Dimension formula for induced maximal faces of separable states and genuine entanglement

0.19 MB·English

by  Lin Chen

Dimension formula for induced maximal faces of separable
states and genuine entanglement
Lin Chen 1,2, Dragomir Zˇ Dokovi´c3,4
¯
1 School of Mathematics and Systems Science, Beihang University, Beijing 100191, China
5
2 International Research Institute for Multidisciplinary Science, Beihang University, Beijing 100191, China
1
0 3 Institute for Quantum Computing, University of Waterloo, Waterloo, Ontario, Canada
2
4 Department of Pure Mathematics, University of Waterloo, Waterloo, Ontario, Canada
n
u
J June 16, 2015

5
1

] Abstract
h
p
- The normalized separable states of a finite-dimensional multipartite quantum system, repre-
t
n sented by its Hilbert space , form a closed convex set . The set has two kinds of faces,
1 1
a H S S
induced and non-induced. An induced face, F, has the form F = Γ(F ), where V is a subspace of
u V
q , FV is the set of ρ 1 whose range is contained in V, and Γ is a partial transposition operator.
H ∈ S
[ Such F is a maximal face if and only if V is a hyperplane. We give a simple formula for the dimen-

2 sion of any induced maximal face. We also prove that the maximum dimension of induced maximal
v faces is equal to d(d 2) where d is the dimension of . The equality DimΓ(F )= d(d 2) holds
V
5 − H −
if and only if V⊥ is spanned by a genuinely entangled vector.
4
7
0
0 Contents
.
1
0
1 Introduction 1
5
1
: 2 Dimension formula 4
v
i
X
3 Action of Θ on induced maximal faces 7
r
a
4 Some maximal faces in the bipartite case 9
4.1 Some 3 3 non-induced maximal faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
×
4.2 Some 2 4 non-induced maximal faces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
×
5 Conclusion and discussion 12
1 Introduction
Let = be the complex Hilbert space of a finite-dimensional n-partite quantum
1 2 n
H H ⊗H ⊗···⊗H
system. We denote by d the dimension of , and so d := d is the dimension of . We assume
i i i
H Q H
that each d > 1. A vector x is normalized if x = 1. We denote by H the space of Hermitian
i
| i ∈ H k k
1
operators ρ on . Note that H is a real vector space of dimension d2. The mixed quantum states of this
H
system are represented by their density matrices, i.e., operators ρ H which are positive semidefinite
∈
(ρ 0) and have unit trace (Trρ = 1). For convenience, we often work with non-normalized states, i.e.,
≥
Hermitian operators ρ such that ρ 0 and ρ = 0. It will be clear from the context whether we require
≥ 6
the states to be normalized. We denote by (ρ) the range of a linear operator ρ.
R
Weassumethatanorthonormalbasisisfixedineach andweusethestandardnotation 0 ,..., d
i i
H | i | −
1 for the corresponding basis vectors. We write EndV for the algebra of linear operators on a
i
complex vector space V. The operation of transposition applied only to the ith tensor factor of
End = n End will be denoted by Γ . (The transposition is with respect to the basis fixed
H ⊗i=1 Hi i
above.) We denote by Θ the abelian group of order 2n generated by the Γ s. We refer to the elements of
i
Θ as the partial transposition operators. Thus if ρ is a state on , then Γ (ρ) is the ith partial transpose
i
H
of ρ. We recall a definition from [5]: an operator ρ H is full if Γ(ρ) has rank d for all Γ Θ.
∈ ∈
Aproduct vectoris a nonzero vector of the form x = x x where x . We shall write
1 n i i
| i | i⊗···⊗| i | i ∈ H
this product vector also as x ,...,x . A pure product state is a state ρ of the form ρ = x x where
1 n
| i | ih |
x is a product vector. The product vectors i ,i ,...,i , 0 i < d , form an orthonormal (o.n.)
1 2 n k k
| i | i ≤ l
basis of . A quantum state ρ is separable if it is a sum of pure product states, i.e., ρ = z z ,
H Pk=1| kih k|
where the z are product vectors. A non-separable state is also called an entangled state. Quantum
k
| i
entanglement has been used to realize many tasks, such as quantum teleportation [3], cryptography
[2] and dense coding, that surpass their classical counterparts both theoretically and experimentally.
Deciding whether a given state is entangled turns out to be hard. This problem has been solved only for
the Hilbert spaces of dimension 2 2 and 2 3 [17]. Thus it is important to understand the properties
× ×
of the set of normalized separable states. In this paper, we will investigate the faces of . The full
1 1
S S
separable states that lie on the boundary, ∂ , of are of special interest.
