Definition of entanglement for pure and mixed states

Definition of entanglement for pure and mixed states seminar talk given by Marius Krumm in the master studies seminar course Selected Topics in Mathematical Physics: Quantum Information Theory at the University of Heidelberg bstract: keypoint in which quantum mechanics differs from classical mechanics is the possibility of (quantum) entanglement. Entanglement is an important resource in quantum information theory and quantum computation and can be used to realize many surprising effects like quantum teleportation. This document is a written elaboration of the seminar lecture Definition of entanglement for pure and mixed states by Marius Krumm as part of the master studies seminar course Selected Topics in Mathematical Physics: Quantum Information Theory at the University of Heidelberg. Its purpose is to give an introduction to entanglement. In the beginning, we define entanglement for pure and mixed states. fterwards, the von Neumann entropy is introduced as an entanglement measure for pure states. t last, there will be a short discussion of entanglement witnesses.

2 1 Introduction fundamental element of quantum information science is quantum entanglement. It can be used to realize surprising effects like quantum teleportation and is considered as an important resource in quantum computation. The aim of this document is to give a first introduction to quantum entanglement and its quantification. This knowledge is collected from the sources found at the end of this document, which are also a good starting point to obtain further knowledge. 2 Quantum Entanglement efore we define entanglement, at first we provide a tool that will prove useful several times. Theorem 1 (Schmidt decomposition). Let H, H be two Hilbert spaces and ψ a normalized state in H H. Then there exist orthonormal sets { j }, { j } in H, H, such that: ψ = pj j j. (2.1) j Here p j are the non-zero eigenvalues of ρ := Tr ( ψ ψ ). Proof. We define ρ := ψ ψ. Furthermore we expand ψ in an orthonormal eigenbasis { j } of ρ (i.e. ρ = j p j j j ) and an arbitrary orthonormal basis { k } of H : ψ = j,k a jk j k. Now we define ĵ := k a jk k to obtain ψ = j j ĵ. We claim, that these vectors are orthogonal: ˆk ĵ δ jk. To prove this claim, we expand the density matrix: ρ = ( j,k j ĵ )( k ˆk ). We take the partial trace: ( ( ρ = Tr j ĵ )( k ˆk )) = ( ( m j ĵ )( k ˆk )) m j,k m = m,j,k = jk [ ] m ĵ ˆk m j k = j,k ˆk ĵ j k j,k [ ( ) ] ˆk m m ĵ j k m We compare to ρ = j p j j j and indeed find ˆk ĵ = p j δ jk. For those j with p j = 0 we find ĵ ĵ = 0 and thus ĵ = 0. For the j with p j 0 we can define j := 1 pj ĵ. Thus we have an orthonormal set ( k j = δ jk) that allows us to write ψ like in eq. (2.1).

3 Corollary 1. Let ψ be a normalized state in H H. Then ρ := Tr ( ψ ψ ) and ρ := Tr ( ψ ψ ) have the same non-zero eigenvalues. Proof. We apply the Schmidt decomposition: ψ = j pj j j. Thus we find ρ = Tr ( j,k pj p k ( j j )( k k )) = m,j,k pj p k m j k m j k = j p j j j This is the spectral decomposition of ρ, which shows that ρ also has the non-zero eigenvalues p j. Definition 1 (Schmidt number). The Schmidt number or Schmidt rank of a pure state ψ H H is defined as the number of non-zero eigenvalues of ρ or ρ : SR( ψ ) := #{j p j 0} (2.2) Definition 2 (Entanglement of pure states). Let ψ H H be a pure state. We call ψ entangled or non-separable if its Schmidt number is larger than 1. Theorem 2. Let ψ H H be a pure state. Then: ψ entangled a H, b H such that ψ = a b Proof. If ψ is entangled, it cannot be written as a b, because that state would have Schmidt number 1. nd vice versa, if ψ cannot be written as a b, then its Schmidt decomposition contains at least 2 summands and thus the Schmidt number is at least 2. Definition 3. Let ψ H H be a pure state, dim(h ) dim(h ). Then ψ maximally entangled : ρ = 1 dim(h ) Example 1. n example for a maximally entangled state (in a composite system of two spin- 1 systems) is the ell state ψ 2 = 1 2 ( + ). It has Schmidt rank 2. It may describe the full composite quantum system. ut only after a measurement we can tell if the spin of system is up or down. ut if have measured the spin of, we also know the spin of without any further measurement.

