A complete criterion for convex-gaussian states detection

A complete criterion for convex-gaussian states detection Anna Vershynina Institute for Quantum Information, RWTH Aachen, Germany joint work with B. Terhal NSF/CBMS conference Quantum Spin Systems The University of Alabama at Birmingham June 17, 2014

Outline of the talk Introduction Fermionic Gaussian states Gaussian-symmetric states A complete criterion for detection of convex-gaussian states Work in progress

Introduction Given m Dirac fermions created by the operator aj, j = 1,.., m, with CAR {a j, ak } = δ jk and {a j, a k } = 0, 2m Majorana fermions are defined as c 2k 1 = a k + ak and c 2k = i(ak a k ). They satisfy c k = ck, and c2 k = 1 c k c j = c j c k. Majorana operators form an algebra C 2m with relations {c j, c k } = 2δ jk.

Introduction Given m Dirac fermions created by the operator a j, j = 1,.., m, with CAR 2m Majorana fermions are defined as They satisfy {a j, a k } = δ jk and {a j, a k } = 0, c 2k 1 = a k + a k and c 2k = i(a k a k ). c k = c k, and c2 k = 1 c k c j = c j c k. Majorana operators form an algebra C 2m with relations {c j, c k } = 2δ jk. The number operator n k = a k a k has eigenvalues either 0 or 1. The fermion parity operator P k = I 2n k = ( 1) n k I has eigenvalue +1 if the number of fermions is even and 1 if it is odd. In terms of Majorana operators P k = ic 2k 1 c 2k.

The total fermion parity operator is P all = k P k = i m c 1 c 2...c 2m. Any even pure state ψ ψ is the eigenstate of P all, i.e. P all ψ ψ = ± ψ ψ the eigenvalues depend on whether the parity of the number of fermions in ψ is even or odd. A Hermitian operator X is called even if it has the form X = α 0 I + m (i) k k=1 1 j 1 <...<j 2k 2m where α 0, α j1,...,j 2k are real. Any even operator commutes with P all, [X, P all ] = 0. α j1,...,j 2k c j1...c j2k,

Fermionic Gaussian states Fermionic Gaussian state is defined as ρ = γ exp{ i i j A ij c i c j } where γ is a normalization and (A ij ) is a real anti-symmetric matrix. Block-diagonalizing A we can re-express ρ is standard form ρ = 1 2 m m (I + iλ k c 2k 1 c 2k ), k=1 where c = R T c with R block-diagonalization of A. Here λ [ 1, 1]. For Gaussian pure states λ { 1, 1}. Check: ρ is a pure state iff ρ 2 = ρ. So Therefore ρ 2 = ρ iff λ 2 k = 1. ρ 2 = 1 2 2m = 1 2 2m = 1 2 m m (I + iλ k c 2k 1 c 2k ) 2 k=1 m (I + λ 2 k I + 2iλ k c 2k 1 c 2k ) k=1 m ( 1 + λ 2 I + iλ k c 2k 1 c 2k ). 2 k=1

Given a state ρ C 2m, the correlation matrix M is M ab = i 2 Tr(ρ[ca, c b]), with a, b = 1,.., 2m. For a = b, we have M aa = 0 and for a b, we have M ab = i 2 Tr(ρcac b ρc b c a) = itr(ρc ac b ). Note that M ab = M ba. It can be block-diagonalized by R SO(2m) M = R m j=1 ( 0 λj λ j 0 ) R T. Fermionic linear optics (FLO) transformation maps Gaussian states into Gaussian states Uc i U = j R ij c j with R SO(2m). The total fermionic parity operator is invariant under FLO UP all = P all U.

Lemma [1] (de Melo, Ćwikliński, Terhal). The correlation matrix M of any even density state ρ C 2m has eigenvalues ±iλ k, with λ k [ 1, 1], k = 1,..., m. Moreover M T M I with equality iff ρ is a Gaussian pure state. Dephasing procedure. Define the FLO transformation U k, k = 1,..., m as U k c 2k U k = c 2k, U k c 2k 1 U = c 2k 1, U k c i U k = c i i 2k 1, 2k. It leaves the correlation matrix of ρ invariant. With ρ 0 = ρ, let ρ k = 1 2 (ρ k 1 + U k ρ k 1 U k ). After m steps, we get ρ m = k p k ψ k ψ k where each ψ k ψ k = 1 2 m m j=1 (I + iβ kj c 2j 1 c 2j ) is a Gaussian state, i.e. β kj = ±1. It is an eigenvector to all ic 2j 1 c 2j, j = 1,.., m, since ic 2j 1 c 2j ψ k ψ k = β kj ψ k ψ k. The correlation matrix of ρ m is M ρm = R m ( j=1 0 k p k β jk k p k β jk 0 ) R T. Therefore M T ρ m M ρm = I iff ρ m is pure Gaussian. Since M ρm = M ρ, ρ is pure Gaussian.

