How many invariant polynomials are needed to decide local unitary equivalence of qubit states?
HHow many invariant polynomials are needed to decide local unitary equivalence ofqubit states?
Tomasz Macia¸ ˙zek,
1, 2
Michał Oszmaniec, and Adam Sawicki
1, 3 Center for Theoretical Physics, Polish Academy of Sciences, Al. Lotników 32/46,02-668 Warszawa, Poland Faculty of Physics, University of Warsaw, ul. Ho˙za 69, 00-681 Warszawa,Poland School of Mathematics, University of Bristol, University Walk, Bristol BS8 1TW,UK
Given L -qubit states with the fixed spectra of reduced one-qubit density matrices, we find aformula for the minimal number of invariant polynomials needed for solving local unitary(LU) equivalence problem, that is, problem of deciding if two states can be connected bylocal unitary operations. Interestingly, this number is not the same for every collection ofthe spectra. Some spectra require less polynomials to solve LU equivalence problem thanothers. The result is obtained using geometric methods, i.e. by calculating the dimensionsof reduced spaces , stemming from the symplectic reduction procedure.1 a r X i v : . [ qu a n t - ph ] M a y . INTRODUCTION We consider a quantum system consisting of L isolated qubits with the Hilbert space H =( C ) ⊗ L . We assume that states are normalized to one and neglect a global phase. In this waythe space of pure states is isomorphic to the complex projective space P ( H ) . In the following,by [ φ ] ∈ P ( H ) we will denote a state corresponding to a vector φ ∈ H . Each qubit is locatedin a different laboratory and the available operations are restricted to the local unitaries describedby the group K = SU (2) × L . Two states are called locally unitary equivalent (LU equivalent)if and only if they can be connected by the action of K , that is, belong to the same K -orbit.The problem of local unitary equivalence of states can be in principle solved by finding the setof K -invariant polynomials, i.e. polynomials that are constant on K -orbits (see for anotherapproach). When the number of qubits is large this, however, becomes hard as the number ofpolynomials grows exponentially with the number of constituents of the system. The problem ofLU equivalence for bipartite and three-qubit pure states was recently studied form the symplecto-geometric perspective (see also ). In particular, the connection with the symplectic reductionwas established. The current paper can be seen as a generalization of these ideas to an arbitrarynumber of qubits.Among the K -invariant polynomials there are L polynomials { tr ( ρ l ([ φ ])) } Ll =1 , where ρ l ([ φ ]) are the reduced one-qubit density matrices. Consequently, for two LU equivalent states [ φ , ] ∈ P ( H ) the spectra of the corresponding reduced one-qubit density matrices are the same. If wedenote by Ψ the map which assigns to [ φ ] ∈ P ( H ) the shifted spectra of its reduced one-qubitdensity matrices, i.e. Ψ([ φ ]) = { diag( − λ , λ ) , . . . , diag( − λ L , λ L ) } , where λ i = − p i and { p i , − p i } is the increasingly ordered spectrum of ρ i ([ φ ]) , then the states satisfying the abovenecessary condition form a fiber of Ψ . Fibers of Ψ are connected collections of K -orbits . More-over, the image, Ψ( P ( H )) , is a convex polytope . The polynomials { tr ( ρ l ([ φ ])) } Ll =1 restrictedto a fiber of Ψ are constant functions. Therefore, typically, for two states [ φ ] and [ φ ] with α := Ψ([ φ ]) = Ψ([ φ ]) , where α denote the collection of spectra of one-qubit density matrices,some additional K -invariant polynomials are needed to decide the LU equivalence. The number ofthese polynomials is given by the dimension of the reduced space M α := Ψ − ( α ) /K (see ).Interestingly, dim M α may not be the same for every α ∈ Ψ( P ( H )) , that is, some collectionsof spectra of reduced one-qubit density matrices require less additional polynomials to solve LUequivalence problem than others. In particular, if the fiber Ψ − ( α ) contains exactly one K -orbit,2.e. dim M α = 0 , no additional information is needed and any two states [ φ , ] ∈ Ψ − ( α ) are LUequivalent.In this paper we find the formula for the dimension of the reduced space M α = Ψ − ( α ) /K ,for any α ∈ Ψ( P ( H )) and for an arbitrary number L of qubits. Our result is obtained in twosteps. First, we consider the points α gen ∈ Ψ( P ( H )) which belong to the interior of the polytope Ψ( P ( H )) . In this case the map Ψ is regular and the calculation is rather straightforward. Thedimension of M α gen does not depend on α gen . Moreover, dim M α gen + L is equal to the cardinalityof the spanning set of K -invariant polynomials. For points α b ∈ Ψ( P ( H )) which belong to theboundary of Ψ( P ( H )) the problem requires more advanced methods and turns out to be moreinteresting. In particular, for a large part of the boundary of Ψ( P ( H )) we have dim M α b = 0 . Wealso observe that for α b ∈ Ψ( P ( H )) corresponding to the { ρ l ([ φ ]) } Ll =1 such that k matrices aremaximally mixed dim M α b = dim M α gen − k . II. HOW MANY INVARIANT POLYNOMIALS ARE NEEDED TO DECIDE LUEQUIVALENCE OF 4-QUBIT STATES?
