Cited by 366Quantum Information and CryptographyAdvanced Operator Algebra ResearchQuantum Mechanics and Applications

An analysis of completely-positive trace-preserving maps on M2

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Beth Ruskai, Mary, Szarek, Stanislaw, Werner, Elisabeth·2002·Linear Algebra and its Applications

Integrity signals

Registry

Crossref · 10.1016/s0024-3795(01)00547-x

Cites

36 of 36 references verified0 citation errors · 0 not checkable

Statistics

Nothing to recompute.

Claims the paper makes

0 of 22 cited sources checked · 0 supported · 0 flagged

  • Findingown data

    The paper provides a new characterization of completely positive trace-preserving maps on M2 that makes it easy to test any trace-preserving map for complete positivity.

    “We give a useful new characterization of the set of all completely positive, trace-preserving maps : M 2 → M 2 from which one can easily check any trace-preserving map for complete positivity.”
    Quote in the paperSubstantial claim
  • Findingreasoning

    Conditions on the diagonal elements and determinant of I - R†R are sufficient to ensure that R is a contraction.

    “Conditions on the diagonal elements and determinant of I -R † R suffice to insure R is a contraction.”
    Quote in the paperSubstantial claim
  • Findingreasoning

    When a map is in the closure of the extreme points, its matrix in (3) can be written in the trigonometric form (17) with u in [0,2π) and v in [0,π).

    “The reformulation of the conditions of Theorem 4 as ( 15) and ( 16) implies that when is in the closure of the extreme points, the matrix in (3) can be written in the useful form”
    Quote in the paperSubstantial claim
  • Findingreasoning

    Requiring the diagonal elements of R†R and RR† to be 1 implies inequalities (9) and (10).

    “The requirement that the diagonal elements of R † R and R R † must be 1 then implies that”
    Quote in the paperSubstantial claim
  • Findingreasoning

    A map belongs to the closure of the extreme points of S exactly when it can be reduced to form (3) with at most one tk nonzero and, taking that index as 3, (λ1 ± λ2)^2 = (1 ± λ3)^2 - t3^2.

    “Theorem 4. A map belongs to the closure of the set of extreme points of S if and only if it can be reduced to the form (3) so that at most one t k is non-zero and (with the convention this is t 3 ) (λ 1 ± λ 2 ) 2 = (1 ± λ 3 ) 2 -t 2 3 .”
    Quote in the paperSubstantial claim
  • May combine two claims. 38
  • Classified as probably not a substantial claim. 9
  • The quote only partly states this claim. 6
  • May combine two claims.

    Findingreasoning

    If λ3 = 0 and |t3| + |λ3| = 1, then |t3| = 1 and the map is not completely positive unless t1 = t2 = λ1 = λ2 = 0; in that case R = σx is consistent and sufficient.

    “If λ 3 = 0 and |t 3 | + |λ 3 | = 1, then |t 3 | = 1 in which case will not be completely positive unless t 1 = t 2 = λ 1 = λ 2 = 0; then it is consistent and sufficient to interpret”
    Quote in the paperSubstantial claim
  • Probably not a substantial claim: may not pass S3 (load-bearing).

    Findingnone given

    A trace-preserving map is positivity-preserving exactly when it sends the Bloch sphere into the Bloch ball.

    “Note that a trace-preserving map is positivity-preserving if and only if it maps the Bloch sphere into the “Bloch ball”, defined as the Bloch sphere plus its interior.”
    Quote in the paperNot a substantial claim
  • May combine two claims.

    Findingreasoning

    Adding and subtracting conditions (14) gives the equivalent conditions λ3 = λ1λ2 and t3^2 = (1 - λ1^2)(1 - λ2^2).

    “Adding and subtracting conditions (14) yields the equivalent conditions λ 3 = λ 1 λ 2 , (15 ) t 2 3 = (1 -λ 2 1 )(1 -λ 2 2 ).”
    Quote in the paperSubstantial claim
  • The quote only partly states this claim.

    Findingreasoning

    The Algoet-Fujiwara condition (11) always holds.

    “This implies that the Algoet-Fujiwara condition [1] (λ 1 ± λ 2 ) 2 (1 ± λ 3 ) 2 -t 2 3 (11 )”
    Quote in the paperNot a substantial claim0 of 1 sources checked
    • cites [1] A. Fujiwara, P. Algoet, Affine parameterization of completely positive maps on a matrix algebra, preprint; A. Fujiwara, P. Algoet, One-to-one parameterization of quantum channels, Phys. Rev. A 59 (1999) 3290–3294 source

      Not checked in the free run.