Proposition [1] Any even pure state ψ ψ C 2m for m = 1, 2, 3 is Gaussian.

Proposition [1] Any even pure state ψ ψ C 2m for m = 1, 2, 3 is Gaussian. Proof. For m = 1, a state can be written as ρ = ψ ψ = αi + βc 1 c 2. Since Trρ = 1, we get α = 1/2. From ρ 2 = ρ we get (α 2 β 2 )I + 2αβc 1 c 2 = αi + βc 1 c 2. So β = i/2. So ψ ψ is Gaussian. For m = 2, block-diagonalize the correlation matrix. Then any state can be written as ρ = ψ ψ = αi + 2 k=1 β k c 2k 1 c 2k + P all. Apply dephasing procedure: ρ 2 = ρ. Since ρ 2 is a convex mixture of pure Gaussian states, ρ = ψ ψ is Gaussian. For m = 3, after block-diagonalization, ψ ψ = αi + βp all + k γ k c 2k 1 c 2k + i<j<k<l η ijkl c i c j c k c l. Note that P all ψ ψ = ± ψ ψ for even pure states. Apply dephasing procedure: ρ 3 = ρ is a convex mixture of pure Gaussian states. So ψ ψ is Gaussian.

A convex-gaussian state is ρ = i p i σ i, where σ i are pure Gaussian states, p i 0 and i p i = 1. Proposition [1] For any even state ρ C 2m there exists ɛ > 0 such that ρ ɛ = ɛρ + (1 ɛ)i/2 m is convex-gaussian. Define 2m Λ = c i c i C 2m C 2m. The operator Λ is invariant under U U for any FLO transformation. i=1 Check: for any FLO U with R SO(2m) we have 2m U U Λ U U = Uc i U Uc i U i=1 = i,j,k R ijcj R ik c k = j c j c j. Here we used that i R ij R ik = i r T ji R ik = (R T R) jk = δ jk.

Gaussian-symmetric states Lemma [2] (Bravyi). An even state ρ C 2m is Gaussian iff [Λ, ρ ρ] = 0.

Gaussian-symmetric states Lemma [2] (Bravyi). An even state ρ C 2m is Gaussian iff [Λ, ρ ρ] = 0. 2m Λ(ρ ρ)λ 1 = Tr[(c a c a)(ρ ρ)(c b c b )] a,b=1 2m = Tr(c aρc a c aρc a) + (Trc aρc b ) 2 a=1 a b = 2m (Tr i c ac b ρ) 2 a b = 2m a,b (M ab ) 2 = 2m a ( b M ab M ba ) = 2m a Mab T M ba = 2m Tr M T M. b For a pure state we have M T M = I, so Λ(ρ ρ) = 0. For a mixed Gaussian state or non-gaussian state M T M < I, so Λ(ρ ρ)λ 1 > 0.

For every Gaussian state ψ, the state ψ, ψ is contained in the null space of Λ. Define a FLO twirl S(ρ) = du U UρU U. FLO Lemma [1], [4] (Terhal, V.) The projector onto the null-space of Λ is Π Λ=0 = ( 2m) m S( 0, 0 0, 0 ). Thus the states ψ, ψ where ψ is Gaussian span the null space of Λ. Proof. To show that Π = ( 2m m ) S( 0, 0 0, 0 ), we need to show that for any X, ( 2m ) ( ) Tr(X Π) = Tr XS( 0, 0 0, 0 ). m Need only to consider S(X) instead of X. The invariant subspace of S is spanned by I I, Λ,..., Λ 2m operators. [4] For any i 0 we have Tr Λ i Π Λ=0 = 0 and Tr Λ i S( 0, 0 0, 0 ) = Tr S(Λ i ) 0, 0 0, 0 = 0.

Overall prefactor: Λ = 2m i=1 c i c i, each term c i c i has eigenvalue µ i = ±1. Then eigenvalues of Λ are 2m i=1 µ i with the projector onto the eigenstates P µ = 1 2m 2 2m (I + µ i c i c i ). i=1 The null space is spanned by ( 2m) m eigenvectors P µ such that 2m i=1 µ i = 0. Therefore Tr Π = ( 2m) m. Thus states ψ, ψ with ψ Gaussian span the null space of Λ. Q.E.D. The null space of Λ is called Gaussian symmetric subspace.

