How many orthonormal bases are needed to distinguish all pure quantum states?
Claudio Carmeli, Teiko Heinosaari, Jussi Schultz, Alessandro Toigo
HHOW MANY ORTHONORMAL BASES ARE NEEDEDTO DISTINGUISH ALL PURE QUANTUM STATES?
CLAUDIO CARMELI, TEIKO HEINOSAARI, JUSSI SCHULTZ,AND ALESSANDRO TOIGO
Abstract.
We collect some recent results that together providean almost complete answer to the question stated in the title. Forthe dimension d = 2 the answer is three. For the dimensions d = 3and d ≥ d = 4 the answeris either three or four. Curiously, the exact number in d = 4 seemsto be an open problem. Introduction
How many different measurement settings are needed in order touniquely determine a pure quantum state, and how should such mea-surements be chosen? This problem goes back to a famous remark byW. Pauli [1], in which he raised the question whether or not the po-sition and the momentum distributions are enough to define the wavefunction uniquely modulo a global phase. The original
Pauli problem has a negative answer [2], but it has evolved into many interesting vari-ants and has been studied from several fruitful perspectives. Discussionof the vast literature lies outside the scope of this work. Instead, wewill concentrate only on a specific form of the Pauli problem, which isconcerned with the minimal number of orthonormal bases, or projec-tive measurements, in a finite dimensional Hilbert space that is neededin order to distinguish all pure quantum states. We require that all pure states are determined, so schemes that allow the determination ofmerely almost all pure states are outside of the scope of this work, evenif they are interesting and important from the practical point of view.The purpose of this paper is to present the essential results related toour question in a comprehensible way.It is quite obvious that a single orthonormal basis cannot distinguishall pure states in a d -dimensional Hilbert space, while it is known thatwith d + 1 bases it is possible to distinguish all states, pure or mixed.The problem of finding the minimal number of orthonormal bases de-termining an unknown pure state has been raised several times in the a r X i v : . [ qu a n t - ph ] A p r past. It is easy to verify that the minimal number is three in dimen-sion 2, but in higher dimensions the problem becomes more difficult. In1978 A. Vogt reported on R. Wright’s conjecture that three orthonor-mal bases are sufficient to identify an unknown pure state in any finitedimension [3]. In 1983 B.Z. Moroz made the same claim and presenteda proof for it [4], but in the erratum he explains that his proof doesnot work for all pure states and credits M. Gromov for pointing outan argument that shows that at least four bases are needed in largedimensions [5]. In 1994 this argument was spelled out in greater detailby Moroz and A.M. Perellomov [6], and they concluded that at leastfour bases are needed for any dimension d ≥
9. We will see that thisconclusion can be extended to all dimensions d ≥ d . Then in Sec. 4 we reviewthe rank criterion first found in [7] and explain how this implies thatthree bases are not enough in dimension 3. In Sec. 5 we give a thoroughpresentation of Jaming’s construction of four bases. We continue inSec. 6 with the most technical part of this paper, which reviews theargument presented in [7] that implies the impossibility of three basesin dimensions d ≥ • three for d = 2 • four for d = 3 and all dimensions d ≥ • either three or four in d = 4.Curiously, the final answer in d = 4 still remains open. In Sec. 7we rule out specific types of triples of bases, namely, those consistingsolely of product vectors with respect to a splitting of the Hilbert spaceinto a tensor product of 2-dimensional spaces. Finally, in Sec. 8 wetreat spin-1 measurements to highlight the fact that even if four basescan distinguish all pure quantum states, these bases must be chosenappropriately and it may happen that some natural choices are not thebest ones. We end this paper with a brief discussion of the problem inan infinite dimensional Hilbert space in Sec. 9. Formulation of the question
Let H be a finite d -dimensional Hilbert space. The quantum statesare described by density matrices, i.e., positive operators (cid:37) on H thatsatisfy tr [ (cid:37) ] = 1. A quantum state (cid:37) is pure if it cannot be writtenas a mixture (cid:37) = (cid:37) + (cid:37) of two different states (cid:37) and (cid:37) . Purequantum states correspond to 1-dimensional projections. They canbe alternatively described as rays of vectors in H , meaning that twovectors ψ and ψ correspond to the same pure state if there is a nonzerocomplex number c such that ψ = cψ . For a unit vector ψ ∈ H , thecorresponding density matrix is (cid:37) = | ψ (cid:105)(cid:104) ψ | .Let { ϕ j } dj =1 be an orthonormal basis of H . (From now on, the term basis means an orthonormal basis.) If we perform a measurement ofthis basis in a state (cid:37) , then we get an outcome j with the probability (cid:104) ϕ j | (cid:37)ϕ j (cid:105) . The probability distribution p ( j ) = (cid:104) ϕ j | (cid:37)ϕ j (cid:105) encodes theinformation that this measurement gives us about the unknown state (cid:37) . It is quite clear that this information is not enough to determine theinput state uniquely. For instance, the pure states corresponding tothe unit vectors 1 / √ ϕ ± ϕ ) are different but they lead to the sameprobability distribution. If our aim is to identify an unknown quantumstate uniquely, we should thus measure more than one orthonormalbasis.Let B = { ϕ j } dj =1 , . . . , B m = { ϕ mj } dj =1 be m orthonormal bases of H .We say that the bases B , . . . , B m distinguish two different states (cid:37) and (cid:37) if (cid:10) ϕ (cid:96)j | (cid:37) ϕ (cid:96)j (cid:11) (cid:54) = (cid:10) ϕ (cid:96)j | (cid:37) ϕ (cid:96)j (cid:11) (1)for some (cid:96) = 1 , . . . , m and j = 1 , . . . , d . This means that if we getthe probability distributions related to all m bases in the states (cid:37) and (cid:37) , then the measurement data corresponding to these two states aredifferent.We say that the orthonormal bases B , . . . , B m distinguish all purestates if they distinguish any pair of different pure states. Following[10], we also say that in this case the bases B , . . . , B m are information-ally complete with respect to pure states .We recall that it is possible to find finite collections of orthonormalbases that can distinguish all pure states. Namely, it is known thatthere exist d + 1 orthonormal bases which distinguish all states, pureor mixed; see e.g. [11] for a construction. Less than d + 1 bases cannotdistinguish all states simply because not enough parameters are deter-mined. However, the pure states form a non-convex subset of all states,and one cannot thus rule out the possibility that less than d + 1 bases can distinguish all pure states. This leads to our main question statedin the title: How many orthonormal bases are needed in order to distinguish allpure states?
One may be tempted to criticize this question on the grounds thatin any real experiment the states are perhaps never completely pure.But as any problem of this type, also this should be considered as aquestion on the fundamental limits of quantum theory. As such, webelieve that it reveals an interesting and even surprising aspect of theduality of states and measurements.3.
