An analysis of completely-positive trace-preserving maps on M2
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- 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 .”
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- Findingnone given
When |t3| + |λ3| = 1, equation (6) has no solution unless t1 = t2 = 0 and either λ1 = λ2 or λ1 = -λ2.
“When |t 3 | + |λ 3 | = 1 no solution to (6) exists unless t 1 = t 2 = 0 and either λ 1 = λ 2 or λ 1 = -λ 2 .”
Quote in the paperSubstantial claim - Findingnone given
The equivalence in (4) preserves the property of being a generalized extreme point of S.
“Moreover, as will be shown in Section 2.3, that “equivalence” preserves the property of being a generalized extreme point of S.”
Quote in the paperNot a substantial claim - Findingnone given
For non-unital maps, the extreme points are those for which the translated ellipsoid touches the boundary of the Bloch sphere at two points, with degenerate cases treated as two coincident points.
“We will see that for non-unital maps the extreme points correspond to those maps for which the translation allows the ellipsoid to touch the boundary of the sphere at two points”
Quote in the paperSubstantial claim - Findingreasoning
Such a map is extreme unless t3 = 0 and |λ3| < 1.
“Moreover, is extreme unless t 3 = 0 and |λ 3 | < 1.”
Quote in the paperSubstantial claim - Findingreasoning
A necessary condition for a map to be an extreme point of S is that it is equivalent in the sense of (4) to a map of the form (3) whose corresponding R is unitary.
“Thus, in particular, a necessary condition for to be an extreme point of S is that it is equivalent [in the sense of (4)] to a map of the form (3) for which the corresponding R is unitary.”
Quote in the paperSubstantial claim - Findingcitation
Not every ellipsoid contained in the Bloch ball is the image of a stochastic map.
“However, not all ellipsoids contained in the “Bloch ball” correspond to images of a stochastic map .”
Quote in the paperSubstantial claim - Findingreasoning
A stochastic map induced by a matrix T of the form (3) belongs to the closure of the extreme points of S exactly when the matrix R given by (7) or (8) is unitary.
“Theorem 3. Let be a stochastic map induced by a matrix T of the form (3). Then belongs to the closure of the set of extreme points of S if and only if the matrix R [as given by (7) or (8) above] is unitary.”
Quote in the paperSubstantial claim - Findingreasoning
The interesting new class of extreme points consists of those satisfying 1 > |λ1| > |λ2| > |λ3| > 0.
“For now, we emphasize that the interesting new class of extreme points are those that also satisfy 1 > |λ 1 | > |λ 2 | > |λ 3 | > 0.”
Quote in the paperSubstantial claim - Findingreasoning
Because a cyclic permutation of indices corresponds to a rotation in R3 that can be absorbed into U and V in (4), treating λ3 and t3 as special does not lose generality.
“Since a cyclic permutation of indices corresponds to a rotation in R 3 which can be incorporated into the maps U and V in (4), stating conditions (for membership of S or for any form of extremality) in a form in which, say, λ 3 and t 3 play a special role does not lead to any real loss of generality.”
Quote in the paperSubstantial claim - Findingnone given
Every map in S is unitarily equivalent, in the sense of (4), to a map of the form (3).
“If ∈ S, then it is unitarily equivalent, in the sense of (4), to a map of the form (3).”
Quote in the paperSubstantial claim - Findingnone given
A map of the form (3) with |t3| + |λ3| ≤ 1 is completely positive exactly when equation (6) has a contraction solution R.
“Theorem 1. A map given by (3) for which |t 3 | + |λ 3 | 1 is completely positive if and only if the equation”
Quote in the paperSubstantial claim - Findingreasoning
A necessary condition for complete positivity of a map of the form (3) is (λ1 ± λ2)^2 ≤ (1 ± λ3)^2.
“Moreover, it follows that a necessary condition for complete positivity of a map of the form (3) is (λ 1 ± λ 2 ) 2 (1 ± λ 3 ) 2 .”
Quote in the paperSubstantial claim - Findingown data
Every stochastic map on M2 can be expressed as a convex combination of two generalized extreme points.
“and show that any stochastic map on M 2 can be written as a convex combination of two “generalized” extreme points.”
Quote in the paperSubstantial claim - Findingnone given
Condition (12) holds exactly when it remains valid under any permutation of the parameters λk.
“Although not obvious from this analysis, (12) holds if and only if it is valid for any permutation of the parameters λ k .”