1 1
S S
We denote by and the set of normalized and non-normalized states, respectively. Thus =
1 1
D D D
ρ : Trρ = 1 is a compact convex subset of the affine hyperplane of H defined by the equation
{ ∈ D } ˜
Trρ = 1. The faces of are parametrized by vector subspaces V [1, section II]. The face F that
1 V
D ⊆ H
corresponds to V consists of all states ρ such that (ρ) V. The intersection
1
∈ D R ⊆
F := F˜ = ρ : (ρ) V (1)
V V 1 1
∩S { ∈ S R ⊆ }
is a face (possibly empty) of . We say that the face F is associated to V. The following basic fact
1 V
S
was proved recently [5, Theorem 15].
Theorem 1 For a subspace V , the associated face F of is maximal if and only if V is a
V 1
⊆ H S
hyperplane of .
H
As in [5] we enlarge the collection of faces of type F by using the group Θ.
V
Definition 2 A face F of is induced if F = Γ(F ) for some vector subspace V and some partial
1 V
S ⊆ H
transposition Γ Θ.
∈
We warn the reader that our general definition above is different from the one adopted (for the
bipartite case) by H.-S. Choi and S.-H, Kye [7].
The boundary of is the union of all maximal faces and in order to describe geometrically the set
1
S
one needs the description of the maximal faces. The induced maximal faces have the form Γ(F )
1 V
S
where V is a hyperplane of and Γ Θ. The non-induced faces are more elusive. It was shown in [5]
H ∈
that there exist maximal faces of which are not induced. Explicit examples of non-induced maximal
1
S
faces of are presently known in several systems, notably in 3 3 and 2 4 systems. A remarkable
1
S ⊗ ⊗
family of non-induced faces of in the case of two qutrits was constructed recently in [13]. Each of
1
S
2
these faces is a 9-dimensional simplex. Moreover, each of them is the intersection of two maximal faces
[19]. For more details see section 4.
It is certainly of interest to know the maximum dimension of maximal faces of . In this regard,
1
S
we propose the following conjecture.
Conjecture 3 For any finite dimensional quantum system we have
max DimF = max DimF . (2)
V
F, maximal face V⊂H, hyperplane
Since DimΓ(F ) = DimF for any Γ Θ and any vector subspace V , the rhs is equal to
V V
∈ ⊆ H
the maximum dimension taken over all induced maximal faces of . The lhs is obviously greater
1
S
than or equal to the rhs. We shall prove that the rhs is equal to d(d 2) (see Corollary 7). The
−
number d(d 2) arises very naturally. Indeed, Dim∂ = d2 2 and induced maximal faces are
1
− S −
parameterized by the complex projective space whose real dimension is 2(d 1). Therefore, the number
−
(d2 2) 2(d 1) = d(d 2) is expected. The genuinely entangled vectors are generic and the maximal
− − − −
faces induced by the hyperplanes orthogonal to such vectors indeed have the expected dimension.
The lhs of (2) is known only when d 6, and in these cases the conjectured equality holds [5]. See
≤
section 4 for additional evidence in support of the conjecture.
If α is a product vector and V = α ⊥ then we have shown in [5, Proposition 17] that
| i ∈ H | i
n
DimF = d2 1 (2d 1). (3)
V − −Y i −
i=1
The main result of the paper, Theorem 6, generalizes this formula. It provides a method for com-
puting the dimension of any induced maximal face in any multipartite quantum system. It turns out
that DimF , V a hyperplane, depends only on the full tensor factorization of the vector α orthogonal
V
| i
to V. Corollary 7 shows that the genuine entanglement [8] from quantum information is linked to the
geometry of faces of . We say that α is genuinely entangled (g.e.) if it is entangled for all bipartitions
1
S | i
of . This corollary states that α is g.e. if and only if DimF = d(d 2). For instance, when α is
V
H | i − | i
the three-qubit Greenberger-Horne-Zeilinger (GHZ) state (which is g.e. [11]) then DimF = 48. Note
V
that, according to the above definition, in the case n = 1 each nonzero vector is g.e. by default since
the system admits no bipartitions. (This makes sense only mathematically, as entanglement physically
exists only when n > 1.)
Fromtheviewpoint ofexperiments, thegenuineentanglement iseasier todetect thanthenon-genuine
one [16]. Most extensively studied quantum states are g.e. such as the multiqubit GHZ, W and Dicke
states, and the 4-qubit L and M states [9]. They indeed play essential role in the implementation of
many quantum-information tasks. In particular, entanglement witnesses detecting genuine multiqubit
entanglement have been realized recently in experiments [21]. Many g.e. states can be constructed by
using combinatorial designs [10].
