Additivity of Entanglement of Formation of Two Three-level-antisymmetric States

6 Pages·0.14 MB·English

by  Toshiyuki Shimono

Additivity of Entanglement of Formation
of Two Three-level-antisymmetric States
Toshiyuki Shimono
Department of Computer Science, University of Tokyo
7-3-1, Hongo, Bunkyo-ku,Tokyo,113-0033, Japan
February 1, 2008
3
0
Abstract 2 New additivity result
0
2

n Quantum entanglement is the quantum information 2.1 Antisymmetric states
a
processingresource. Thusitisofimportancetounder-
J
stand how much of entanglement particular quantum Let us start with an introduction of our notations
6
states have, and what kinds of laws entanglement and concepts. will stand for an antisymmetric

1 and also transformation between entanglement states Hilbertspace,whHic−hisasubspaceofabipartiteHilbert
v subject to. Therefore, it is essentialy important to space := , where both and are
AB A B A B
1 use proper measures of entanglement which have 3 dimHensionalHHilb⊗erHt spaces, spanneHd by basHic vec-
1 −
nice properties. One of the major candidates of such tors i 3 . is three-dimensional Hilbert space,
0 measuresis”entanglementofformation”,andwhether spann{e|di}bi=y1stHat−es i,j , where the state
1 ij=23,31,12
{| i}
0 this measurement is additive or not is an important i,j is defined as |ii|ji−|ji|ii. The space is called
3 open problem. We aim at certain states so-called | i √2 H−
antisymmetricbecausebyswappingthepositionoftwo
0 ”antisymmetric states” for which the additivity are
qubitsinanyofitsstates ψ wegetthestate ψ . Let
h/ not solved as far as we know, and show the additivity ⊗n be the tensor produ|ctiof n copies of −|. Ti hese
p for two of them. cHo−pieswillbediscriminatedbytheupperindHe−xas (j),
-
nt forj =1...n. (j) willthenbeanantisymmetricHsu−b-
a Keywords: quantum entanglement, entanglement of (j) H−(j)
space of .
u formation, additivity of entanglement measures, anti- HA ⊗HB
q symmetric states.
:
v
i 2.2 The result and proof sketch
X
r 1 Introduction
a
It has been shown in [1] that E (ρ) = 1 for any
f
Concerning the additivity of entanglement of for- mixed state ρ ( ). This result will play the key
∈ S H−
mation, only a few results have been known. Vidal role in our proof. We prove now that :
et al. [1] showed that additivity holds for some mix-
ture of Bell states and other examples by reducing the Theorem.
argument of additivity of the Holevo capacity of so-
called ”entanglement breaking quantum channels” [2]
E (ρ ρ )=E (ρ )+E (ρ ) (=2) (1)
andtheyarethenon-trivialfirstexamples. Matsumoto f 1⊗ 2 f 1 f 2
etal. [3]showedthatadditivityofentanglementoffor-
mation holds for a family of mixed states by utilizing for any ρ1,ρ2 ( ).
∈S H−
theadditivityofHolevocapacityforunitalqubitchan-
nels [4] via Stinespring dilation [5].
Proof. To prove this theorem, it is sufficient to show
In this extended abstract we prove that entangle- that
mentofformationisadditivefortensorproductoftwo
three-dimensional bipartite antisymmetric states with Ef(ρ1 ρ2) 2 (2)
⊗ ≥
a sketch of the proof. We proved by combination of
elaborate calculations. since the subadditivity E (ρ ρ ) E (ρ ) +
f 1 2 f 1
⊗ ≤
E (ρ )=2 is trivial. Indeed, it holds (We proved this lemma by solving a cubic equa-
f 2
tion and bounding the Shannon entropy function with
Ef (ρ1 ρ2)=inf piE(ψi ψi ) polynomial functions.) Local unitary operators do not
⊗ | ih |
change von Neumann reduced entropy, and therefore
inf p(1)p(2)EX(ψ(1) ψ(1) ψ(2) ψ(2) )
≤ i i | i ih i |⊗| i ih i | E(ψ ψ ) = E(ψ′ ψ′ ) 2. That is, the claim (4) is
| ih | | ih | ≥
= infXp(1)E(ψ(1) ψ(1) ) proven.