Quote in the paperNot a substantial claim - Findingown data
All extreme points of the set of completely positive trace-preserving maps on M2 are determined explicitly.
“We also determine explicitly all extreme points of this set”
Quote in the paperSubstantial claim - Findingnone given
The condition t1 = t2 = 0 is equivalent to R being skew diagonal.
“However, it should be noted that the requirement t 1 = t 2 = 0 is equivalent to requiring that R is skew diagonal.”
Quote in the paperNot a substantial claim - Findingreasoning
A map lies in the closure of the extreme points of S exactly when, up to a permutation of indices, equality holds in (9), (10), and (13).
“Thus, is in the closure of the extreme points of S if and only if, up to a permutation of indices, equality holds in ( 9), ( 10) and ( 13).”
Quote in the paperSubstantial claim - Findingreasoning
An important class of non-unital extreme points is characterized by having exactly two images on the Bloch sphere.
“This allows a detailed examination of an important class of non-unital extreme points that can be characterized as having exactly two images on the Bloch sphere.”
Quote in the paperSubstantial claim - Findingreasoning
After reduction to the diagonal form (3), there are three equivalent sets of conditions for the closure of the extreme points of S.
“We conclude this summary by noting that we have obtained three equivalent sets of conditions on the closure of the set of extreme points of S after appropriate reduction to the “diagonal” form (3).”
Quote in the paperSubstantial claim - Findingnone given
For maps of the form (3), a necessary condition for positivity preservation is |tk| + |λk| ≤ 1 for each k = 1,2,3.
“For such maps, it is easy to verify that a necessary condition for to be positivity-preserving is that |t k | + |λ k | 1 for k = 1, 2, 3.”
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When |tk| + |λk| < 1 for all k, equation (6) has the unique solution R given by (7).
“When |t k | + |λ k | < 1 (6) has the unique solution R =”
Quote in the paperSubstantial claim - Findingreasoning
If one of the λk is zero, condition (15) implies that at least two of the λk are zero.
“It may be worth noting that when one of the λ k = 0, (15) implies that at least two λ k are zero.”
Quote in the paperSubstantial claim - Findingreasoning
In the trigonometric parameterization, the map can be realized with the Kraus operators A± given by (18).
“It is straightforward to verify that in this parameterization can be realized with the Kraus operators”
Quote in the paperNot a substantial claim - Findingown data
In a fixed basis, the convex set of stochastic maps with t = 0 forms a tetrahedron.
“In a fixed basis, the convex set of stochastic maps with t = 0 forms a tetrahedron (see Fig. 2) that we can describe using the λ k in (3) which we write as”
Quote in the paperSubstantial claim - Findingreasoning
Depending on the sign of λ3, one of the pair inequalities in (39) holds.
“one can see that, depending on the sign of λ 3 one of the pair inequalities in (39) holds.”
Quote in the paperSubstantial claim - Findingreasoning
When t3 = 0, the linearity in λ3 yields a line segment, so there are no longer true extreme points.
“When t 3 = 0, the linearity in λ 3 yields a line segment so that we no longer have”
Quote in the paperSubstantial claim - Findingnone given
Theorem 16: If a map is an extreme point of S, then at least one point in its image (D) is a pure state.
“Theorem 16. If is an extreme point of S, then at least one point in the image (D) is a pure state.”
Quote in the paperSubstantial claim - Findingreasoning
In the special case where the map is represented by a matrix T of form (3), the first three conditions of Theorem 12 are equivalent.
“In the preceding section we showed that, in the special case when is represented by a matrix T of the form (3), the first three conditions are equivalent.”
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Maps of type (I) and (II) correspond respectively to the non-unital and unital extreme points of S.
“Maps of type (I) and (II) correspond to the non-unital and unital extreme points of S, respectively.”
Quote in the paperSubstantial claim - Findingnone given
The eigenvalues [1/3, 1/3, 1/3] and [1/3, 1/3, -1/3] both yield the same ellipsoid, a sphere of radius 1/3.
“The eigenvalues [ 1 3 , 1 3 , 1 3 ] and [ 1 3 , 1 3 , -1 3 ] both yield the same ellipsoid (actually, a sphere of radius 1 3 ).”
Quote in the paperSubstantial claim - Findingreasoning
If R is a contraction, its singular value decomposition can be written in the form of equation (36).