4 Definition 4 (General definition of entanglement). state ρ on H H is called non-entangled/separable if there exist density operators ρ (j), ρ(j) and p j 0 with j p j = 1 such that: ρ = p j ρ (j) ρ(j) (2.3) j The idea behind that definition is: systems described by ρ (j) ρ(j) are completely independent of each other. Now assume two physicists lice (system ) and ob (system ) use classical communication: lice rolls a dice, which has the result j with probability p j, and tells ob the result; Then lice produces ρ (j) and ob ρ(j). They repeat this procedure many times and forget, which j belongs to the samples. Then the resulting ensemble is described by eq. (2.3). Theorem 3. For a pure state ψ H H, definition (4) reduces to definition (2). Proof. We first assume ρ = j p jρ (j) ρ(j) = ψ ψ. We define ρ j := ρ (j) ρ(j) (again a density operator). Complete ψ to an orthonormal basis { ψ, φ 1, φ 2...}. Furthermore we perform an eigen-decomposition: ρ j = k a(j) k 0. Then we find 0 = φ x ρ φ x = j,k p ja (j) k a(j) k a(j) k with a(j) k > φ x a (j) k a(j) k φ x and thus φ x a (j) k 2 = ψ we find a (j) ψ. We thus found 0. s a (j) k k ρ j = ρ and thus ρ = ρ ρ for some ρ and ρ. Now we notice ρ 2 = ρ2 ρ2 = x φ x a j k φ x + ψ a (j) k and thus 1 = Tr(ρ 2 ) = Tr (ρ 2 ) Tr (ρ 2 ) and thus Tr(ρ2 ) = 1, which means that ρ is pure and thus Schmidt number = 1. Now we assume ψ = φ α. Then ρ = φ φ α α. Theorem 4. ρ separable sets of normalized states { a (j) j p j = 1 such that ρ = j p j a (j) a(j) b(j) b(j) Proof. : We simply set ρ (j) := a(j) a(j) and ρ(j) := b(j) b(j). and ρ(j) }, { b(j) }, p j 0 with : We perform an eigenvalue decomposition of all ρ (j). y relabeling and change of numeration we achieve the desired form. full proof is found in ppendix. So far we have only defined entanglement. Now we want to find a notion for how strongly a state is entangled. For a pure state, this is done by the von Neumann entropy. Definition 5. Let ρ be an arbitrary density operator. Then the von Neumann entropy is defined as: S(ρ) = Tr(ρ log 2 ρ) (2.4)