i | i ih i |
We are now almost done. Indeed, the entanglement
+Xinf p(2)E(ψ(2) ψ(2) )
i | i ih i | of formation is defined as
= Ef(ρ1)X+Ef(ρ2) (3) E (ρ)= inf p E(ψ ψ ) (8)
f i i i
where(p(j), ψ(j) )aresubjecttothe conditionofρ = [(pi,ψi)]i∈∆(ρ)Xi | ih |
p(j) ψi(j)|ψi(j)i. To prove (2), we first show thajt where
i i | i ih i | p =1,p >0 i
P E(|ψihψ|)≥2, for any pure state |ψi∈H−⊗2. (4) ∆(ρ)=(cid:26)[(pi,ψi)]i (cid:12) Piipii|ψiihψii|=ρ∀,hψi|ψii=1∀i(cid:27)
UsingtheSchmidtdecomposition,thestate ψ canbe andit is knowntha(cid:12)(cid:12)taPll ψi induced from∆(ρ) satisfy
decomposed as follows: | i ψi Range(ρ), where|Rainge(ρ) is sometimes called
| i ∈
the image space of the matrix ρ, which is the set of
3 ρψ with ψ running over the domain of ρ. Hence
ψ = √p ψ(1) ψ(2) , (5) | i | i
| i i| i i⊗| i i E (ρ) inf E(ψ ψ ) ψ Range(ρ), ψ ψ =1 .
Xi=1 f ≥ | ih | | i∈ h | i
(9)
wishaenreoprt1h,po2n,opr3m>al0b,aps1is+opf2th+epH3i=lbe1r,tasnpdac{e|ψi(j()j)i,}3if=o1r Sfoirntche(ρ21)⊗isρp2r(cid:8)∈ovSen(H. T−⊗h2)e,r(cid:12)(cid:12)eRfoarneg(e1(ρ)1h⊗avρe2b)e⊆enHs−⊗ho2,whne.n(cid:9)ce-
j =1,2. NotethatthisSchmidtdecompositioHn−iswith
respect to (1) : (2), or, it could be said that with 3 Conclusions and discussion
H− H−
respectto (1) (1) : (2) (2) , notwithre-
HA ⊗HB HA ⊗HB
Additivity of the entanglement of formation for two
spect to (cid:16)(1) (2) (cid:17): (cid:16)(1) (2) ,(cid:17)where “:” in-
HA ⊗HA HB ⊗HB three-dimensional bipartite antisymmetric states has
dicatesho(cid:16)wtoseparat(cid:17)eth(cid:16)esysteminto(cid:17)twosubsystems been proven in this paper. The next goal could be
for the decomposition. to prove additivity for more than two antisymmetric
states. Perhapsthe proofcanutilize the valueoflower
First, we will use the following fact.
boundofthereducedvonNeumannentropy. Ofcourse,
the main goal is to show that entanglement of forma-
Lemma. If ψ 3 is an orthonormal basis of ,
then there ex{i|stisi}ai=n1unitary operator U, actingHo−n tion is additive, in general. However, this seems to be
a very hard task.
both and , such that U U maps the states
A B
H H ⊗
ψ , ψ , ψ into the states 2,3 , 3,1 , 1,2 , respec-
1 2 3
| i | i | i | i | i | i References
tively.
[1] G. Vidal, W. Du¨r, J. I. Cirac, Entanglement Cost
Therefore, by this Lemma, there exist unitary oper-
of Bipartite Mixed States (2002), Physical Review
ators U(1),U(2) such that
Letters, 89, 027901.
