Common Resource State for Preparing Multipartite Quantum Systems via Local Operations and Classical Communication

0.16 MB·

by  Cheng Guo

Common Resource State for Preparing Multipartite Quantum
Systems via Local Operations and Classical Communication
6
Cheng Guo1, Eric Chitambar2, and Runyao Duan1
1
0 1Centre for Quantum Computation and Intelligent Systems (QCIS),
2 Faculty of Engineering and Information Technology,
University of Technology, Sydney (UTS), NSW 2007, Australia and
n 2Department of Physics and Astronomy,
a
Southern Illinois University, Carbondale, Illinois 62901, USA
J

3 Given a set of multipartite entangled states, can we find a common state to prepare them by
2 local operations and classical communication? Such a state, if exists, will be a common resource

for the given set of states. We completely solve this problem for bipartite pure states case by
]
h explicitly constructing a unique optimal common resource state for any given set of states. In
p the multipartite setting, the general problem becomes quite complicated, and we focus on finding
- nontrivial common resources for the whole multipartite state space of given dimensions. We show
nt that|GHZ3i=1/√3(|000i+|111i+|222i)isanontrivialcommonresourceforthree-qubitsystems.
a
u
The problem of transforming one entangled state to another by local operations and
q
[ classical communication (LOCC) is of central importance in quantum entanglement the-
ory. This problem has been studied extensively in last two decades, and many interesting
1
results havebeen reported. Notably,Nielsen pointed out thatlocaltransformationsamong
v
bipartite pure states can be completely characterized by an algebraic relation of majoriza-
0
2 tion between their Schmidt coefficient vectors [1]. The majorization characterization can
2 be extended to a class of multipartite pure states having Schmidt decompositions [2]. Un-
6 fortunately, the Schmidt decomposition for a generic multipartite pure state doesn’t exist,
0 and it is still an open problem to determine whether a general multipartite pure state can
.
1 be transformed into another one by LOCC.
0 Despitetheoverallcomplexityofmultipartiteentanglementtransformations,wecanoften
6 findentangledstatesthatcanbetransformedintoanyotherstateinthesame statespaceby
1 LOCC.These kind of states arecalled maximally entangled states, and they existin spaces
:
v if and only if the dimension of one subsystem is no less than the product of dimensions of
i all other subsystems [5]. Bell states are one such example, which can be transformed into
X
any pure state in two-qubit systems. As a simple corollary of the dimensionality bound,
r there is no maximally entangledstate in three-qubit systems. In fact, it is well-knownthat
a
three-qubit states can be entangled in two different ways, one class consisting of so-called
W-type states and other consisting of Greenberge–Horn–Zeilinger (GHZ)-type states [3].
While one copyof GHZ cannotbe stochasticallytransformedinto W , interestingly this
| i | i
transformation becomes possible at an asymptotic rate approaching unity if multi-copy
transformationis allowed [4].
In this paper, we generalize the notion of maximally entangled states with respect to
LOCCtransformations. Theproblemwestudycanbebestdescribedthroughthefollowing
scenario. Assume that Alice and Bob are going to implement a series of quantum informa-
tion tasks, each one requiring a different entangled state to perform. However, instead of
sharing a multitude of different states, they wish to share only one type of entangled state
and then transform this state into a different form as needed. Thus, the question is: for a
givenset ofpure entangledstates,is there a certainstate whichcanbe locally transformed
2
intoallofthembyLOCC?Below,we giveacomplete solutiontothis problemforbipartite
systems. In addition, we study the case when the set of target states is the entire state
spaceforsomegivendimensions. Whenthedimensionofonesubsysteminthistargetspace
is not smaller than the product of all the other subsystem dimensions, there always exists
some (perhaps higher-dimensional) state that can obtain all states in the target space [5].
However,this dimensionalityconditionturnsoutnottobe necessary. Interestingly,wefind
a non-trivial state GHZ = 1 (000 + 111 + 222 ) which can be transformed into any
| 3i √3 | i | i | i
three-qubit pure state by LOCC.