5 If p j are the eigenvalues of ρ, then we can rewrite the entropy as 1 S(ρ) = j p j log 2 p j (2.5) Example 2. 1. ρ = ( ) 1 0 has eigenvalues 0 and 1 S(ρ) = 0 0 0 2. ψ = 1 2 ( + ). Then ρ = ψ ψ ˆ= 1 2 again S(ρ) = 0. ( 1 ) 0 3. ρ = 2 S(ρ) = log 2 (2) 0 1 2 ( ) 1 1. s ρ has eigenvalues 0, 1, 1 1 Theorem 5 (General properties of the von Neumann entropy). 1. S(ρ) 0 and S(ρ) = 0 ρ pure state 2. dim(h) = d S(ρ) log 2 (d) and S(ρ) = log(d) ρ = 1 d 3. ρ on H H pure S(ρ ) = S(ρ ). 4. Let p j 0, j p j = 1, ρ j density operators which have support on orthogonal subspaces 2. Then S( j p jρ j ) = H({p j }) + j p js(ρ j ), where H({p j }) = j p j log 2 p j is the Shannon entropy. 5. Joint Entropy Theorem: Let p j 0, j p j = 1, { j } H orthonormal, ρ j density operators on H. Then S( j p j j j ρ j ) = H({p j })+ j p js(ρ j ) 6. ρ, ρ density operators on H, H. Then S(ρ ρ ) = S(ρ ) + S(ρ ). Proof. 1. s all eigenvalues p j [0, 1], we find S(ρ) = j p j log 2 p j 0 because p j log 2 p j 0. Furthermore p j log 2 p j = 0 p j {0, 1}. s the p j form a probability distribution, we find S(ρ) = 0 j : p j = 1 state pure 1 For a hermitian operator with eigenvalue-decomposition = j a j a j a j (where { a j } is an orthonormal basis) and a function f : C C, we define f() := j f(a j) a j a j. Especially this implies that ρ log 2 ρ has the eigenvalues p j log 2 p j. Furthermore we define 0 log 2 0 := 0. 2 For a hermitian operator = j a j a j a j (where { a j } is an orthonormal basis), the support is defined as span{ a j a j 0}. If and have orthogonal support, then a j = 0 for a j 0.

6 2. We prove this claim by induction. t first assume d = 1. Then we have p 1 = 1 and thus S(ρ) = 0 = log 2 1. Now assume we already know: d 1 j=1 p j log 2 p j log 2 (d 1) for all {p j j {1, 2,..., d 1}} with p j 0, d 1 j=1 p j = 1. We consider now {p j j {1, 2,..., d}} with p j 0, d j=1 p j = 1. We impose this last constraint by identifying p d as function of the other eigenvalues: p d = p d (p 1,..., p d 1 ) = 1 d 1 j=1 p j. This function is defined for p := (p 1,...p d 1 ) D := [0, 1] d 1 { q d 1 j=1 q j 1} (see figure 2.1). Here, we demand d 1 j=1 p j 1 because we want p d [0, 1]. s D is compact, there must exist a global maximum. To find the border of this set, we notice that [0, 1] d 1 is a hypercube, while { q d 1 j=1 q j 1} constrains us to one half of that cube, because { q d 1 j=1 q j = 1} is a hyperplane (compare ppendix ) which splits R d 1 and [0, 1] d 1 in two parts. Then the border is given by the sides of the hypercube (with j : p j = 0) and the part of the hyperplane, that lies in the cube (here p d = 0). Thus on D, we always have at least one p j = 0, wlog we assume p d = 0. ut we already know d 1 j=1 p j log 2 p j log 2 (d 1). However, S( 1 d d ) = log 2 d, so we have to look for our maximum in the interior D 0 = D \ D. Here, every global maximum is also a local maximum. Such maxima must satisfy p ( d j=1 p j log 2 p j ) = 0. For k {1, 2,...d 1} we find p k ( d j=1 p j log 2 (e) ln(p j )) = 0 0 = ln p k 1 + ln p d + 1 p k = p d. Thus our maximum is indeed found for ρ = 1 d d and S(ρ) = log 2 d. Figure 2.1: The domain of p = (p 1,...p d 1 ) for d = 4 dimensions (left) and d = 3 dimensions (right). 3. We know that ρ and ρ have the same non-zero eigenvalues. S(ρ ) and S(ρ ) depend only on the eigenvalues. 4. We perform eigenvalue decompositions: ρ i = j p(i) j orthogonal supports, the p (i) j with p(i) p (i) j p(i) j. ecause of the j 0 are also eigenstates of the other ρ k with ρ k p (i) j = 0. This implies that j p jρ j also has the eigenstates p (i) j, with eigenvalues p i p (i) j. Thus we find S( j p jρ j ) = i,j p ip (i) j log 2 (p i p (i) j ) = i p i log 2 p i + i p ( i j p(i) j log 2 p (i) ) j = H({pj }) + i p is(ρ i ). 5. The j j ρ j obviously have orthogonal supports, and S( j j ρ j ) = S(ρ j ) because j j ρ j and ρ j have the same non-zero eigenvalues.