U(1) U(1) U(2) U(2) ψ [2] Peter W. Shor (2002), Additivity of the Classi-
⊗ ⊗ ⊗ | i
(cid:0) = √pij i,j(cid:1) i,j =: ψ′ , (6) cal Capacity of Entanglement-Breaking Quantum
| i⊗| i | i Channels, quant-ph/0201149.
i,j
ij=X23,31,12
[3] K. Matsumoto, A. Winter, T. Shimono (2002),
where p23 :=p1, p31 :=p2, p12 :=p3. Remarks on additivity of the Holevo channel ca-
pacity and of the entanglement of formation,
As is written in the following, we use the following
quant-ph/0206148.
fact.
[4] ChristopherKing,Additivityforunitalqubitchan-
Lemma. nels (2001), quant-ph/0103156.
[5] W.F.Stinespring(1955),PositivefunctionsonC –
p ,p ,p 0 ∗
E(ψ′ ψ′ ) 2, if 23 31 12 ≥ . algebras,Proceedingsofthe AmericanMathemati-
| ih | ≥ ( p23+p31+p12 =1 cal Society, 6, 211-216.
(7)
A. Appendix
We provide here proofs of two facts used in the proof of our main result.
Lemma. If ψ 3 is an orthonormal basis, there exists an unitary operator U, acting on both
and , such{|thiait}iU=1 ⊂UHm−aps the states ψ , ψ , ψ into the states 2,3 , 3,1 , 1,2 , respectively. HA
B 1 2 3
H ⊗ | i | i | i | i | i | i
Proof. Let us start with notational conventions. In the following, T2 stands for the transpose of a matrix, 2
∗
stands for taking complex conjugate of eachelementof a matrix, 2Θ denotes the transformationdefined later.
LetU berepresentedas uu1211 uu1222 uu1233 withrespecttothebasis 1 , 2 , 3 .Formathematicians,anoperatorand
u31 u32 u33 | i | i | i
its matrix representationm(cid:16)ightbe diff(cid:17)erentobjects,but for convenience,we identify U with uu1211 uu1222 uu1233 here.
u31 u32 u33
Lengthy calculationsshowthat whena 9 9dimensionalmatrix U U is consideredas map(cid:16)ping from (cid:17) into
× ⊗ H−
,itcanberepresentedbythe following3 3dimensionalmatrix,withrespecttothebasis 2,3 , 3,1 , 1,2 ,
H− × | i | i | i
u u u u u u u u u u u u
22 33 23 32 23 31 21 33 21 32 22 31
− − −
UΘ := u u u u u u u u u u u u .
32 13 33 12 33 11 31 13 31 12 32 11
 − − − 
u u u u u u u u u u u u
12 23 13 22 13 21 11 23 11 22 12 21
− − −
 
One can then show that
UΘ TU =(detU) 100100 ,
· 001
(cid:16) (cid:17)
and by multiplying with U from the right in the above equation, one obtain UΘ =(detU) U , since U is an
∗ ∗
·
unitary matrix, and TU U is equal to the identity matrix.
∗
·
Since ψ is an orthonormal basis of , there exists an unitary operator on such that
i i=1,2,3
{| i} H− H−
ψ 2,3 , ψ 3,1 , ψ 1,2 , and let Θ be the corresponding matrix with respect to the basis
1 2 3 ψ
| i 7→ | i | i 7→ | i | i 7→ | i
i,j .
ij=23,31,12
{| i}
Let Uψ := (detΘψ)12 ·Θ∗ψ .1 It holds UψΘ = Θψ. 2 Therefore Uψ ⊗Uψ = Uψ′. The operator Uψ is the one
needed to satisfy the statement of Lemma.