The rest of the paper is organized as follows. In Sec. 2 we prove our main result, Theorem 6, and
derive two important corollaries 7 and 8. In Sec. 3, we study the action of Θ on induced maximal faces.
The main technical result is Theorem 11. The two corollaries 12 and 13 describe the action of Θ in more
details. In particular, it is shown that if V := α ⊥ then the face F and the state α α have the same
V
| i | ih |
stabilizer in Θ. In section 4 we examine two known infinite families of non-induced maximal faces (in
3 3 and 2 4) and show that their dimensions are less than d(d 2). We conclude in Sec. 5.
⊗ ⊗ −
3
2 Dimension formula
We shall compute the dimension of the maximal face F where V is any hyperplane. We shall
V
⊆ H
need the following elementary lemma and its corollary.
Lemma 4 Let n f g∗ = 0 where the f and the g are polynomials in independent complex variables
Pi=1 i i i i
z ,...,z . If the f are linearly independent, then all g = 0.
1 m i i
Proof. We use induction on n. The case n = 1 is trivial. Assume that n > 1. Let µ be a monomial
in z ,...,z which occurs in f . Choose c C such that µ does not occur in f cf . We can replace
1 m 1 2 1
f with f cf and g with g +c∗g whil∈e preserving the equality n f g∗ =−0. Therefore we may
2 2 − 1 1 1 2 Pi=1 i i
assume that µ does not occur in f . Similarly, we may assume that it does not occur in any other f .
2 i
By [5, Lemma 7] we must have g = 0. By the induction hypothesis, all other g = 0.
1 i
⊔⊓
Corollary 5 Let f ,...,f and g ,...,g be polynomials in independent complex variables z ,...,z .
1 p 1 q 1 m
If both f ,...,f and g ,...,g are linearly independent, then also the products f g∗ are linearly inde-
1 p 1 q i j
pendent.
∗
Proof. Assume that c f g∗ = 0 for some c C. Thus f c∗ g = 0 and the lemma
Pi,j ij i j ij ∈ Pi i(cid:16)Pj ij j(cid:17)
implies that c∗ g = 0 for each i. As the g are linearly independent, all c = 0.
Pj ij j j ij ⊔⊓
We recall another important fact which we use in our proofs. Let H be the real subspace of the
algebra End consisting of all Hermitian operators. As usual we identify End with the space M of
H H
d d complex matrices. Let T be the real subspace of M consisting of all upper triangular matrices
×
with real diagonal elements. The map H T which sends a Hermitian matrix to its upper triangular
→
part is an isomorphism of real vector spaces. For any subspace L of H, we have DimL = DimL′ where
L′ denotes the image of L by this isomorphism.
It is immediate from our definition of genuinely entangled states that each nonzero vector α ,
| i ∈ H
after a permutation of the parties, can be written as the tensor product
α = α′ α′ α′ = α′,α′,...,α′ , (4)
| i | 1i⊗| 2i⊗···⊗| mi | 1 2 mi
where each α′ is g.e. The tensor factors α′ are unique up to scalar factors and a permutation, and
| ii | ii
we shall refer to them as the g.e. components of α . We also say that (4) is a g.e. decomposition of α .
| i | i
Using this terminology, we state our main theorem which generalizes our previous result [5, Proposition
17].
Theorem 6 Let V be any hyperplane and let α V⊥ be a nonzero vector. Let (4) be the g.e.
⊂ H | i ∈
decomposition of α with α′ ′ := and n = n + n + + n ,
| i | ii ∈ Hi Hn1+···+ni−1+1 ⊗ ··· ⊗ Hn1+···+ni 1 2 ··· m
n 1. Then
i
≥
m
DimF = d2 1 (2d′ 1), (5)
V − −Y i −
i=1
where d′ = Dim ′.
i Hi
Proof. We can view also as a quantum system consisting of m parties ′, i = 1,...,m. In that
H Hi
case we shall denote it by ′ and its set of (normalized) separable states by ′. Note that ′, and
H S1 S1 ⊆ S1
if m < n then the inclusion is proper. Thus, if F′ is the face of ′ associated to V, then (1) implies
V S1
4
that F F′ and so DimF DimF′ . Since α is a product vector in ′, by applying the formula
V ⊆ V V ≤ V | i H
(3), we obtain that
m
DimF′ = d2 1 (2d′ 1). (6)
V − −Y i −
i=1
It remains to prove that DimF = DimF′ . We use the induction on m. The crucial case is m = 1,
V V
i.e., when α is genuinely entangled. In that case we know that DimF DimF′ = d(d 2).
