The geometric measure of multipartite entanglement

Transcription

1 The geometric measure of multipartite entanglement Anthony Sudbery Department of Mathematics University of York International Confrence on Quantum Information Bhubaneshwar, December 2011

2 Collaborators Joe Hilling (York) Sayatnova Tamaryan (Yerevan) Levon Tamaryan (Yerevan)

3 MULTIPARTITE ENTANGLEMENT Ψ H 1 H n is entangled if Ψ φ 1 φ n φ k H k How entangled? How far from nearest product state?

4 MULTIPARTITE ENTANGLEMENT Ψ H 1 H n is entangled if Ψ φ 1 φ n φ k H k How entangled? How far from nearest product state? Fubini-Study metric: d( Ψ, Φ ) 2 = 1 Φ Ψ 2

5 MULTIPARTITE ENTANGLEMENT Ψ H 1 H n is entangled if Ψ φ 1 φ n φ k H k How entangled? How far from nearest product state? Fubini-Study metric: d( Ψ, Φ ) 2 = 1 Φ Ψ 2 Geometric measure of entanglement 1 g( Ψ ) g( Ψ ) = ( ) 2 max Ψ φ 1 φ n φ 1 φ n φ k φ k =1 Shimony 1995, Wei & Goldbart 2003

6 MULTIPARTITE ENTANGLEMENT Ψ H 1 H n is entangled if Ψ φ 1 φ n φ k H k How entangled? How far from nearest product state? Fubini-Study metric: d( Ψ, Φ ) 2 = 1 Φ Ψ 2 Geometric measure of entanglement 1 g( Ψ ) g( Ψ ) = ( ) 2 max Ψ φ 1 φ n φ 1 φ n φ k φ k =1 Shimony 1995, Wei & Goldbart 2003 Extend to mixed states by the usual convex roof definition

7 APPLICATIONS Constructing optimal entanglement witnesses Experimental estimate of entanglement Quantifies difficulty of distinguishing multipartite states by local means Identifying multipartite states for perfect quantum teleportation and superdense coding Identifying multipartite states for perfect quantum teleportation and superdense coding Detecting quantum phase transitions in spin models

8 PROPERTIES 1. The geometric measure is an entanglement monotone. Barnum & Linden, Wei and Goldbart

9 PROPERTIES 1. The geometric measure is an entanglement monotone. Barnum & Linden, Wei and Goldbart 2. In the form G( Ψ ) = log g( Ψ ), the geometric measure is additive: G ( Ψ Φ ) = G ( Ψ ) + G ( Φ ) for certain symmetric states Ψ. Dhen, Zhu and Wei

10 PROPERTIES 1. The geometric measure is an entanglement monotone. Barnum & Linden, Wei and Goldbart 2. In the form G( Ψ ) = log g( Ψ ), the geometric measure is additive: G ( Ψ Φ ) = G ( Ψ ) + G ( Φ ) for certain symmetric states Ψ. Dhen, Zhu and Wei 3. The geometric measure is a lower bound for the relative entropy relative to separable states: G ( Ψ ) min tr[ρ log ρ ρ log σ] separable σ (ρ = ψ ψ ). Dhen, Zhu and Wei

11 SCHMIDT DECOMPOSITION Schmidt decomposition of bipartite state (n = 2): Given Ψ H 1 H 2, bases 1 1,..., d 1 1 of H 1 and 1 2,..., d 2 2 of H 2 such that Ψ = k λ k k 1 k 2

12 SCHMIDT DECOMPOSITION Schmidt decomposition of bipartite state (n = 2): Given Ψ H 1 H 2, bases 1 1,..., d 1 1 of H 1 and 1 2,..., d 2 2 of H 2 such that Ψ = k λ k k 1 k 2 λ k = singular values of (c ij ) where Ψ = c ij i j Geometric measure g( Ψ ) = max λ 2 k

13 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion Ψ = c i1 i n i 1 i n has the minimum number of terms, with coefficients satisfying: 1. c ji i = c iji i = = c i ij = 0 if 1 i < j d; This reduces the number of non-zero coefficients by nd(d + 1)/2.

14 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion Ψ = c i1 i n i 1 i n has the minimum number of terms, with coefficients satisfying: 1. c ji i = c iji i = = c i ij = 0 if 1 i < j d; 2. c jd d, c djd d,..., c d dj are real and non-negative; This reduces the number of non-zero coefficients by nd(d + 1)/2.

15 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion Ψ = c i1 i n i 1 i n has the minimum number of terms, with coefficients satisfying: 1. c ji i = c iji i = = c i ij = 0 if 1 i < j d; 2. c jd d, c djd d,..., c d dj are real and non-negative; 3. c i i c j1 j n if i j r, r = 1,..., n. This reduces the number of non-zero coefficients by nd(d + 1)/2.

