: First, from (9),we have xij xjk = xik xjj , i.e, xim−(i+j) xjm−(i+j) xjm−(j+k) xkm−(j+k) = xim−(i+k) xkm−(i+k) xjm−2j xjm−2j . Let j − i = k − j. We obtain xim−(i+j) xjm−(i+j) xjm−(j+k) xkm−(j+k) = xim−2j xkm−2j xjm−2j xjm−2j , which
gives
rise
to
xim−i+j xjm−i+j xjm+j−k xkm+j−k
=
xim xkm xjm xjm .
Namely, ij
xim+s xjm+s xjm−s xkm−s = xim xkm xjm xjm . However from (10) we have xim+s xjm−s = xim xjm eiθm and kj
xkm+s xjm−s = xkm xjm ei(−θm ) . Therefore, ij kj θm = θm
if j − i = k − j,
(12)
ij and θm depends on the difference of i and j. ij s 1 Set s = |i − j|. In the following, we denote θm as θm . In particular, we denote θm as θm . s There are [ n2 ] angles {θm } for given m, with s = 1, 2, · · · , [ n2 ], [x] denoting the integer that
is less or equal to x. Second, from (11) and (12), for any integer j ≤ [ n2 ] and given m, we can get the following P Q Q i sl=0 θm+l = sl=0 xjm+l+1 xj+1 equation: sl=0 xjm+l xj+1 m+l−1 . That is m+l e i xjm xj+1 m+s e
Following (13), j+2 i xj+1 m−1 x(m+s−1)−1 e
Ps
l=0 θm+l
= xjm+s+1 xj+1 m−1 .
we can get s equations: Ps−1 l=0
θm−1+l
=
j+2 xj+1 m+s−1 xm−1−1 ,
i xjm xj+1 m+s−1 e
···,
(13) Ps−1 l=0
θm+l
xjm+s xj+1 m−1 ,
=
j+s i xj+s−1 m−(s−1) xm e
Ps−1 l=0
θm−(s−1)+l
=
j+s xj+s−1 m+s−(s−1) xm−1−(s−1) . Multiplying these equations together, we get s
j+s j+s i(sθm +(s−1)(θm−1 +θm+1 )+···+(θm+s−1 +θm−s+1 )) j+s i(θm ) xjm xm e = xjm+s xm−1−(s−1) = xjm xm e , s i.e. any θm , s ≥ 2, m = 1, 2, · · · , n can be expressed according to the angles θm , m =
1, 2, · · · , n. Furthermore, Qn−1
for
fixed
j, k
or
s,
jk iΣn−1 m=0 θm
we
have
Qn−1
m=0
xjm−2j xkm−2k
=
xjm−(j+k) xkm−(j+k) e . On the other hand, by direct computation, we have Pn−1 s Qn−1 j Qn−1 j n−1 jk k k Hence Σm=0 θm = 0, or m=0 θm = 0, m=0 xm−2j xm−2k = m=0 xm−(j+k) xm−(j+k) . m=0
s = 1, 2, · · · , [ n2 ]. Therefore, there are in fact only n − 1 independent angles θm . 6
For example, using (11) we have x11 x21 eiθ1 = x12 x2n , x12 x22 eiθ2 = x13 x21 . Hence x11 x22 ei(θ1 +θ2 ) = x13 x2n . From x21 x31 eiθ1 = x22 x3n and x2n x3n eiθn = x21 x3n−1 , we get x31 x2n ei(θ1 +θn ) = x22 x3n−1 , which 13
12 +θ 12 +θ 23 +θ 23 ) n 2 1
give rise to x11 x31 ei(θ1 ) = x11 x31 ei(θ1
= x13 x3n−1 , i,e.
θ113 = θ12 = (2θ1 + θ2 + θn ).
(14)
We are now ready to construct pure separable state decompositions of ρ when ρ is PPT. Let U be a unitary transformation, with its entries given by ukl = ( √1n ei((k−1)(l−1)ω+δk ) ), where P δk k = 1, 2, · · · , n is an angle, ω is the n-th unit root, ω n = 1. Then ρ = nl=1 |Xl ihXl | = Pn l=1 |Zl ihZl |, where n n X X ukl |Xl i = blrs |rsi. (15) |Zl i = i,j=1
k=1
Denoting Bl = (blrs ), one has Bl = (blrs ) = ((ei((s−r)(l−1)ω+δs−r+1 ) xrs−r+1 )rs ). For example, when n = 5, one has u1l x11 u x2 5l 5 3 Bl = u4l x4 u3l x4 3 u2l x52
u2l x12 u3l x13 u4l x14 u5l x15
u1l x21 u2l x22 u3l x23 u4l x24 u5l x35 u1l x31 u2l x32 u3l x33 . u4l x44 u5l x45 u1l x41 u2l x42 u3l x53 u4l x54 u5l x55 u1l x51
(16)
Theorem 3: There exist δk such that every order two minors {(m, k), (α, β)} in Bl is zero, P and so that ρ = nl=1 |Zl ihZl | is a pure separable state decomposition for ρ that is PPT. Proof That any order two minors {(m, k), (α, β)} of Bl are zero implies: k m k ei(δα−m+1 +δβ−k+1−δα−k+1 −δβ−m+1 ) xm α−m+1 xβ−k+1 = xβ−m+1 xα−k+1 .
(17)
Namely, any order two minors {(m, m + 1), (α, α + 1)} should be zero, m+1 m m+1 ei(2δα−m+1 −δα−m+2 −δα−m ) xm α−m+1 xα−m+1 = xα−m+2 xα−m .
(18)
iθm From the PPT conditions, we have: xjm xj+1 = xjm+1 xj+1 m e m−1 . Applying Theorem 2, we have i(2δm −δm+1 −δm−1 ) j j+1 iθm xjm xj+1 = xjm+1 xj+1 . Therefore m e m−1 = xm xm e
2δi − δi+1 − δi−1 = θi , i = 1, 2, · · · , n. 7
(19)
Eq. (19) has always solutions for δi with the relationship
Pn
i=1 θi
= 0. As every order two
minors {(m, k), (α, β)} of Bl is zero, the rank of Bl is one. Therefore, |Zl i is separable. In fact the solutions of Eq. (19) are not unique. By calculating, we know that there is a free variable of the parameters δm , m = 1, 2, · · · , n, therefore exist many different separable pure state decompositions for such ρ. We have investigated a class of n ⊗ n bipartite mixed states for which the PPT criterion is both sufficient and necessary for separability. The PPT conditions for these states are derived. We have presented a general approach to find the separable pure state decompositions of this class, and the separable pure state decompositions have been explicitly constructed. Acknowledgments Huang gratefully acknowledges the grant support from the Natural Sciences Foundation for Hainan Province 111006. Jing is grateful for the support of Simons Foundation grant 198129 and NSFC grant 11271138 during the work. This work is also supported by the NSFC under number 11105226. We would like to acknowledge helpful discussions with Bobo Hua.
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