| i V ≤ V −
Assume that DimF < d(d 2) and let
V
−
d1−1d2−1 dn−1
α = a j ,j ,...,j . (7)
| i X X ··· X j1,j2,...,jn| 1 2 ni
j1=0j2=0 jn=0
Without any loss of generality we may assume that the coefficient a = 0. Let ζ :=
d1−1,d2−1,...,dn−1 6 | i
ζ ,ζ ,...,ζ be any product vector with
1 2 n
| i ∈ H
di−1
ζ = z j , i = 1,...,n. (8)
| ii X i,ji| ii
ji=0
The inner product P := α ζ is a polynomial of degree n which is multilinear in the n sets of complex
h | i
variables z : j = 0,1,...,d 1 , i = 1,...,n. In order to write this polynomial explicitly (and for
{ i,ji i i− }
later use) it is convenient to introduce the abbreviations
f = z z z , 1 k n. (9)
j1,j2,...,jk 1,j1 2,j2··· k,jk ≤ ≤
Then we have
d1−1d2−1 dn−1
P = a∗ f . (10)
X X ··· X j1,j2,...,jn j1,j2,...,jn
j1=0j2=0 jn=0
Let L H be the real span of all pure product states
⊆
ζ ζ = f f∗ j ,j ,...,j k ,k ,...,k , (11)
| ih | X X j1,j2,...,jn k1,k2,...,kn| 1 2 nih 1 2 n|
(j1,j2,...,jn)(k1,k2,...,kn)
where the complex varables z are subject only to the constraint P = 0 (which guarantees that
i,ji
ζ V). Let us denote by the hypersurface in defined by the equation P = 0.
1 2 n
| i ∈ X H ×H ×···×H
Here we consider H as the space of d d Hermitian matrices whose rows and columns are indexed
×
by the n-tuples (j ,j ,...,j ) with j 0,1,...,d 1 . These n-tuples are linearly ordered so that
1 2 n k k
∈ { − }
(j ,j ,...,j ) < (k ,k ,...,k ) holds if and only if j = k for i < s and j < k for some index s.
1 2 n 1 2 n i i s s
Since a = 0, the monomial f (when restricted to ) is a linear combi-
d1−1,d2−1,...,dn−1 6 d1−1,d2−1,...,dn−1 X
nation with constant coefficients of d2 1 other monomials. Thus, we can eliminate this monomial (and
−
its complex conjugate) from (11). Then this equation can be rewritten as
ζ ζ = f f∗ ρ , (12)
| ih | X j1,...,jn k1,...,kn j1,...,jn;k1,...,kn
whereρ† = ρ andneither(j ,...,j )nor(k ,...,k )isequalto(d 1,...,d 1).
j1,...,jn;k1,...,kn k1,...,kn;j1,...,jn 1 n 1 n 1− n−
Thus this sum has (d 1)2 terms. Moreover, the (d 1)2 matrices ρ are linearly independent.
− − j1,...,jn;k1,...,kn
5
The coefficient of ρ is real (and nonnegative) if (j ,...,j ) = (k ,...,k ) and the other
j1,...,jn;k1,...,kn 1 n 1 n
coefficients occur in complex conjugate pairs.
As DimF < d(d 2), we have DimL < (d 1)2. We have P = P z +P , where
V − − 0 n,dn−1 1
d1−1 dn−1−1
P := a∗ f , (13)
0 X ··· X j1,...,jn−1,dn−1 j1,...,jn−1
j1=0 jn−1=0
d1−1 dn−1−1dn−2
P := a∗ f . (14)
1 X ··· X X j1,...,jn−1,jn j1,...,jn−1,jn
j1=0 jn−1=0jn=0
As in the previous paragraph, we know that L is spanned by the matrices (12) with the restriction on
the indices j and k specified there. Let L′ be the subspace of L spanned by all matrices (12) with the
i i
additional constraint P = 0 on the variables z . Obviously, we have DimL′ DimL < (d 1)2.
0 6 i,ji ≤ −
Assume that the polynomial P is irreducible. By multiplying (12) with P 2, we obtain the equation
0
| |
P 2 ζ ζ = (P f )(P f )∗ρ . (15)
| 0| ·| ih | X 0 j1,...,jn 0 k1,...,kn j1,...,jn;k1,...,kn
The variablez canbe eliminated fromtherhsby using theequation P z = P . Forinstance,
n,dn−1 0 n,dn−1 − 1
if j = d 1 we can replace P z with P . After this elimination, the coefficients of ρ
n n − 0 n,jn − 1 j1,...,jn;k1,...,kn
on the rhs will have the form gh∗, where
g,h Φ := P f : k = d 1
∈ { 0 k1,...,kn−1,kn n 6 n − }∪
P f : i < n, k < d 1 . (16)
{− 1 k1,...,kn−1 ∃ i i − }
We claim that the d 1 polynomials in Φ are linearly independent. Otherwise we have an identity
−
P Q = P Q where Q is a nonzero multilinear polynomial of degree n and the monomial f
0 1 1 0 1 d1−1,...dn−1−1
does not occur in Q . Since P is irreducible, P and P are relatively prime. Hence, we may assume
0 0 1
that P = Q . But this is impossible since f occurs in P . Thus, our claim is proved.