Let us now formulate our problem more precisely. Let S = ψ , ψ , be a set of
1 2
{| i | i ···}
(multipartite) states, possibly infinite. A common resource state ψ to S can be trans-
| i
formed into any state in S by LOCC. We say ψ an optimal common resource (OCR)
| i
if for any other common resource φ we have either φ can be transformed into ψ by
| i | i | i
LOCC,or φ and ψ are not comparableunder LOCC.In general,it is a hardproblemto
| i | i
find OCR for a set of multipartite states. For bipartite pure states, majorization charac-
terizes the LOCC transformation between two pure state [1]. Given a bipartite pure state
ψ , λψ denotes a probability vector whose entries are in descending order of the Schmidt
|coiefficients of ψ . For instance, if ψ = 1 0 0 + 1 1 2 + 1 2 1 , λψ = (1,1,1).
| i | i √2| i| i √6| i| i √3| i| i 2 3 6
k k
If λψ = (x , ,x ) and λφ = (y , ,y ) satisfy i) k,1 k d, x y , ii)
1 d 1 d j j
··· ··· ∀ ≤ ≤ ≤
jP=1 jP=1
d d
x = y , we say that λψ is majorized by λφ and write λψ λφ. Nielsen established
j j
≺
jP=1 jP=1
the following fundamental result: A bipartite pure state ψ canbe transformedto another
pure state φ by LOCC if and only if λψ λφ [1]. | i
| i ≺
Nielsen’s result together with the properties of majorization leads us to an explicit con-
struction of the unique OCR of a set of bipartite pure states.
Theorem 1. Let S = φ ,i I be a set of d d pure states, where I is an index set
i
{| i ∈ } ⊗
(finite or infinite). Assume that the Schmidt coefficient vector of φ is given by λ =
| ii φi
(x(i), ,x(i)).
1 ··· d
Then if I is a finite set, the OCR for S always exists and is unique. The OCR state ψ
is given by λψ =(y , ,y ), where | i
1 d
···
k k 1
y =min x(i) min − x(i).
k i I j − i I j
∈ Xj=1 ∈ Xj=1
Furthermore, if I is infinite, the min sign in above equations should be replaced with inf.
The proof is given in the Appendix.
Let us consider an example to demonstrate the application of the above theorem. Let
a d dimensional bipartite target set be S = φ λφ a , where a 1/d. Then an
OC−R ψ for S can be chosen as ψ such tahat λ{ψ| =i| (1a,≥1 a},1 a, ,1 ≥a). The maximal
| i a | i d−1 d−1 ··· d−1
d 1 − − −
entangled state 1 − kk is always a common resource but usually not optimal.
√d | i
kP=0
Weshallnowmovetomultipartitesetting,andconsidertheproblem“whatisthecommon
resource of the whole system”. For n partite quantum system d d where d
1 n 1
− ⊗···⊗ ≥
3
d d , the maximal entangled state exists, in the sense that all other states in the
2 n
≥···≥ n
systemcanbeobtainedfromthestatebyLOCC,ifandonlyifd d [5]. Forinstance,
1 i
≥
iQ=2
the state 1(000 + 101 + 210 + 311 ) is an OCR of tripartite system.
2 | i | i | i | i H4 ⊗H2 ⊗H2
Interestingly, the OCR exists even if any sub-system’s dimension is less than the product
of other sub-system’s dimensions.
What is the connection between ψ being an OCR for S and S being a possible multi-
| i
outcome image for some LOCC map performed on ψ ? The latter question was first
| i
posed by Jonathan and Plenio [6], and is more precisely stated as follows: If p(i) is an
arbitrary distribution over the φ S, when does there exist an LOCC transformation
i
| i ∈
thattransforms ψ into φ withprobabilityp(i)foreach ψ ? If ψ isanOCRforS,then
i i
| i | i | i | i
clearly ψ can generate the ensemble ψ ,p(i) ; Alice and Bob simply sample from I
i i I
with dis|triibution p(i) and then determ{in|istiically}p∈repare ψ given i. We can also see the
i
| i
possibility of generating ψ ,p(i) directly from Theorem1. For any l 1, ,d we
i i I
have {| i }∈ ∈{ ··· }
l l k k 1 l l
yk = mi iIn x(ji)−mi iIn − xj(i)=mi iIn xj(i) ≤ p(i) xj(i).
Xk=1 Xk=1 ∈ Xj=1 ∈ Xj=1 ∈ Xj=1 Xi I Xj=1
  ∈
Satisfying this inequality is both necessary and sufficient for transforming ψ into the
| i
ensemble ψ ,p(i) [6].
i i I
{| i }∈
Theorem 2. GHZ = 1 (000 + 111 + 222 ) is a common resource of three-qubit
| 3i √3 | i | i | i
system.