  • May combine two claims. The quote only partly states this claim.

    Findingreasoning

    For a map of the form (3), the image of the Bloch sphere is an ellipsoid whose axis lengths are given by the eigenvalues and whose center is the translation vector.

    “The image of the Bloch sphere of pure states under a map of the form (3) is the ellipsoid”
    Quote in the paperSubstantial claim
  • The quote only partly states this claim.

    Findingown data

    After reduction to a canonical form, the paper gives a useful parameterization of the stochastic maps and their extreme points.

    “and give a useful parameterization after reduction to a certain canonical form.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    The requirement R†R ≤ I also implies det(I - R†R) ≥ 0, which leads to condition (13).

    “In addition to a condition on the diagonal, the requirement R † R I also implies that det(I -R † R ) 0. This leads to the condition”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingcitation

    In general, the capacity is achieved neither with a pair of orthogonal input states nor with a pair of minimal entropy states.

    “Moreover, the capacity is, in general, achieved neither with a pair of orthogonal input states, which would yield h(sin u sin v)h( 1sin 2 u cos 2 v), nor with a pair of minimal entropy states, which would yield h(sin u/sin v), where”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    In that case the non-zero Kraus operator is one of I, σx, σy, σz, and either none or exactly two of the λk can be negative.

    “In fact, the non-zero Kraus operator is simply one of {I, σ x , σ y , σ z } and accordingly, either none or exactly two of the λ k can be negative.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingnone given

    The expression log 2 - h(sin u) is the capacity of the unshifted channel and in this case is also the classical Shannon capacity.

    “The expression log 2h(sin u) is the capacity of the unshifted channel.In this case, it is also the classical Shannon capacity.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    For non-unital maps, two extreme points suffice, and they can be chosen so that an arbitrary non-unital map is the midpoint of a line connecting two true extreme points.

    “For non-unital maps, not only do two extreme points suffice, but they can be chosen so that an arbitrary non-unital map is the”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    In that case the map is unitary and extreme, and λk = ±1 for k = 1, 2, 3.

    “In this case, is unitary and obviously extreme, and λ k = ±1 for k = 1, 2, 3.”
    Quote in the paperSubstantial claim
  • Classified as probably not a substantial claim.

    Findingreasoning

    Lemma 10 shows that (iv) ⇔ (v).

    “Similarly, Lemma 10 shows that (iv) ⇔ (v).”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    Although this tetrahedron lies in a three-dimensional space in a fixed basis, an arbitrary bistochastic map requires nine parameters, which could be chosen as the nine elements of the real 3×3 matrix T, or as the three λk and two sets of Euler angles.

    “Although this tetrahedron lies in a three-dimensional space in a fixed basis, an arbitrary bistochastic map requires nine parameters that could be chosen either as the nine elements of the real 3 × 3 matrix T, or as the three λ k and two sets of Euler angles corresponding to the two rotations required to reduce to diagonal form.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    Under the conditions that λ3 and λ1λ2 have the same sign and t1 = t2 = 0, R is always a contraction, but is unitary only for |λ3| = μ^2.

    “This is possible if, in addition, λ 3 and λ 1 λ 2 have the same sign, and t 1 = t 2 = 0.Under these conditions R , is always a contraction, but is unitary only for |λ 3 | = µ 2 .”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    Lemma 8 gives (ii) ⇔ rank β()^2, and from (25) rank β() = rank β(), so (ii) ⇔ (v).

    “Concerning the other conditions in the general case, Lemma 8 says that (ii) ⇔ rank β( ) 2, while it follows from (25) that rank β( ) = rank β( ); thus (ii) ⇔ (v).”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    A bistochastic map in diagonal form has a unique decomposition as a convex combination of the four unitary maps at the tetrahedron corners, but can also be written non-uniquely as a convex combination of two maps on the tetrahedron edges.

    “If a bistochastic map is written in diagonal form, then it has a unique decomposition as a convex combination of the four unitary maps corresponding to the corners of a tetrahedron; however, it can also be written non-uniquely as a convex combination of two maps on the”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    The squared norm |x|^2 equals 1 - [sin u cos v - sin θ cos u sin v]^2, so |x| = 1 if and only if sin θ = tan u / tan v.