0 0 d1−1,...dn−1−1 0
ByCorollary 5, the(d 1)2 products gh∗ with g,h Φ arealso linearly independent. As thematrices
− ∈
ρ are linearly independent, it follows from (15) that DimL′ = (d 1)2. This contradiction
j1,...,jn;k1,...,kn −
implies that P is reducible, i.e., it has a nontrivial factorization P = P′P′′, where P′ and P′′ are
homogeneous polynomials of degree m and n m, respectively. Since P is multilinear, the same is true
−
for P′ and P′′. Hence, after permuting the , we may assume that P′ depends only on the variables
i
H
z with i m, and P′′ depends only on the z with i > m. This means that α = α′ α′′ with
i,ji ≤ i,ji | i | i⊗| i
α′ and α′′ , i.e., α is not genuinely entangled. Thus, we have
1 m m+1 n
| i ∈ H ⊗···⊗H | i ∈ H ⊗···⊗H | i
finally reached a contradiction. Hence, we conclude that DimF = DimF′ = d(d 2) when m = 1.
Now let m > 1. Let us introduce the quantum system ˆ = V′ V′ , wh−ich we consider as a
H H1⊗···⊗Hm−1
subsystem of ′. We set αˆ = α′,...,α′ , Vˆ = αˆ ⊥ ˆ and V′ = α′ ⊥ ′. Denote the space
H | i | 1 m−1i | i ⊂ H i | ii ⊂ Hi
of Hermitian operators on ˆ and ′ by Hˆ and H′, respectively. By the induction hypothesis we have
DimF = DimF′ and, byH(3), weHhaive DimF′ =i(d/d′ )2 1 m−1(2d′ 1).
Vˆ Vˆ Vˆ m − −Qi=1 i −
Let ζ be any product vector in V, and note that ζ ˆ V′ or ζ Vˆ ′ . Let L H be
| i | i ∈ H ⊗ m | i ∈ ⊗Hm ⊆
the real span of all ζ ζ . Similarly, let Lˆ Hˆ be the real span of ξ ξ over all product vectors ξ Lˆ
| ih | ⊂ | ih | | i ∈
orthogonal to αˆ . Finally, let L′ be the real span of η η over all product vectors η ′ orthogonal
| i i | ih | | i ∈ Hi
to α′ . Since we have already handled the case m = 1, we know that DimL′ = (d′ 1)2.
| ii m m −
sDpiamcTeLˆhLeˆ=|⊗ζDiRhiζmH|Fmw′ ′.it+hCo1|ζn=ise∈q(duH/eˆdn′t⊗l)y2V, mw′ esphmaan−v1e(t2hLde′=spaLˆ1c)e⊗, iRHtˆfH⊗olm′lRo+wLs′mHˆt,h⊗aantRdLt′mho.seSinwciethD|iζmi L∈′mVˆ=⊗(dH′mm′ −sp1a)2natnhde
Vˆ m −Qi=1 i −
6
DimL = DimLˆ ⊗R Hm′ +DimHˆ ⊗R L′m −DimLˆ ⊗R L′m
= (d′ )2DimLˆ +(d/d′ )2(d′ 1)2 (d′ 1)2DimLˆ
m m m − − m −
m
= d2 (2d′ 1). (17)
−Y i −
i=1
This completes the proof of the induction step, and of the theorem.
We remark that m (2d′ 1) 2d 1 and that equality holds only if m = 1. Thus we have th⊔⊓e
Qi=1 i − ≥ −
following corollary.
Corollary 7 For any hyperplane V = α ⊥ we have DimF d(d 2) and equality holds if and
V
| i ⊂ H ≤ −
only if α is genuinely entangled.
| i
Similarly one can prove the following corollary.
Corollary 8 If we drop in Theorem 6 the hypothesis that each α′ is g.e., then
| ii
m
DimF d2 1 (2d′ 1), (18)
V ≤ − −Y i −
i=1
and the equality holds if and only if each α′ is g.e.
| ii
Example 9 Let us consider the bipartite case d d (n = 2) and let V = α ⊥ where α is a
1 2
⊗ | i | i ∈ H
vector of Schmidt rank s > 0. If s > 1 then the pure state α is genuinely entangled and so m = 1 and
| i
d′ = d. Thus (18) gives that DimF = d(d 2). On the other hand, if s = 1 then m = 2, d′ = d and
1 V − 1 1
d′ = d and so DimF = d2 1 (2d 1)(2d 1).