In this section we briefly discuss the considered problem and present the main results of thepaper on the 4 qubits example.Recently, the problem of finding dim M α was considered for three qubits . In particular it wasshown that for points in the interior of the polytope Ψ( P ( H )) , dim M α = 2 , whereas for points onthe boundary dim M α = 0 . The uniform behaviour of dim M α on the boundary of Ψ( P ( H )) in caseof three qubits is, as already indicated in ref. , a low dimensional phenomenon. As we explain insection III, for an arbitrary number of L qubits the boundary consists of three parts characterizedby a different behaviour of dim M α . The first part is the polytope Ψ( P ( ˜ H )) that corresponds to asystem with one qubit less, that is, ˜ H = ( C ) ⊗ ( L − . The second corresponds to changing one ofthe non-trivial inequalities (6) into an equality. The third represents situations when k one-qubitdensity matrices are maximally mixed. The clear distinction between these parts of the boundarycan be seen already in the four qubits case.The four-qubit polytope Ψ( P ( H )) is a -dimensional convex polytope spanned by 12 vertices(see appendix for the proof and the list of vertcies). The dimension of the reduced space in theinterior of Ψ( P ( H )) is dim M α gen = 14 (see formula (8)). In figure 1 the above mentioned threedifferent parts of the boundary are shown. In particular, in figure 1(a) we see that for -dimensional3ace of Ψ( P ( H )) corresponding to three qubits, dim M α = 2 in the interior and dim M α = 0 onthe boundary which agrees with results of ref. . On the other hand, inside the -dimensional faceshown in figure 1(b) corresponding to one of { ρ i } i =1 being maximally mixed we have dim M α =12 . The boundary of this face contains: -dimensional faces corresponding to two of { ρ i } i =1 being maximally mixed - dim M α = 10 , -dimensional faces - three of { ρ i } i =1 are maximallymixed and dim M α = 8 , and finally, the vertex denoted by v GHZ when all one-particle reduceddensity matrices are maximally mixed - dim M α = 6 . Therefore, as mentioned in the introduction, dim M α b = dim M α gen − k . Finally, in figure 1(c) we see the -dimensional face of Ψ( P ( H )) with dim M α = 0 . (a) (b) (c)
88 81010 10
FIG. 1. Three parts of the boundary of Ψ( P ( H )) . The numbers denote dim M α . If the number is missing, dim M α = 0 . The vertices are defined in appendix. In the next sections we show how to calculate dim M α for any point of Ψ( P ( H )) for an arbitrarynumber of qubits. III. LU EQUIVALENCE OF QUBITS AND THE REDUCED SPACES M α We start with a rigorous statement of the problem and the solution of its easy part. For a detaileddescription of symplecto-geometric methods in quantum information theory see for example .Let H = ( C ) ⊗ L be the L -qubit Hilbert space and denote by P ( H ) the corresponding complexprojective space. It is known that P ( H ) is a symplectic manifold with the Fubiny-Study sym-plectic form ω F S . The action of K = SU (2) × L on P ( H ) is symplectic, i.e. it preserves ω F S .Consequently, there is the momentum map for this action. In the considered setting this map isgiven by µ : P ( H ) → i k ([ φ ]) = (cid:26) ρ ([ φ ]) − I, . . . , ρ L ([ φ ]) − I (cid:27) , where k is the Lie algebra of K , { ρ l ([ φ ]) } Ll =1 are the reduced one-qubit density matrices, I is × identity matrix and i = − . The map µ is equivariant, i.e. for any g ∈ Kµ ([ gφ ]) = (cid:26) ρ ([ gφ ]) − I, . . . , ρ L ([ gφ ]) − I (cid:27) = (1) = (cid:26) g (cid:18) ρ ([ φ ]) − I (cid:19) g ∗ , . . . , g (cid:18) ρ L ([ φ ]) − I (cid:19) g ∗ (cid:27) = gµ ([ φ ]) g ∗ = Ad g µ ([ φ ]) , (2)where g ∗ is the Hermitian conjugate of g and Ad g X := gXg ∗ , for any X ∈ i k . The equivariance of µ implies that K -orbits in P ( H ) are mapped onto adjoint orbits in i k . Consequently, the necessarycondition for two states [ φ , ] ∈ H to be on the same K -orbit in P ( H ) is µ ( K. [ φ ]) = µ ( K. [ φ ]) .On the other hand, adjoint orbits in i k are determined by the spectra of matrices { ρ l ([ φ ]) } Ll =1 . Thecharacteristic polynomial w l ( ν ) for ρ l ([ φ ]) reads w l ( ν ) = ν − tr (cid:18) ρ l ([ φ ]) − I (cid:19) ν + (cid:32)(cid:18) tr (cid:18) ρ l ([ φ ]) − I (cid:19)(cid:19) − tr (cid:18) ρ l ([ φ ]) − I (cid:19) (cid:33) . (3)Using tr ( ρ l ([ φ ])) = 1 , equation (3) reduces to w l ( ν ) = ν − tr (cid:0) ρ l ([ φ ]) (cid:1) − . (4)One can thus say that the necessary condition for two states [ φ , ] ∈ P ( H ) to be on the same K -orbit in P ( H ) is that ∀ l tr (cid:0) ρ l ([ φ ]) (cid:1) = tr (cid:0) ρ l ([ φ ]) (cid:1) . Let i t + be the positive Weyl chamber in i k , i.e. i t + = − λ λ , . . . , − λ L λ L : λ i ∈ R + . (5)We define the map Ψ : P ( H ) → i t + to be Ψ([ φ ]) := µ ( K. [ φ ]) ∩ i t + = (cid:26) ˜ ρ ([ φ ]) − I, . . . , ˜ ρ L ([ φ ]) − I (cid:27) , ˜ ρ l ([ φ ]) is a diagonal × matrix whose diagonal elements are given by the increasinglyordered spectrum σ ( ρ l ([ φ ]) = { p l , − p l } of ρ l ([ φ ]) , that is, ≥ p l ≥ . The shifted spectrum,i.e. the spectrum of ρ l ([ φ ]) − I is given by {− λ l , λ l } , where ≤ λ l = 12 − p l ≤ . Under these assumptions the image Ψ( P ( H )) is known to be a convex polytope, defined by thefollowing set of inequalities ∀ l ≤ λ l ≤ , and (cid:18) − λ l (cid:19) ≤ (cid:88) j (cid:54) = l (cid:18) − λ j (cid:19) . (6)Note that the polytope Ψ( P ( H )) gives 1-1 parametrization of µ ( P ( H )) , that is, of all available ad-joint K -orbits. Moreover, for each point α ∈ Ψ( P ( H )) the set Ψ − ( α ) is a connected K -invariantstratified symplectic space . We will next briefly describe the structure of this space . Notethat isotropy groups of points belonging to a fixed K -orbit are conjugated. First, one can decom-pose P ( H ) according to the conjugacy classes of isotropy groups, i.e. into sets of points, whoseisotropies are conjugated. This decomposition divides P ( H ) into sets of orbits characterized by thesame isotropy type. These sets are in general non-connected, thus performing the decompositionfurther into connected components, one obtains the stratification P ( H ) = ∪ S ν . There exists an unique highest dimensional stratum S max , which is open and dense. Moreover, itcan be shown that for each α ∈ Ψ( P ( H )) the set µ − ( α ) ∩ S ν is smooth and that the decomposition µ − ( α ) = ∪ ν µ − ( α ) ∩ S ν , (7)gives the stratification of µ − ( α ) . Let K α be the isotropy subgroup of α with respect to the adjointaction, i.e. K α = { g ∈ K : Ad g α = α } . Because µ − ( α ) is invariant to the action of K α , one hasthat the reduced space M α = Ψ − ( α ) /K ∼ = µ − ( α ) /K α . The stratification given by (7) inducesthe stratification of M α in a natural way. What is more, every M α also has an unique highestdimensional stratum M maxα . If µ − ( α ) ∩ S max (cid:54) = ∅ then M maxα = ( µ − ( α ) ∩ S max ) /K. Therefore it is clear that the dimension of the highest dimensional stratum in M α , or more precisely dim M α + L is the number of K -invariant polynomials needed to decide LU equivalence of states6atisfying Ψ([ φ ]) = α . The problem of finding dim M α is essentially different for points α in theinterior and on the boundary of Ψ( P ( H )) . Using (6) the boundary of Ψ( P ( H )) can be divided intothree parts Case 1: k of λ l s are equal to . Case 2:
At least one of the inequalities (cid:0) − λ l (cid:1) ≤ (cid:80) j (cid:54) = l (cid:0) − λ j (cid:1) is an equality. Case 3: k of λ l s are equal to .By the permutation symmetry of inequalities (6), it is enough to consider one example for each ofthese cases. In the remaining two paragraphs of this section we find dim M α for the interior andthe boundary points satisfying the condition of the first case. A. Interior of Ψ( P ( H )) For the points α in the interior of the polytope Ψ( P ( H )) the calculation of dim M α turns out tobe rather straightforward.To begin, let us denote by P ( H ) max the union of K -orbits of maximal dimension in P ( H ) . Theprincipal isotropy theorem implies that this set is connected, open and dense. In order to calculatethe maximal dimension of K -orbits in P ( H ) we use the following fact (see ref. ) Proposition 1.
Assume that dimΨ( P ( H )) = dim t + . Then a generic K -orbit has dimension of thegroup K . For L -qubits, L ≥ , one easily checks that the polytope Ψ( P ( H )) given by inequalities (6), is L -dimensional. On the other hand, for the Weyl chamber defined by (5), we have dim t + = L .Therefore, by proposition 1 the K -orbits belonging to P ( H ) max have dimension of the group K = SU (2) × L , i.e L . Moreover, Ψ( P ( H ) max ) contains the interior of the polytope Ψ( P ( H )) (see ref. ) and for α in the interior of the polytope Ψ( P ( H )) the set Ψ − ( α ) ∩ P ( H ) max is thehighest dimensional stratum of Ψ − ( α ) (see ref. ). Hence dim M α = dim(Ψ − ( α ) /K ) = (dim P ( H ) − dimΨ( H )) − dim K == (cid:0)(cid:0) L +1 − (cid:1) − L (cid:1) − L = 2 L +1 − L − . (8)Note that already in the case of three qubits we have dim M α = 2 , that is, one needs , K -invariant polynomials to decide LU equivalence of states whose spectra of reduced density7atrices are the same and belong to the interior of Ψ( P ( H )) . The exponential growth of dim M α ,which for large L is of order L , can be seen as a usual statement that the number of K -invariantpolynomials needed to distinguish between generic K -orbits grows exponentially with the numberof particles. B. Calculation of dim M α for case 1 The Case 1 is once again straightforward. To see this note that when the first k of λ l ’s are equal to inequalities (6) reduce to the inequalities for the polytope of L − k qubits. Moreover, a genericstate belonging to Ψ − ( α ) is of the form φ ⊗ φ where φ is a k -qubit separable state and φ is anarbitrary state of L − k qubits. Therefore, using (8), we get dim M α = (cid:0)(cid:0) L − k +1 − (cid:1) − ( L − k ) (cid:1) − L − k ) = 2 L − k +1 − L − k ) − . In the next two sections we find dim M α for cases 2 and 3. IV. CALCULATION OF dim M α FOR CASE 2
In this section we show that dim M α = 0 for α ’s satisfying assumptions of case 2.Assume that one of the inequalities (cid:0) − λ l (cid:1) ≤ (cid:80) j (cid:54) = l (cid:0) − λ j (cid:1) is an equality, e.g. − λ + L (cid:88) i =2 λ i = 12 L − . (9)Using the fact that states mapped by Ψ onto α are K -orbits through states mapped by µ onto α ,i.e. Ψ − ( α ) /K = µ − ( α ) /K α , one can, if more convenient, use map µ instead of Ψ to calculate dim M α . It turns out that this is the case for the considered α ’s, as we have the following: Proposition 2.