LOCC
Proof. If ψ can be transformed into φ via LOCC, we denote this as ψ φ .
| i | i | i −→ | i
TheproofdetailsoftheprotocolswillbegivenintheAppendix. Hereisanoutlineofthe
proof ideas. We divide the proof into two cases according to the target states are W-type
orGHZ-type states. The case ofGHZ can be further divide into two sub-cases: orthogonal
GHZ and non-orthogonalGHZ.
Ifthe targetstate φ is SLOCC equivalentto a W state, φ canbe writtenas x 000 +
0
| i | i | i
3
x 100 +x 010 +x 001 , where x are all positive real numbers and x2 = 1 [13].
1| i 2| i 3| i k k
kP=0
Then, we can do the following transforms:
i) GHZ LOCC x2+x2 000 +x 111 +x 222
| 3i −→ 0 1| i 2| i 3| i
ii) x2+x2 000p+x 111 +x 222 LOCC x2+x2 100 +x 010 +x 001
0 1| i 2| i 3| i −→ 0 1| i 2| i 3| i
iii)px2+x2 100 +x 010 +x 001 LOCCpφ . The stepi)hasbeenshownin[2],andthe
0 1| i 2| i 3| i −→ | i
steppiii) was given in [13]. For completeness we provided simple proofs in appendix.
If the targetstate φ is SLOCC equivalentto GHZ state, φ can be written as x000 +
| i | i | i
y φ φ φ . When 000φ φ φ = 0, we name it orthogonal GHZ case, otherwise non-
A B C A B C
| i h | i
orthogonalGHZ case [14].
For the orthogonal GHZ case, GHZ = 1 (000 + 111 ) LOCC φ [14]. Notice that,
| 2i √2 | i | i −→ | i
LOCC LOCC
GHZ GHZ [2]. Hence, GHZ φ .
3 2 3
| i −→ | i | i −→ | i
For the non-orthogonal GHZ case, suppose φ = a 0 + a 1 , φ = b 0 +
A 0 1 B 0
| i | i | i | i | i
b 1 and φ = c 0 + c 1 , and a ,a ,b ,b ,c and c are all real num-
1 C 0 1 0 1 0 1 0 1
| i | i | i | i
4
bers. Ψ = xc +ya b 2+ ya b 2 000 + xc 111 + yb 222 and Ψ =
1 0 0 0 1 0 1 1 2
| i | | | | | i | || i | || i | i
xc +ya b 2+pya b 2 000 + xc 101 + yb 210
0 0 0 1 0 1 1
| | | | | i | || i | || i
p
LOCC
i) GHZ Ψ
3 1
| i −→ | i
LOCC
ii) Ψ Ψ
1 2
| i −→ | i
LOCC
iii) Ψ x00φ +y φ φ 0
2 C A B
| i −→ | i | i
LOCC
iv) x00φ +y φ φ 0 x000 +y φ φ φ
C A B A B C
| i | i −→ | i | i
Again the step i) is from [2]. The step iii) is very technical, and is given in appendix.
The step iv): Charlie takes a local unitary operation φ 0 +(c 0 c 1 ) 1.
C 1 0
| ih | | i− | i h |
In conclusion, we introduce a notion of optimal common resource for a set of entangled
states, and explicitly construct it for any bipartite pure state set. We also show that
GHZ state is a nontrivial common resource for three-qubit system, and conjecture its
3
| i
optimality. We hope this problem will stimulate further research interest in entanglement
transformationtheory.
CG and RD were supported in part by the Australian Research Council (Grants No.
DP120103776 and FT120100449) and the National Natural Science Foundation of China
(Grant No. 61179030). CG and EC were supported in part by the National Science Foun-
dation (NSF) Early CAREER Award No. 1352326.
[1] M. Nielsen, Phys. Rev. Lett. 83, 436 (1999).
[2] Y. Xin and R. Duan,Phys. Rev. A 76, 044301 (2007).
[3] W. Du¨r, G. Vidal, and J. I. Cirac, Phys. Rev. A 62, 062314 (2000).
[4] N. Yu,C. Guo and R. Duan,Phys. Rev. Lett. 112, 160401 (2014).
[5] R. Duan and Y.Shi, Quantum Information and Computation 10 (11): 925–935 (2010).
[6] D. Jonathan and M.B. Plenio, Phys. Rev. Lett. 83, 1455 (1999).
[7] D. Li, X. Li, H.Huang and X.Li, Phys. Lett. A 359, 428- 437 (2006).
[8] L. Chen, Y. Chen,and Y.Mei, Phys. Rev. A 74, 052331 (2006).
[9] E. Chitmbar, C. Miller and Y.Shi, J. Math. Phys. 51, 072205 (2010) .