2 2 V − − 1 − 2−
In general, if m = n in Theorem 6 then the formula (5) reduces to (3).
3 Action of Θ on induced maximal faces
For any subset S 1,2,...,n we set Γ = Γ . Moreover, for any Γ Θ there is a unique S such
⊆ { } S Qi∈S i ∈
that Γ = Γ . Note that if S = then Γ is the identity map, and if S = 1,2,...,n then Γ is the
S S S
∅ { }
transposition map on End . For any vector subspace V we denote by P the set of all product
V
H ⊆ H
vectors in V.
Let us introduce the general notion of partial conjugates.
Definition 10 If ζ = ζ ,ζ ,...,ζ is a product vector and S 1,2,...,n , then its S-partial
1 2 n
| i | i ∈ H ⊆ { }
conjugate, ζ∗S , is the product vector
| i
ζ∗S = z ,z ,...,z , (19)
1 2 n
| i | i
where z = ζ∗ if i S and z = ζ otherwise. (Note that this conjugate is well defined only up to
| ii | ii ∈ | ii | ii
a phase factor.)
7
For any subset X of H we set (X) = (ρ) and we define the rank of X, r(X), to be the
R Pρ∈X R
dimension of (X). This rank can be used to distinguish the induced from the non-induced maximal
R
faces of . Indeed, if F is a maximal face of , it was shown in [5, Proposition 13] that r(F) d 1
1 1
S S ≥ −
and that F is induced if and only if r(Γ(F)) = d 1 for some Γ Θ. In particular, for any hyperplane
− ∈
V and any Γ Θ, the rank of Γ(F ) is either d or d 1. The two cases can be distinguished by
V
⊂ H ∈ −
the first corollary of the following theorem.
Theorem 11 Let α be a nonzero vector, V = α ⊥, and let S = 1,2,...,k with 1 k < n. If
α is not separable|foir∈thHe bipartition S S¯, then r(Γ |(Fi )) = d. { } ≤
S V
| i |
Proof. Let us define P∗S := ζ∗S : ζ P . It is easy to verify that P∗S (Γ (F )). Hence, in
V {| i | i ∈ V} V ⊆ R S V
order to prove the theorem it suffices to prove that P∗S spans . We shall prove it by contradiction.
V H
Assume that P∗S does not span . A product vector ζ = ζ ,ζ ,...,ζ belongs to V if and only if
V H | i | 1 2 ni
α ζ = 0. We shall write α as in (7) and the vectors ζ as in (8). We shall also use the abbreviations
i
h | i | i | i
(9). The inner product P := α ζ is given explicitly by the formula (10). By our assumption, the set
h | i
P∗S is contained in some hyperplane of . Hence, there exist complex constants b (not all 0)
V H j1,j2,...,jn
such that
d1−1d2−1 dn−1
b f∗ f = 0 (20)
X X ··· X j1,j2,...,jn j1,j2,...,jk jk+1,...,jn
j1=0j2=0 jn=0
holds whenever P = 0. Let us denote the lhs of the above equation by Q, it is a polynomial in the
variables z∗ , i k, and the variables z , i > k. This means that P = 0 = Q = 0. Equivalently,
i,ji ≤ i,ji ⇒
P 2 = 0 = Q 2 = 0. (21)
| | ⇒ | |
Let P = P P P and Q = Q Q Q be the factorizations into the product of irreducible
1 2 r 1 2 s
··· ···
polynomials. Then we have
r s
P 2 = 0 = Q 2 = 0. (22)
Y| i| ⇒ Y| j|
i=1 j=1
We view the polynomials P 2 and Q 2 as polynomials in the real and imaginary parts of the complex
i j
| | | |
variables z . As such they have real coefficients and are irreducible over R. The factors P 2 are
p,qp | i|
pairwise non-proportional since they depend on different and disjoint sets of variables. Consequently,
we must have r = s and we may assume that P 2 = Q 2 for each i. Moreover, we may also assume
i i
| | | |
that Q = P or Q = P∗ for i = 1,...,r.
i i i i
Notethat P is a multilinear polynomial of degree n in the coordinates of thevectors ζ , i = 1,...,n,
i
| i
and Q is a multilinear polynomial of degree n in the coordinates of the vectors ζ∗ , i k and ζ ,
| ii ≤ | ii
i > k. If n is the degree of P , it follows that P is a multilinear polynomial in the coordinates of n
i i i i
of the vectors ζ ,..., ζ . Since α is not separable for the bipartition S S¯, at least one of the P ,
1 n i
| i | i | i |
say P , depends on the coordinates of at least one ζ with p k and at least one ζ with q > k.