Let φ ∈ H be such that µ ([ φ ]) satisfies (9). Let ξ = [ ξ , . . . , ξ L ] be a vectorperpendicular to the plane given by (9). Then φ is an eigenvector of X = X ⊗ I ⊗ . . . ⊗ I + . . . + I ⊗ I ⊗ . . . ⊗ X L , where X l = diag { ξ l , − ξ l } . In order to prove this, we will use the fact characterizing the image of the differential of themomentum map : 8 roposition 3. The image of dµ | [ φ ] : T [ φ ] M → i k is equal to the annihilator of k [ φ ] , the Lie algebraof the isotropy subgroup K [ φ ] ⊂ K .Proof. (Proposition 2) The Lie algebra of K is a real vector space equipped with the inner productgiven by (cid:104) A | B (cid:105) = − tr( AB ) . The matrices X k = i I ⊗ I ⊗ ... ⊗ σ x ⊗ I ⊗ ... ⊗ I , Y k = i I ⊗ I ⊗ ... ⊗ σ y ⊗ I ⊗ ... ⊗ I , Z k = i I ⊗ I ⊗ ... ⊗ σ z ⊗ I ⊗ ... ⊗ I , where σ x,y,z are Pauli matrices, form an orthogonal basis of k . The matrix representing the collec-tion (cid:8) ρ ([ φ ]) − I, . . . , ρ L ([ φ ]) − I (cid:9) is of the form: ( ρ ([ φ ]) − I ) ⊗ I ⊗ . . . ⊗ I + . . . + I ⊗ . . . ⊗ I ⊗ ( ρ L ([ φ ]) − I ) . Note that dµ | [ φ ] is the map that transports vectors tangent to P ( H ) at the point [ φ ] to the tangentspace to k , T µ ([ φ ]) k ∼ = k . Assume that α := µ ([ φ ]) belongs to Ψ( P ( H )) and that spectra of thecorresponding matrices { ρ l ([ φ ]) } Li =1 are nondegenerate.In this case the Lie algebra k α of the isotropy subgroup K α ⊂ K is given by the diagonalmatrices in k , the set of which we denote by t . On the other hand by the equivariance of themomentum map (1) we have k [ φ ] ⊂ k α = t . In other words, states [ φ ] mapped by µ to α ∈ Ψ( P ( H )) with non-degenerate spectra of { ρ l ([ φ ]) } Li =1 can have isotropy given by at most t . As the off-diagonal matrices in the image of dµ | [ φ ] are orthogonal to t , by proposition 3, in order to find k [ φ ] we need to find matrices in t orthogonal to (annihilated by) diagonal matrices from dµ | [ φ ] . Notethat since Ψ( P ( H )) ⊂ t and dimΨ( P ( H )) = dim t , for α inside Ψ( P ( H )) the diagonal matrices in T µ ([ φ ]) k , for [ φ ] ∈ P ( H ) max , span the space t and hence the isotropy k [ φ ] = 0 . On the other hand,for α satisfying (9) the space T µ ([ φ ]) k ∩ t (cid:54) = t . In order to find matrices in t orthogonal to T µ ([ φ ]) k ∩ t note that any element of t can be written as (cid:80) Lk =1 a k Z k . The inner product of two matrices of thistype reads (cid:104) L (cid:88) k =1 a k Z k | L (cid:88) k = l b l Z l (cid:105) = − L (cid:88) k =1 a k L (cid:88) k = l b l tr ( Z k Z l ) = a · b i.e. is equal to the standard inner product of vectors a = [ a , . . . , a L ] and b = [ b , . . . , b L ] in R L . The vectors a corresponding to the diagonal matrices from the image of dµ | [ φ ] are tangent to9 ( P ( H )) at the point µ ([ φ ]) satisfying (9). Therefore, if ξ = [ ξ , . . . , ξ L ] is a vector perpendicularto the plane given by (9) then the corresponding operator X = X ⊗ I ⊗ . . . ⊗ I + . . . + I ⊗ I ⊗ . . . ⊗ X L , where X l = diag { ξ l , − ξ l } is the element of the Lie algebra of the isotropy group k [ φ ] .Consequently Xφ = λφ , for some λ .The vector v = [ − , , . . . , is perpendicular to the plane given by (9). The correspondingoperator X reads X = X ⊗ I ⊗ . . . ⊗ I + . . . + I ⊗ . . . ⊗ I ⊗ X L , (10)where X = diag {− , } , X = . . . = X L = diag { , − } . By proposition 2 we need to considereigenspaces of X . We have the following: Proposition 4.
The matrix X , defined by (10), is a diagonal L × L matrix. The eigenvalues of X are the integers chosen from − L to L with the step , that is σ ( X ) = {− L, − L + 2 , . . . , L − , L } .The multiplicity of eigenspace H − L +2 k is dim H − L +2 k = (cid:0) Lk (cid:1) .Proof. The matrices X l are diagonal and their spectra are σ ( X l ) = {− , } . Consequently, thematrix X is also diagonal and its eigenvalues are sums of eigenvalues of X l ’s. One can easilyverify that the eigenvalues of X belong to the set σ ( X ) = {− L, − L + 2 , . . . , L − , L } . To seethis note that the eigenvalue − L + 2 k arises as a sum of eigenvalues of { X l } Ll =1 , where k out of L eigenvalues of X l ’s are positive ( +1 ) and L − k negative ( − ). Therefore the multiplicity of H − L +2 k is dim H − L +2 k = (cid:0) Lk (cid:1) .As a direct consequence of proposition 4 we need to consider L + 1 eigenspaces of X . In thefollowing we describe the structure of these spaces and show that only one of them, that is, H − L +2 contains states [ φ ] for which µ ([ φ ]) consists of diagonal matrices whose diagonal elements satisfy(9). The result is obtained in two steps. First in proposition 5 we determine H − L +2 k and show thatcondition (9) is not satisfied for k ∈ { , . . . , L } \ { } . Then in proposition 6 we prove that for H − L +2 condition (9) is satisfied.Denote by D Lk the subspace of ( C ) ⊗ L spanned by separable states of L qubits such that k outof L qubits are in the ground state | (cid:105) and the remaining L − k qubits are in the excited state | (cid:105) ,for example, D = Span C {| (cid:105) , | (cid:105) , | (cid:105)} . Assume that D Lk = { } if k > L . Proposition 5.