16 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion Ψ = c i1 i n i 1 i n has the minimum number of terms, with coefficients satisfying: 1. c ji i = c iji i = = c i ij = 0 if 1 i < j d; 2. c jd d, c djd d,..., c d dj are real and non-negative; 3. c i i c j1 j n if i j r, r = 1,..., n. This reduces the number of non-zero coefficients by nd(d + 1)/2.

17 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion Ψ = c i1 i n i 1 i n has the minimum number of terms, with coefficients satisfying: 1. c ji i = c iji i = = c i ij = 0 if 1 i < j d; 2. c jd d, c djd d,..., c d dj are real and non-negative; 3. c i i c j1 j n if i j r, r = 1,..., n. This reduces the number of non-zero coefficients by nd(d + 1)/2. Three qubits Ψ = a b c d f 111 a, b, c, d real, a b c d 0.

18 GENERALISED SCHMIDT DECOMPOSITION H. Carteret, A. Higuchi, AS Given Ψ H 1 H n, there are bases of H 1,..., H n such that the expansion of Ψ has the minimum number of terms. Geometric measure g( Ψ ) = coeff. of Three qubits Ψ = a b c d f 111 a, b, c, d real, a b c d 0.

19 GENERALISED SINGULAR-VALUE EQUATIONS Problem (n = 3): Given Ψ = a ijk i j k, find θ = x i i, φ = y j j, ψ = z k k to maximise Ψ ( θ φ ψ ) 2 subject to θ θ = φ φ = ψ ψ = 1.

20 GENERALISED SINGULAR-VALUE EQUATIONS Problem (n = 3): Given Ψ = a ijk i j k, find θ = x i i, φ = y j j, ψ = z k k to maximise Ψ ( θ φ ψ ) 2 subject to θ θ = φ φ = ψ ψ = 1. Lagrange multipliers λ 1, λ 2, λ 3 = a ijk y j z k = λ 1 x i a ijk x i z k = λ 2 y j a ijk x i y j = λ 3 z k x = y = z = 1 λ 1 = λ 2 = λ 3 = required maximum λ.

21 SINGULAR VALUES OF A TENSOR Lagrange multipliers λ 1, λ 2, λ 3 = a ijk y j z k = λ 1 x i a ijk x i z k = λ 2 y j a ijk x i y j = λ 3 z k x = y = z = 1 λ 1 = λ 2 = λ 3 = required maximum λ.

22 SINGULAR VALUES OF A TENSOR Lagrange multipliers λ 1, λ 2, λ 3 = a ijk y j z k = λ 1 x i a ijk x i z k = λ 2 y j a ijk x i y j = λ 3 z k x = y = z = 1 λ 1 = λ 2 = λ 3 = required maximum λ. Compare n = 2 : a ij y j = λx i Ay = λx a ij x i = λy j A x = λy Choose phases to make λ real and 0: λ is a singular value of A (λ 2 is an eigenvalue of A A).

23 DISCRIMINANTS AND HYPERDETERMINANTS Homogeneous polynomial f (z 1,..., z N ). The discriminant f is a polynomial in the coefficients of f such that f = 0 z s. t. f z k (z) = 0, k = 1,..., N. Examples 1. f (z) = a ij z i z j = z T Az: f = det A.

24 DISCRIMINANTS AND HYPERDETERMINANTS Homogeneous polynomial f (z 1,..., z N ). The discriminant f is a polynomial in the coefficients of f such that f = 0 z s. t. f z k (z) = 0, k = 1,..., N. Examples 2. N = d 1 + d 2, z = (x, y), f (x, y) = a ij x i x j : f = det A T A.

25 DISCRIMINANTS AND HYPERDETERMINANTS Homogeneous polynomial f (z 1,..., z N ). The discriminant f is a polynomial in the coefficients of f such that f = 0 z s. t. f z k (z) = 0, k = 1,..., N. Examples 3. N = d 1 + d 2 + d 3, z = (u, v, w), f = a ijk u i v j w k : f = hyperdeterminant of the hypermatrix a ijk = 0 if a ijk has singular value 0. d 1 = d 2 = d 3 (three qubits): f 2 = 3-tangle (al. et Wootters)

26 THE 3 x 3 HYPERDETERMINANT

27 CHARACTERISTIC POLYNOMIAL OF A TENSOR Theorem (Hilling, AS) ( α α : C d 1 C dn C multilinear: u (1),..., u (dn)) = a i1 i n u (1) i 1 u (n) i n Given λ R, define α(λ) : R 2d 3 R 2dn R by ( α(λ) u (3),..., u (n)) [ ] = det A A u (3) 2 u (n) 2 I, If λ 0, the equations A i1 i 2 = a i1 i 2 i 3 i n u (3) i 3 u (n) i n. a i1 i n u (1) i 1 û(r) i r u (n) i n = λu (r) i r have a solution with all u (r) non-zero if and only if α(λ) has a real critical point. If this so, λ satisfies the polynomial equation discriminant of α(λ) = 0.