[10] M. Nielsen and I. Chuang, QuantumComputation and QuantumInformation (2000).
[11] E. Chitambar, R.Duan and Y.Shi, Phys. Rev. Lett. 101, 140502 (2008).
[12] N. Yu,E. Chitambar, C. Guo and R.Duan, Phys. Rev. A 81, 014301 (2010).
[13] S. Kintas and S.Turgut, J. Math. Phys. 51, 092202 (2010).
[14] S. Turgut,Y. Gul and N. K.Pak, Phys. Rev. A 81, 012317 (2010).
5
Appendix
I. PROOF OF THEOREM 1
According to Nielsen’s majorizationcriterionfor entanglement transformation,Theorem
1 is essentially due to the following lemma:
Lemma 3. Suppose that X(k) = (x(k),x(k), ,x(k) are a set of d–dimensional vectors
1 2 ··· d
where x(k) x(k) x(k). There always exists an optimal vector Y = (y , ,y )
1 ≥ 2 ≥ ··· ≥ d 1 ··· d
such that Y X(k) for any k. Furthermore, Y = (y , ,y ) can be chosen as y =
1 d k
≺ ···
k k k k 1 k 1 k 1
min( x(1), x(2), , x(n)) min( − x(1), − x(2), , − x(n)). Furthermore, if n
j j ··· j − j j ··· j
jP=1 jP=1 jP=1 jP=1 jP=1 jP=1
k k 1
is infinite, y =inf( x(h) h=1,2 ) inf( − x(h) h=1,2 ).
k j | ··· − j | ···
jP=1 jP=1
Proof. Let us first consider the case that n is finite. We will complete the proof via the
following steps:
Step 1: y y .
k k+1
≥
k k k k
Suppose min( x(1), x(2), , x(n)) is just x(t).
j j ··· j j
jP=1 jP=1 jP=1 jP=1
k+1 k+1 k+1 k k k
y =min( x(1), x(2), , x(n)) min( x(1), x(2), , x(n))
k+1 j j ··· j − j j ··· j
jP=1 jP=1 jP=1 jP=1 jP=1 jP=1
x(t) x(t)
≤ j+1 ≤ j
k k k k 1 k 1 k 1
min( x(1), x(2), , x(n)) min( − x(1), − x(2), , − x(n))=y .
≤ j j ··· j − j j ··· j k
jP=1 jP=1 jP=1 jP=1 jP=1 jP=1
k k k k k
Notice that: y = min( x(1), x(2), , x(n)), so k, f(k), y =
j j j ··· j ∀ ∃ j
jP=1 jP=1 jP=1 jP=1 jP=1
k
x(f(k)).
j
jP=1
Step 2: m, Y X(m).
∀ ≺
k k
It is obvious to see that y x(m).
j ≤ j
jP=1 jP=1
Step 3: If k, Z X(k), then Z Y.
∀ ≺ ≺
k0 k0
Otherwise, if k , z > y .
0 j j
∃
jP=1 jP=1
k0 z > k0 y = k0 x(f(k0)). This is a contradiction against with Z X(f(k0)).
j j j ≺
jP=1 jP=1 jP=1
If n is infinite, we modify our proof as follows:
k k k
(h) (m)
Notice that: k,m, y =inf( x ) x .
∀ j j ≤ j
jP=1 jP=1 jP=1
Step 1: y y .
k k+1
≥
6
k k k
Suppose inf( x(h))= lim x(g(h)), where x(g(h)) is in descending order. We can
j j j
jP=1 h→∞jP=1 jP=1
k k
alsofindasub–sequence(f(h)) sequence(g(h)),suchthatinf( x(h))= lim x(f(h)),
⊂ jP=1 j h→∞jP=1 j
(f(h)) (f(h))
also lim x and lim x exist. Thisisbecauseanyinfinite boundedsequencemust
k k+1
h h
have a→m∞onotonic conv→e∞rgentsub–sequence with a limit.
k+1 k k+1 k
y =inf( x(h)) inf( x(h))=inf( x(h)) lim xf(h)
k+1 jP=1 j − jP=1 j jP=1 j −h→∞jP=1 j
k+1 k
lim xf(h) lim xf(h)
≤h→∞jP=1 j −h→∞jP=1 j
lim xf(h) lim xf(h)
≤h k+1 ≤h k
→∞ k →∞ k 1 k k 1
lim xf(h) inf( − x(h))=inf( x(h)) inf( − x(h))=y .
≤h→∞jP=1 j − jP=1 j jP=1 j − jP=1 j k
Step 2: m, Y X(m).