“If R is a contraction, its singular value decomposition can be written in the form R =V cos θ 1 0 0 cosθ 2 W † = 1 2 V e iθ 1 0 0 e iθ 2 W † + 1 2 V e -iθ 1 0 0 e -iθ 2 W † , (36 )”
Quote in the paperSubstantial claim - Findingreasoning
A general extreme point is the composition of a map of type (I) or (II) with unitary maps, as in (4).
“A general extreme point is then the composition of a map of type (I) or (II) with unitary maps, as in (4).”
Quote in the paperSubstantial claim - Findingnone given
In general, it is not possible to translate the image ellipsoid to the boundary of the unit ball.
“In general, however, it is not possible to translate the image ellipsoid to the boundary of the unit ball.”
Quote in the paperSubstantial claim - Findingreasoning
The matrix β( ) is positive semidefinite exactly when β( ) is positive semidefinite.
“In particular, (25) shows that β( ) is positive semi-definite if and only if β( ) is.”
Quote in the paperSubstantial claim - Findingcitation
The minimal number of Kraus operators needed to represent a completely positive map in form (1) equals rank[β( )].
“Lemma 10.The minimal number of Kraus operators A k needed to represent a completely positive map in the form (1) is rank[β( )].”
Quote in the paperSubstantial claim0 of 1 sources checked- cites [4] Choi, 1975, Completely positive linear maps on complex matrices, Linear Algebra Appl. source
Not checked in the free run.
- Findingreasoning
Such a map is a convex combination of two extreme maps of type (II).
“However, it is readily verified that such a is a convex combination of two extreme maps of type (II).”
Quote in the paperSubstantial claim - Findingreasoning
Theorem 14: Any stochastic map on M2 can be written as the convex combination of two maps in the closure of the extreme-point set.
“Theorem 14. Any stochastic map on M 2 can be written as the convex combination of two maps in the closure of the set of extreme points.”
Quote in the paperSubstantial claim - Findingreasoning
For each pair (j,k), one obtains a line between two extreme points that form the tetrahedron of bistochastic maps.
“For each pair (j, k) we obtain a line between two of the extreme points that form the tetrahedron of bistochastic maps.”
Quote in the paperNot a substantial claim - Findingreasoning
Theorem 13: A stochastic map that requires two Kraus operators but not more is either an extreme point of S or bistochastic.
“Theorem 13. A stochastic map that, when written in the form (1), requires two Kraus operators A k but not more, is either an extreme point of S or bistochastic.”
Quote in the paperSubstantial claim - Findingreasoning
For a positive semidefinite matrix M in M4, M admits a factorization (29) with R unitary exactly when rank(M) ≤ 2.
“Lemma 8. Let M ∈ M 4 with M 0. Then M admits a factorization (29) with R unitary if and only if rank(M) 2.”
Quote in the paperSubstantial claim - Findingreasoning
When t3 is non-zero, the corners depend non-linearly on λ3, yielding a curve of true extreme points.
“When t 3 / = 0, the corners depend non-linearly on λ 3 , yielding a curve of true extreme points.”
Quote in the paperSubstantial claim - Findingnone given
If (D) contains two or more pure states, then the map must be in the closure of the extreme points of S.
“If (D) contains two or more pure states, it must be in the closure of the extreme points of S.”
Quote in the paperSubstantial claim - Findingreasoning
If the intersection of (D) with the Bloch sphere consists of exactly two non-orthogonal pure states, then the map must be a non-unital extreme point.
“Moreover, if the intersection of (D) with the Bloch sphere consists of exactly two non-orthogonal pure states, then must be a non-unital extreme point;”
Quote in the paperSubstantial claim - Findingreasoning
If = α1 + (1-α)2 with 0 < α < 1 is not extreme and has an image point x = (u) on the boundary, then x = 1(u) = 2(u).
“Indeed, if = α 1 + (1α) 2 with 0 < α < 1 is not extreme and has an image point x = (u) on the boundary of the Bloch sphere, then it follows immediately from the strict convexity of the Euclidean unit ball that x = (u) = 1 (u) = 2 (u).”
Quote in the paperSubstantial claim - Findingreasoning
There are many situations in which the translation is severely limited despite contraction of the ellipsoid.
“On the contrary, there are many situations in which the translation is severely limited despite contraction of the ellipsoid.”
Quote in the paperSubstantial claim - Findingreasoning
Such a translation would have to be representable in the form [[1,0],[t,I]], which does not satisfy the conditions for complete positivity.
“Such a translation would have to be representable in the form 1 0 t I , which does not satisfy the conditions for complete positivity.”