Proof Let M τ be a correlation matrix of state τ. m ( 0 λj M τ = R λ j 0 j=1 Then ) R T. Λτ τλ 1 = 2m Tr M T τ Mτ = 2m j λ 2 j ɛ. So j λ2 j 2m ɛ. Since every λ 2 j 1, for every j, λ 2 j 1 ɛ. A pure Gaussian state ψ is such that β j = sign λ j. Consider the fidelity M ψ = R m j=1 ( 0 βj β j 0 F (τ, ψ ) = ψ τ ψ. ) R T, Apply a dephasing procedure to τ. Since the correlation matrix M τ stays invariant, the fidelity F (τ, ψ ) stays the same. After dephasing the state has the form τ m =: k p k φ k φ k, where φ k φ k = 1 2 m m j=1 (I + iβk j c 2j 1 c 2j ) and φ 0 φ 0 := ψ ψ.

Then F (τ, ψ ) = p 0 and so For every j = 1,..., m we have τ ψ ψ 1 2 1 p 0. λ j = M τ (2j 1, 2j) = Tr(ic 2j 1 c 2j τ m) = k β k j p k. Note that λ j = 1 2 p k = 2 p k 1. k:β j k = 1 k:β j k =1

Then F (τ, ψ ) = p 0 and so For every j = 1,..., m we have τ ψ ψ 1 2 1 p 0. λ j = M τ (2j 1, 2j) = Tr(ic 2j 1 c 2j τ m) = k β k j p k. Note that λ j = 1 2 p k = 2 p k 1. k:β j k = 1 k:β j k =1 For every j = 1,..., m we have 1 ɛ λ 2 j 1. Consider two possible cases: 1 If 1 ɛ λ j 1, we have β j k = 1 p k 1 2 (1 1 ɛ). Note that p 0 is not in the sum, since βj 0 = 1. 2 If 1 ɛ λ j 1, we have β j k =1 p k 1 2 (1 1 ɛ). Note that p 0 is not in the sum, since βj 0 = 1. Summing all inequalities (for every j) we obtain m j=1 k:β j k = sign λ p j k m 2 (1 1 ɛ). There is no p 0 in the sum and the only p k absent in the sum is p 0, so 1 p 0 = k 0 p k m 2 (1 1 ɛ), therefore Q.E.D. τ ψ ψ 1 2 1 p 0 2m(1 1 ɛ).

Finding convex-gaussian states Let ρ = i p i σ i with σ i pure Gaussian. Then there exists a symmetric extension ρ ext = i p i σ n C n i 2m, which is annihilated by Λk,l and Tr 2,..,n ρ ext = ρ. Program. Input: ρ C 2m and an integer n 2. Question: Is there a ρ ext C n 2m s.t. Trρ ext = 1 ρ ext 0 Tr 2,...,n ρ ext = ρ Λ k,l ρ ext = 0, k l Output: Yes, then provide ρ ext, or No. Since the null space of Λ k,l is spanned by ψ, ψ k,l where ψ is Gaussian, the intersection of all null-spaces of Λ k,l is spanned by vectors ψ n. The program can be done in the standard form of semi-definite program: minimize c T x subject to F 0 + i x i F i 0 Ax = b, here x R d, c R d is a given vector, {F i } i=0,...,d are given symmetric matrices and A R p d with rank(a) = p and b R p are given.

Finding convex-gaussian states Let ρ = i p i σ i with σ i pure Gaussian. Then there exists a symmetric extension ρ ext = i p i σ n C n i 2m, which is annihilated by Λk,l and Tr 2,..,n ρ ext = ρ. Program. Input: ρ C 2m and an integer n 2. Question: Is there a ρ ext C n 2m s.t. Trρ ext = 1 ρ ext 0 Tr 2,...,n ρ ext = ρ Λ k,l ρ ext = 0, k l Output: Yes, then provide ρ ext, or No. Since the null space of Λ k,l is spanned by ψ, ψ k,l where ψ is Gaussian, the intersection of all null-spaces of Λ k,l is spanned by vectors ψ n. Theorem [1] If an even state ρ C 2m has an n-gaussian-symmetric extension for all n, then ρ is convex-gaussian, in other words, there exists a sequence of convex-gaussian states ω 1, ω 2,... C 2m s.t. lim n ρ ωn 1 = 0.