    “After some straightforward trigonometry, we find |x| 2 =cos 2 θ cos 2 u + sin 2 θ cos 2 u cos 2 v + sin 2 u sin 2 v + 2 sin θ [sin u sin v cos u cos v] =1 -[sin u cos v -sin θ cos u sin v] 2(37) so that |x| = 1 if (and only if) sin θ = tan u/tan v.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    If R is unitary but not of special form (7), it still defines a stochastic map, though not one of diagonal form (3).

    “If R is unitary, but does not have the special form (7), it will still define a stochastic map, albeit not one of the”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    In a fixed basis, an arbitrary bistochastic map can be written uniquely as a convex combination of four true extreme points, or non-uniquely as a convex combination of two quasi-extreme points on the edges.

    “In a fixed basis, an arbitrary bistochastic map can be written (uniquely) as a convex combination of four true extreme points, or non-uniquely as a convex combination of two quasi-extreme points on the”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    If W is diagonal, then A1†A1 = V2† D V2 and A1†A1 + A2†A2 = I, so the map is bistochastic.

    “We first note that the fact that W is diagonal implies A † 1 A 1 = V † 2 W † DW V 2 = V † 2 DV 2 . Thus A † 1 A 1 + A † 2 A 2 = V † 2 DV 2 + V † 2 (I -D)V 2 = I so that is a bistochastic map.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    These maps have two non-zero Kraus operators and hence have the form s j + (1-s) k, where j,k ∈ {0,1,2,3} and j(ρ) = σj ρ σj, with σ0 = I.

    “These maps have two non-zero Kraus operators and hence, the form s j + (1s) k , where j, k ∈ {0, 1, 2, 3} and j (ρ) = σ j ρσ j (and σ 0 = I ).”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    Unlike the case μ > 0, such maps cannot be translated to the boundary; for μ = -1/3, one must have t = 0 so no translation is possible.

    “However, unlike the case with µ > 0 such maps cannot be translated to the boundary; in fact, for µ = - 1 3 , one must have t = 0 so that no translation is possible.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    If exactly one of sin v and sin u is zero, then t3 = 0 and the map is unital.

    “If exactly one of {sin v, sin u} is zero, then it follows from (17) that t 3 = sin v sin u = 0, and is unital.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    Under the assumptions 0 < ω < θ < 1/2 and 0 < u < v < 1/2, the conditions cos u = cosω / cosθ and sin v = sin u / sinω follow from (38) and imply that the solution is unique.

    “The conditions cos u = cos ω cos θ , sin v = sin u sin ω follow from (38) and imply that the solution is unique.”
    Quote in the paperSubstantial claim
    • The authors state this as an assumption: For solving the inverse problem, the authors assume 0 < ω < θ < 1/2 and seek solutions with 0 < u < v < 1/2, corresponding to λ1 > λ2 > 0.
  • May combine two claims.

    Findingreasoning

    Example (2a) includes maps with λ1 = λ2 = μ and λ3 = 2μ - 1, but the condition 2μ - 1 = λ3 > λ2^2 = μ^2 is never satisfied, so the conditions for Examples (2a) and (1b) do not overlap.

    “Example (2a) includes maps with λ 1 = λ 2 = µ, λ 3 = 2µ -1, which may seem to contradict Example (1b).However, the condition 2µ -1 = λ 3 > λ 2 1 = µ 2 is never satisfied since if it would imply -µ 2 + 2µ -1 = -(1µ) 2 > 0. Hence the conditions for these two examples do not overlap.”
    Quote in the paperNot a substantial claim
  • May combine two claims.

    Findingreasoning

    The former can be translated in any direction with |t| ≤ 2/3; the latter cannot be translated at all without losing complete positivity.

    “The former can be translated in any direction with |t| 2/3; however, the latter cannot be translated at all without losing complete positivity.”
    Quote in the paperSubstantial claim
  • The quote only partly states this claim.

    Findingnone given

    If both sin v and sin u vanish, then u and v are both of the form nπ for some integer n.

    “If both of {sin v, sin u} are zero, then u, v are both of the form n for some integer n.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    For 0 < μ < 1/2, translation is possible, but it never reaches the boundary of pure states.

    “For 0 < µ < 1 2 translation is possible, but clearly never reaches the boundary of pure states.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingnone given

    For extreme points of types (II) and (III), the Holevo capacity equals log 2, so the capacity attains its maximal value; for the completely noisy channel in Case (IC), the Holevo capacity is 0.