Quote in the paperSubstantial claim - Findingreasoning
The operators in (32) are linearly independent if and only if sin v sin u is non-zero.
“Because the set {I, σ x , σ y , σ z } forms an orthonormal basis (with respect to the Hilbert-Schmidt inner product) for M 2 , the operators in (32) are linearly independent if and only if sin v sin u / = 0.”
Quote in the paperSubstantial claim - Findingreasoning
Lemma 11 implies that condition (v) is invariant under a change of basis.
“In addition, Lemma 11 above implies that (v) is invariant under a change of basis.”
Quote in the paperSubstantial claim - Findingnone given
A map with λ1 = λ2 = λ3 = μ < 0 is not completely positive unless |μ| ≤ 1/3.
“(b) A map with λ 1 = λ 2 = λ 3 = µ < 0 is not completely positive unless |µ| 1 3 .”
Quote in the paperSubstantial claim - Findingreasoning
The map sends the pair of states corresponding to (±cosθ, 0, sinθ) to the pair corresponding to (±cosω, 0, sinω) when equations (38) hold.
“In particular, maps the pair of states corresponding to (± cos θ, 0, sin θ) on the Bloch sphere to the pair of states corresponding to (± cos ω, 0, sin ω) when cos ω = cos θ cos u = cos 2 u sin 2 v -sin 2 u cos 2 v sin v , (38 ) sin ω = sin u sin v + sin θ cos u cos v = sin u sin v .”
Quote in the paperSubstantial claim - Findingreasoning
Lemma 15: Any contraction in M2 can be written as the convex combination of two unitary matrices.
“Lemma 15. Any contraction in M 2 can be written as the convex combination of two unitary matrices.”
Quote in the paperNot a substantial claim - Findingreasoning
A map is a generalized extreme point of S exactly when its corresponding matrix M = β( ) has the form (28) with 0 ≤ A ≤ I and U unitary.
“Theorem 7. A map is a generalized extreme point of S if and only if the corresponding matrix M = β( ) defined via (23) is of the form”
Quote in the paperSubstantial claim - Findingreasoning
A non-extreme point can have an image point on the boundary of the Bloch sphere only if all its extreme components have the same pre-image for that point.
“A non-extreme point can have an image point on the boundary of the Bloch sphere only if all its extreme components have the same pre-image for that point.”
Quote in the paperSubstantial claim - Findingreasoning
Lemma 11: For a positive map on M2 and unitary U, V, the equality = U • D • V holds if and only if equation (34) holds.
“Lemma 11.Let be a positive map on M 2 and U, V unitary.Then = U • D • V if and only if ( ⊗ I )(B 0 ) = [U ⊗ V T ] [( D ⊗ I )(B 0 )] [U † ⊗ V ], (34 )”
Quote in the paperSubstantial claim - Findingreasoning
By Lemma 9, the map is extreme if and only if the set {A1A1†, A2A2†, A1A2†, A2A1†} is linearly independent.
“By Lemma 9, is extreme if and only if the set {A 1 A † 1 , A 2 A † 2 , A 1 A † 2 , A 2 A † 1 } is linearly independent.”
Quote in the paperSubstantial claim - Findingnone given
When λ1 = λ2 = μ and λ3 = 0, which implies |μ| ≤ 1/2, the image of the map is a circle in a plane parallel to the equator.
“(c) When λ 1 = λ 2 = µ, λ 3 = 0 which implies |µ| 1 2 and the image of is a circle in a plane parallel to the equator.”
Quote in the paperNot a substantial claim - Findingreasoning
For a pure state ρ = 1/2[I + w•σ] with w = (±cosθ, 0, sinθ), the image is 1/2[I + x•σ] with x = (±cosθ cos u, 0, sin u sin v + sinθ cos u cos v).
“We consider a pure state of the form ρ = 1 2 [I + w• ] with w = (± cos θ, 0, sin θ) so that (ρ) = 1 2 [I + x• ] with x = (± cos θ cos u, 0, sin u sin v + sin θ cos u cos v).”
Quote in the paperSubstantial claim - Findingown data
The non-unital situation (I) includes channels for which the Holevo capacity is strictly larger than the classical Shannon capacity, demonstrating a definite quantum advantage.
“By contrast, the non-unital situation (I) is more interesting because it includes channels for which capacity (42) is strictly bigger than the classical Shannon capacity, i.e., the capacity for communication restricted to product input and measurements.Such channels demonstrate a definite quantum advantage.”