Qubit and insufficiency of two bases
As a warm up, let us consider the case of a qubit, i.e., d = 2. Itis well known and explained also in almost any textbook that one canchoose three orthonormal bases such that the related measurement out-come distributions determine an unknown qubit state uniquely. This israther obvious if one looks at the Bloch representation (cid:37) (cid:126)r = ( + (cid:126)r · (cid:126)σ )of qubit states, where (cid:126)r is a vector in R satisfying (cid:107) (cid:126)r (cid:107) ≤ (cid:126)σ = ( σ x , σ y , σ z ) consists of Pauli matrices. Since (cid:126)r = (tr [ (cid:37)σ x ] , tr [ (cid:37)σ y ] , tr [ (cid:37)σ z ]) , we conclude that measurements of the eigenbases of σ x , σ y and σ z specify the vector (cid:126)r and hence also the state (cid:37) (cid:126)r . More generally, if wefix three linearly independent unit vectors (cid:126)a , (cid:126)b and (cid:126)c in R , then themeasurements of the eigenbases of (cid:126)a · (cid:126)σ , (cid:126)b · (cid:126)σ and (cid:126)c · (cid:126)σ distinguish allstates.A qubit state (cid:37) (cid:126)r is pure exactly when (cid:107) (cid:126)r (cid:107) = 1. The direction of aunit vector (cid:126)r depends on two parameters only, so one may wonder iftwo orthonormal bases can suffice to determine any pure qubit state.This is not true, in fact, in any dimension: Proposition 1.
In any dimension d ≥ , two orthonormal bases can-not distinguish all pure states.Proof. Our proof of this statement has been motivated by Theorem 1in [12]. Let B = { ϕ j } dj =1 and B = { φ k } dk =1 be two orthonormal basesof a d -dimensional Hilbert space. We need to find two nonparallel unitvectors ψ + and ψ − such that |(cid:104) ξ | ψ + (cid:105)| = |(cid:104) ξ | ψ − (cid:105)| (2)for all vectors ξ ∈ B ∪ B . Let η ∈ H be a unit vector orthogonal to ϕ . We set ψ ± = √ ( ϕ ± η ).Then |(cid:104) ψ ± | ξ (cid:105)| = (cid:0) |(cid:104) ϕ | ξ (cid:105)| + |(cid:104) η | ξ (cid:105)| (cid:1) ± Re ( (cid:104) ϕ | ξ (cid:105) (cid:104) ξ | η (cid:105) ) , so that |(cid:104) ψ + | ξ (cid:105)| − |(cid:104) ψ − | ξ (cid:105)| = 2 Re ( (cid:104) ϕ | ξ (cid:105) (cid:104) ξ | η (cid:105) ) . Since (cid:104) ϕ | ϕ j (cid:105) (cid:104) ϕ j | η (cid:105) = 0 for all ϕ j ∈ B , the condition (2) holds forall ξ ∈ B ∪ B if Re ( (cid:104) ϕ | φ k (cid:105) (cid:104) φ k | η (cid:105) ) = 0 (3)for all φ k ∈ B . The remaining thing is to show that it is possible tochoose a unit vector η ∈ H such that η is orthogonal to ϕ and (3)holds for all φ k ∈ B .Firstly, suppose that (cid:104) ϕ | φ (cid:105) = 0 or (cid:104) ϕ | φ (cid:105) = 0. Then thecorresponding choice η = φ or η = φ implies that (3) holds for all φ k ∈ B . If otherwise (cid:104) ϕ | φ (cid:105) (cid:54) = 0 and (cid:104) ϕ | φ (cid:105) (cid:54) = 0, we then set η = i | (cid:104) ϕ | φ (cid:105) (cid:104) ϕ | φ (cid:105) | (cid:112) | (cid:104) ϕ | φ (cid:105) | + | (cid:104) ϕ | φ (cid:105) | ( (cid:104) ϕ | φ (cid:105) − φ − (cid:104) ϕ | φ (cid:105) − φ ) . It is easy to verify that η is orthogonal to ϕ . Furthermore, we get (cid:104) ϕ | φ (cid:105) (cid:104) φ | η (cid:105) = − (cid:104) ϕ | φ (cid:105) (cid:104) φ | η (cid:105) = i | (cid:104) ϕ | φ (cid:105) (cid:104) ϕ | φ (cid:105) | (cid:112) | (cid:104) ϕ | φ (cid:105) | + | (cid:104) ϕ | φ (cid:105) | (cid:104) ϕ | φ k (cid:105) (cid:104) φ k | η (cid:105) = 0 for k ≥ , hence (3) holds for all φ k ∈ B . (cid:3) Rank criterion and qutrit
Let B = { ϕ j } dj =1 , . . . , B m = { ϕ mj } dj =1 be m orthonormal bases of H . For each vector ϕ (cid:96)j , we denote P (cid:96)j = | ϕ (cid:96)j (cid:105)(cid:104) ϕ (cid:96)j | . Each P (cid:96)j is thusa 1-dimensional projection. Using this notation we observe that theorthonormal bases B , . . . , B m cannot distinguish two different states (cid:37) and (cid:37) if and only iftr (cid:2) P (cid:96)j ( (cid:37) − (cid:37) ) (cid:3) = 0 for all (cid:96) = 1 , . . . , m and j = 1 , . . . , d . (4)This condition means that the operator (cid:37) − (cid:37) is orthogonal to allthe projections P (cid:96)j in the Hilbert-Schmidt inner product. (We recallthat the Hilbert-Schmidt inner product of two operators A and B is (cid:104) A | B (cid:105) HS = tr [ A ∗ B ]). Note that the operator (cid:37) − (cid:37) is selfadjointand traceless. Moreover, if (cid:37) and (cid:37) are pure states, then (cid:37) − (cid:37) hasrank 2. The previous observation can be developed into a useful criterionwhen we look at all selfadjoint operators that are orthogonal to theprojections P (cid:96)j . Suppose T is a nonzero selfadjoint operator satisfyingtr (cid:2) P (cid:96)j T (cid:3) = 0 for all (cid:96) = 1 , . . . , m and j = 1 , . . . , d . First of all, as (cid:80) j P (cid:96)j = , the operator T satisfies tr [ T ] = 0. To derivesome other properties of T , we write T in the spectral decomposition T = p (cid:88) i =1 λ + i | ψ + i (cid:105)(cid:104) ψ + i | − n (cid:88) i =1 λ − i | ψ − i (cid:105)(cid:104) ψ − i | , where λ +1 , . . . , λ + p and − λ − , . . . , − λ − n are the strictly positive and strictlynegative eigenvalues of T , respectively, and ψ +1 , . . . , ψ + p , ψ − , . . . , ψ − n areorthogonal unit vectors. From tr [ T ] = 0 it follows that p (cid:88) i =1 λ + i = n (cid:88) i =1 λ − i ≡ λ . The rank of T is p + n , the sum of its nonzero eigenvalues counted bytheir multiplicities. As T (cid:54) = 0 and tr [ T ] = 0, T must have both positiveand negative eigenvalues, meaning that n ≥ p ≥
1. Therefore,the rank of T is at least 2. If the rank of T is 2, then n = p = 1 andthus λ − T = | ψ +1 (cid:105)(cid:104) ψ +1 | − | ψ − (cid:105)(cid:104) ψ − | . This implies that the orthonormal bases B , . . . , B m cannot distinguishthe pure states (cid:37) = | ψ +1 (cid:105)(cid:104) ψ +1 | and (cid:37) = | ψ − (cid:105)(cid:104) ψ − | Our previous discussion can be summarized in the form of the fol-lowing criterion.