    “It is easy to see that for extreme points of the types (II) and (III), C Holv ( ) = log 2 so that the capacity attains its maximal value, and for the completely noisy channel in Case (IC), C Holv ( ) = 0.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    If D = μI, then both μ^{-1/2} A1 and (1-μ)^{-1/2} A2 are unitary, so the map is a convex combination of unitary maps.

    “If D = µI , then both µ -1/2 A 1 and (1µ) -1/2 A 2 are unitary so that is a convex combination of unitary maps.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Findingreasoning

    For type (D), the image is an ellipsoid whose major axis has length one, and it has one pair of orthogonal states on the unit sphere that are also fixed points.

    “(D) is an ellipsoid whose major axis has length one.Thus, (D) has one pair of orthogonal states on the unit sphere, which are also fixed points.”
    Quote in the paperSubstantial claim
  • Probably not a substantial claim: may not pass S3 (load-bearing).

    Findingreasoning

    This curve forms the boundary of Fig. 3.

    “This curve forms the boundary of Fig. 3.”
    Quote in the paperNot a substantial claim
  • The quote only partly states this claim.

    Findingreasoning

    In place of the remaining two edges, there appear two segments connecting pairs of the four points for which λ3 has the same sign.

    “In place of the remained two edges there appear two segments (still on the surface of the”
    Quote in the paperNot a substantial claim
  • May combine two claims. The quote only partly states this claim.

    Findingreasoning

    This yields a map of type (IC) with u = 2x - π/2 and v = 1/2, but the Kraus operators do not have form (18).

    “One can easily check that this yields a map of type (IC) with u = 2x -2 , v = 1 2 but the Kraus operators do not have the form (18).”
    Quote in the paperNot a substantial claim
  • May combine two claims.

    Premisecitation

    Choi's theorem states that a linear map on Mn is completely positive exactly when its amplification by the identity on Mn is positivity-preserving, equivalently when the block matrix (20) is positive semidefinite.

    “Theorem 5. A linear map : M n → M n is completely positive if and only if ⊗ I M n is positivity-preserving on M n ⊗ M n or equivalently, if and only if the matrix”
    Quote in the paperSubstantial claim0 of 2 sources checked
    • cites [4] Choi, 1975, Completely positive linear maps on complex matrices, Linear Algebra Appl. source

      Not checked in the free run.

    • cites [19] Lindblad, 1975, Completely positive maps and entropy inequalities, Commun. Math. Phys. source

      Not checked in the free run.

  • Probably not a substantial claim: may not pass S3 (load-bearing).

    Premisecitation

    Rieffel and Zalka independently obtained the same parameterization while considering maps arising from interactions between a pair of qubits.

    “This parameterization was also obtained independently by Rieffel and Zalka [21] who, like Niu and Griffiths, considered maps that arise via interactions between a pair of qubits”
    Quote in the paperNot a substantial claim0 of 1 sources checked
    • cites [21] M.A. Nielsen, I.L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge, MA (in press)

      Not checked in the free run.

  • Probably not a substantial claim: may not pass S3 (load-bearing).

    Premisecitation

    Niu and Griffiths obtained a similar parameterization in their work on two-bit copying devices.

    “A similar parameterization was obtained by Niu and Griffiths [18] in their work on two-bit copying devices.”
    Quote in the paperNot a substantial claim0 of 1 sources checked
    • cites [18] Lieb, 1974, Some operator inequalities of the Schwarz type, Adv. Math. source

      Not checked in the free run.

  • Probably not a substantial claim: may not pass S3 (load-bearing). May combine two claims.

    Premisecitation

    For bistochastic maps, the extreme points are known to be unitary conjugations and to correspond to the four corners of a tetrahedron in the λk representation.

    “For bistochastic maps, the extreme points are known [1,11] to consist of the maps that conjugate by a unitary matrix and in the λ k”
    Quote in the paperNot a substantial claim0 of 2 sources checked
    • cites [1] A. Fujiwara, P. Algoet, Affine parameterization of completely positive maps on a matrix algebra, preprint; A. Fujiwara, P. Algoet, One-to-one parameterization of quantum channels, Phys. Rev. A 59 (1999) 3290–3294 source

      Not checked in the free run.

    • cites [11] Horn, 1991

      Not checked in the free run.