Quote in the paperSubstantial claim - Findingreasoning
Unitarity of R implies condition (14) in every case.
“Thus, we have verified that unitarity of R implies (14) in all cases.”
Quote in the paperSubstantial claim - Findingreasoning
In general, it is not possible to write a non-unital map as the composition of a unital map and a translation.
“In general, it is not possible to write a non-unital map as the composition of a unital map (which contracts the Bloch sphere to an ellipsoid) and a translation.”
Quote in the paperSubstantial claim - Findingreasoning
When μ = 1/2, no translation of the image (D) is possible despite the ellipsoid having shrunk to a flat disk of radius 1/2.
“Thus, when µ = 1 2 no translation the image (D) is possible despite the fact that the ellipsoid has shrunk to a flat disk of radius 1 2 .”
Quote in the paperSubstantial claim - Findingreasoning
The implication (i) ⇒ (ii) of Theorem 12 holds in full generality.
“Moreover, the implication (i) ⇒ (ii) holds in full generality.”
Quote in the paperSubstantial claim - Findingreasoning
Even when equality holds with only one sign, (11) implies t = 0 so no translation is possible.
“However, even when equality holds with only one sign, (11) implies that t = 0 so that no translation is possible.”
Quote in the paperSubstantial claim - Findingreasoning
Case (IC) is the only situation in which one can have extreme points with more than one t_k non-zero.
“It is worth pointing out that Case (IC) is the only situation in which one can have extreme points with more than one t k non-zero.”
Quote in the paperSubstantial claim - Findingreasoning
(D) contains the entire Bloch sphere if and only if the map is a unital extreme point.
“while (D) contains the entire Bloch sphere if and only if it is a unital extreme point.”
Quote in the paperSubstantial claim - Findingreasoning
The unitarity of R implies condition (14) in Theorem 4.
“We now show that this unitarity implies condition (14) in Theorem 4.”
Quote in the paperSubstantial claim - Findingreasoning
The convex combination in Theorem 14 can be chosen to be a midpoint.
“Note that in both cases we have proven the somewhat stronger result that the convex combination can be chosen to be a midpoint.”
Quote in the paperNot a substantial claim - Findingnone given
The cases t = (√(1-α^2), 0, 0) and t = (0, √(1-α^2), 0) are similar, differing only in orientation and displacement.
“The cases t = ( √ 1α 2 , 0, 0) and t = (0, √ 1α 2 , 0) are similar.The only difference is the orientation and displacement, e.g., whether the point [1, 1, 1] is replaced by [α 2 , α, α] or to [α, α 2 , α].”
Quote in the paperNot a substantial claim - Findingreasoning
If t1 = t2 = 0 and condition (14) holds, then R as given by (7) or (8) is always unitary.
“Conversely, one can easily check that under the hypotheses t 1 = t 2 = 0 and ( 14), R as given by ( 7) or ( 8) is always unitary.”
Quote in the paperSubstantial claim - Findingreasoning
The argument yields the extreme points of S represented by a matrix T of form (3).
“The argument presented thus far yields the extreme points of S that are represented by a matrix T of the form (3).”
Quote in the paperSubstantial claim - Findingreasoning
This set is linearly independent unless D is a multiple of the identity or W is diagonal.
“One can readily verify that this set is linearly independent unless D is a multiple of the identity or W is diagonal.”
Quote in the paperSubstantial claim - Findingnone given
If (λ1 ± λ2)^2 = (1 ± λ3)^2 holds with exactly one sign, then λ1, λ2, λ3 lie on the surface of the tetrahedron defined by (II), but on the interior of one of its four faces.
“(a) If (λ 1 ± λ 2 ) 2 = (1 ± λ 3 ) 2 holds with exactly one sign, then λ 1 , λ 2 , λ 3 lie on the surface of the tetrahedron defined by (II), but on the interior of one of the four faces.”
Quote in the paperSubstantial claim - Findingreasoning
This is equivalent to linear independence of the set D, I-D, √D W √(I-D), √(I-D) W† √D, where W = V1 V2†.
“This is equivalent to linear independence of the set D, I -D, √ DW √ I -D, √ I -DW † √ D , where W = V 1 V † 2 .”
Quote in the paperSubstantial claim - Findingreasoning
In the case where both sin v and sin u vanish, exactly one of the two Kraus operators in (18) is non-zero and hence unitary.