Proposition 2.
Orthonormal bases B , . . . , B m can distinguish all purestates if and only if every nonzero selfadjoint operator T that satisfies tr (cid:2) P (cid:96)j T (cid:3) = 0 for all (cid:96) = 1 , . . . , m and j = 1 , . . . , d (5) has rank at least . Using this criterion we can prove the following statement.
Proposition 3.
In dimension d = 3 ,(i) three orthonormal bases cannot distinguish all pure states;(ii) four orthonormal bases can distinguish all pure states if and onlyif the Hilbert-Schmidt orthogonal complement of the projections { P (cid:96)j | (cid:96) = 1 , , , , j = 1 , , } is either { } or is the linear spanof a single nonzero and invertible selfadjoint operator. Proof.
Our proof is adapted from the analogous one of [7, Proposi-tion 5]. By Proposition 2, m orthonormal bases B , . . . , B m distin-guish all pure states if and only if every nonzero selfadjoint operator T ∈ { P (cid:96)j | (cid:96) = 1 , . . . , m, j = 1 , , } ⊥ is invertible. We claim that inthis case there cannot exist two linearly independent selfadjoint oper-ators T , T ∈ { P (cid:96)j | (cid:96) = 1 , . . . , m, j = 1 , , } ⊥ .Indeed, suppose on the contrary that T and T are two such opera-tors. Since det( T ) and det( T ) are nonzero, there are real numbers α , α such that det ( α T ) > α T ) <
0. By linear indepen-dence, the convex combination λα T + (1 − λ ) α T is nonzero for all λ ∈ [0 , λ α T + (1 − λ ) α T ] = 0for some λ ∈ (0 , T = λ α T + (1 − λ ) α T is not invertible. But T satisfies (5), which thencontradicts Proposition 2.We now come to the proof of (i). Three orthonormal bases B , B , B give 9 projections P (cid:96)j . However, as (cid:80) j =1 P (cid:96)j = for each (cid:96) = 1 , , P (cid:96)j . The previous claim then implies that B , B , B cannot distinguish all pure states.To prove item (ii), observe that, if the linear space { P (cid:96)j | (cid:96) = 1 , , , , j =1 , , } ⊥ has dimension k , then we can find a basis of it consisting ofselfadjoint operators. Indeed, if T , . . . , T k is any linear basis, then theselfadjoint operators T +1 , . . . , T + k , T − , . . . , T − k given by T + h = T + T ∗ , T − h = i ( T − T ∗ )still satisfy (5) and generate the linear space { P (cid:96)j | (cid:96) = 1 , , , , j =1 , , } ⊥ . Extracting k linearly independed elements from these opera-tors, we get a basis of selfadjoint operators. Therefore, by our earlierclaim the four bases B , . . . , B can distinguish all pure states only ifeither k = 0 or k = 1. In the latter case, { P (cid:96)j | (cid:96) = 1 , , , , j =1 , , } ⊥ = C T for some selfadjoint operator T , which must then beinvertible by Proposition 2. Conversely, the sufficiency of these twoconditions is clear by Proposition 2. (cid:3) In the next section, we will see that in dimension d = 3 actuallythere exist four orthonormal bases distinguishing all pure states. Thecondition in item (ii) of Proposition 3 is then very useful to explicitelyconstruct such bases. As an example, Section 8 below will provide an application to the measurement of the orthonormal bases correspondingto four different spin directions in a spin-1 system.5. Four bases that distinguish all pure states
Up to now we have seen that already in dimension 3, we can neverfind three orthonormal bases which would yield unique determinationof all pure states. Therefore, the minimal number of bases in that caseis at least four. In this section we show that, perhaps surprisingly,four properly chosen orthonormal bases are sufficient regardless of thedimension of the Hilbert space.
Proposition 4.
In any finite dimension, there exist four orthonormalbases B , B , B , B that can distinguish all pure states. The proof is based on an explicit construction of the bases in theHilbert space H = C d , as presented by Jaming in [9]. His constructionuses properties of the Hermite polynomials, but a similar constructionworks also for any other sequence of orthogonal polynomials. Differentpolynomials will lead to different bases, so this freedom in choosing thepolynomials may be sometimes useful.The construction begins by fixing a sequence of orthogonal polyno-mials. By a sequence of orthogonal polynomials we mean a sequence p , p , p , . . . of real polynomials such that the degree of p n is n and (cid:90) ba p n ( x ) p (cid:96) ( x ) w ( x ) d x = δ n(cid:96) for a nonnegative weight function w and either finite or infinite interval[ a, b ]. The most common sequences of orthogonal polynomials are the(normalized versions of) Chebyshev, Hermite, Laguerre and Legendrepolynomials. For instance, the n th Hermite polynomial H n is definedby the formula H n ( x ) = ( − n √ n n ! e x d n d x n e − x . The Hermite polynomials form a sequence of orthogonal polynomialswith respect to the weight function w ( x ) = √ π e − x and the infiniteinterval ( −∞ , ∞ ).The following construction uses three basic properties shared by anysequence of orthogonal polynomials. Let c n denote the highest coeffi-cient of a polynomial p n . It can be shown (see e.g. [13, pp. 43-46])that the following properties hold:(a) The roots of p n are all real and distinct.(b) p n and p n +1 have no common roots. (c) For all x (cid:54) = y , the Christoffel-Darboux formula holds: n (cid:88) j =0 p j ( x ) p j ( y ) = c n c n +1 p n +1 ( x ) p n ( y ) − p n ( x ) p n +1 ( y ) x − y . In dimension d , only the first d + 1 polynomials of the sequence will beneeded.To construct the first basis, let x , . . . , x d be the roots of the poly-nomial p d , and define (cid:101) ϕ j = ( p ( x j ) , p ( x j ) , . . . , p d − ( x j )) T (6)for j = 1 , . . . , d . Then each vector (cid:101) ϕ j is nonzero since p ( x j ) (cid:54) = 0 ( p is a nonzero constant polynomial), and since the zeros are all distinct,we may apply the Christoffel-Darboux formula to get (cid:104) (cid:101) ϕ i | (cid:101) ϕ j (cid:105) = d − (cid:88) k =0 p k ( x i ) p k ( x j )= c d − c d p d ( x i ) p d − ( x j ) − p d − ( x i ) p d ( x j ) x i − x j = 0for i (cid:54) = j . Thus, the vectors are orthogonal. By normalizing ϕ j = (cid:107) (cid:101) ϕ j (cid:107) − (cid:101) ϕ j we obtain an orthonormal basis B = { ϕ j } dj =1 of C d .Let then y , . . . , y d − be the roots of the polynomial p d − and define (cid:101) ϕ j = ( p ( y j ) , p ( y j ) , . . . , p d − ( y j )) T . (7)for j = 1 , . . . , d −
1. These vectors are again nonzero and orthogonal,and since the y j ’s are the roots of p d − , the last component is p d − ( y j ) =0. Hence, we can again normalize ϕ j = (cid:107) (cid:101) ϕ j (cid:107) − (cid:101) ϕ j and define ϕ d =(0 , . . . , , T to obtain another orthonormal basis B = { ϕ j } dj =1 .For the two remaining bases, we first pick a number α ∈ R which isnot a rational multiple of π . We then define (cid:101) ϕ j = ( p ( x j ) , e iα p ( x j ) , . . . , e i ( d − α p d − ( x j )) T , (8)which, after normalization, gives the third basis B = { ϕ j } dj =1 . Finally,we set (cid:101) ϕ j = ( p ( y j ) , e iα p ( y j ) , . . . , e i ( d − α p d − ( y j )) T (9)which, after normalizing and adding the vector ϕ d = (0 , . . . , , T ,gives the last basis B = { ϕ j } dj =1 .Using the basis B , . . . , B , we can now prove our main result. Proof of Proposition 4.