28 THREE-QUBIT STATES Ψ = a ijk i j k (i, j, k = 0, 1) Geometric measure of entanglement g( Ψ ) = λ given by a ijk y j z k = λx i a ijk x i z k = λy j a ijk x i y j = λz k Let A(z) = 2 2 matrix a ijk z k ; then so A(z)y = λx, A(z) x = λy, [ ] α(λ)(z, z) = det A(z) A(z) λ 2 z z = 0. Characteristic equation of a ijk : (λ) = discriminant z,z α(λ) = z ( z α(λ)) = 0 Degree 12 in λ 2.

29 THE ART OF THE SOLUBLE Tetrahedral state Ψ = a b c d 110

30 THE ART OF THE SOLUBLE Tetrahedral state Ψ = a b c d 110 (λ) = 16a 2 b 2 c 2 d 2 (λ 2 a 2 )(λ 2 b 2 )(λ 2 c 2 )(λ 2 d 2 )Q(λ) 2 Q(λ) = λ 4 (4S 2 λ 2 L 2 )(4S 2 λ 2 L 2 ) L 2 = (ab + cd)(ac + bd)(ad + bc) S 2 = (s a)(s b)(s c)(s d) s = 1 2 (a + b + c + d) S = area of cyclic quadrilateral with sides a, b, c, d (Brahmagupta) S, L obtained from S, L by changing sign of d.

31 SOLUTIONS OF THE CHARACTERISTIC EQUATION Ψ = a b c d 110 λ = 0 (twice), a, b, c, d, L 2S (twice), L 2S (twice). L = 2R = diameter of circumcircle of cyclic quadrilateral 2S L 2S = 2R = diameter of circumcircle of self-intersecting quadrilateral

32 GEOMETRY OF THE GEOMETRIC MEASURE Tamaryan, Park, Son & Tamaryan S S } = WX YX + WZ YZ = { area of WXYZ TWZ TXY

33 GETTING THE RIGHT SOLUTION S 2 = 1 ( a+b +c +d)(a b +c +d)(a+b c +d)(a+b +c d) 16 S 2 = S 2 abcd If S 2 < 0, there is no cyclic quadrilateral and no solution 2R. In this case, g( Ψ ) = max(a, b, c, d). If 0 < S 2 < abcd, 2R = L/2S is the largest solution. If S 2 > abcd, the largest solution is either 2R or 2R. But only 2R gives a real critical point.

34 GETTING THE RIGHT SOLUTION S 2 = 1 ( a+b +c +d)(a b +c +d)(a+b c +d)(a+b +c d) 16 S 2 = S 2 abcd If S 2 < 0, there is no cyclic quadrilateral and no solution 2R. In this case, g( Ψ ) = max(a, b, c, d). If 0 < S 2 < abcd, 2R = L/2S is the largest solution. If S 2 > abcd, the largest solution is either 2R or 2R. But only 2R gives a real critical point. g( Ψ ) = { max(a, b, c, d) if S 2 < 0; L/2S if S 2 > 0.

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36 MANY QUBITS Generalised W state Ψ = c c c n (0 c 1 c n R) Slightly entangled (max c k 2 > 1/2): g( Ψ ) = c 2 n Ψ has many different nearest product states.

37 Highly entangled (c 2 n 1/2): Ψ has nearest product state u 1... u n where and the geometric measure is u k = sin θ k 0 + e iφ cos θ k 1 g( Ψ = 2r sin θ 1 sin θ 2... sin θ n where r is the unique solution of and Then 1 c2 1 r c2 n 1 r 2 ± sin 2θ k = c k r. cos 2 θ cos 2 θ n = 1, 1 c2 n r 2 = n 2 so the set of highly entangled states has the form S n 1 S 1.

38 Highly entangled (c 2 n 1/2): Ψ has nearest product state u 1... u n where and the geometric measure is u k = sin θ k 0 + e iφ cos θ k 1 g( Ψ = 2r sin θ 1 sin θ 2... sin θ n where r is the unique solution of 1 c2 1 r c2 n 1 r 2 ± 1 c2 n r 2 = n 2 (+ if c n < r 0 ; if r 0 < c n < c c2 n 1 ); r 0 satisfies the same equation with the last term removed; φ is arbitrary; and sin 2θ k = c k r.