Step 3: I∀f k, Z≺ X(k), then Z Y.
∀ ≺ ≺
k0 k0
Otherwise, if k , z > y .
0 j j
∃
jP=1 jP=1
k0 z > k0 y =inf( k x(h)). This is a contradiction against with k, Z X(k).
j j j ∀ ≺
jP=1 jP=1 jP=1
II. LOCC TRANSFORMATION PROTOCOLS FROM GHZ3 TO ANY
THREE-QUBIT PURE STATE
Thissectionisthedetailsoftheprotocoltransforming GHZ intoanythree-qubitpure
3
| i
state. Theproofisdividedinto2parts: WandGHZ.GHZparthavetwocases: orthogonal
GHZ and non-orthogonalGHZ case.
The following lemma has been shown in [2]. This also is our step i) in all the cases.
2
Lemma 4. z ,z ,z , z 2 =1,
0 1 2 k
∀ | |
kP=0
LOCC
GHZ z 000 +z 111 +z 222 .
3 0 1 2
| i −→ | i | i | i
Proof. Alice takes the following measurement and sends Bob and Charlie the result,
M =z 0 0 +z 1 1 +z 2 2,
1 0 1 2
{ | ih | | ih | | ih |
M =z 0 1 +z 1 2 +z 2 0,
2 0 1 2
| ih | | ih | | ih |
M =z 0 2 +z 1 0 +z 2 1 .
3 0 1 2
| ih | | ih | | ih |}
The, Bob and Charlie make some unitary operations based on Alice’s measurement out-
come, which transform the state to z 000 +z 111 +z 222 .
0 1 2
| i | i | i
7
A. Protocol of entanglement transformation from GHZ3 to W type states
Ifthetargetstate φ isstochasticlocaloperationsandclassicalcommunication(SLOCC)
| i
equivalent to W state, φ can be written as x 000 +x 100 +x 010 +x 001 , where
0 1 2 3
| i | i | i | i | i
3
x are all positive real numbers and x2 =1 [13].
k k
kP=0
i) GHZ LOCC x2+x2 000 +x 111 +x 222
| 3i −→ 0 1| i 2| i 3| i
ii) x2+x2 000p+x 111 +x 222 LOCC x2+x2 100 +x 010 +x 001
0 1| i 2| i 3| i −→ 0 1| i 2| i 3| i
iii)p x2+x2 100 +x 010 +x 001 LOCCpφ .
0 1| i 2| i 3| i −→ | i
Stpep i) is lemma 4 which z = x2+x2, z =x and z =x .
0 0 1 1 2 2 3
Step ii): x2+x2 000 +x 1p11 +x 222 LOCC x2+x2 100 +x 010 +x 001
0 1| i 2| i 3| i −→ 0 1| i 2| i 3| i
Alice takeps the measurement: p
M =(1 0 + 0 1 + 0 2)/√2, M =(1 0 + 0 2 0 1)/√2 .
1 2
{ | ih | | ih | | ih | | ih | | ih |−| ih | }
Bob takes the measurement:
M =(1 1 + 0 0 + 0 2)/√2, M =(1 1 + 0 0 0 2)/√2 .
1 2
{ | ih | | ih | | ih | | ih | | ih |−| ih | }
Charlie takes the measurement
iii) x2+x2 100 +x 010 +x 001 LOCC φ .
0 1| i 2| i 3| i −→ | i
p
M =(1 2 + 0 1 + 0 0)/√2, M =(1 2 + 0 1 0 0)/√2 .
1 2
{ | ih | | ih | | ih | | ih | | ih |−| ih | }
Alice transmits her result to Bob. Bobtransmits his resultto Charlie. Charlie transmits
its result to Alice.
If the result is 2, he or she should take a Z operation, Z = 0 0 1 1.
− | ih |−| ih |
Now, the state is x2+x2 100 +x 010 +x 001 .
0 1| i 2| i 3| i
Step iii): x2+px2 100 +x 010 +x 001 LOCC φ .
0 1| i 2| i 3| i −→ | i
Alice takespthe measurement:
M = 1 (0 0 + x1 1 1 + x0 0 1),
{ 1 √2 | ih | √x20+x21| ih | √x20+x21| ih |
M = 1 (0 0 + x1 1 1 x0 0 1) .