States which belong to eigenspace H − L +2 k are of the form φ = p | (cid:105) ⊗ ψ + p | (cid:105) ⊗ ψ , where ψ ∈ D L − k and ψ ∈ D L − k − . The reduced one-qubit density matrices for any ∈ H − L +2 k are diagonal. For any φ ∈ H − L +2 k condition (9) is equivalent to ( L − k − || φ || = L − .Proof. We first determine vectors spanning eigenspace H − L +2 k of X . The eigenvalue − L + 2 k arises as the sum of eigenvalues of { X l } Ll =1 with k out of L eigenvalues equal to (+1) and L − k equal to ( − . Note, however, that matrix X = diag {− , } , whereas X = . . . = X L =diag { , − } . Therefore, the eigenvalue − L + 2 k corresponds to separable states with either thefirst qubit in the ground state | (cid:105) and the remaining L − qubits in a state from D L − k or with thefirst qubit in the excited state | (cid:105) and the remaining L − qubits in a state from D L − k − . Thus thegeneric state belonging to H − L +2 k can be written as φ = ( L − k ) (cid:88) l =1 a l | (cid:105) ⊗ e l + ( L − k − ) (cid:88) l =1 b l | (cid:105) ⊗ f l , where { e i } and { f i } are the separable states spanning D L − k and D L − k − respectively. It is straight-forward to see that the reduced one-qubit density matrices of φ are diagonal. The first one is of theform: ρ ([ φ ]) − I = (cid:80) ( L − k ) l =1 | a l | − (cid:80) ( L − k − ) l =1 | b l | − , and hence λ = ( L − k − ) (cid:88) l =1 | b l | − . (11)We now show that (9) is equivalent to ( L − k − || φ || = L − . As matrices ρ l ([ φ ]) are diagonal,the ( ρ l ([ φ ])) entry of each ρ l , which is equal to λ l + , is the sum of | b j | and | a j | coefficientscorresponding to vectors with the l -th qubit in the excited state | (cid:105) . Consequently, in the sum (cid:80) Ll =2 ( ρ l ([ φ ])) , each | b j | coefficient occurs L − k times and each | a j | coefficient L − k − times. Therefore, L (cid:88) l =2 λ l = ( L − k −
1) ( L − k ) (cid:88) i =1 | a i | + ( L − k ) ( L − k − ) (cid:88) i =1 | b i | −
12 ( L −
1) = (12) = ( L − k ) || φ || − ( L − k ) (cid:88) i =1 | a i | −
12 ( L − . − λ + L (cid:88) i =2 λ i = ( L − k − || φ || − L + 1 . Hence equation (9) reads ( L − k − || φ || = L − . Using fact 5 one easily finds that for normalized state, i.e. when || φ || = 1 condition (9) can besatisfied only when k = 1 . The following proposition ensures that indeed this is the case. Proposition 6.
The reduced one-qubit density matrices of states φ ∈ H − L +2 are diagonal andsatisfy condition (9).Proof. The eigenspace H − L +2 is L -dimensional. Any vector φ ∈ H − L +2 can be written as φ = c | (cid:105) ⊗ | . . . (cid:105) + c | (cid:105) ⊗ | . . . (cid:105) + c | (cid:105) ⊗ | . . . (cid:105) + . . . + c L | (cid:105) ⊗ | . . . (cid:105) , (13)that is, φ is a linear combination of separable state where all qubits are in the excited state and thestates for which the first and one additional qubits are in the ground state (while other are in theexcited state). Assume that φ is normalized, i.e. (cid:80) Lk =1 | c i | = 1 . It is straightforward to calculate ρ l ([ φ ]) − I = | c l | − − + (cid:80) Lk (cid:54) = l | c k | , i ∈ { , . . . , L } ,ρ ([ φ ]) − I = − + (cid:80) Lk =2 | c k | | c | − . Note that we can assume that | c | ≥ (cid:80) Lk =2 | c k | . This means that for all i ∈ { , . . . , L } wehave (cid:0) ρ l ([ φ ]) − I (cid:1) ≤ as required. It is also easy to see that condition (9) is equivalent to L (cid:88) k =2 | c k | = L (cid:88) k =2 | c k | , and is satisfied. 12y proposition 6, states mapped by Ψ onto α ’s satisfying (9) belong to K -orbits through φ ∈H − L +2 . On the other hand states φ ∈ H − L +2 are K C -equivalent to L -qubit W state, where K C = SL (2 , C ) × L is complexification of K = SU (2) × L and [ W ] = | . . . (cid:105) + | . . . (cid:105) + . . . + | . . . (cid:105) . This can be easily seen by changing | (cid:105) ↔ | (cid:105) on the first qubit of (13). It was shown in ref. thatthe variety K C . [ W ] is spherical , i.e. reduced spaces stemming from the restriction Ψ | K C . [ W ] arezero-dimensional. Therefore: Theorem 7.
Let α ∈ Ψ( P ( H )) be such that at least one of the inequalities (cid:0) − λ l (cid:1) ≤ (cid:80) j (cid:54) = l (cid:0) − λ j (cid:1) is equality. Then dim M α = 0 . V. CALCULATION OF dim M α FOR CASE 3
As we showed in section III, for points α in the interior of Ψ( P ( H )) the dimension of thereduced space is dim M α gen = 2 L +1 − L − . In the following we show that for α = ( α , . . . , α L ) ∈ Ψ( P ( H )) with matrices α = . . . = α k = 0dim M α = (cid:0) L +1 − L − (cid:1) − k = dim M α gen − k. (14)This means that the dimension dim M α drops by every time one of the reduced density matricesbecomes maximally mixed. The argument for this is based on the existence of stable states whichwe discuss first. A. Stable states
In the following we assume that ˜ K is any compact semisimple group acting in the symplecticway on the complex projective space P ( H ) , where H can be, for example, the Hilbert space of L qubits. We will denote by ˜ µ the corresponding momentum map and assume that ˜ µ − (0) (cid:54) = ∅ . Let ˜ G = ˜ K C be the complexification of ˜ K . Following ref. we denote X (˜ µ ) = { [ φ ] ∈ P ( H ) : ˜ G. [ φ ] ∩ ˜ µ − (0) (cid:54) = ∅} .