As usual, we denote P (cid:96)j = | ϕ (cid:96)j (cid:105)(cid:104) ϕ (cid:96)j | for (cid:96) =1 , . . . , j = 1 , . . . , d . By (4), in order to prove that the bases B , . . . , B can distinguish all pure states, we need to show that, forany two pure states (cid:37) = | ξ (cid:105)(cid:104) ξ | and (cid:37) = | η (cid:105)(cid:104) η | , the conditiontr (cid:2) P (cid:96)j ( (cid:37) − (cid:37) ) (cid:3) = 0 for all (cid:96) = 1 , . . . , m and j = 1 , . . . , d (10)implies ξ = e iθ η for some θ ∈ R . To see this, let T = (cid:37) − (cid:37) , andwrite T = ( t ij ) di,j =1 with respect to the standard basis of C d . Theselfadjointness of T implies that t ii ∈ R and t ji = t ij . Using the explicitform of the vectors from (6)–(9), the orhogonality condition (10) thenyields d − (cid:88) k,l =0 t k +1 ,l +1 p k ( z ) p l ( z ) = 0 (11) d − (cid:88) k,l =0 t k +1 ,l +1 e i ( l − k ) α p k ( z ) p l ( z ) = 0 (12)for every z ∈ { x , . . . , x d , y , . . . , y d − } . The degree of p n is n , so thepolynomials in (11) and (12) have degree at most 2 d −
2. But the aboveequations state that these polynomials have 2 d − z ∈ R .We can now look at the highest order terms in (11) and (12). Thiscorresponds to k = l = d − , z, z , . . . , z d − we have t d,d = 0. Since t ij = ξ i ξ j − η i η j for all i, j , it follows that | ξ d | = | η d | so that ξ d = e iθ η d for some θ ∈ R . Assumefor the moment that ξ d (cid:54) = 0.We can now consider the terms of order 2 d −
3. Since they appear onlyfor k = d − l = d − t d,d − + t d − ,d = t d,d − e − iα + t d − ,d e iα = 0 . By substituting t d − ,d = t d,d − we then getRe ( t d,d − ) = Re ( t d,d − e − iα ) = 0 . Since e − iα / ∈ R , we have t d,d − = 0 which implies that ξ d ξ d − = η d η d − .By our assumption, ξ d − = e iθ η d − .We now proceed by induction. Suppose that ξ d − p = e iθ η d − p for p =0 , . . . , n −
1. It follows that t d − p,d − q = 0 for p, q ≤ n −
1, so that thehighest order terms in (11) and (12) are of order 2 d − n −
2, and theyappear only with k = d − l = d − n − us Re ( t d,d − n ) = Re ( t d,d − n e − iα ) = 0so that t d,d − n = 0. Hence, ξ d − n = e iθ η d − n .Finally, if we have ξ d = 0 so that also η d = 0, then t d,n = 0 and t n,d = 0for all n , and the summations in (11) and (12) terminate at d −
2. Hence,we can repeat the above process starting from the highest order termwhich is now of order 2 d −
4. By induction, if ξ d = . . . = ξ d − ( n − = 0but ξ d − n (cid:54) = 0, then also η d = . . . = η d − ( n − = 0 so that the processcan be started from the terms of order 2( d − n ) − ξ d − n = e iθ η d − n and then proceed as before. This completes the proofof Proposition 4. (cid:3) Insufficiency of three bases in dimension and higher We have seen that in every finite dimension d = 2 , , . . . , it is possibleto choose a set of four orthonormal bases that distinguish all purestates, but no pair of orthonormal bases can have this property. Cana set of three orthonormal bases distinguish all pure states? As weknow already, this question has a positive answer in d = 2 and anegative answer in d = 3. In the following, we explain how topologicalconsiderations lead to a negative answer for all dimensions d ≥
5. Formore details on the topological aspects of state determination we referto [7] and [14].First, it is useful to generalize the property of distinguishing purestates to arbitrary sets A = { A , . . . , A n } of selfadjoint operators on H . We say that such a set A distinguishes all pure states if for twodifferent pure states (cid:37) and (cid:37) , there exists A i ∈ A such thattr [ A i (cid:37) ] (cid:54) = tr [ A i (cid:37) ] . (13)Clearly, if A consists of the orthogonal projections defined by a collec-tion of orthonormal bases, we obtain again (1).In what follows, we use the notation below: for each dimension d =2 , , ... , we denote by • s d the minimal number of selfadjoint operators which distin-guish all pure states; • b d the minimal number of orthonormal bases which distinguishall pure states.It is easy to see that s d gives a lower bound for b d . Namely, if we have m bases, they give m · d projections. All the projections correspondingto a basis sum up to the identity operator ; since tr [ (cid:37) ] = 1 for all states (cid:37) , one projection for each basis can then be left out withoutlosing any information. We thus conclude that( d − · b d ≥ s d . (14)Therefore, lower bounds for s d translate into lower bounds for b d .Let us denote by P the set of pure states. Saying that A distinguishesall pure states means that the map f A : P → R n , (cid:37) (cid:55)→ (tr [ A (cid:37) ] , . . . , tr [ A n (cid:37) ]) (15)is injective. Hence, roughly speaking, the selfadjoint operators A dis-tinguish all pure states if and only if the map (15) identifies P witha subset of the Euclidean space R n . This is a very useful observation:indeed, suppose, for instance, that it is possible to prove that there is anatural number n such that no injective map P → R n exists if n < n ;then one can conclude that s d ≥ n .Up to this point our considerations were purely set theoretical. How-ever, it can be proved that, if the map f A is injective, then it is actuallya smooth embedding (see Proposition 7 of Appendix A; we refer to [15]for the necessary notions from differential geometry). Therefore, theselfadjoint operators A can distinguish all pure states only if the man-ifold of pure states P can be smoothly embedded in R n .The minimal n for which P can be smoothly embedded in R n , iscalled the embedding dimension of P . From the heuristic point ofview, if M and M are two smooth manifolds of the same dimension,we expect that the embedding dimension of M is greater than theembedding dimension of M if the shape of M is more involved thanthe shape of M . For instance, a 2-dimensional torus can be smoothlyembedded in R whereas the Klein bottle requires R .The problem of determining the embedding dimension of complexprojective spaces has been studied extensively in the mathematicalliterature and the best bounds are, up to our knowledge, those obtainedin [16]. They lead to the conclusion that s d ≥ d − α − d ≥ d − α − d odd, and α = 2 mod 44 d − α − d odd, and α = 3 mod 4 (16)where α is the number of 1’s in the binary expansion of d − d − (cid:80) nj =0 a j j , we have that 2 n ≤ d − n ≤ log ( d − α ≤ n + 1 ≤ log ( d −
1) + 1 Figure 1.