2 √2 | ih | √x20+x21| ih |− √x20+x21| ih | }
Now, the state is ( x 000 + x 100 + x 010 + x 001 ). If Alice’s result is 1, we
0 1 2 3
± | i | i | i | i
already get the target. If it is 2, Alice transmits 2 to Bob and Charlie. Bob and Charlie
makeunotaryoperationZ = 0 0 1 1 ontheirownpart. Finally,Alicemakesanunitary
| ih |−| ih |
operation Z = 0 0 + 1 1. Now, the state is x 000 +x 100 +x 010 +x 001 .
0 1 2 3
− −| ih | | ih | | i | i | i | i
B. Protocol of entanglement transformation from GHZ3 to GHZ type states
Withoutlossofgenerality,aGHZ typepurestatescanbedenotedasx000 +y φ φ φ ,
A B C
| i | i
where a = 0φ ,a = 1φ ,b = 0φ ,b = 0φ ,c = 0φ ,c = 0φ .
0 A 1 A 0 B 0 B 0 C 0 C
h | i h | i h | i h | i h | i h | i
a ,a ,b ,b ,c and c are all real numbers.
0 1 0 1 0 1
The LOCC protocols from GHZ to orthogonal GHZ state (suppose c =0, φ = 1 .
3 0 C
Thus, x2+ y 2 =1): | i | i
| | | |
8
LOCC
i) GHZ GHZ .
3 2
| i −→ | i
LOCC
ii) GHZ φ .
2
| i −→ | i
Stepi)isforlemma4whichz =z =1/√2andz =0.Thisisaprotocol GHZ LOCC
0 1 2 3
| i −→
GHZ .
2
| i
Step ii):
LOCC LOCC LOCC
GHZ x000 +y 111 x000 +y φ 11 x000 +y φ φ 1 = φ .
2 A A B
| i −→ | i | i −→ | i | i −→ | i | i | i
LOCC
For GHZ (x000 +y 111 ): Alice takes the measurement:
2
| i −→ | i | i
M =x0 0 +y 1 1, M =x1 1 +y 0 0 .
1 2
{ | ih | | ih | | ih | | ih |}
If output is “1”, finish. If output is “2”, Alice take an X operation, X = 1 0 + 0 1.
− | ih | | ih |
LOCC
The protocol for x000 +y 111 x000 +y φ 11 is as following. Alice takes the
A
| i | i −→ | i | i
measurement:
M = √2(0 0 + φ 1), M = √2(0 0 φ 1) .
{ 1 2 | ih | | Aih | 2 2 | ih |−| Aih | }
If output is “1”, finish. If output is “2”, Alice transmits the result to Charlie and Charlie
takes an Z operation, Z = 0 0 1 1.
− | ih |−| ih |
LOCC
The protocol for x000 +y φ 11 x000 +y φ φ 1 is similar. Bob takes the
A A B
| i | i −→ | i | i
measurement:
M = √2(0 0 + φ 1), M = √2(0 0 φ 1) .
{ 1 2 | ih | | Bih | 2 2 | ih |−| Bih | }
If output is “1”, finish. If output is “2”, Bob transmits the result to Charlie and Charlie
takes an Z operation, Z = 0 0 1 1.
− | ih |−| ih |
The LOCC protocols from GHZ to non-orthogonalGHZ state:
3
LOCC
i) GHZ Ψ
3 1
| i −→ | i
LOCC
ii) Ψ Ψ
1 2
| i −→ | i
LOCC
iii) Ψ x00φ +y φ φ 0
2 C A B
| i −→ | i | i
LOCC
iv) x00φ +y φ φ 0 x000 +y φ φ φ
C A B A B C
| i | i −→ | i | i
where Ψ = xc +ya b 2+ ya b 2 000 + xc 111 + yb 222 and
1 0 0 0 1 0 1 1
| i | | | | | i | || i | || i
Ψ = xc +pya b 2+ ya b 2 000 + xc 101 + yb 210 .
2 0 0 0 1 0 1 1
| i | | | | | i | || i | || i
Steppi) is for lemma 4 which z = xc +ya b 2+ ya b 2, z = xc and z = yb .
0 0 0 0 1 0 1 1 2 1
| | | | | | | |
Step ii) Ψ LOCC Ψ : p
1 2
| i −→ | i
Bob takes the measurement:
M = √2(1 2 + 0 0 + 0 1),M = √2(1 2 + 0 0 0 1) .
{ 1 2 | ih | | ih | | ih | 2 2 | ih | | ih |−| ih | }
Charlie takes the measurement:
M = √2(1 1 + 0 2 + 0 0),M = √2(1 1 0 2 + 0 0) .