1 p q
| i ≤ | i
Consequently, Q depends on the vectors ζ∗ and ζ . Hence, Q is not a scalar multiple of P or P∗.
1 | pi | qi 1 1 1
Thus we have a contradiction.
⊔⊓
The following two corollaries describe the action of Θ on the set of induced maximal faces of .
1
S
Corollary 12 For any induced maximal face F of the following are equivalent:
1
S
(i) r(F) = d 1;
−
(ii) F = F for some hyperplane V ;
V
⊂ H
(iii) F = F .
R(F)
8
Proof. As F is a maximal face, we have r(F) d 1.
≥ −
(i) = (ii) Since F is an induced maximal face, we have F = Γ (F ) for some hyperplane V and
S V
⇒
some S 1,2,...,n . Without any loss of generality, we may assume that S = 1,2,...,k for
⊆ { } { }
some k. Let α be a nonzero vector orthogonal to V. By Theorem 11, we have α = β,γ with
| i | i | i
β 1 k and γ k+1 n. Then F = FV′ where V′ is the hyperplane orthogonal
| i ∈ H ⊗···⊗H | i ∈ H ⊗···⊗H
to β∗,γ .
| i
(ii) = (iii) By the definition of F we have (F) = (F ) V. As r(F) d 1, it follows that
V V
⇒ R R ⊆ ≥ −
(F) = V and so (iii) holds.
R
(iii) = (i) Since F = , we have (F) = . Hence, we must have r(F) = d 1.
1
⇒ 6 S R 6 H − ⊔⊓
Corollary 13 Let (4) be the g.e. decomposition of a nonzero vector α , V = α ⊥, and let ′ =
| i | i H
′ ′ ′ be the m-partite quantum system obtained from by splitting the n parties of
H1 ⊗H2 ⊗···⊗Hm H H
into m groups as in Theorem 6. Denote by Θ′ the group of partial transposition operators of the quantum
system ′. For Γ Θ we have
H ∈
(i) r(Γ(F )) = d 1 if and only if Γ Θ′;
V
− ∈
(ii) Γ fixes α α if and only if Γ Θ′ and α′ is real up to a phase factor whenever Γ acts on
| ih | ∈ | ii
End ′ non-trivially (i.e., as the transposition map).
Hi
(iii) Γ fixes α α if and only if it fixes F ;
V
| ih |
Proof. As (i) and (ii) follow easily from the theorem, we shall prove only (iii). By (i) and (ii) we may
assume that Γ Θ′. After permuting the ′, we may assume that Γ = Γ acts as the transposition
∈ Hi S
operator on ′ ′ and as identity on ′ ′ for some k.
H1 ⊗···⊗Hk Hk+1 ⊗···⊗Hm
Suppose that Γ fixes α α . By (ii) r(Γ(FV)) = d 1. By Corollary 12, Γ(FV) = FV′ for some
| ih | −
hyperplane V′. Moreover, each α′ , i k, must be real up to a phase factor. Thus, α∗S is proportional
| ii ≤ | i
to α (see Definition 10). This implies that V′ = V and so Γ(FV) = FV′ = FV.
| i
For the converse, suppose that Γ fixes F . Then V must be orthogonal to α∗S . It follows that α∗S
V
| i | i
is equal to α up to a phase factor. Hence, Γ( α α ) = α∗S α∗S = α α .
| i | ih | | ih | | ih | ⊔⊓
If α is a product vector, then obviously the hyperplane α ⊥ has an orthogonal basis consisting
| i ∈ H | i
of product vectors. We shall prove the converse.
Lemma 14 Let α be a nonzero vector and let V = α ⊥. If V has an orthogonal basis consisting
| i ∈ H | i
of product vectors, then α is a product vector.
| i
Proof. The case n = 1 is trivial. The bipartite case, n = 2, is well known. Assume that n > 2. By
applying the assertion to the bipartition := , := with 1 k < n,
1 k k+1 n
P H ⊗···⊗H Q H ⊗···⊗H ≤
we deduce that α = β γ with β and γ . Since this holds for each k, α is a product
| i | i⊗| i | i ∈ P | i ∈ Q | i
vector.
⊔⊓
4 Some maximal faces in the bipartite case
It is easy to see that there exist maximal faces of which are not induced provided that d > 6. Indeed,
1
S
there exist full states ρ ∂ , and any maximal face which contains such ρ is non-induced. In this
1
∈ S
section we consider only bipartite systems and we set m = d and l = d .