“In this case one and only one of the two Kraus operators given by ( 18) is non-zero and hence, unitary.”
Quote in the paperSubstantial claim - Findingnone given
The condition |t3| = (1 ± λ3) implies that the North or South pole is a fixed point.
“However, the condition |t 3 | = (1 ± λ 3 ) implies that the North or South pole is a fixed point.”
Quote in the paperSubstantial claim - Findingreasoning
For non-zero t with t fixed, the extreme points are given by the curve [cos u, cos v, cos u cos v] subject to sin u sin v = t3.
“It follows from (17) that the extreme points are given by the curve [cos u, cos v, cos u cos v] subject to the constraint sin u sin v = t 3 .”
Quote in the paperSubstantial claim - Findingreasoning
The four points split the curve into four pieces corresponding to four of the six edges of the tetrahedron.
“The four points above split the curve into four pieces, corresponding to four of the six edges of the tetrahedron.”
Quote in the paperNot a substantial claim - Findingreasoning
Letting α^2 = 1 - t3, the curve (41) passes through the four points [α, α, α^2], [-α, -α, α^2], [α, -α, -α^2], and [-α, α, -α^2] when sin u = ± sin v.
“Letting α 2 = 1t 3 , we see that the curve (41) passes through the four points [α, α, α 2 ], [-α, -α, α 2 ], [α, -α, -α 2 ], [-α, α, -α 2 ] when sin u = ± sin v.”
Quote in the paperNot a substantial claim - Findingnone given
Unlike the case t3 = 0, these lines are not extreme points, even in a generalized sense.
“Unlike the case t 3 = 0 these lines are not extreme points, even in a generalized sense.”
Quote in the paperSubstantial claim - Findingreasoning
This curve can be written in parametric form as a pair of curves (41) with sin^{-1}(t3) ≤ u ≤ π - sin^{-1}(t3).
“This can be written in parametric form as a pair of curves cos u, ± cos sin -1 t 3 sin u , ± cos u cos sin -1 t 3 sin u (41) with sin -1 (t 3 ) u sin -1 (t 3 ).”
Quote in the paperNot a substantial claim - Premisecitation
The parameters are constrained by the Algoet-Fujiwara inequalities (λ1 ± λ2)^2 ≤ (1 ± λ3)^2.
“It was shown in [1,11] that the λ k 's are limited by the inequalities (λ 1 ± λ 2 ) 2 (1 ± λ 3 ) 2 .”
Quote in the paperSubstantial 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.
- Premisecitation
Any positivity-preserving map on M2 can be reduced by changes of basis to the diagonal form (3).
“King and Ruskai [11] showed that any such map can be reduced, via changes of basis in C 2 , to the form”
Quote in the paperSubstantial claim0 of 1 sources checked- cites King and Ruskai [11]
Not checked in the free run.
- Premisecitation
For n = 2, Choi's condition is equivalent to (⊗ I)(B0) ≥ 0, where B0 projects onto the maximally entangled Bell state.
“We are interested in n = 2, in which case this condition is equivalent [10] to ( ⊗ I ) (B 0 ) 0,”
Quote in the paperSubstantial claim0 of 1 sources checked- cites [10] Horn, 1985
Not checked in the free run.
- Interpretationnone given
Under the reduction (4), a map is completely positive exactly when its reduced map is completely positive.
“The crucial point about the reduction (4) is that is completely positive if and only if t, is.”
Quote in the paperSubstantial claim - Premisecitation
Niu and Griffiths's construction showed that taking the partial trace T1 instead of T2 is equivalent to swapping sin and cos in (17).
“Moreover, Niu and Griffiths's construction showed that taking the partial trace T 1 rather than T 2 is equivalent to switching sin and cos in (17).”
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.
- Interpretationreasoning
It is not accurate to view a stochastic map as the composition of a rotation, a contraction to an ellipsoid, a translation, and another rotation, because the individual maps in that process are no longer stochastic.
“It is tempting to think of a stochastic map as the composition of a rotation, contraction (to an ellipsoid), translation, and another rotation.However, this is not accurate in the sense that the individual maps in this process will no longer be stochastic.”
Quote in the paperSubstantial claim - Interpretationreasoning
Even when |λ3| > μ^2 so the map is not extreme, the ellipsoid is shifted to the boundary in a manner analogous to an amplitude damping channel.
“Hence, even when |λ 3 | > µ 2 so that is not extreme, the ellipsoid is shifted to the boundary in a manner analogous to an amplitude damping channel.”