{ 1 2 | ih | | ih | | ih | 2 2 | ih |−| ih | | ih | }
They transmit their results (1 or 2) to Alice. Suppose Bob gets β and Charlie gets α.
Alice takes the unitary operation M = 0 0 ( 1)β 1 1 ( 1)α 2 2.
| ih |− − | ih |− − | ih |
Now, the state is Ψ .
2
| i
Step iii): Alice does the measurement
{
M1 =( √(xc|x0+c0y+ay0ab00)b|00|i2++y|yaa11bb00|1|2ih0|+ |xxcc11||0ih1|+ |yybb11|(a0|0i+a1|1i)h2| )/2,
M2 =( √(xc|x0+c0y+ay0ab00)b|00|i2++y|yaa11bb00|1|2ih0|+ −|xxcc11||0ih1|+ |yybb11|(a0|0i+a1|1i)h2| )/2,
M3 =( √(xc|x0+c0y+ay0ab00)b|00|i2++y|yaa11bb00|1|2ih0|+ |xxcc11||0ih1|+ −|yybb11|(a0|0i+a1|1i)h2| )/2,
9
M4 =( √(xc|x0+c0y+ay0ab00)b|00|i2++y|yaa11bb00|1|2ih0|+ −|xxcc11||0ih1|+ −|yybb11|(a0|0i+a1|1i)h2| )/2}
The resulting states are all LOCC-equivalent to ((xc + ya b )0 + ya b 1 )00 +
0 0 0 1 0
| i | i | i
xc 001 + yb (a 0 + a 1 )10 = x00φ + y φ φ 0 . It is LOCC-equivalent to
1 1 0 1 C A B
| i | i | i | i | i | i
x000 +y φ φ φ , which can be done in next step.
A B C
| i | i
Step iv): Charlie takes a local unitary operation φ 0 +(c 0 c 1 ) 1.
C 1 0
| ih | | i− | i h |
C. The common resource of three-qubit system cannot be a 3 2 2 pure state
⊗ ⊗
We try to prove GHZ is an optimal common resource. We still need more time for
3
| i
furtherresearch. Thissectionshowsthatthereisnocommonresourceofthree-qubitsystem
in 3 2 2 systems. (We research 3 3 2 system nowadays.)
⊗ ⊗ ⊗ ⊗
TherearetwoSLOCCequivalentclassesin3 2 2system. Theyare 000 + 101 + 210
⊗ ⊗ | i | i | i
and 000 + 111 + 201 + 210 . The front one is with rank 3, and another one is with
| i | i | i | i
rank 4 [9]. A distinguish difference between them is in their Bob’s and Charlie’s (BC)
subsystem. BC subsystem’s dimension is 4 and Alice’s dimension is 3. Hence, a 3 2 2
⊗ ⊗
purestatealwayshasamissing dimensionintheirBCsubsystem. Themissingdimension
ofa 3 2 2 pure state ψ canbe defined as the pure state in the orthogonalcomplement
⊗ ⊗ | i
of ψ’s BC support.
If ψ’s missing dimension is a product state, (without loss of generality, suppose it is
11 ,) ψ is SLOCC equivalent to 000 + 101 + 210 . If it is entangled, (without loss of
| i | i | i | i | i
generality,supposeitis 01 10 ,) ψ isSLOCCequivalentto 000 + 111 + 201 + 210 .
| i−| i | i | i | i | i | i
The common resource of three-qubit system cannot be a 3 2 2 pure state, because
⊗ ⊗
i) a pure state ψ , which is SLOCC equivalent to 000 + 111 + 012 + 102 , cannot be
| i | i | i | i | i
transformedintoaWkindofpurestatewithnon-symmetricinBCsubsystemsviaLOCC;
ii)apurestate ψ ,whichisSLOCCequivalentto 000 +111 +012 ,cannotbetransformed
| i | i | i | i
into GHZ via LOCC.
2
| i
Let
1
ϕ = (1 00 + 0 τ ), (1)
λ ABC A BC A λ BC
| i √2 | i ⊗| i | i ⊗| i
where
τ =√1 λ10 +√λ01 . (2)
λ
| i − | i | i
Theorem 5. Let ψ be any 3 2 2 state with tensor rank four. Then ψ
ABC ABC
| i ⊗ ⊗ | i →
ϕ by finite-round LOCC only if λ = 1/2; i.e. only if ϕ is symmetric w.r.t.
λ ABC ABC
| i | i
Bob and Charlie.