13t is known that the set X (˜ µ ) is an open dense subset of P ( H ) . Moreover the set G. ˜ µ − (0) ⊂ X (˜ µ ) is also an open dense subset of P ( H ) (see ref. ). We will use the following terminology, typicalfor geometric invariant theory:1. [ φ ] is unstable iff [ φ ] / ∈ X (˜ µ ) ,2. [ φ ] is semistable iff [ φ ] ∈ X (˜ µ ) .Among semistable states we distinguish the class of stable states. By definition, a semistable state [ φ ] is stable if and only if ˜ µ ([ φ ]) = 0 and dim ˜ K. [ φ ] = dim ˜ K . Note that since for [ φ ] ∈ ˜ µ − (0) one has dim ˜ K C . [ φ ] = 2dim ˜ K. [ φ ] (see ref. ), the condition dim ˜ K. [ φ ] = dim ˜ K can be phrased as dim ˜ G. [ φ ] = dim ˜ G . Remarkably, the existence of a stable state implies that almost all semistablestates are stable, in particular almost all states in ˜ µ − (0) are stable . Note that since ˜ G. ˜ µ − (0) isopen and dense in P ( H ) and a generic ˜ G -orbit in ˜ G. ˜ µ − (0) has dimension dim ˜ G we get dim P ( H ) = dim ˜ G. ˜ µ − (0) = dim ˜ G + dim (cid:16) ˜ G. ˜ µ − (0) (cid:17) / ˜ G. (15)One of the central results in the geometric invariant theory reads (cid:16) ˜ G. ˜ µ − (0) (cid:17) / ˜ G = ˜ µ − (0) / ˜ K. (16)Hence, under the assumption of stable states existence and using (15) and (16) we get ˜ µ − (0) / ˜ K = dim P ( H ) − K. (17)As we will see formula (17) plays a major role in showing (14). B. The strategy for showing (14)
Let K = K × K , where K = SU (2) × k and K = SU (2) × ( L − k ) . We first consider thenatural action of K on the first k qubits in P ( H ) , where H = ( C ) ⊗ L . The momentum map µ forthis action gives the first k reduced density matrices. Therefore, µ − (0) consists of all states withthe first k reduced density matrices maximally mixed, but no assumption is made on the spectraof the remaining ( L − k ) matrices. In the following we assume that there exists a stable state for14 -action on P ( H ) (see lemma 8 for proof). Under this assumption and using formula (17) thedimension of dim µ − (0) /K is dim µ − (0) /K = dim P ( H ) − K = 2 L +1 − k − . Recall that µ − (0) /K is a stratified symplectic space and we consider the highest dimensionalstratum which is a symplectic manifold. Removing K freedom does not affect K action, i.e. theactions of K and K commute. Therefore, we can consider action of K on the highest dimen-sional stratum of µ − (0) /K . The momentum map µ for K action on µ − (0) /K gives the re-maining L − k reduced density matrices. Moreover, using inequalities (6) with λ = . . . = λ k = 0 it is straightforward to see that the image of the corresponding map Ψ is L − k dimensional poly-tope. Using fact 1 and formula (8), for a point inside of this polytope, e.g. when λ k +1 , . . . , λ L (cid:54) = 0 ,the dimension of Ψ -fiber is (cid:0)(cid:0) dim µ − (0) /K (cid:1) − ( L − k ) (cid:1) − dim K = (cid:0)(cid:0) L +1 − k − (cid:1) − ( L − k ) (cid:1) − L − k )= 2 L +1 − L − k − . But the Ψ -fiber is exactly the reduced space we look for, i.e. the one which corresponds to λ = . . . = λ k = 0 and λ k +1 , . . . , λ L (cid:54) = 0 . Therefore, as promised dim M α = dim M α gen − k = 2 L +1 − L − k − . In order to complete the above reasoning we now show that an appropriate stable state indeedexists.
Lemma 8.
Let K = SU (2) × k , k ≤ L . If L ≥ then the L -qubit state [ φ ] = ( | . . . (cid:105) + | . . . (cid:105) ) + ( | . . . (cid:105) + | . . . (cid:105) ) + ( | . . . (cid:105) + | . . . (cid:105) ) ++ . . . + ( | . . . (cid:105) + | . . . (cid:105) ) , is K -stable. For L = 4[ φ ] = α ( | (cid:105) + | (cid:105) ) + ( | (cid:105) + | (cid:105) ) + ( | (cid:105) + | (cid:105) ) ++ ( | (cid:105) + | (cid:105) ) , α ∈ R \ { , − } , is K -stable. roof. We need to show that the first k reduced density matrices of [ φ ] are maximally mixed andthat dim K . [ φ ] = dim K . Note first that if the state [ φ ] is stable with respect to K = SU (2) × L action it is also stable with respect to K ⊂ K action. Therefore, we will show that [ φ ] is K -stable.The state [ φ ] consists of L ≥ pairs of separable states. In each pair the second vector is thefirst vector with the swap | (cid:105) ↔ | (cid:105) performed on every qubit. The first pair is GHZ state. In theremaining pairs the first vector is such that the first and one additional qubits are in the excitedstate | (cid:105) while the remaining L − qubits are in the ground state | (cid:105) . The construction ensures that µ ([ φ ]) = 0 . What is left is to calculate the dimension dim G. [ φ ] , where G = K C = SL (2 , C ) × L .This is equivalent to calculating the dimension of the tangent space T [ φ ] G. [ φ ] which is generatedby the action of Lie algebra g = sl (2 , C ) × L on [ φ ] . More precisely let E = , E = , H = − , be the basis of sl (2 , C ) . Let E ( l )12 = I ⊗ I ⊗ . . . ⊗ E ⊗ I ⊗ . . . ⊗ I ,E ( l )21 = I ⊗ I ⊗ . . . ⊗ E ⊗ I ⊗ . . . ⊗ I ,H ( l ) = I ⊗ I ⊗ . . . ⊗ H ⊗ I ⊗ . . . ⊗ I . Our goal is to show that vectors { E ( l )12 φ, E ( l )21 φ, H ( l ) φ } Ll =1 are linearly independent and orthogonalto φ . Denote by | ... (cid:105) ⊗ | (cid:105) l a separable state whose l -th qubit is in the excited state | (cid:105) andremaining qubits are in the ground state | (cid:105) (e.g. a 5 - qubit state | . . . (cid:105) ⊗ | (cid:105) = | (cid:105) anda 6 - qubit state | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) = | (cid:105) ). It is straightforward to verify that E (1)21 φ = | . . . (cid:105) ⊗ | (cid:105) + L (cid:88) l (cid:48) =2 | . . . (cid:105) ⊗ | (cid:105) l (cid:48) ,E ( l )21 φ = | . . . (cid:105) ⊗ | (cid:105) l + | . . . (cid:105) ⊗ | (cid:105) + (cid:88) l (cid:48) (cid:54) = l | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) , l ≥ . Moreover, the action of E gives vectors that can be obtained from the above set of vectors bypreforming the swap operation on every qubit. The vectors obtained by the action of E arelinearly independent, because each vector from this group contains a unique separable state of theform | ... (cid:105) ⊗ | (cid:105) l (an analogous argument can be applied to the set of vectors obtained by theaction of E ). We next examine the linear independence of vectors from groups E ( l )12 and E ( l )21 .16irst, note that E ( l )12 φ, l ≥ are linearly independent from all vectors E ( l )21 φ , because they consistof separable states of the form | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) that do not appear in the vectors E ( l )21 φ (it is not true in the 4 - qubit case, but we will return to this problem later). The last thing to showis the linear independence of the vector E (1)12 φ = | . . . (cid:105) ⊗ | (cid:105) + (cid:80) Ll (cid:48) =2 | . . . (cid:105) ⊗ | (cid:105) l (cid:48) fromthe vectors E ( l )21 φ, l ≥ . Note that this vector is orthogonal to E (1)21 φ . Assume that E (1)12 φ can beexpressed as the linear combination of the remaining vectors E (1)12 φ = L (cid:88) l =2 λ l E ( l )21 φ. (18)We will show that this leads to a contradiction. To this end, let us calculate the sum L (cid:88) l =2 E ( l )21 φ = L (cid:88) l =2 | . . . (cid:105) ⊗ | (cid:105) l + ( L − | . . . (cid:105) ⊗ | (cid:105) + (cid:88) l (cid:88) l (cid:48) (cid:54) = l | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) == E (1)12 φ + ( L − | . . . (cid:105) ⊗ | (cid:105) + (cid:88) l (cid:88) l (cid:48) (cid:54) = l | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) . Using (18) we get L (cid:88) l =2 (1 − λ l ) E ( l )21 φ = ( L − | . . . (cid:105) ⊗ | (cid:105) + (cid:88) l (cid:88) l (cid:48) (cid:54) = l | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) (19)On the other hand, L (cid:88) l =2 (1 − λ l ) E ( l )21 φ = L (cid:88) l =2 (1 − λ l ) | . . . (cid:105) ⊗ | (cid:105) l + L (cid:88) l =2 (1 − λ l ) | . . . (cid:105) ⊗ | (cid:105) ++ (cid:88) l (1 − λ l ) (cid:88) l (cid:48) (cid:54) = l | . . . (cid:105) ⊗ | (cid:105) ⊗ | (cid:105) l ⊗ | (cid:105) l (cid:48) . (20)Thus, comparing the coefficients by | . . . (cid:105) ⊗ | (cid:105) and | . . . (cid:105) ⊗ | (cid:105) l in equations (19) and(20), we get L (cid:88) l =2 (1 − λ l ) = L − , ∀ l − λ l = 0 (21)This is a contradiction, because the first equation (21) is equivalent to (cid:80) Ll =2 λ l = 1 , which cannotbe satistied by λ l = 1 for all l (which is implied by the second equation (21)). Clearly for any l ∈ { , . . . , L } we also have (cid:104) E ( l )12 | φ (cid:105) = 0 = (cid:104) E ( l )21 φ | φ (cid:105) . We are left with vectors H ( l ) φ . It17s straightforward to see that (cid:104) H ( l ) φ | φ (cid:105) = (cid:104) H ( l ) φ | E ( l )21 φ (cid:105) = (cid:104) H ( l ) φ | E ( l )12 φ (cid:105) = 0 . The matrix ofcoefficients for { H ( l ) φ } is given by C (cid:48) = . . . .
11 1 − . . . − − − ... − − − − ... − . . .. . . − − . . − in the basis { ( | ... (cid:105)−| ... (cid:105) ) , ( −| ... (cid:105) + | ... (cid:105) ) , ( −| ... (cid:105) + | ... (cid:105) ) , . . . , ( −| ... (cid:105) + | ... (cid:105) ) } . By direct calculation one checks that det C (cid:48) (cid:54) = 0 . Therefore the dimension of G. [ φ ] is equal to the dimension of G , which is L and φ is K -stable. In the case of 4 qubits we need toconsider a slightly different state [ φ ] = α ( | (cid:105) + | (cid:105) ) + ( | (cid:105) + | (cid:105) ) + ( | (cid:105) + | (cid:105) ) ++ ( | (cid:105) + | (cid:105) ) , α ∈ R \ { , − } . It can be shown by similar calculation that this state is K - stable. VI. SUMMARY
Given spectra of one-qubit reduced density matrices, we found the formula for the minimalnumber of polynomials needed to decide LU equivalence of L -qubit pure states. As we showedthis number is the same for spectra belonging to the interior of the polytope Ψ( P ( H )) . This isnot the case on the boundary where the behaviour of dim M α is not uniform. In particular, for alarge part of the boundary of Ψ( P ( H )) we have dim M α b = 0 . We also observed that for α b ∈ Ψ( P ( H )) corresponding to the { ρ l ([ φ ]) } Ll =1 such that k matrices are maximally mixed dim M α b =dim M α gen − k .The methods used in this paper can be in principle applied to L -particle systems with an ar-bitrary finite-dimensional one-particle Hilbert spaces. The argument for points in the interior of Ψ( P ( H )) can be used mutatis mutandis in this case. We note, however, that inequalities describing18he polytope Ψ( P ( H )) are much more complicated when H (cid:54) = ( C ) ⊗ L (see ref. ) and thereforethe problem for boundary points is of higher computational complexity. Nevertheless, one shouldexpect that, similarly to the qubit case, there is a large part of the boundary characterized by dim M α = 0 . ACKNOWLEDGMENTS
We would like to thank Marek Ku´s for many discussions on the geometric aspects of quantumcorrelations and his continuous encouragement. This work is supported by Polish Ministry ofScience and Higher Education Iuventus Plus grant no. IP2011048471.
