We provide a proof that entanglement witnesses considered recently in [D.
Chru\'sci\'nski, F.A. Wudarski, arXiv:1105.4821] are optimal.Comment: 4 page

Chruściński, Dariusz

Sarbicki, Gniewomir

English

arXiv

We provide a proof that entanglement witnesses considered recently in [D. Chru\'sci\'nski, F.A. Wudarski, arXiv:1105.4821] are optimal

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Optimal entanglement witnesses for two qutritsDariusz Chruściński and Gniewomir SarbickiInstitute of Physics, Nicolaus Copernicus University,Grudzia̧dzka 5/7, 87–100 Toruń, PolandAbstractWe provide a proof that entanglement witnesses considered recently in [1] are optimal.In a recent paper [1] we analyzed z class of entanglement witnesses (EW) given byW [a, b, c] =a · · · −1 · · · −1· b · · · · · · ·· · c · · · · · ·· · · c · · · · ·−1 · · · a · · · −1· · · · · b · · ·· · · · · · b · ·· · · · · · · c ·−1 · · · −1 · · · a, (1)where to make the picture more transparent we replaced zeros by dots (for simplicity we skipped thenormalization factor which is not essential). One proves the following result [2]Theorem 1. W [a, b, c] defines an entanglement witness if and only if1. 0 ≤ a < 2 ,2. a+ b+ c ≥ 2 ,3. if a ≤ 1 , then bc ≥ (1− a)2.Moreover, being EW it is indecomposable if and only if bc < (2− a)2/4.In particular we analyzed [1] a subclass of EWs defined by0 ≤ a ≤ 1 , a+ b+ c = 2 , bc = (1− a)2 . (2)The corresponding EWs W [b, c] := W [2− b− c, b, c] belong to the ellipse on bc-plane – see Fig. 1. Itwas conjectured [1] that W [b, c] are optimal. In the present paper we show that this conjecture is true.Theorem 2. EWs W [b, c] defined by (2) are optimal.1arXiv:1108.0513v1  [quant-ph]  2 Aug 2011Boundaryiiiiiiivv0.2 0.4 0.6 0.8 1.0 1.20.20.40.60.81.01.2Figure 1: A convex set of EWs W [b, c]. A line b = c corresponds to decomposable EW. Special points:(i) and (ii) Choi EWs, (iii) EW corresponding to reduction map, (v) positive operator with b = c = 0,(iv) decomposable EW with b = c = 1/3.Proof: let us definePbc = {x⊗ y ∈ C3⊗C3 | 〈x⊗ y|W [b, c]|x⊗ y〉 = 0} . (3)It is well known [3] that if the set Pbc spans the entire Hilbert space C3⊗C3, then W [b, c] is an optimalEW. If we find a set of vectors y ∈ C3 such that the 3× 3 matrixWy[b, c] := Tr2(W [b, c] · I3⊗ |y〉〈y|) , (4)is singular, then for each vector xy belonging to the kernel of Wy[b, c] the product vector xy ⊗ y belongsto Pbc (Tr2 denotes a partial trace over the second factor in C3⊗C3). The matrix Wy[b, c] is given bythe formulaWy[b, c] = a|y1|2 + b|y2|2 + c|y3|2 y∗1y2 y∗1y3y∗2y1 c|y1|2 + a|y2|2 + b|y3|2 y∗2y3y∗3y1 y∗3y2 b|y1|2 + c|y2|2 + a|y3|2= (a+ 1)|y1|2 + b|y2|2 + c|y3|2 0 00 c|y1|2 + (a+ 1)|y2|2 + b|y3|2 00 0 b|y1|2 + c|y2|2 + (a+ 1)|y3|2− |y∗〉〈y∗|Let us observe, that for any a, b, c satisfying Theorem 1 and y = [eiα, eiβ, eiγ ] one findsWy[b, c] = diag[e−iα, e−iβ, e−iγ ] 2 −1 −1−1 2 −1−1 −1 2diag[eiα, eiβ, eiγ ] .This matrix has rank 2 and its 1-dim. kernel is spanned by the vector xy = [e−iα, e−iβ, e−iγ ]. Hencewe have the following continuous family of product vectorsxy ⊗ y = [1, ei(β−α), ei(γ−α), ei(α−β), 1, ei(γ−β), ei(α−γ), ei(β−γ), 1] (5)2Note that this family spans at most 7-dimensional subspace of C3 ⊗ C3. To show, that this subspaceis exactly 7-dimensional, it suffices to consider the following set of (α, β, γ)(0, 0, 0), (0, 0, π), (0, π, 0), (0, π, π), (0, 0, π/2), (0, π/2, 0), (0, π/2,−π/2) . (6)Consider now y = (0, y2, y3). One hasWy[b, c] = b|y2|2 + c|y3|2 0 00 a|y2|2 + b|y3|2 −y∗2y30 −y∗3y2 c|y2|2 + a|y3|2 .Its determinant is given by the formula:detWy[b, c] = (b|y2|2 + c|y3|2)(ab|y2|4 + (a2 + ac− 1)|y2|2|y23|+ bc|y3|4) .We are looking for y ∈ C3, that the determinant vanishes.Case 1: b, c 6= 0.Now, the first term is always positive and so the second term has to vanish. Taking ||y|| = 1, one canreplace |y3|2 by 1− |y2|2. The second term reads as followsa (4− 3a)|y2|4 + 2 a (a− b− 1)|y2|2 + ab = 0 . (7)We use here relations bc = (a− 1)2 and a = 2− b+ c. One also assume that b < c (the case c < b maybe treated in the same way using a symmetry b ←→ c [1]). One obtains the following formulae for band cb =12(2− a−√4a− 3a2 ) , c = 12(2− a+√4a− 3a2 )The discriminant of the quadratic equation (for |y2|2) vanishes (it can not be positive due to the factthat W [b, c] is an EW) and one easily solves (7) to get|y2|2 =1 + b− a4− 3a.The vector y is then equal (after calculating |y3|2, we drop the normalization):y = [0,√1 + b− a,√3− b− 2aeiφ] =: [0, p, qeiφ1 ] . (8)For such y, the kernel of Wy[b, c] is spanned by the vectorxy = [0, y∗2 · y3, a|y2|2 + b|y3|2] =: [0, reiφ1 , s] (9)The numbers p, q, r, s are nonzero and depend only on parameters a, b, c. LetΨ(1) := xy ⊗ y = [0, 0, 0, 0, preiφ1 , ps, 0, qre2iφ1 , qseiφ1 ] .3Because of the cyclic symmetry of the problem, one can find the similar product vectors for y2 = 0and y3 = 0:Ψ(2) = [qseiφ2 , 0, qre2iφ2 , 0, 0, 0, ps, 0, preiφ2 ] ,Ψ(3) = [preiφ3 , ps, 0, qre2iφ3 , qseiφ3 , 0, 0, 0, 0] .Now, it turns out that 7 vectors from the family (5) generated by a set (6) plus two arbitrary vectorsfrom the family (Ψ(1),Ψ(2),Ψ(3)) defines a basis in C3⊗C3. Indeed, taking 7 vectors from (5) andΨ(1), Ψ(2) one obtains the following 9× 9 matrix:1 1 1 1 1 1 1 1 11 1 −1 1 1 −1 −1 −1 11 −1 1 −1 1 −1 1 −1 11 −1 −1 −1 1 −1 −1 1 11 1 i 1 1 i −i −i 11 i 1 −i 1 −i 1 i 11 i −i −i 1 −1 i i 10 0 0 0 preiφ1 ps 0 qre2iφ1 qseiφ1qseiφ2 0 qre2iφ2 0 0 0 ps 0 preiφ2. (10)Its determinant reads(−32 + 160i)ei(φ1+φ2)[ (qs)2 + (pr)2 − qspr ] ,and is different from zero except qs = pr = 0. Note, however, that for b, c 6= 0 one has qs, pr 6= 0.Case 2: b = 0, c = 1.Now, the determinant readsdetWy[b, c] = |y1|2|y2|4 + |y2|2|y3|4 + |y3|2|y1|4 − 3|y1|2|y2|2|y3|2 .If one of coordinates, say y1 is zero, then the determinant is equal |y2|2|y3|4 and vanishes only if y2 ory3 vanishes, so the only vectors y with at least one zero coordinate for which Wy[b, c] vanishes areΦ(1) := [1, 0, 0]⊗ [0, 0, 1] , Φ(2) := [0, 1, 0]⊗ [1, 0, 0] , Φ(3) := [0, 0, 1]⊗ [0, 1, 0] . (11)Now we will look for the remaining vectors and we assume that all coordinates are non-zero. Dividingthe determinant by |y1|2|y2|2|y3|2 and gets the following equation|y2||y3|+|y3||y1|+|y1||y2|− 3 = 0 .Its LHS is nonnegative and vanishes only for |y1| = |y2| = |y3|, and hencey = [eiα, eiβ, eiγ ] , xy = [e−iα, e−iβ, e−iγ ] ,and one gets again the 7-dimensional family of vectors (5). However, vectors Φ(k) are not linearlyindependent from (5). Therefore, P01 spans only 7-dim. subspace of C3⊗C3.4Actually, one obtains Φ(k) from Ψ(k) in the limit b→ 0. Let us recall that the determinant of (10)vanishes only when qs = pr = 0. Now, p = s = 0 when b = 0 and c = 1, whereas q = r = 0 whenb = 1 and c = 0. Hence, apart from two witnesses corresponding to Choi maps W [1, 0] and W [0, 1],the remaining EWs have spanning property, i.e. Pbc spans C3⊗C3, and hence they are optimal. 2As this paper was completed we were informed by professors Kil-Chan Ha and Seung-Hyeok Kyethat they provided an independent proof of optimality [4]. Moreover, they proved [5] that all witnessesW [b, c] are exposed (and hence extremal) except W [1, 1], W [1, 0] and W [0, 1].References[1] D. Chruściński and F.A. Wudarski, Geometry of entanglement witnesses for two qutrits,arXiv:1105.4821.[2] S. J. Cho, S.-H. Kye, and S. G. Lee, Linear Algebr. Appl. 171, 213 (1992).[3] M. Lewenstein, B. Kraus, J. I. Cirac, and P. Horodecki, Phys. Rev. A 62, 052310 (2000).[4] K.-C. Ha and S.-H. Kye, One parameter family of indecomposable optimal entanglement witnessesarising from generalized Choi maps, arXiv:1107.2720.[5] K.-C. Ha and S.-H. Kye, Entanglement witnesses arising from exposed positive linear maps,arXiv:1108.0130.5

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