Geometry of Entanglement

Transcription

1 Geometry of Entanglement Bachelor Thesis Group of Quantum Optics, Quantum Nanophysics and Quantum Information University of Vienna WS SE Seminar Quantenphysik I Supervising Professor: Ao. Univ.-Prof. i.r. Dr. Reinhold Bertlmann Handed in by: Bogdan Pammer 8158 Vienna, June 29, 211

2 Contents 1 Introduction What is entanglement? entangled non local Free entanglement and bound entanglement Mathematical foundations 2.1 Positive operators and density matrices Positive Maps How to draw the line - necessary theorems 5.1 Pure states The PPT criterion A necessary condition for separability Entanglement witness theorem Three final steps to proof the PPT criterion Hilbert-Schmidt measure and entanglement witness Drawing a map for two-parameter entangled qubits 1.1 The Werner states A two parameter family Making a good guess? Finding the closest separable state in region I Finding the closest separable state in region II

3 1 Introduction 1.1 What is entanglement? Phenomenologically (experimentally) Entanglement is the correlation of results from mesurements in two different degrees of freedom within a physical system. (Alice, Bob,...) Mathematically (theoretically) Two degrees of freedom are entangled, iff they are not separable. Def. Pure or mixed states are separable in the Hilbertspaces A and B (bipartite system), if they can be written in the following way: ψ = ψ A ψ B (1) whereas ψ H A H B and ψ x H x ρ = i ρ mµnν = i p i ρ i A ρ i B p i ρ i Amn ρ i Bµν (2) whereas ρ x = i ψ i x ψ i, ψ x H x, p i 1 and i p i = 1. A separable state s eigenvalues therefore are invariant under separate unitary transformations of the bases used by the two observers. (U A U B )ρ(u A U B ) = i p i U A ρ i AU A U Bρ i BU B () Def. If a state H A H B is not separable in the way defined above, it is bipartite entangled. In this paper we will exclusively deal with bipartite entanglement. As example we can take a look at the (pure) spin-state ψ. ψ = 1 2 ( ) =

4 This vector cannot be written as tensor product of two state-vectors ( ) ( ) ac a c = ad b d bc. bd From (2) naturally follows that the set of separable states S is convex, meaning: ρ z = p ρ x + (1 p)ρ x S () ρ x, ρ y S and p entangled non local There are entangled states that can be described by a LRT (local realistic theory), that means due to hidden variables. Not all entangled states violate the Bell inequality. [1] So entangled states can be further classified in such that violate the bell inequality and those which don t. Def. An entangled state, that violates the Bell inequality maximally is defined as maximally entangled. 1. Free entanglement and bound entanglement Def. Fidelity F of an entangled state ρ ent in relation to an maximally entangled (and in this case pure) state ψ max is defined as: F ψmax = ψ max ρ ent ψ max, F ψmax 1 (5) Def. Def. Free entangled states ρ free are those states, which through repeated application of LOOC-Protocols (local operations and classical communication) can be distilled - reach an aribitrarily high fidelity.[2] Bound entangled states ρ bound are those states, which cannot be distilled. Bound entanglement is matter of intensive research in quantum information.

5 2 Mathematical foundations Dealing with bipartite entanglement in two qu-dit Hilbert spaces, we want to find out which states are entangled and which are not - draw the line. Qudits are the d-dimensional analogon to the two dimensional qu-bits. Before we can draw the line and analyse the geometry of entanglement in those two qu-dit Hilbert spaces we first need a criterion to distinguish entangled from not entangled states. Since this is a high dimensional problems this task is not as easy as it appears at first sight. 2.1 Positive operators and density matrices Operators acting on a Hilbert Space H of a qunatum mechanical system make up themselves a Hilbert Space called Hilbert-Schmidt Space A. Since we are dealing with Hilbert spaces of finit dimension the Elements of A written as matrices. The density matrices ρ from above are in strict mathematical terms elements of the positive semi-definite sub-space with a trace equal to 1.(In the case of pure states they are projectors on the respective ψ.) The respective scalar product of two operators A, B, A is defined as: A B = T ra B (6) Def. An operator P A on a Hilbert Space H is positive semi-definite iff P ψ ψ ψ H. Short: P A positive operator is necessarily Hermitian. 2.2 Positive Maps A linear map Λ : A 1 A 2 (7) maps an operator from A 1 into A 2. Def. A linear map Λ is positive iff Λ(A) A A 1. Short: Λ A postive map maps positive operators into positive operators. For a map Λ to be completely positive also a the map 1 d Λ : M d A 1 M d A 2 (8) needs to be positive d = 2,,,...; 1 d is the identity on the matrix space M d of all d d matrices. This is an advanced mathematical structure, since