Lemma.
i,j
ψ′ = √pij i,j i,j
| i | i| i
E(ψ′ ψ′ ) 2 if  ij=23,31,12 .
| ih | ≥  p23,p31,p12P≥0
p +p +p =1
23 31 12
Proof. Let p :=p ,p :=p ,p :=p . Thenit holds,
32 23 13 31 21 12
i,j
ψ′ = √pij i,j i,j
| i | i| i
1 i<j 3
≤X≤
i,j
1
= √pij ii;jj ij;ji ji;ij + jj;ii
2 {| i−| i−| i | i}
1 i<j 3
≤X≤
i,j
1
= √pij ii;jj ij;ji ,
2 {| i−| i}
1 i=j 3
≤X6 ≤
where i i ;i i denotes the tensor product i i i i , i (1), i (2), i (1) and
| 1 2 3 4i | 1i⊗| 2i⊗| 3i⊗| 4i | 1i ∈ HA | 2i ∈ HA | 3i ∈ HB
i (2) , and the condition 1 i = j 3 actually means ”1 i 3 and 1 j 3 and i = j”. This
| 4i ∈ HB ≤ 6 ≤ ≤ ≤ ≤ ≤ 6
convention will be used also in the following.
1Intheabovedefinitionitdoesnotmatter whichoftworootsof detΘψ aretaken
2Indeed, UψΘ = (detUψ)Uψ∗ = (detΘψ)32 detΘ∗ψ · ((detΘψ)12)∗Θψ = Θψ . Note that detUψ = det(det(Θψ)12 Θ∗ψ) =
(detΘψ)32 detΘ∗ψ becauseΘ∗ψ isa3×3matrix.
We are now going to calculate the reduced matrix of ψ ψ , which we will denote as Ξ, and it will be
′ ′
| ih |
decomposed into the direct sum as follows.
Ξ := Tr ψ ψ
′ ′
(1) (2)| ih |
HB ⊗HB
i,j,k,l
1 ii;jj kk;ll ii;jj kl;lk
= √pijpkl Tr | ih |−| ih |
4 (1) (2) ij;ji kk;ll + ij;ji kl;lk
11≤≤Xki6=6=jl≤≤33 HB ⊗HB (cid:18) −| ih | | ih |(cid:19)
i,j,k,l
1
= √pijpkl Tr ii;jj kk;ll + ij;ji kl;lk
4 (1) (2) | ih | | ih |
11≤≤Xki6=6=jl≤≤33 HB ⊗HB (cid:0) (cid:1)
i,j,k i,j
1 1
= √pikpjk ii jj + pij ij ij
4 | ih | 4 | ih |
1≤i6=k≤3 1 i=j 3
1≤Xj6=k≤3 ≤X6 ≤
1 p12+p13 √p13p23 √p12p23 1 1 1
∼= 4(cid:18)√√pp1132pp2233 √p1p21+2pp2133 √p1p31+2pp2133(cid:19)⊕ 4(p12)⊕2⊕ 4(p13)⊕2⊕ 4(p23)⊕2, (10)
where denotes the direct sum of matrices, and 2 n denotes the direct sum of n copies of the same matrix.
⊕
⊕
We need to get eigenvalues of Ξ in order to calculate reduced von Neumann entropy
E(ψ ψ )= Tr(Ξlog Ξ)= λ log λ.
| ′ih ′| − 2 − 2
λ:e.v.ofΞ
X
In this case, fortunately, the eigenvalues can be determined explicitly from the expression (10). They are the
following ones:
1 cosθ 1 cos(θ+ 2π) 1 cos(θ+ 4π) p p p p p p
− , − 3 , − 3 , 12, 12, 13, 13, 23, 23 (11)
6 6 6 4 4 4 4 4 4
(cid:18) (cid:19)
for a certain π <θ π.3 These eigenvaluesare denotedas (λ ,λ ,...,λ ), respectively. Although λ ,...,λ
−3 ≤ 3 1 2 9 4 9
are trivial, λ ,λ ,λ are the roots of the cubic polynomial
1 2 3
1 1 p p p
g(λ):=λ3 λ2+ λ 12 13 23 (12)
− 2 16 − 16
that is the characteristic polynomial function of the cubic matrix that appeared in the expression (10). We
must solve this cubic equation to obtain (11). The cubic equation g(λ) = 0 is in Cardan’s irreducible form,4
because Ξ is the density matrix. In such a case, the roots of the cubic equation are
2π 4π
α+βcosθ,α+βcos(θ+ ),α+βcos(θ+ ). (13)
3 3
One can easily show that λ +λ +λ =3α, and λ2+λ2+λ2 =3α2+ 3β2. If λ ,λ ,λ are equal to the roots
1 2 3 1 2 3 2 1 2 3
of the cubic equation λ3 +a λ2 +a λ+a = 0, then λ +λ +λ = a , λ2+λ2+λ2 = a2 2a . Taking
1 2 3 1 2 3 − 1 1 2 3 1− 2
a = 1,a = 1 from the expression (12), we get the following system of equations 3α = 1,3α2+ 3β2 = 1,
1 −2 2 16 2 2 8
and (α,β)= 1, 1 is sufficient. Applying this argument into (13), we complete (11).