The function f ( x ) = 4 − ( x − x − givesa lower bound for the mimimal number of orthonormalbases. Since b d ≥ f ( d ) for each integer d ≥
2, we con-clude that b d ≥ d ≥ s d ≥ d − α − ≥ d − ( d − − d ≥
2. Inserting this into (14) we obtain b d ≥ d − ( d − − d − − ( d −
1) + 1 d − . (17)To see the consequences of the derived lower bound (17), we examinethe function f : [2 , ∞ ) → R , f ( x ) = 4 − ( x −
1) + 1 x − f (2) = 3 and lim x →∞ f ( x ) = 4.Secondly, the derivative f (cid:48) ( x ) has a single zero at x = 1 + e √ (cid:39) . f (cid:48) ( x ) < x ∈ (2 , x ) and f (cid:48) ( x ) > x > x .Finally, f (8) (cid:39) .
055 so that f ( x ) > x ≥
8. Since b d is aninteger and b d ≥ f ( d ), we thus have b d ≥ d ≥ d = 2 , . . . , s d given in (16). We thus observe that unless d = 2 or4, three orthonormal bases cannot be sufficient. d s d d − d = 3 and d ≥ d = 2 the minimal number is three and inthe case d = 4 it is, based on our knowledge, either three or four.7. Insufficiency of four product bases in dimension d = 4In our search for the minimal number of orthonormal bases that candistinguish all pure states, the remaining question is: Is it possible to find three orthonormal bases in dimension that candistinguish all pure states? Unfortunately, we are able to provide only a partial answer to thisquestion. Namely, in the following we will see that if we considerthe splitting of the 4-dimensional Hilbert space into a tensor product H = C ⊗ C , then for any three bases consisting solely of productvectors, the answer is negative. In fact, we will show that even fourproduct bases are not enough.Before we concentrate on dimension 4, we slightly elaborate the statedistinction criterion used in earlier sections. As we have seen, a set A = { A , . . . , A n } of selfadjoint operators cannot distinguish two states (cid:37) and (cid:37) if and only iftr [ A j ( (cid:37) − (cid:37) )] = 0 for all j = 1 , . . . , n . Since the trace is a linear functional, this is equivalent totr (cid:34)(cid:16) n (cid:88) j =1 α j A j (cid:17) ( (cid:37) − (cid:37) ) (cid:35) = 0 for all α , . . . , α n ∈ C . Moreover, as tr [ (cid:37) − (cid:37) ] = 0, we can rewrite the previous condition astr (cid:34)(cid:16) α + n (cid:88) j =1 α j A j (cid:17) ( (cid:37) − (cid:37) ) (cid:35) = 0 for all α , α , . . . , α n ∈ C . This equivalent formulation shows that, for the purpose of state dis-tinction, we can always switch from A to the linear space spanned by A and . We denote this subspace of operators by R ( A ), i.e., R ( A ) = (cid:40) α + n (cid:88) j =1 α j A j (cid:12)(cid:12)(cid:12)(cid:12) α j ∈ C (cid:41) (18)and call it the (complex) operator system generated by the selfadjointoperators A . If the set A consists of the projections corresponding to m orthonormal bases B = { ϕ j } dj =1 , . . . , B m = { ϕ mj } dj =1 , we write also R ( B , . . . , B m ) = R ( { P (cid:96)j | (cid:96) = 1 , . . . , m, j = 1 , . . . , d } ) , where as usual P (cid:96)j = | ϕ (cid:96)j (cid:105)(cid:104) ϕ (cid:96)j | . Our discussion then yields the followingconclusion. Proposition 5.
Let A and A (cid:48) be two sets of selfadjoint operators. If R ( A ) = R ( A (cid:48) ) , then A and A (cid:48) distinguish the same pairs of states. Let us make use of this fact to show that in the Hilbert space H = C ⊗ C four product bases cannot distinguish all pure states. Byproduct basis, we mean an orthonormal basis of H that is constructedfrom two orthonormal bases of C by taking their tensor product. Moreprecisely, if { ϕ , ϕ } and { φ , φ } are two orthonormal bases of C ,then { ϕ i ⊗ φ j | i, j = 1 , } is an orthonormal basis of C . From thephysics point of view this corresponds to a scheme where two partiestry to determine the unknown pure state of a composite system byperforming only local measurements.Using the Bloch representation, any 1-dimensional projection on C can be written as P (cid:126)n = ( + (cid:126)n · (cid:126)σ ), where (cid:126)n ∈ R is a unit vector.Furthermore, tr [ P (cid:126)m P (cid:126)n ] = 12 (1 + (cid:126)m · (cid:126)n ) . Therefore, the 1-dimensional projections corresponding to an orthonor-mal basis of C are P (cid:126)n and P − (cid:126)n where (cid:126)n is fixed by the choice of thebasis.Suppose that we have two quadruples of (not necessarily distinct)orthonormal bases B (cid:48) , B (cid:48) , B (cid:48) , B (cid:48) and B (cid:48)(cid:48) , B (cid:48)(cid:48) , B (cid:48)(cid:48) , B (cid:48)(cid:48) of C with the cor-responding quadruples of unit vectors (cid:126)m , (cid:126)m , (cid:126)m , (cid:126)m and (cid:126)n , (cid:126)n , (cid:126)n , (cid:126)n .We define four product bases of C via B j = { ϕ ⊗ φ | ϕ ∈ B (cid:48) j , φ ∈ B (cid:48)(cid:48) j } . The four projections P (cid:126)m j ⊗ P (cid:126)n j , P (cid:126)m j ⊗ P − (cid:126)n j , P − (cid:126)m j ⊗ P (cid:126)n j , P − (cid:126)m j ⊗ P − (cid:126)n j corresponding to an orthonormal basis B j have the same linear span asthe selfadjoint operators ⊗ , (cid:126)m j · (cid:126)σ ⊗ , ⊗ (cid:126)n j · (cid:126)σ , (cid:126)m j · (cid:126)σ ⊗ (cid:126)n j · (cid:126)σ . Therefore, by introducing the set A = { (cid:126)m j · (cid:126)σ ⊗ , ⊗ (cid:126)n j · (cid:126)σ, (cid:126)m j · (cid:126)σ ⊗ (cid:126)n j · (cid:126)σ | j = 1 , . . . , } , we have the