6 not all positive maps remain positive when -expanded by the identity. A positive map therefore preserves hermiticity. How to draw the line - necessary theorems.1 Pure states It s rather easy to determine whether a pure states is separable. As we know from (1) ψ = ψ A ψ B ψ ψ = ( ψ A ψ B )( ψ A ψ B ) = ψ A ψ A ψ B ψ B (9) = ρ A ρ B For a separable state ψ ρ A and ρ B are density matrices of a pure state. The reduced density matrices ρ A = T r B ( ψ ψ ) and ρ B = T r A ( ψ ψ ) therefore need to be density matrices of a pure state, meaning T r(ρ 2 A ) = T r(ρ2 B ) = 1. If this is not the case the state is entangled..2 The PPT criterion The PPT (partial positive transposition) criterion is a necessary and sufficient criterion for C d C e (for d e 6) Hilbertspaces to determine whether a state is separable. For higher dimensional bipartite Hilbertspaces it is sufficient but not necessary. If a state ρ S A S B is separable, then ρ T B := (1 A T B )ρ (1) where T B is the transposition on the second subsystem of S A S B..2.1 A necessary condition for separability ρ T B := (1 A T B )ρ ρ S A S B is a necessary condition for separability. This is shown by Peres in [7]. Per definition (2) a separable state can be written as: 5

7 ρ = i p iρ i A ρi B ρ mµnν = i p iρ i Amn ρ i Bµν For the partial transposition T B we transpose the elements of the B-space, which is equivalent with switching the respective (greek) indices. ρ T B = ρ mνnµ = = i = i p i ρ i Amn ρ i Bνµ = p i ρ i A ρ i B T (11) Under separate unitary basis transformations, which are defined above (see ), the partial transposition s eigenvalues are conserved. This means the eigenvalues of ρ T B = (1 A T B )ρ are equal to the eigenvalues of (1 A T B )(U A U B ρ (U A U B ) ). This can be seen after a few manipulation steps: (1 A T B )(U A U B ρ U A U B ) = =(1 A T B ) p i U A ρ i AU A U Bρ i BU B = i = i = i = i p i U A ρ i AU A (U Bρ i BU B )T = p i U A ρ i AU A (U B )T ρ i B T U T B = p i U A ρ i AU A U Bρ i B T (U B) = (12) =(U A U B) ρ T B (U A U B) ρ T B (U A U B ) ρt B (U A U B ) is an unitary transformation und therefore conserves eigenvalues. For this reason it doesn t matter in which basis A and B view their system, the partial transposition s eigenvalues remain and are therefore able to conduct a useful criterion. ρ i B T is a density matrix as well, since a transposition keeps the matrix eigenvalues. ρ T B is therefore necceserily a legitimate (separable) density matrix and non-negative. The opposite direction 6

8 ρ T B := (1 A T B )ρ ρ S A S B is true only in the special cases of C d C e (for d + e 6). The Horodecki s showed this in the following steps [6]. The first step is the entanglement witness theorem..2.2 Entanglement witness theorem First it is to show, that a state ρ ent is entangled iff there exists an Hermitian operator A A, called entanglement witness [, 6, 9, 1] so that: ρ ent, A = T raρ ent <, σ, A = T raσ, σ S (1) The proof is based on the Hahn-Banach Theorem of functional analysis, saying that if W 1 and W 2 are convex closed sets in a real Banach space and one of them is compact, then there exists a continuous functional f and α R such that for all pairs w 1 W 1 and w 2 W 2 we have f(w 1 ) < α f(w 2 ) (1) We will operate on a subspace of the Hilbert-Schmidt space A 1 A 2 consisting only of hermitian operators. (It can be easily shown they generate a subspace). Therefore and since the separable states S are a closed set and a set consisting of one element is compact we can use the Hahn-Banach Theorem in the following way: W 1 ρ ent (the set consisting only of the not-separable density matrix ρ ent ) W 2 S (σ S) α β f g(ρ) = T raρ with A = A, we can obtain any continous linear functional g this way, since any continous functional in a Hilbert space is represented by a vector from this space. [6] Further we see that g is real. According to Hahn-Banach we therefore know, that there exists an A so that If we substitute: g(ρ ent ) < β g(σ) (15) A A = A β1 We obtain the desired entanglement witness, since 7