1 2
Given a linear map Φ : End End , the so called Choi matrix, C End , of Φ is defined
1 2 Φ
H → H ∈ H
by
m−1
C := i j Φ( i j ). (23)
Φ X | ih |⊗ | ih |
i,j=0
9
The map Φ is positive if Φ(ρ) 0 for all ρ 0. We denote by the closed convex cone in
≥ ≥ P
Hom(End ,End ) consisting of all positive maps. We say that a point φ is exposed if it
1 2
H H ∈ P
lies on an exposed ray of the cone .
P
Define the bilinear pairing , : End Hom(End ,End ) C by
1 2
h· ·i H× H H →
ρ,Φ := Tr(ρTC ), (24)
Φ
h i
where T is the transposition map. For any subset P , the set
⊆ P
P′ = ρ : ρ,φ = 0, φ P (25)
1
{ ∈ S h i ∀ ∈ }
is a face of (which may be empty). If F is a face of then we say that F′ is the dual face of F. One
1
S P
defines similarly the dual of a face of , which is a face of .
1
S P
Let φ and φ = 0. To simplify notation, we shall denote the face φ ′ also by φ′. We denote
∈ P 6 { }
by P the set of product vectors z = x,y such that z z ,φ = 0 or, equivalently, φ( x x ) y∗ = 0.
φ
| i | i h| ih | i | ih | | i
This set is important because the extreme points of the face φ′ are exactly the pure product states z z
| ih |
with z P and z = 1. We say that φ has the spanning property if P spans . If φ is an exposed
φ φ
| i ∈ || || H
point of then φ′ is a maximal face of (see [18, Proposition 5.3]). Moreover, it follows from Corollary
1
P S
12 that this maximal face is non-induced if and only if φ has the spanning property.
Let L : be a nonzero linear map. Then the linear map φ : End End defined by
2 1 L 1 2
H → H H → H
φ (X) = L†XL is a positive map. It has been proved recently [20] that φ is an exposed point of .
L L
P
Hence, φ′ is a maximal face of . We shall prove below that it is also induced.
L S1
Let us recall that there is a natural isomorphism, Ψ, of complex vector spaces =
1 2
H H ⊗ H →
Hom( , ) which sends x y x y∗ for each x and y . While is already
2 1 1 2
H H | i ⊗ | i → | ih | | i ∈ H | i ∈ H H
a Hilbert space, we make also Hom( , ) into a Hilbert space by using the standard inner product
2 1
H H
(X Y) := Tr(X†Y). Then Ψ becomes an isometry.
|
Proposition 15 If L : is a nonzero linear map, then the maximal face φ′ of is induced.
H2 → H1 L S1
More precisely, if α := Ψ−1(L) and V := α ⊥ then φ′ = F .
| i | i L V
Proof. A product vector z = x,y belongs to P if and only if φ ( x x ) y∗ = 0. This condition
| i | i φL L | ih | | i
is equivalent to x L y∗ = 0, and also to Tr(L† x y∗ ) = 0, i.e., to (L x y∗ ) = 0. By applying the
h | | i | ih | | | ih |
isometry Ψ−1, we conclude that x,y P if and only if α x,y = 0. Thus, P is the set of all
| i ∈ φL h | i φL
product vectors in the hyperplane V. Consequently, φ′ = F .
L V ⊔⊓
4.1 Some 3 3 non-induced maximal faces
×
K.-C. Ha and S.-H, Kye have constructed recently [13] concrete examples of non-induced maximal faces
in 3 3. Let us mention first that they have constructed a 1-parameter family ∆ , b > 0, b = 1, of faces
b
⊗ 6
of . Each ∆ is a 9-dimensional simplex, the convex hull of pure product states z z , i = 1,...,10.
1 b i i
S | ih |
The product vectors z are the normalizations of the six vectors listed in [13, Eq. 10] and the additional
i
| i
four vectors listed in [13, Eq. 12]. They have shown that the face ∆ is the intersection of two faces
b
Φ(1/b)′ and (Φ(1/b) t)′ (see their paper for more details). Further, they have shown that
◦
(i) each interior point of ∆ is a full state;
b
(ii) (Φ(1/b) t)′ = Γ (Φ(1/b)′);
1
◦
(iii) any product vector z for which z z Φ(1/b)′ has one of the four forms listed in [13, Eq. 13].
| i | ih | ∈
To simplify the notation we set F = Φ(1/b)′, and so by (ii) we have (Φ(1/b) t)′ = Γ (F ). It follows
b 1 b
◦
from (i) that all three faces ∆ , F and Γ (F ) are non-induced.
b b 1 b
10