equality R ( B , . . . , B ) = R ( A ) . We will next show that there exist two maximally entangled pure states (cid:37) (cid:54) = (cid:37) which are not distinguished by the set A , thus implying byProposition 5 that the bases B , . . . , B cannot distinguish all purestates.Recall that a unit vector Ω ∈ C ⊗ C is called maximally entangled if Ω = √ ( ϕ ⊗ φ + ϕ ⊗ φ ) for some orthonormal bases { ϕ , ϕ } and { φ , φ } of C . If { e , e } denotes the canonical basis of C , and Ω = √ ( e ⊗ e + e ⊗ e ), then there always exists a unitary operator U on C such that Ω = ( U ⊗ )Ω (see, e.g., [17, Lemma 2]). Since the globalphase of Ω is irrelevant for our purposes, we see that the maximallyentangled pure states are in one-to-one correspondence with elementsof the quotient group SU (2) / {± I } , which in turn is diffeomorphic to SO (3) through mapping U (cid:55)→ R U given by the equality U ∗ (cid:126)x · (cid:126)σU = R U (cid:126)x · (cid:126)σ for all (cid:126)x ∈ R . Suppose that we try to determine the pure state (cid:37) U = | Ω (cid:105)(cid:104) Ω | usingthe selfadjoint operators A . Sincetr [( (cid:126)m j · (cid:126)σ ⊗ ) (cid:37) U ] = tr [( U ∗ (cid:126)m j · (cid:126)σU ⊗ ) | Ω (cid:105)(cid:104) Ω | ]= 12 tr [ U ∗ (cid:126)m j · (cid:126)σU ] = 0and similarlytr [( ⊗ (cid:126)n j · (cid:126)σ ) (cid:37) U ] = tr [( ⊗ (cid:126)n j · (cid:126)σ ) | Ω (cid:105)(cid:104) Ω | ] = 12 tr [ (cid:126)n j · (cid:126)σ ] = 0 , we find that the only relevant information that can be extracted istr [( (cid:126)m j · (cid:126)σ ⊗ (cid:126)n j · (cid:126)σ ) (cid:37) U ] = tr [( U ∗ (cid:126)m j · (cid:126)σU ⊗ (cid:126)n j · (cid:126)σ ) | Ω (cid:105)(cid:104) Ω | ]= tr [( R U (cid:126)m j · (cid:126)σ ⊗ (cid:126)n j · (cid:126)σ ) | Ω (cid:105)(cid:104) Ω | ]= h j ( R U ) where h j is defined on the linear space M ( R ) of real 3 × h j ( A ) = tr [( A (cid:126)m j · (cid:126)σ ⊗ (cid:126)n j · (cid:126)σ ) | Ω (cid:105)(cid:104) Ω | ] . If the selfadjoint operators A could distinguish all pure states, then,in particular, they could distinguish all maximally entangled pure states.Therefore, the linear map f : M ( R ) → R given by f ( A ) = ( h ( A ) , h ( A ) , h ( A ) , h ( A )) T would restrict to an injective map ˜ f : SO (3) → R . By Proposition8 in Appendix 9, such a map would then be a smooth embedding of SO (3) into R . Since SO (3) is diffeomorphic to the real projective3-dimensional space RP [18, Proposition 5.2.10], and RP cannot beembedded into R by [19, 20], we then obtain a contradiction. We thusconclude that the selfadjoint operators A cannot distinguish all purestates.8. Spin-1 – It is not only about number of bases
As we have now seen, it is enough to measure four bases in order todistinguish all pure states. However, not all sets of four bases do thisas was implied by our consideration of product bases in dimension 4.In this section we demonstrate this further by giving another examplewhere some natural candidates for the bases fail to distinguish all purestates. We consider the problem of determining all pure states of aspin-1 system by measuring four orthonormal bases corresponding todifferent spin directions.The Hilbert space of the spin-1 quantum system is H = C , and theusual spin operators along the three axes are L x = 1 √ , L y = 1 √ − i i − i i , L z = − . If (cid:126)n ∈ R is any unit vector, the spin operator along the direction (cid:126)n is (cid:126)n · (cid:126)L = n x L x + n y L y + n z L z , which is selfadjoint and has eigenvalues { +1 , , − } . We denote the corresponding eigenprojections as P (cid:126)nj , andthey can be written as P (cid:126)n +1 = ( (cid:126)n · (cid:126)L ) + (cid:126)n · (cid:126)L P (cid:126)n − = ( (cid:126)n · (cid:126)L ) − (cid:126)n · (cid:126)L P (cid:126)n = − ( (cid:126)n · (cid:126)L ) . Note that these projections span the same linear space as the operators , (cid:126)n · (cid:126)L , and ( (cid:126)n · (cid:126)L ) . The next result shows that, if the four directions (cid:126)n , . . . , (cid:126)n are suit-ably chosen, then the corresponding spin measurements actually deter-mine all pure states. This refines the lower bound of [21], where theauthors prove that all pure states of a spin-1 system are uniquely deter-mined by six spin components, and should be compared with [22, 23],where it is shown that, for any spin- s system, the spin observablesalong two infinitesimally near directions (cid:126)n and (cid:126)n (cid:48) together with theexpectation value of the spin observable orthogonal to (cid:126)n and (cid:126)n (cid:48) areenough to determine all pure states up to a set of measure zero. Proposition 6.
Let (cid:126)n , . . . , (cid:126)n ∈ R be four directions with the com-ponents (cid:126)n k = ( n kx , n ky , n kz ) T , and denote by B k the eigenbases of thespin operator along (cid:126)n k for a spin- quantum system. Then the bases B , . . . , B can distinguish all pure states if and only if the followingconditions hold:(a) the × real matrix M with entries M k, = 2 √ n kx n kz M k, = − √ n ky n kz M k, = n kx − n ky M k, = − n kx n ky M k, = 3 n kz − has rank ;(b) there exists a nonzero real solution x = ( x , . . . , x ) T ∈ R of thelinear system M x = 0 such that x x x + x ( x − x ) + x ( x + x + x − x − x ) (cid:54) = 0 . (20) Proof.