  • May combine two claims.

    Premisecitation

    Terhal et al. showed that Lloyd's conjecture, that every map in S can be represented in form (19) with a mixed state ρ2, is false; another counterexample was obtained by Rieffel and Zalka.

    “However, Terhal et al. [22] showed that this is false and another counterexample was obtained in [21].”
    Quote in the paperSubstantial claim0 of 2 sources checked
    • cites [22] Niu, Griffiths, Two qubit copying machine for economical quantum eavesdropping, quant-ph/9810008

      Not checked in the free run.

    • cites [21] M.A. Nielsen, I.L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, Cambridge, MA (in press)

      Not checked in the free run.

  • May combine two claims.

    Interpretationreasoning

    When the map has form (17), more than one non-zero t_k can occur only when λ3 is degenerate, equal to λ1 or λ2; this is precluded by (15) unless λ3 = 0, ±1, and in the latter case the channel is unital with all t_k = 0, so two non-zero t_k occur only in channels with at least two λ_k = 0.

    “When has the form (17) this can only happen when λ 3 is degenerate [i.e., equal to λ 1 or λ 2 ].However, this is precluded by (15) unless λ 3 = 0, ±1.In the latter case we must have a unital channel with all t k = 0. Thus, two non-zero t k occur only in channels that are so noisy that at least two λ k = 0.”
    Quote in the paperSubstantial claim
  • Probably not a substantial claim: may not pass S3 (load-bearing). May combine two claims.

    Premisecitation

    This behavior was first observed by Fuchs for a non-extreme channel, and the amplitude-damping channel was later studied by Schumacher and Westmoreland.

    “This behavior was observed first by Fuchs for a non-extreme channel [5]; subsequently, the amplitude-damping channel was studied by Schumacher and Westmoreland [23].”
    Quote in the paperNot a substantial claim0 of 2 sources checked
    • cites [5] Choi, 1974, A schwarz inequality for positive linear maps on C* algebras, Illinois J. Math. source

      Not checked in the free run.

    • cites [23] V.I. Paulsen, Completely Bounded Maps and Dilations, Pitman Research Notes in Mathematics Series, vol. 146, Longman, Harlow, Wiley, New York, 1986

      Not checked in the free run.

  • Probably not a substantial claim: may not pass S3 (load-bearing). May combine two claims.

    Premisecitation

    Holevo introduced such channels and showed that they suffice to demonstrate the quantum advantage, although both minimal entropy and capacity are achieved with orthogonal inputs.

    “Holevo [7,8] introduced such channels and showed that they suffice to demonstrate the”
    Quote in the paperNot a substantial claim0 of 2 sources checked
    • cites [7] Haag, 1964, An algebraic approach to quantum field theory, J. Math. Phys. source

      Not checked in the free run.

    • cites [8] Holevo, 1979, On the capacity of quantum communication channel, Probl. Peredachi Inform.

      Not checked in the free run.

  • May combine two claims.

    Interpretationreasoning

    As changes of basis rotate the tetrahedron, one obtains a much larger set, so by Theorem 14 any non-extreme map can be written as the midpoint of two generalized extreme points in different bases.

    “As the changes of basis rotate the tetrahedron (considered as a set of diagonal 3 × 3 matrices in R 9 ), we obtain a much larger set so that (by Theorem 14) any non-extreme map can be written as the midpoint of two”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Interpretationreasoning

    Finding a map that takes the pair (±cosω, 0, sinω) to the pair (±cosθ, 0, sinθ) is possible only if |sinθ| > |sinω| and |λ1| > |λ2|.

    “Geometric considerations and the arguments leading to (38) imply that this will be possible only if | sin θ| > | sin ω| and |λ 1 | > |λ 2 |.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Interpretationreasoning

    Because every map in S is equivalent via changes of bases to a map of form (3), this describes all extreme points of S and all points in the closure of the extreme-point set.

    “Since every ∈ S is equivalent in the sense of (4) (i.e., after changes of bases in the two-dimensional Hilbert spaces corresponding to the domain and the range of ) to a map of the form (3), this describes in principle all extreme points of S and all points in the closure of the set of extreme points.”
    Quote in the paperSubstantial claim
  • May combine two claims.

    Interpretationcitation

    In both cited cases the work was numerical, but it suggests that this situation is generic for channels translated orthogonal to the major axis of the ellipse.