7 6. Use the joint entropy theorem with ρ j := ρ j and j p j j j = ρ. Then H({p j }) = S(ρ ) and j p js(ρ ) = S(ρ ). Theorem 6 (Von Neumann entropy as measure for pure state entanglement). Let ρ = ψ ψ be a pure state on H H, dim(h ) dim(h ). Then: 1. S(ρ ) = 0 ψ separable, S(ρ ) > 0 ψ entangled 2. S(ρ ) = log(dimh ) ( = maximal ) ψ maximally entangled Proof. 1. S(ρ ) = 0 ρ pure Schmidt number = 1 separable 2. S(ρ ) maximal ρ = 1 dimh maximally entangled The last theorem shows that the von Neumann entropy is a good measure for entanglement of pure states. For mixed states, it is much harder to find good entanglement measures, because one has additional classical correlations that need to be distinguished from quantum entanglement. There exist many measures, but none of them is perfect. s a simple example to show why the von Neumann entropy is a bad measure for entanglement of mixed states we consider ρ = ρ ρ. This mixed stated is non-entangled. However, we are completely free to choose ρ. We could 1 take ρ = ψ ψ with S(ρ ) = 0 or ρ = dim(h ) 1 with S(ρ ) = log 2 (dim(h )). Thus even for non-entangled states, S(ρ ) can take values between its minimum and maximum and is thus a bad measure. n idea to detect entanglement of mixed states is given by entanglement witnesses. efore we state the Entanglement Witness Theorem, at first we need to prove a few statements: Theorem 7. 1. The density operators form a convex set 3 : D := {ρ ψ ρ ψ 0 ψ, ρ = ρ, Tr(ρ) = 1} (2.6) 2. The separable density operators form a convex set: S := {ρ D ρ separable } (2.7) Proof. 1. Let, D, p [0, 1]. Then Tr(p + (1 p)) = ptr() + (1 p)tr() = 1. lso (p + (1 p)) = (p + (1 p)) and ψ [p + (1 p)] ψ = p ψ ψ + (1 p) ψ ψ 0 3 set is called convex, if for all,, p [0, 1] also p + (1 p)

8 2. ρ, σ D separable density operators ρ (j) j p j = j q j = 1 such that: ρ = m j=1 p jρ (j), ρ(j), σ(j), σ(j) ρ(j) For j {1,..., m} and a [0, 1] we define r j := ap j and τ (j) For j {m + 1,..., m + n} we define r j τ (j) := σ(j m) n+m j=1 r jτ (j). Then we find r j [0, 1], n+m τ (j) S and p j, q j 0 with and σ = n j=1 q jσ (j) := ρ(j), τ (j) σ(j). := ρ(j). := σ(j m), := (1 a)q j m and τ (j) j=1 r j = 1 and aρ + (1 a)σ = Theorem 8 (Entanglement witness theorem (EWT)). Let ρ be an arbitrary density operator on a finite-dimensional Hilbert space H H. We define M := { linear operator on H H with = }. Then: ρ entangled W M : Tr(ρW ) < 0 and Tr(σW ) 0 for all separable density operators σ. Proof(sketch). We assume that ρ D is entangled. M is a real vector space with scalar product, := Tr( ) = Tr() (see ppendix C for more information). s we have learned from theorem 7, we have the situation shown in figure 2.2 (left image). It is intuitive, that there exists a hyperplane which separates ρ and S. 4 Figure 2.2: The set of density operators D is convex and contains the convex set of separable density operators S. lso shown is the hyperplane that separates S and the entangled density operator ρ. Let this hyperplane be described by Tr(Ŵ τ) = Ŵ, τ = c (see also ppendix ), where Ŵ = Ŵ is the normal vector of the hyperplane, chosen such that Tr(Ŵ ρ) = Ŵ, ρ < c and Tr(Ŵ σ) = Ŵ, σ c σ S. Now instead of Ŵ, we take W := Ŵ c1. Then we find (τ D): W, τ < 0 Tr(W τ) < 0 Tr(Ŵ τ) ctr(τ) < 0 Tr(Ŵ τ) < c Ŵ, τ < c We also find Ŵ, τ = c W, τ = 0 and Ŵ, τ > c W, τ > 0. Thus we are now in the situation shown in figure 2.2(right image) and W fulfills the EWT. 4 The existence of this hyperplane follows from the Hahn-anach theorem of functional analysis.