We will show that conditions (a) and (b) in the previous state-ment are equivalent to the condition in item (ii) of Proposition 3.The operator system R ( B , . . . , B ) is spanned by and the set of 8selfadjoint operators A = { (cid:126)n · (cid:126)L, . . . , (cid:126)n · (cid:126)L, ( (cid:126)n · (cid:126)L ) , . . . , ( (cid:126)n · (cid:126)L ) } . Since the operators { (cid:126)n · (cid:126)L, . . . , (cid:126)n · (cid:126)L } are linearly dependent, we havedim R ( B , . . . , B ) ≤
8, that is, dim R ( B , . . . , B ) ⊥ ≥
1. Hence byProposition 3.(ii) we actually need to show that conditions (a) and (b)are equivalent to(a’) dim R ( B , . . . , B ) ⊥ = 1;(b’) there exists an invertible selfadjoint operator T ∈ R ( B , . . . , B ) ⊥ .We begin by showing that conditions (a) and (a’) are equivalent, and,when (a) holds, R ( B , . . . , B ) ⊥ = Φ(ker( M )) , (21) where Φ : C → M ( C ) is the linear map given byΦ x ... x = x x + ix x + ix x − ix − x − x − ix x − ix − x + ix x . Note that Φ is injective, and Φ( C ) = { , L x , L y , L z } ⊥ . Moreover, it iseasy to check that tr (cid:104) ( (cid:126)n i · (cid:126)L ) Φ( x ) (cid:105) = m i x for all x ∈ C , where m i is the i th row of the matrix M defined in (19), and m i x is the usualmatrix product. Hence,Φ(ker( M )) = { , L x , L y , L z } ⊥ ∩ { ( (cid:126)n · (cid:126)L ) , . . . , ( (cid:126)n · (cid:126)L ) } ⊥ = R ( A (cid:48) ) ⊥ (22)where A (cid:48) = { L x , L y , L z , ( (cid:126)n · (cid:126)L ) , . . . , ( (cid:126)n · (cid:126)L ) } . Now, suppose that rank( M ) <
4. Then, dim ker( M ) >
1, from which itfollows that dim R ( A (cid:48) ) ⊥ > R ( B , . . . , B ) = R ( A ) and A ⊂ A (cid:48) , we have R ( A (cid:48) ) ⊥ ⊂ R ( B , . . . , B ) ⊥ , and this impliesthat condition (a’) does not hold.Conversely, assume that rank( M ) = 4. We claim that in this casethe four unit vectors (cid:126)n , . . . , (cid:126)n span the whole space R . Indeed, ife.g. (cid:126)n i = α i (cid:126)n + β i (cid:126)n for i = 3 ,
4, then we would have( (cid:126)n i · (cid:126)L ) = α i ( (cid:126)n · (cid:126)L ) + β i ( (cid:126)n · (cid:126)L ) + α i β i (cid:110) (cid:126)n · (cid:126)L, (cid:126)n · (cid:126)L (cid:111) , where {· , ·} is the anticommutator. Hence, m i = α i m + β i m + α i β i u ,where u T ∈ C (actually, u T ∈ R ) is defined by ux = tr (cid:104)(cid:110) (cid:126)n · (cid:126)L, (cid:126)n · (cid:126)L (cid:111) Φ( x ) (cid:105) for all x ∈ C . Thus, rank( M ) ≤
3, which is a contradiction. Ourclaim then implies R ( A ) = R ( A (cid:48) ). Taking the orthogonal comple-ment, we have R ( A ) ⊥ = R ( A (cid:48) ) ⊥ = Φ(ker( M )) by (22), and (21) fol-lows since R ( B , . . . , B ) = R ( A ). In particular, by injectivity of Φ,dim R ( B , . . . , B ) ⊥ = dim ker( M ) = 1, that is, condition (a’).Finally, assuming (a), we come to the proof of (b) ⇔ (b’). First of all,observe that the operator Φ( x ) is selfadjoint if and only if x ∈ R . By(21), condition (b’) then amounts todet x x + ix x + ix x − ix − x − x − ix x − ix − x + ix x (cid:54) = 0for some nonzero x = ( x , . . . , x ) T ∈ R such that M x = 0, that is,condition (b). (cid:3) By Proposition 6, it is easy to check that the three orthogonal spindirections (cid:126)e , (cid:126)e , and (cid:126)e cannot be completed to a set of four directionswhich would allow unique determination of pure states. Indeed, if (cid:126)n = (cid:126)e , (cid:126)n = (cid:126)e , (cid:126)n = (cid:126)e and (cid:126)n is any direction, thenrank( M ) = rank −
10 0 − −
10 0 0 0 2 ∗ ∗ ∗ ∗ ∗ ≤ (cid:126)n = (0 , , T (cid:126)n = (1 / √ , / √ , T (cid:126)n = (1 / √ , , / √ T (cid:126)n = (0 , √ / , / T satisfy both conditions (a) and (b) of Proposition 6, hence the corre-sponding bases can distinguish all pure states.Finally, we remark that the property of distinguishing all pure statesis robust against small perturbations of the unit vectors (cid:126)n , . . . , (cid:126)n . In-deed, suppose that conditions (a) and (b) of Proposition 6 are satisfiedby the four directions (cid:126)n , . . . , (cid:126)n . The matrix M = M ( (cid:126)n , . . . , (cid:126)n ) de-fined in (19) is a continuous function of ( (cid:126)n , . . . , (cid:126)n ), hence so are thediagonalizable matrix-valued functions M ∗ M and M M ∗ . By condition(a), the 4 × M M ∗ ( (cid:126)n , . . . , (cid:126)n ) is invertible, hence M M ∗ isinvertible in a neighborhood of ( (cid:126)n , . . . , (cid:126)n ), that is, rank( M ) = 4 inthat neighborhood. Thus, condition (a) still holds around ( (cid:126)n , . . . , (cid:126)n ).Now, let x ∈ R be a nonzero real solution of M ( (cid:126)n , . . . , (cid:126)n ) x = 0which satisfies (20). Moreover, let Q be the orthogonal projection ontoker( M ∗ M ), and define ˜ x ( (cid:126)n , . . . , (cid:126)n ) = Q ( (cid:126)n , . . . , (cid:126)n ) x . By [24, Theo-rem II.5.1], Q is a continuous function of ( (cid:126)n , . . . , (cid:126)n ) in a neighborhoodof ( (cid:126)n , . . . , (cid:126)n ). As M ∗ M is a real matrix, also Q is real. Combiningthese two facts, the map ˜ x is a nonzero continuous R -valued functionsuch that ˜ x ∈ ker( M ∗ M ) = ker( M ) and ˜ x ( (cid:126)n , . . . , (cid:126)n ) = x . By continu-ity, ˜ x satisfies (20) around ( (cid:126)n , . . . , (cid:126)n ). Therefore, also condition (b) ofProposition 6 remains fulfilled for small perturbations of ( (cid:126)n , . . . , (cid:126)n ).We conclude that, if the eigenbases of the spin operators along the di-rections (cid:126)n , . . . , (cid:126)n distinguish all pure states, then this still holds truein a neighborood of these directions. Remarks on the question in infinite dimensionalHilbert space
The question of the title is meaningful also in the case of an infinitedimensional Hilbert space. The proof of Proposition 1 works withoutchanges also in that case, so we conclude that two orthonormal basescannot distinguish all pure states even when d = ∞ . However, theconstruction of four orthonormal bases in Sec. 5 has no direct general-ization to the infinite dimensional Hilbert space. We are, in fact, notaware of a construction in infinite dimension that would give a finitenumber of bases that can distinguish all pure states.In the infinite dimensional case it is natural to allow also measure-ments of continuous observables such as position Q and momentum P . In fact, the determination of pure states from the statistics of suchmeasurements was precisely what was addressed in the original Pauliproblem. Since it is known that Q and P alone are not sufficient for thispurpose, we can ask if this set can be suitably completed to make it ableto distinguish all pure states. One natural attempt to obtain such acompletion would be to add rotated quadratures Q θ = cos θ Q +sin θ P .It is known that by measuring all of the quadratures, it is possible todetermine an arbitrary state, pure or mixed [25], but it is not knownif a smaller subset is sufficient for pure state determination. It wasrecently shown that no triple { Q θ , Q θ , Q θ } of quadratures is enough[26]. In particular, it is not sufficient to measure position, momentum,and a single additional rotated quadrature.For larger, but still finite, sets of rotated quadratures it is only knownthat if such a set can distinguish all pure states, then the choice ofthe angles θ is a delicate issue. In [26], it was shown that if a finiteset of angles θ , . . . , θ n satisfies θ i − θ j ∈ Q π for all i, j = 1 , . . . , n ,then the corresponding observables are not sufficient for pure statedetermination. In [27], a similar result was proved in the case thatcot θ j ∈ Q for all j = 1 , . . . , n . Acknowledgements.