9 T ra ρ ent = g(ρ ent ) β < and T ra σ = g(σ) β Later we will see that the entanglement witness theorem in its own is a very useful tool in higher dimensions to draw the line..2. Three final steps to proof the PPT criterion 1. T raρ ρ S. (16) A A 1 A 2 with A = A. : If A = A then T ra( ψ ψ φ φ ). This can be easily shown by decomposing the basis of A 1 A 2 in the eigenbasis of the Hermitian operator A. Then from the definition of separability (see (2)) obviously follows the proof s first direction. : Indirect proof: Suppose for all Hermitian A T raρ, but ρ / S. Since ρ / S according to the EWT (1) there exists a Hermitian A so that T ra ρ <, which is a contradiction to the assumption A T raρ. Qed. 2. (1 Λ) ρ ρ S. (17) for all positive Λ s. In this second step as it is shown in [6] the Horodecki s translate (16) in the language of positive maps. For this purpose we use the following isomorphism, Λ S(Λ) = i E i Λ(E i) A 1 A 2. (18) Whereas Λ : A 1 A 2 is a linear map and E i is a arbitrary orthonormal basis for A 1 and A 2 respectivly. We can now substitute (16) A S(Λ) and gain: T r( i E i Λ(E i))ρ. (19) A theorem by Jamiolkowski [8] states, that for a map Λ to be positive it is necessary and sufficient that S(Λ) is Hermitian and T r(s(λ) P Q). Whereby P and Q are elements of A 1 and A 2 respectivly. As mentioned above a density matrix is a sum of tensored pairs of projectors. In this notation ρ could be written as 8

10 ρ = p n P n Q n (2) Therefore iff Λ is an arbitrary positive map (Λ ) S(Λ) is Hermitian, (19) fulfilled and the substitution allowed. Now we choose the specific projector basis {P ij } dimax i,j=1, with P ije l = δ jl e i P ij projects the basis vector e j on e i and the others on. (19) turns into: T r( i,j (1 Λ)P ji P ij )ρ T r((1 ΛT ) i,j P ij P ij )ρ (21) T r(1 ΛT P )ρ T is a positive map, since it conserves matrices eigenvalues. Λ := ΛT is accordingly a positive form. Written as scalar product: ρ (1 Λ P ) (22) Positive maps preserve hermiticity and hence the tensor product 1 Λ also does. Then, as P O is Hermitian we obtain: ρ 1 Λ P (2) Since we deal with finite operators we can pass to adjoint maps and gain: 1 Λ ρ P (2) for an arbitrary map Λ : A 2 A 1. For separable ρ this condition is obviously fulfilled. Conversely, since P is a sum of projectors, if 1 Λ ρ conditions (2) and accordingly (16) are fulfilled. whereas ρ T B := (1 A T B )ρ see (1).. ρ T B > ρ S. (25) According to the Strømer-Woronowicz Theorem [11, 12] for C d C e (d e 6) a positive operator Λ can be written as whereas Λ CP A and ΛCP B Λ = Λ CP A + Λ CP B T (26) are completely positive maps. 9