9 Comment. s shown in ppendix C, the density operators can all be found in a hyperplane T := { M Tr() = 1} M. This means that T and the other hyperplane orthogonal to W intersect in an even lower dimensional line. This gives us the freedom to chose another W to effectively tilt the hyperplane such that the origin is also found in that plane. This is what we have done by replacing Ŵ with W. The whole situation is visualized in figure 2.3. Figure 2.3: The density operators can be found in an affine hyperplane T. The plane that we obtain from the Hahn-anach theorem can be tilted such that it contains the origin. 3 Conclusion In this document we have introduced entanglement for pure and mixed states. We have discussed the von Neumann entropy as measure for entanglement of pure states. However, many aspects of entanglement have not been adressed in this introduction, e.g. what entanglement can be used for and what its application for information science is. We also have not discussed the problems with the interpretation of entanglement, e.g. why it does not allow for faster-than-light communication. lso a delicate topic is how to define entanglement measures for mixed states. Here one typically defines some properties that such an measure should satisfy and defines measures constructed such that the constraints are fulfilled. There exist a lot of such measures; but many of them are hard to calculate. ppendix: Extended Proof Of Theorem 4 We start with a separable (mixed) state: ρ = a j=1 p jρ (j) expansion of all density operators and obtain: ρ(j). We perform a spectral ρ = a p j ( j=1 b q k q k q k ) ( k=1 c r l r l r l ) = l=1 a,b,c j,k,l=1 (p j q k r l ) q k q k r l r l

10 Now we define a bijection f : N 3 N and := f({1, 2,...a} {1, 2,...b} {1, 2,...c}) (see e.g. figure.1). Then f : {1, 2,...a} {1, 2,...b} {1, 2,...c} is also a bijection. Now we define t f(j,k,l) := p j q k r l, α f(j,k,l) := q k and β f(j,k,l) := r l. Thus: ρ = a,b,c j,k,l=1 t f(j,k,l) α f(j,k,l) α f(j,k,l) β f(j,k,l) β f(j,k,l) = f t f α f α f β f β f t f(j,k,l) = p j q k r l 0 and f t f = a,b,c j,k,l=1 t f(j,k,l) = ( j p j)( k q k)( l r l) = 1. Figure.1: The figure illustrates how to define a bijection g : N 2 N. With this function a bijection f : N 3 N, f(j, k, l) = g(j, g(k, l)) can be defined. ppendix: Euclidean space Here we want to recall some simple facts about planes in Euclidean space (or isomorphic spaces). Figure.1: hyperplane in 2 dimensions. The difference of two vectors pointing on the hyperplane is orthogonal to the normal vector. Figure.2: The vector p points above the plane. Let q be an arbitrary vector pointing on the hyperplane with normal vector n. Then for any vector p also pointing on the plane, we find ( p q) n = 0 (see figure.1). This last equation defines the plane. We note that the exact choice of q does not matter, because the different choices only differ in a vector orthogonal to n. Defining c := q n, we find the hyperplane to be defined by p n = c.