Proof. Considerany finite-roundLOCC protocolperformedon ψ . Since ϕ is a W-class
λ
| i | i
state, it follows that Alice must be the last party to perform a non-trivial measurement
along each branch of the protocol. For one of the branches, let ψ be the final state
′
| i
before Alice performs her last non-trivial measurement. At this point, all Kraus operators
performed along this branch are invertible and ψ is still a 3 2 2 state.
′
| i ⊗ ⊗
We can always decompose Alice’s final measurement into finite-depth tree generated
by binary-outcome measurements. Let ψ denote one of the 3 2 2 states obtained
′′
| i ⊗ ⊗
10
immediately before Alice performs her final measurement along one branch of this binary-
outcome tree. Thus, Alice measures M ,M on ψ and both outcomes must be LU
0 1 ′′
{ } | i
equivalent to |ϕλi. Let {|0i,|1i,|2i} be a basis in which M0†M0 is diagonal. The condition
M0†M0+M1†M1 =IimpliesthatM1†M1 isalsodiagonalinthisbasis,andthereforewehave
the general forms
α 0 0 1 α 0 0
−
M0†M0 =0 1 0, M1†M1 = 0 0 0. (3)
0 0 0 0 0 1
   
The polar decomposition of Mi is given by Mi = Wi Mi†Mi for unitary Wi. We next
q
expand
ψ′′ =c0 0 A χ0 BC +c1 1 A χ1 BC +c2 2 A χ2 BC. (4)
| i | i | i | i | i | i | i
LetU andV bethelocalunitariesofBobandCharlieperformedafteroutcomeiofAlice’s
i i
measurement. Since M U V ψ ϕ , we must have that either (i) U V χ =
0 0 0 ′′ λ 0 0 0
eiθ0 00 andU V χ =⊗eiφ0⊗τ ,|ori(i≈i)U| iV χ =eiφ0 τ andU V χ ⊗=ei|θ0 0i0 .
0 0 1 λ 0 0 0 λ 0 0 1
C|onsiider firs⊗t cas|e (ii). Let| χi = αβ ⊗so t|hati U V| αiβ = e⊗iθ0 0|0 .i The o|their
0 0 0
| i | i ⊗ | i | i
measurementoutcomeisM U V ψ ϕ ,andsowemustalsohaveU V αβ =
1 1 1 ′′ λ 1 1
eiθ1 00 and U V χ =ei⊗φ1 τ ⊗. It|thein≈fo|llowis that there exists a local unit⊗ary|U iV
1 1 2 λ
| i ⊗ | i | i ⊗
suchthatU V =(Z(η ,ζ ) Z(µ ,ν ))(U V),whereZ(a,b)= eia 0 . Thuswehave
i⊗ i i i ⊗ i i ⊗ (cid:16) 0 eib(cid:17)
ψ =c 0 αβ +I U V c eiφ0I Z (η ,ζ ) Z (µ ,ν )1 τ
′′ 0 † † 1 † 0 0 † 0 0 λ
| i | i| i ⊗ ⊗ (cid:20) ⊗ ⊗ | i| i
+c eiφ1I Z (η ,ζ ) Z (µ ,ν )2 τ . (5)
2 † 1 1 † 1 1 λ
⊗ ⊗ | i| i(cid:21)
This state has a tensor rank of three.
Now consider case (ii). The argument is very similar to case (i). We have the two
conditions U V χ = eiφ0 τ and U V χ = eiφ1 τ . The key observation is
0 0 0 λ 1 1 0 λ
⊗ | i | i ⊗ | i | i
that if λ = 1/2, then the respective Schmidt bases of χ and τ are unique, and so
0 λ
6 | i | i
the actions of U and V are fixed up to phases. That is, we can again write U V =
i i i i
⊗
(Z(η ,ζ ) Z(µ ,ν ))(U V), and ψ will have a rank three decomposition:
i i i i ′′
⊗ ⊗ | i
ψ =c 0 χ +I U V c eiφ0I Z (η ,ζ ) Z (µ ,ν )1 00
′′ 0 0 † † 1 † 0 0 † 0 0
| i | i| i ⊗ ⊗ (cid:20) ⊗ ⊗ | i| i
+c eiφ1I Z (η ,ζ ) Z (µ ,ν )2 00 . (6)
2 † 1 1 † 1 1
⊗ ⊗ | i| i(cid:21)
This meansany 3 2 2state withtensor rankfour isnot acommonresourcefor three-
⊗ ⊗
qubit systems. The following theorem shows any 3 2 2 state with tensor rank three is
⊗ ⊗
not a common resource for three-qubit systems neither.