    “In both cases the work was numerical, but suggests that this situation is generic for channels that are translated orthogonal to the major axis of the ellipse.”
    Quote in the paperNot a substantial claim0 of 2 sources checked
    • cites [5] Choi, 1974, A schwarz inequality for positive linear maps on C* algebras, Illinois J. Math. source

      Not checked in the free run.

    • cites [23] V.I. Paulsen, Completely Bounded Maps and Dilations, Pitman Research Notes in Mathematics Series, vol. 146, Longman, Harlow, Wiley, New York, 1986

      Not checked in the free run.

  • Probably not a substantial claim: may not pass S3 (load-bearing).

    Premisenone given

    Holevo called channels of form (IC) binary because their Kraus operators can be represented as A1 = |e1><ψ+| and A2 = |e2><ψ-|, where the |e_k> are orthonormal and ψ± can be any pair of states.

    “Holevo called channels of the form (IC)”
    Quote in the paperNot a substantial claim
48 of 48
  • The authors state this as an assumption.

    For describing the subclasses of extreme points, the authors assume |λ1| ≤ |λ2|, noting that permutations of indices cover all maps of form (3).

    “We make the additional assumption that |λ 1 | |λ 2 | since, in any case, all permutations of indices must be considered to obtain the extreme points of all maps of the form (3).”
  • The authors state this as an assumption.

    For solving the inverse problem, the authors assume 0 < ω < θ < 1/2 and seek solutions with 0 < u < v < 1/2, corresponding to λ1 > λ2 > 0.

    “For simplicity, we assume 0 < ω < θ < 1 2 [corresponding to an upward translation and u ∈ (0, )] and seek solutions satisfying 0 < u < v < 1 2 (corresponding to λ 1 > λ 2 > 0).”
2 of 2

Claims other papers make about this one

Each is checked against this paper’s own text: does it say what the citing paper says it does?

Self-check

Percentages, sums and reported differences re-computed from the paper’s own numbers.

Each contradiction between two passages listed here was confirmed by an independent check.

  • Some edge channels swap their pure boundary states rather than fix them

    Minor

    The listed tetrahedron edges include channels with λ1 = −1. Such a channel sends each pure state on the x-axis to the other one, contradicting the claim that the boundary states are fixed points.

  • The stated Pauli basis is not orthonormal

    Minor

    The paper defines the Hilbert-Schmidt inner product without a normalization factor. Under that definition, each listed basis matrix has squared norm 2, not 1, so the basis is orthogonal but not orthonormal.

2 of 2

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Authors
Beth Ruskai, M., Szarek, S., & Werner, E.
Affiliations
Université de Lille (FR)
Venue
Linear Algebra and its Applications
Publisher
Elsevier BV
Published
2002-05
Volume
347
Issue
1-3
Pages
159-187
ISSN
0024-3795 (ISSN-L)
OpenAlex
W2160212841
Cited by
366
References
29
Citations by year
2026: 20 · 2025: 15 · 2024: 17 · 2023: 22 · 2022: 10 · 2021: 14 · 2020: 20 · 2019: 20 · 2018: 26 · 2017: 10 · 2016: 16 · 2015: 14 · 2014: 13 · 2013: 17 · 2012: 13
FWCI
13.77
Type
journal article
Language
en
Open access
Locations
  • Linear Algebra and its Applications (publishedVersion)
  • arXiv (Cornell University) (acceptedVersion)
  • arXiv (Cornell University) (submittedVersion)
  • CERN Document Server (European Organization for Nuclear Research) (submittedVersion)
  • arXiv (Cornell University)
Funders
  • National Science Foundation (9706981, 0074566)
  • National Security Agency
  • Agricultural Research Development Agency
  • Technische Universität Berlin
  • Army Research Office
Topics
  • Physical Sciences › Computer Science › Artificial Intelligence › Quantum Information and Cryptography
  • Physical Sciences › Mathematics › Mathematical Physics › Advanced Operator Algebra Research
  • Physical Sciences › Physics and Astronomy › Atomic and Molecular Physics, and Optics › Quantum Mechanics and Applications
Keywords
completely positive maps, trace-preserving maps, extreme points, stochastic maps, Bloch sphere, convex combination, canonical form, quantum channels

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Confidence bands are shown only for fields calibrated on our annotated test set. None is calibrated yet, so no band is shown.

What counts as a claim ›How we measure confidence ›How we check this ›

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