11 <[1 Λρ] = [1 (Λ CP A + Λ CP B T )]ρ = [1 Λ CP A ]ρ + [1 Λ CP B T ]ρ = [1 Λ CP A ]ρ + [1 Λ CP B ]ρ T B (27) Since Λ CP A and Λ CP B are completely positive maps is ρ T B > a sufficent condition for separability.. Hilbert-Schmidt measure and entanglement witness In (2.2.) it was shown that a state ρ ent is entangled iff there exists an Hermitian operator A A, called entanglement witness, so that: ρ ent, A = T raρ ent <, σ, A = T raσ, σ S (28) An entanglement witness A therefore divides the operator space in two spaces. Both spaces are convex due to the linearity of the scalar product. One of them contains the entangled state, the other S, the (convex) set of all separable states. An entanglement witness A is optimal, if additionally to the criterions mentioned in (28) there exists a separable state ρ S for which A ρ =. (29) A = ρ ρ ent ρ ρ ρ ent 1 ρ ρ ent A as defined above is an optimal entanglement witness. The states fulfilling A ρ = ρ ρ ent ρ ρ ρ ent 1 ρ ρ ρ ent = ρ ρ ρ ent ρ ρ ρ ρ ent 1 ρ ρ ρ ent = () (1) A ρ = (2) 1

12 can be illustrated as tangent plane to S, due to the linearity of the scalar product. If two different ρ 1 and ρ 2 fulfill (2) also their linear combinations do so. If σ 1 and σ 1 S fulfill equation (2) also those states on the straight line connecting them fulfill (2) and are in S, since S is convex. All separable states are either on the tangent plane or on the side of it. The situation can therfore be illustrated in the following way: Figure 1: Illustration of an optimal entanglement witness The Hilbert-Schmidt measure of entanglement is defined as the minimal Hilbert-Schmidt distance of an entangled state ρ ent to any separable state ρ S. D(ρ ent ) := min ρ S ρ ρ ent = ρ ρ ent () The entanglement witness theorem can help to identify the respective ρ. A separable state σ equals the separable state ρ closest to ρ ent iff is an entanglement witness. C = σ ρ ent σ σ ρ ent σ ρ ent () To show this we make use of the Bertlmann-Narnhofer-Thirring Theorem (BNT Theorem) []. Firstly we define the maximal violation of the entaglement witness inequality: B(ρ ent ) = max (min ρ A ρ ent A ) (5) A, A a1 1 ρ S 11

13 The BNT Theorem then states: Theorem 1 (BNT Theorem) Considering an abritary entangled state ρ ent, then its closest distance to the set of separable states D(ρ ent ) () equals the maximal violation of the entanglement witness inequality (5) The following graph illustrates the BNT Theorem: D(ρ ent ) = B(ρ ent ). (6) Figure 2: Illufation of the Bertlmann-Narnhofer-Thirring Theorem Due to the definition A from above and the BNT Theorem the very intuitive from above is justified: (1) The closest separable state ρ ent is on the straight line representing A (2) A is tangent to the convex set S () the line illustrated representing A can be illustrated as being orthogonal to the line connecting ρ ent and ρ, since for A the entanglement witness inequality gets violate maximally by ρ ent, the closest separable state. From the BNT Theorem we can conclude that if ρ is the closest separable state, the respective A is an entanglement witness. So we can guess closest separable states σ and proof them as such by analysing whether is an entanglement witness. C = σ ρ ent σ σ ρ ent σ ρ ent (7) 12

14 Drawing a map for two-parameter entangled qubits We will now explicitly calculate Hilbert-Schmidt measure of two-parameter entangled qubits. We will first guess states and then proof that they are the closest separable states by the criterion mentioned above..1 The Werner states The Werner State is the 2 2 dimensional case of an isotropic state. An isotropic state is a d d dimensional bipartite quantum state that is invariant under any unitary of the form U U, where * denotes complex conjugate. That is, any state with the property that for any unitary U on one part of the system, ρ = (U U )ρ(u (U ) ). (8) The isotropic states is a one-parameter familiy of states and can be written as ρ α = α φ + φ + (1 α)1 + (9) d 2 The Werner States therefore can be written as: ρ α = 1+α α 2 1 α 1 α α 1+α 2 Since ρ α is a density matrix all eigenvalues need to be positive, therefore: 1 < α 1. ρt B α can be written as ρ T B α = 1+α 1 α α 2 α 2 1 α 1+α The eigenvalues of ρ T B α are: λ 1,2, = 1+α and λ = 1 α. λ 1,2, are positiv for all α in its domain. λ is not negative for α 1. According to the PPT criterion therefore ρ α is separable for 1 < α 1 and entangled for 1 < α 1. 1