11 Now we consider a vector p that does not point on the hyperplane (see figure.2). We decompose p = q + α n, where α R and q points on the plane. Then we define p to point above the plane, if α > 0. This implies p n = α n 2 + q n = c + α > c. Thus we say that p points above the hyperplane if p n > c, and equivalently we say that p points below the plane if p n < c. Of course the assignment of above and below is arbitrary and can be exchanged. C Hermitian operators form a real vector space Theorem 9. The set M := { linear operator on H H with = }, where dim(h H ) <, forms a real vector space with a scalar product, := Tr( ) = Tr(). Proof. Let, M, a, b R. ddition of matrices of course is associative and + = +, the neutral element 0 is hermitian, and the inverse is hermitian if is. Furthermore ( + ) = + = +, thus (M, +) is an abelian group. Of course we also have a (+) = a +a, (a+b) = a +b, a (b) = (ab) and 1 = ; (a ) = a = a. Thus M is a real vector space 5., is linear in both arguments because matrix multiplication and Tr() are linear. ecause of Tr() = Tr(),, is also symmetric. Furthermore Tr() = Tr(() ) = Tr(), thus, : M M R is well-defined. Furthermore, for = j a j a j a j (eigenvalue decomposition) we find, = j a j 2, which means, 0 and = 0 exactly for = 0. Theorem 10. The set T := { M Tr() = 1} is a hyperplane in M given by T = 1 + ˆT, where ˆT := { M Tr() = 0}. Proof. t first we show T = 1 + ˆT : Every T we can decompose into = M + D where M ˆT contains all offdiagonal elements of, and D contains all diagonal elements of. s also Tr(D) = 1, we find Tr(D 1) = 0 and thus = 1 + ( (D 1) + M ) 1 + ˆT. Now we argue that ˆT M is a sub-vectorspace: 0 ˆT,, ˆT + ˆT because Tr( + ) = Tr() + Tr() = 0 and + hermitian too. Furthermore for ˆT, a R: a also hermitian and Tr(a ) = a Tr() = 0. Note that ˆT has one dimension less than M, because the additional condition Tr() = 0 puts exactly one constraint on the elements of. 6 5 For some a C, we find (a ) = a a. Thus M is not a complex vector space. 6 basis for ˆT could be given by E (jk) := (δ ja δ kb +δ ka δ jb ) ab for j k, Ê(jk) := (iδ ja δ kb iδ jb δ ka ) ab for j k and D (j) := (δ ja δ jb δ (j+1),a δ (j+1),b ) ab for j {1, 2,..., dim 1}. Here, E (jk) and Ê(jk) are used to obtain the off-diagonal elements. D (j) are used to obtain the diagonal elements: First we use D (1) to choose the first diagonal element; then we use D (2) to choose the second diagonal element,... The last diagonal element is uniquely determined by the condition Tr( ) = 0, which is fulfilled because all basis elements have vanishing trace. To get a basis for M, we can add (δ a,dim δ b,dim ) ab such that we can also choose the last diagonal element freely. Thus ˆT is indeed a hyperplane in M.

12 References [1] Quantum Computation and Quantum Information, M. Nielsen, I. Chuang, Cambridge University Press, tenth printing [2] lecture notes Physics 219 by John Preskill [3] D. ruß, G. Leuchs Lectures on Quantum Information, Wiley-VCH(2007) [4] V. Vedral, Introduction to Quantum Information Science, Oxford University Press [5] M. Piani, Lecture Notes on Entanglement Theory, IQC Waterloo, 2012 [6] bachelor thesis by Lukas Schneiderbauer, Entanglement or Separability-an introduction [7] P. Krammer Characterizing entanglement with geometric entanglement witnesses, arxiv: 0807.4830v2[quant-ph] [8] Diploma thesis by Philipp Krammer, Quantum Entanglement: Detection, Classification, and Quantification