Theorem [4] Suppose that an even state ρ C 2m has an extension to n number of parties, where n is sufficiently large. Then there exists a convex-gaussian state σ(n) such that ρ σ(n) 1 ɛ(n) := 4 2m ( n + 2m 1 1 4 Λ 2 2 m 1 n ). Denote G the set of convex-gaussian states and G n ρ iff ρ n ext C n 2m. Then G 1 G 2... G and lim n Gn = G. Proof Denote ρ n := ρ ext to n parties. The state ρ n is symmetric, therefore use quantum de Finetti theorem: Theorem [3] (Christandl, König, Mitchison, Renner). Let Ψ be a symmetric state on n parties and let ζ k = Tr k+1,...,n Ψ Ψ. Then there exists a probability distribution {α j } and states {τ j } such that ζ k j α j τ j τ j 1 ɛ

Therefore there exists α j 0, j α j = 1, and states τ j C 2m, s.t. ρ n 1,2 j α j τ j τ j 1 ɛ = 4 2m n. Applying Λ 1,2 we obtain Λ 2 1 ɛ =: ˆɛ Λ j α j τ j (n) τ j (n)λ 1 = j α j Λ τ j (n) τ j (n) Λ 1. Therefore for every j = 1,...m we obtain Λτ j (n) τ j (n)λ 1 ɛ. For every j the state τ j (n) is ɛ -close to a pure Gaussian state ψj (n). Therefore the state ρ is close to the convex-gaussian state: ρ j α j ψj (n) ψ j (n) 1 ρ j α j τ j (n) 1 + j α j τ j (n) ψj (n) ψ j (n) 1 ɛ + ɛ. Q.E.D.

Complementary criterion Separability criteria Let H = H A H B be a finite dimensional Hilbert space. Denote S the set of separable operators, i.e. the conical combination of all pure product states { ψ A ψ A ψ B ψ B }. ρ AB S N iff ρ AB N B(H AB N ) s.t. 1. ρ AB N 0 2. Tr B N 1 (ρ AB N ) = ρ AB 3. ρ AB N is Bose symmetric in HB N, i.e. ρ AB N (1l PN sym) = ρ AB N. The sequence {S N } N=1 converges to S from outside [5] (Doherty et al.): lim N SN = S.

Program. Given a state ρ AB B(H) determine whether it is separable. 1. Check whether ρ AB S k. If NO, then ρ is entangled. Done. If YES, then go to 2. 2. Check whether ρ AB S k. If YES, then it is separable. Done. If NO, repeat steps for k = k + 1. We have that G 1 G 2... G. Define G n = {ɛ nρ + (1 ɛ n)i/2 m : ρ G n } G. Problem. Show that lim G n = G. n Proposition [1] For any even state ρ C 2m there exists ɛ > 0 such that ρ ɛ = ɛρ + (1 ɛ)i/2 m is convex-gaussian. I.e. need to show that ɛ n 1 as n, for ρ n G n.

Extension to C 2mn The isomorphism J between C 2mn and C n 2m is the following: J(c j ) = I... c j P all... P all, for 2m(k 1) + 1 j 2mk, where c j stands on the k-th component and P all acts on each C 2m. J is the isomorphism since J(XY ) = J(X)J(Y ). For n = 2, for any state µ = k α k Z 1 k Z 2 k C 4m, where Z 1 acts on c 1,..., c 2m and Z 2 acts on c 2m+1,..., c 4m. Applying J this state gets mapped to: J(µ) = α k Zk 1 Pɛ k all Z k 2 C 2m C 2m, k where ɛ k = 0, if Zk 1 is even and ɛ k = 1, if Zk 1 is odd.

The operator Λ k,l is equivalent to Γ acting on C 2mn : for any 1 k l n, 2m Γ k,l = ( 1) m j c 2m(k 1)+j c 2m(l 1)+1...ĉ 2m(l 1)+j...c 2ml. j=1

The operator Λ k,l is equivalent to Γ acting on C 2mn : for any 1 k l n, 2m Γ k,l = ( 1) m j c 2m(k 1)+j c 2m(l 1)+1...ĉ 2m(l 1)+j...c 2ml. j=1 For an even state ρ = j α j Z 1 j C 2m the extension ρ ext C 2mn has the following form ρ ext = 2 m(n 1) ρ + Corr. Every term in the correlations Corr = j β j Z 1 j...z n j acts nontrivially on the last (n 1) spaces spanned by c 2m+1... c 2mn. Here Z k acts on the space spanned by c 2m(k 1)+1... c 2mk.

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