15 .2 A two parameter family We know now for this one parameter family, where the boarder between entangled and separable lies. From here we want to expand this by analysing the following state: ρ α,β = 1 α β 1 + α φ + φ + + β 2 ( ψ+ ψ + + ψ ψ ) () ρ α,β = 1+α β α 2 1 α+β 1 α+β α 2 1+α β The eigenvalues of ρ α,β are: λ 1,2 = 1 α+β, λ = 1 α β and λ = 1+α β. We see that the eigenvalues sum equals 1. This is basically the Werner State mixed with an additional parameter β. Again ρ α,β is only a density matrix if all its eigenvalues are positive. This constrains the possible values for α and β, namely: α β + 1, α 1 β 1, α β + 1, (1) We can now apply the partial transposition on ρ α,β eigenvalues and separability conditions for ρ α,β. and obtain following ρ T B α = 1+α β α 1 α+β α α+β 1+α The eigenvalues of ρ T B α,β are: λ 1,2 = 1+α β, λ = 1 α+β and λ = 1+α+β. α β 1, α 1 β + 1, α β 1, (2). Making a good guess? Now we need some intuition or a good guess for the closest separable state to progress on our endeavor. Since the states ρ α,β compile a plane in the Hilbert-Schmidt space Bertlmann and Krammer decided to make use of it s inner product to introduce a coordinate system and display it. After illustrating the situation and the lines (conditions) discerning separable and 1

16 entangle states, we will guess the closest state (in the Euclidian sense) as the closest separable state ρ (minimal Hilbert-Schmidt distance). We will proof that this guess was right, but as we will see this is by no means trivial. To find a helpful coordinate system to display the plane containing our states from (). We check first which angle the α- and β-axis include and where we have right angles (where the scalar product diminishes). Let us again take a look at the states we want to display: ρ α,β = 1 α β 1 + α φ + φ + + β 2 ( ψ+ ψ + + ψ ψ ) For α = β = we get the identity, which shall be at the origin of our coordinate system. Now our coordinate system needs axis and angles. A general vector (in the plane) is of the form ρ α,β 1: ρ α,β 1 = We calculate a general angle: α β α 2 α+β α+β α 2 +α β ρ α1,β 1 1 ρ α2,β 2 1 = T r((ρ α2,β 2 1) (ρ α1,β 1 1)) = 1 (α 1α 2 α 2 β 1 α 1 β 2 + β 1 β 2 ) () We can calculate for which vectors (α 2,β 2 ) the α-axis (β 1 = ) includes a right angle. We obtain: α 1 α 2 α 1 β 2 = α 2 = 1 β 2 () We see that the α-axis includes a right angle with those vectors fulfilling α = 1 β. Looking at the conditions constraining the separable states in (1) and (2), we see that in both cases the direction vector of those straight lines resembling the second conditions (α 1β 1 and α 1β + 1 ) fulfil the conditions and therefore include a right angle with the α-axis. For the β axis we optain through equivalent calculations, that the straight lines resembling the first conditions of (1) and (2) (α β+1 and α β 1) include a right angle with it. 15

17 The two parameter family of states and the positivity conditions constraining ρ α,β and ρ T B α,β can be displayed in the following way: Figure : Illustration of the qu-bit states ρ α,β and their partial transposition It is to mention that this illustration makes a significant compromise. If we calculate the angle between the α- and β-axis with equation we realize that their angle is a function of the position you are on the respective axis. Therefore the image above is a helpful illustration, but one needs to be careful with drawing conclusions from it. Three straight lines resembling the positivity conditions meet at the points (α =, β = 1) and (α =, β = 1). This can be easily seen, by inserting in equation 1 and 2. From this we can draw, that there are two sectors (I and II) of entangled states in ρ α,β. Those states fulfill the positivity criterion, but their partial transposition ρ T B α,β has negative eigenvalues.. Finding the closest separable state in region I The previous section s illustration helps us to make educated guesses, for the closest separable states. In section I α 1β + 1 is the constraining condi- 16