JS and AT acknowledge financial support from the Italian Ministryof Education, University and Research (FIRB project RBFR10COAQ).
Appendix A. Linear smooth embeddings in R n If M is a real differentiable manifold and x ∈ M , we denote by T x ( M ) the (real) tangent space of M at x . If N is another differentiablemanifold and f : M → N is a differentiable map, we let d f x : T x ( M ) → T f ( x ) ( N ) be the differential of f at x . In this section we assume that M is a submanifold of a real linearspace V , and we prove that, if a linear map f : V → R n restricts toan injective map ˜ f : M → R n , then ˜ f is a smooth embedding in thefollowing two cases:(1) V = S d ( C ) is the linear space of complex d × d selfadjointmatrices and M = P is the submanifold of pure states (see [7,Section 4.2]);(2) V = M ( R ) is the linear space of real 3 × M = SO (3) is the submanifold of orthogonal matrices withunit determinant.The next two results are Theorem 5 and a particular case of Lemma1 in [7]. Up to our knowledge, Proposition 8 below is new. Lemma 1.
Let V , W be two real linear spaces, and M a compactsubmanifold of V . Suppose that T x ( M ) ⊆ R ( M − M ) for all x ∈ M .Then, if a linear map f : V → W restricts to an injective map ˜ f : M → W , the restriction ˜ f is a smooth embedding of M in V .Proof. Since M is compact and ˜ f is continuous, injectivity implies that˜ f is a homeomorphism of M onto ˜ f ( M ) [28, Proposition 1.6.8]. In orderto show that it is a smooth embedding, it remains to prove that d ˜ f x isinjective for all x ∈ M .Note that by linearity d ˜ f x = f | T x ( M ) for all x ∈ M . Thus, if u ∈ T x ( M )with u = λ ( x − x ) for some λ ∈ R and x , x ∈ M , thend ˜ f x u = f ( u ) = λ ( f ( x ) − f ( x )) = λ ( ˜ f ( x ) − ˜ f ( x )) ≡ λ = 0 or ˜ f ( x ) = ˜ f ( x ), that is, x = x by injectivity of˜ f . In both cases, we have u = 0, hence d ˜ f x is injective as claimed. (cid:3) Proposition 7. If ˜ f : P → R n is injective and it is the restriction ofa linear map f : S d ( C ) → R n , then ˜ f is a smooth embedding of P in R n .Proof. By Lemma 1, it suffices to prove that T (cid:37) ( P ) ⊆ R ( P − P ) for allpure states (cid:37) ∈ P . Indeed, P is an orbit for the adjoint action of thegroup SU ( d ) on the linear space S d ( C ), hence T (cid:37) ( P ) = (cid:26) dd t e itH (cid:37) e − itH | H ∈ S d ( C ) (cid:27) = { i [ H, (cid:37) ] | H ∈ S d ( C ) } . If (cid:37) is a pure state and H ∈ S d ( C ), then i [ H, (cid:37) ] is a selfadjoint trace-less matrix with rank at most 2. Therefore, i [ H, (cid:37) ] = λ ( | ψ + (cid:105)(cid:104) ψ + | −| ψ − (cid:105)(cid:104) ψ − | ) for some λ ∈ R and unit vectors ψ + , ψ − ∈ C d , which provesthe claim. (cid:3) Proposition 8. If ˜ f : SO (3) → R n is injective and it is the restrictionof a linear map f : M ( R ) → R n , then ˜ f is a smooth embedding of SO (3) in R n .Proof. Again, by Lemma 1 it is enough to prove that T R ( SO (3)) = R ( SO (3) − SO (3)) for all R ∈ SO (3).Denote by M − ( R ) the linear subspace of antisymmetric matrices in M ( R ). Then, T R ( SO (3)) = RM − ( R ) for all R ∈ SO (3). We claimthat any X ∈ M − ( R ) can be written X = λ ( R − R T ) for some R ∈ SO (3) and λ ∈ R . Indeed, the map g : SO (3) → M − ( R ) with g ( R ) = R − R T is a diffeomorphism of an open neighborhood U of the identity I onto a neighborhood g ( U ) of 0, since its differentiald g I ( X ) = dd t (exp( tX ) − exp( − tX )) (cid:12)(cid:12)(cid:12)(cid:12) t =0 = 2 X is bijective. It follows that R g ( U ) = M − ( R ), hence the claim. (cid:3) References [1] W. Pauli.
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Analysis now . Springer-Verlag, New York, 1989. Claudio Carmeli ; DIME, Universit`a di Genova, Via Magliotto 2, I-17100 Savona, Italy
E-mail address : [email protected] Teiko Heinosaari ; Turku Centre for Quantum Physics, Departmentof Physics and Astronomy, University of Turku, Finland
E-mail address : [email protected] Jussi Schultz ; Dipartimento di Matematica, Politecnico di Milano,Piazza Leonardo da Vinci 32, I-20133 Milano, Italy, and Turku Centrefor Quantum Physics, Department of Physics and Astronomy, Univer-sity of Turku, Finland
E-mail address : [email protected] Alessandro Toigo ; Dipartimento di Matematica, Politecnico di Mi-lano, Piazza Leonardo da Vinci 32, I-20133 Milano, Italy, and I.N.F.N.,Sezione di Milano, Via Celoria 16, I-20133 Milano, Italy
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