6 −6
(cid:0) (cid:1)
Our idea is now to show that
9
E(ψ ψ )= ( λ log λ ) 2. (14)
| ′ih ′| − i 2 i ≥
i=1
X
This will be shown if we prove that it holds
3 9
( λ log λ ) 1 and ( λ log λ ) 1. (15)
− i 2 i ≥ − i 2 i ≥
i=1 i=4
X X
3Theexactvalueofθ willbenoimportanceforus.
4AcubicequationissaidtobeinCardan’sirreducibleformifitsthreerootsarereal.
0.5
0.4
0.3
0.2
0.1
0
0.05 0.1 0.15 0.2 0.25 0.3
z
Figure 1: By virture of Shannon entropy function being lower-bounded by the polynomial functions, we can
evaluate the von Neumann entropy of Ξ.
The second inequalities is easy to verify by simple calculations. To finish the proof of the lemma we therefore
need to show that
3
( λ log λ ) 1. (16)
− i 2 i ≥
i=1
X
Withoutlossofgenerality,weassumeθ ∈ 0,π3 .5 Clearly,λ1 ∈ 0,112 andλ2,λ3 = 41−λ1±√λ21−3λ21 ∈ 112,31
(λ ,λ canbe regardedasthesolutionofthe followingsystemsofequations: λ +λ +λ = 1,λ2+λ2+λ2 = 1
2 3 (cid:2) (cid:3) (cid:2) (cid:3) 1 2 3 2 1 2 (cid:2)3 8(cid:3)
) . You can also show that
(log 12)z if z 0, 1
zlog z 2 ∈ 12 (17)
− 2 ≥( 21 + lolgoege4−21(z− 14)−4(z− 41)2 if z ∈(cid:2)112,13(cid:3)
(cid:2) (cid:3)
(see Fig.1). The first inequality of (17) is easily confirmed. On the other hand, one way of the proof of the
secondinequalityisasfollows: Letf(z):= −zlog2z − 21 + lolgoege4−21(z− 14)−4(z− 41)2 . Differentiatingthis
expression by z once and twice, we can get the increasing(cid:16)and decreasing table as follows(cid:17).
(cid:0) (cid:1)
z 1 1 1 1
12 8log 2 4 3
e
f(z) + y + 0
ց ր
f (z) 0 +
′
−
f (z) 0 + + +
′′
−
The table indicates f(z) 0 for z 1 ,1 . Now we indeed get the lower bounds by polynomial functions.
≥ ∈ 12 3
Combining all of the above inequalities we get (16) as
(cid:2) (cid:3)
3
log 3+2
λ log λ 1+ e 2 λ +4λ2 1 .
− i 2 i ≥ log 2 − 1 1 ≥
i=1 (cid:18) e (cid:19)
X
5The sequence of {λi}3i=1 doesn’t change if θ is replaced by −θ. Thus we can change the assumption θ ∈ −π3,π3 , into
θ∈ 0,π .
3 (cid:0) (cid:3)
(cid:2) (cid:3)
so that (15) and (14) are successively shown and that our proof is finished.