18 tion. The nearest states in the Euclidian sense (according to our coordinate system) to a state ρ ent α,β =(α,β) in area I is therefore ρ β =( 1 β + 1,β). Let us now proof whether C = ρ β ρent α,β ρ β ρ β ρent α,β 1 ρ ent α,β ρ β is an entanglement witness and thereby ρ β the closest separable state (). To calculate C we use a different but equivalent expression for ρ α,β : ρ α,β = 1 (1 + α(σ 1 σ 1 σ 2 σ 2 ) + (α β) σ σ ) (5) Whereby σ i represents the respective Pauli matrix. We gain with ρ β ρ ent α,β = 1 (1 + β α)σ (6) Σ = σ 1 σ 1 σ 2 σ 2 + σ σ. (7) Since Σ = T r(σσ) = 2 we optain, that ρ β ρ ent α,β = For the scale product we obtain 2 (1 + β α) (8) ρ β ρ ent α,β ρ β = T rρ β(ρ β ρ ent α,β = (α 1 β ). (9) Using (6),(8) and (9) we obtain C = 1 2 (1 Σ) (5) having calculated C we need, check whether it is an entanglement witness. For this to be the case it needs to fulfill the conditions pointed out in (). By construction we obtain for ρ ent C = ρ ρ ent <. The mathematically more interesting test to to prove whether all separable states σ fulfill σ C >. In [] this is done by proofing the following Lemma: 17

19 Lemma 1 For any Hermitian operator C on a Hilbert space of dimension 2 2 that is of the form C = a(1 + c i σ i σ i ), a R +, c i R (51) the expectation value for all separable states is positive, i=1 σ C σ S if c i 1 i. (52) Proof: Any separable state ρ is a convex combination of product states and thus a separable two qubit states can be written as σ = k p k (1 1 + i n k i σ i 1 + j m k j σ j 1 + i,j n k i m k j σ i σ j ), with n k i, m k j R, n k 1, m k 1, p k 1, k p k = 1, (5) where n k = i (nk i ) 2. Performing the trace, we obtain σ C = T rσc = k p k a(1 + i c i n k i m k i ), (5) and usig the restriction c i 1 i we have i c i n k i m k i i n k i m k i 1, (55) and since the convex sum of positive terms stays positive we get σ C σ S. Qed. Since (5) is of the form (51), we can use this Lemma to very that C from (5) as an entanglement witness. ρ β are therefore the closest separable states in area I. D I (ρ ent α,β) = ρ β ρ ent α,β = 2 (α 1 β ) (56) is therefore the Hilbert-Schmidt measure for area I and equal to the maximal violation of the entanglement witness inequality as mentioned above in (5). 18

20 .5 Finding the closest separable state in region II The calculation for region II follows the same pattern as in region I. α < β 1 is here the constraining condition separating entangled from separable states. In order to calculate the nearest states in the Euclidian sense (according to our coordinate system) to a state ρ ent α,β =(α,β) in area II we need to find the line including a right angle with those states fulfilling α = β 1. As in () we apply the scalar product and obtain ρ 1+2α β α,β =(, 2 2α+β ). To calculate C we need so that ρ α,β ρ ent α,β = 1 12 (α β)σ 1 σ 1 σ 2 σ 2 σ σ, (57) ρ α,β ρ ent α,β = 1 2 ( α 1 β) (58) ρ α,β ρ ent α,β ρ α,β = 1 (α β), (59) 12 C = 1 2 (1 + σ 1 σ 1 σ 2 σ 2 σ σ ). (6) The first condition of (1) is again fulfilled by the constuction of C. To poof the second condition we again apply the condition from (.) and obtain Therefore σ C σ S. (61) D II (ρ ent α,β) = ρ α,β ρ ent α,β = 1 2 ( α 1 β) (62) is the Hilbert-Schmidt measure for region II and equal to the maximal violation of the entanglement witness inequality as mentioned above in (5). With this we mapped this two parameter family of two entangled qu-bits properly and calculated the Hilbert-Schmidt measure. 19

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