Quote in the paperSubstantial claim - Interpretationreasoning
A map of form (3) with all three t_k non-zero does not necessarily require three extreme points; in general two maps in the generalized-point set suffice.
“One might think that a map of the form (3) with all three t k non-zero would require three extreme points.However, this is not the case.In general, two maps in the set of generalized points suffice.”
Quote in the paperSubstantial claim - Premisecitation
A stochastic map is an extreme point of S exactly when it can be written in Kraus form with the identity trace-preserving condition and the set of products Aj Ak† is linearly independent.
“Lemma 9. A stochastic map is an extreme point of S if and only if it can be written in the form (1) (necessarily with k A k A † k = I ), so that the set of matrices {A j A † k } is linearly independent.”
Quote in the paperSubstantial claim0 of 1 sources checked- cites [4] Choi, 1975, Completely positive linear maps on complex matrices, Linear Algebra Appl. source
Not checked in the free run.
- Interpretationreasoning
When two distinct extreme points have a common pair of points on the image of the Bloch sphere, their pre-images must be distinct.
“Hence, it follows that when two distinct extreme points have a common pair of points on the image of the Bloch sphere, their pre-images must be distinct.”
Quote in the paperSubstantial claim - Interpretationreasoning
By Lemma 9, the corresponding stochastic map is extreme when both sin v and sin u are non-zero, equivalently when t3 is non-zero and all |λk| are less than 1.
“Thus, by Lemma 9, the corresponding stochastic map is extreme if both of {sin v, sin u} are non-zero or, equivalently, t 3 / = 0 and all |λ k | < 1.”
Quote in the paperSubstantial claim - Interpretationown data
Example (2b) shows that the ellipsoid picture, while useful, is incomplete.
“Example (2b) illustrates that the ellipsoid picture, while useful, is incomplete.”
Quote in the paperSubstantial claim - Interpretationreasoning
The statement that extreme points correspond to maps for which two or more image points lie on the Bloch sphere is correct only if a single point at a pole is interpreted as a pair of degenerate images when |λ1| = |λ2|.
“The discussion at the beginning of Section 3.1 suggests that, roughly speaking, extreme points correspond to maps for which two (or more) points in the image (D) lie on the Bloch sphere as shown in Fig. 1.However, this statement is correct only if we interpret the single point at a pole as a pair of degenerate images when |λ 1 | = |λ 2 |.”
Quote in the paperSubstantial claim - Interpretationreasoning
The implications (ii) ⇒ (iii) ⇒ (i) in full generality follow from reduction of every stochastic map to form (3) and invariance of the conditions under changes of bases.
“It thus remains to prove that (ii) ⇒ (iii) ⇒ (i) in full generality rather than just for maps of the form (3).But this follows from the facts that every stochastic map can be reduced to the form (3) via changes of bases (4) and that all these conditions are invariant under such changes of bases.”
Quote in the paperSubstantial claim - Interpretationreasoning
One can find maps with all |λk| < 1 for which the image (D) is an ellipsoid strictly contained within the unit ball but no translation in any direction is possible.
“Thus, one can find maps with all |λ k | < 1 for which the image (D) is an ellipsoid strictly contained within the unit ball but no translation in any direction is possible.”
Quote in the paperSubstantial claim - Interpretationreasoning
Maps for which exactly one of sin v and sin u is zero cannot be extreme.
“Hence maps for which exactly one of {sin v, sin u} is zero cannot be extreme.”
Quote in the paperSubstantial claim - Interpretationreasoning
As t3 moves away from zero, the corners of the tetrahedron are replaced by these points, forming an asymmetric rounded tetrahedron.
“Thus, as t 3 moves away from zero, the corners of the tetrahedron are replaced by these points as shown in Fig. 3 which might be described as an asymmetric rounded tetrahedron.”
Quote in the paperNot a substantial 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.
FindingreasoningIf 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 claimProbably not a substantial claim: may not pass S3 (load-bearing).
Findingnone givenA 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 claimMay combine two claims.
FindingreasoningAdding 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 claimThe quote only partly states this claim.
FindingreasoningThe 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.
FindingreasoningFor 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 claimThe quote only partly states this claim.
Findingown dataAfter 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 claimMay combine two claims.
FindingreasoningThe 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 claimMay combine two claims.
FindingcitationIn 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 claimMay combine two claims.
FindingreasoningIn 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 claimMay combine two claims.