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Appendix: Vertices of the polytope Ψ( P ( H )) for L qubits In order to find vertices of the polytope Ψ( P ( H )) , it is more convenient to view the inequalitiesdescribing the polytope in terms of the minimal eigenvalues of the reduced one-qubit densitymatrices (remembering that the shifted spectra are given by λ i = − p i ). As proven in theseinequalities read: p i ≤ (cid:88) j (cid:54) = i p j , i = 1 , . . . , L , (A.1a) p i ≥ , i = 1 , . . . , L , (A.1b) p i ≤ , i = 1 , . . . , L , (A.1c)where each p i denotes the minimum eigenvalue of the reduced one-qubit density matrix describing i -th qubit, ρ i . The general algorithm of finding the vertices of the polytope given by a set ofinequalities is to choose the set of L of the inequalities, write them as a set of equations and checkwhether there exists a unique solution . If obtained solution satisfies the remaining L inequalitiesthen it defines a vertex of Ψ( P ( H )) . The following fact describes the structure of vertices of Ψ( P ( H )) roposition 9. Vertices of the polytope Ψ( P ( H )) in the case of L qubits are given by equations p i = 0 or p i = 12 , i = 1 , . . . , L (A.2) such that the number of indecies i for which p i = belongs to { , , , ..., L } , that is, |{ i : p i = }| ∈ { , , , ..., L } .Proof. First note that each vertex of this form can be obtained by picking |{ i : p i = }| equationsfrom (A.1c) and L − |{ i : p i = }| linearly independent equations from (A.1b). One easilychecks that in each case L remaining inequalities are trivially satisfied. Thus, points described byconditions (A.2) are indeed vertices of Ψ( P ( H )) . A single exceptional case occurs when |{ i : p i = }| = 1 , e.g. p = . Then by (A.1a) we get p = ≤ , which is a contradiction. We will nowshow that these are all solutions. We achieve our goal by considering all remaining possibilities ofchoosing L out of L inequalities. Before we proceed let us introduce some useful notation. Anychoice of L out of L inequalities (and turning them into equalities) is uniquely given when thefollowing three auxiliary sets are specified: I = { i : i th inequality of the form (A.1a) was chosen } ,I = { i : i th inequality of the form (A.1b) was chosen } ,I = { i : i th inequality of the form (A.1c) was chosen } . We have | I | + | I | + | I | = L , where | I | denotes the number of elements of the finite set I . Wehave already covered cases when I = ∅ . What remains to be checked are three possibilities: I = ∅ or I = ∅ and the case when each I i is non-empty. Firstly, let us consider the case when I = ∅ . If additionally I = ∅ , one can check by direct calculations that the set of L equationsfrom I gives p i = 0 for all i . Next, if the set I is non-empty, i.e. we choose some number of p i ’s equal to zero and combine these conditions with the equations from I , our problem eitherreduces to a problem analogous to the previous case with I = ∅ , or there exists such i ∈ I thatreads p i = (cid:80) j (cid:54) = i p j with p i = 0 . Because all p i ’s are positive or equal to zero, we obtain in bothcases that p j = 0 for all j . One of the last things to check is the case when the only empty set is I , i.e. I = ∅ , I (cid:54) = ∅ and I (cid:54) = ∅ . If, in addition | I | = k > , there exists such i ∈ I that reads p i = (cid:80) j (cid:54) = i p j = (cid:80) j / ∈ I p j + k . This implies that p i > , which is a contradiction. Now, if | I | = 1 from the same equation we get that for all j / ∈ I p j = 0 and p i = . This is a contradiction,because we assumed that | I | = 1 . This argument remains also true in the case when all sets arenon-empty. 21y the above fact, the number of the vertices of the polytope Ψ( P ( H )) for L qubits is thenumber of ways to place k out of L p i ’s equal to and the remaining p i s equal to zero on L places: V = (cid:18) L (cid:19) + (cid:18) L (cid:19) + ... + (cid:18) LL (cid:19) = N (cid:88) k =0 (cid:18) Lk (cid:19) − (cid:18) L (cid:19) = 2 L − L. Moreover, to find the L − dimensional faces of the polytope, one has to change one of theinequalities (A.1a), (A.1b), (A.1c) into equality and find vertices that satisfy this condition (theminimal number of vertices sufficient to span such face is L ). In this way, for L ≥ qubits, oneobtains that there are L such faces.
1. The 4 qubits example
The four-qubit polytope is a 4-dimensional convex polytope spanned by 12 vertices (in termsof λ ’s): |{ i : p i = }| Vertices in terms of λ i = − p i v SEP = { diag ( − , ) , diag ( − , ) , diag ( − , ) , diag ( − , ) } v B1 = { diag (0 , , diag (0 , , diag ( − , ) , diag ( − , ) } v B2 = { diag (0 , , diag ( − , ) , diag (0 , , diag ( − , ) } v B3 = { diag (0 , , diag ( − , ) , diag ( − , ) , diag (0 , } v B4 = { diag ( − , ) , diag (0 , , diag (0 , , diag ( − , ) } v B5 = { diag ( − , ) , diag (0 , , diag ( − , ) , diag (0 , } v B6 = { diag ( − , ) , diag ( − , ) , diag (0 , , diag (0 , } v = { diag (0 , , diag (0 , , diag (0 , , diag ( − , ) } v = { diag (0 , , diag (0 , , diag ( − , ) , diag (0 , } v = { diag (0 , , diag ( − , ) , diag (0 , , diag (0 , } v = { diag ( − , ) , diag (0 , , diag (0 , , diag (0 , } v GHZ = { diag (0 , , diag (0 , , diag (0 , , diag (0 , } TABLE I. The vertices of the four-qubit polytope Ψ( P ( H )) . The 3-dimensional faces can be divided into three groups obtained by: (1) changing one in-equality from (A.1a) into equality: the face is spanned by v SEP and three v B j vertices, (2) choosing22ne λ i = 0 : the face is spanned by three v B j vertices, three v j vertices and v GHZ , (3) choosing one λ i = : the face is spanned by v SEP , three v B j vertices and one v jj