Findingnone givenThe 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 claimMay combine two claims.
FindingreasoningFor 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 claimMay combine two claims.
FindingreasoningIn 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 claimClassified as probably not a substantial claim.
FindingreasoningLemma 10 shows that (iv) ⇔ (v).
“Similarly, Lemma 10 shows that (iv) ⇔ (v).”
Quote in the paperSubstantial claimMay combine two claims.
FindingreasoningAlthough 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 claimMay combine two claims.
FindingreasoningUnder 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 claimMay combine two claims.
FindingreasoningLemma 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 claimMay combine two claims.
FindingreasoningA 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 claimMay combine two claims.
FindingreasoningThe 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 claimMay combine two claims.
FindingreasoningIf 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 claimMay combine two claims.
FindingreasoningIn 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 claimMay combine two claims.
FindingreasoningIf 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 claimMay combine two claims.
FindingreasoningThese 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 claimMay combine two claims.
FindingreasoningUnlike 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 claimMay combine two claims.
FindingreasoningIf 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 claimMay combine two claims.
FindingreasoningUnder 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.
FindingreasoningExample (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 claimMay combine two claims.
FindingreasoningThe 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 claimThe quote only partly states this claim.
Findingnone givenIf 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 claimMay combine two claims.
FindingreasoningFor 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 claimMay combine two claims.
Findingnone givenFor 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 claimMay combine two claims.
FindingreasoningIf 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 claimMay combine two claims.
FindingreasoningFor 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 claimProbably not a substantial claim: may not pass S3 (load-bearing).
FindingreasoningThis curve forms the boundary of Fig. 3.
“This curve forms the boundary of Fig. 3.”
Quote in the paperNot a substantial claimThe quote only partly states this claim.
FindingreasoningIn 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 claimMay combine two claims. The quote only partly states this claim.
FindingreasoningThis 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 claimMay combine two claims.
PremisecitationChoi'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 checkedProbably not a substantial claim: may not pass S3 (load-bearing).
PremisecitationRieffel 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).
PremisecitationNiu 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.
PremisecitationFor 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.
PremisecitationTerhal 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.
InterpretationreasoningWhen 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 claimProbably not a substantial claim: may not pass S3 (load-bearing). May combine two claims.
PremisecitationThis 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.
PremisecitationHolevo 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.
InterpretationreasoningAs 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 claimMay combine two claims.
InterpretationreasoningFinding 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 claimMay combine two claims.
InterpretationreasoningBecause 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 claimMay combine two claims.
InterpretationcitationIn 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 givenHolevo 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
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).”
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?
For all η, the spontaneous emission channel is extremal.
“Interestingly, for all η this channel is extremal48, which means that the CS is not applicable.”
From The elusive Heisenberg limit in quantum-enhanced metrology (2012) · cites this paper as [48]
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
MinorThe 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.
“Thus, (D) has one pair of orthogonal states on the unit sphere, which are also fixed points.”
“The extreme points of the tetrahedron are [1, 1, 1], [1, -1, -1], [-1, 1, -1], [-1, -1, 1].The edges connecting any two of these extreme points have the form [±1, s, ±s] up to permutation and correspond to quasi-extreme points of type (III).”
The stated Pauli basis is not orthonormal
MinorThe 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.
“Because the set {I, σ x , σ y , σ z } forms an orthonormal basis (with respect to the Hilbert-Schmidt inner product) for M 2”
“the Hilbert-Schmidt inner product A, B = TrA † B”
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- Authors
- Beth Ruskai, M., Szarek, S., & Werner, E.
- Affiliations
- Université de Lille (FR)
- ORCID
- Werner, Elisabeth
- 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
- Semantic Scholar
- b5eb07634f6099ceec94304193355fa8b3fe6acd
- 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
- bronze
- doi.org
- Licence
- 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
About these checks
Registry, retraction, Cites and statistics results are computed, not judged, with one exception: a printed reference that no index lists is read once by a model before it counts as a citation error. Claims are extracted by a model and kept only when their quote appears verbatim in the text; each cited source is then checked, with a model’s help, for whether it says what the claim relies on. Those verdicts can be wrong — read the claim’s quote and the cited source before relying on one.
Claims read by deepseek-v4.1-flash under definition version 3 (extract-v3) on 2026-10-06.
Classified by jev-1.13-20260917 (gold-guideline-v0.2) on 2026-10-06.
Statistics and arithmetic are recomputed, not judged.
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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