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arXiv:1001.0004v1 [quant-ph] 31 Dec 2009The Lie Algebraic Significance of Symmetric Informationally Complete Measurements D.M. Appleby, Steven T. Flammia and Christopher A. Fuchs Perimeter Institute for Theoretical Physics Waterloo, Ontario N2L 2Y5, Canada December 30, 2009 Abstract Examplesofsymmetric informationallycomplete positiveoperatorva lued mea- sures (SIC-POVMs) have been constructed in every dimension ≤67. However, it remains an open question whether they exist in all finite dimensions. A SIC- POVM is usually thought of as a highly symmetric structure in quantum state space. However, its elements can equally well be regarded as a basis for the Lie algebra gl(d,C). In this paper we examine the resulting structure constants, which are calculated from the traces of the triple products of the S IC-POVM elements and which, it turns out, characterize the SIC-POVM up to unitary equivalence. We show that the structure constants have numero us remarkable properties. In particular we show that the existence of a SIC-POV M in di- mensiondis equivalent to the existence of a certain structure in the adjoint representation of gl( d,C). We hope that transforming the problem in this way, from a question about quantum state space to a question about Lie algebras, may help to make the existence problem tractable. Contents 1. Introduction 1 2. The Angle Tensors 7 3. Spectral Decompositions 14 4. TheQ-QTProperty 18 5. Lie Algebraic Formulation of the Existence Problem 21 6. The Algebra sl( d,C) 31 7. Further Identities 33 8. Geometrical Considerations 36 9. TheP-PTProperty 49 10. Conclusion 52 11. Acknowledgements 53 References 531 1.Introduction Symmetric informationally complete positive operator-valued measu res (SIC- POVMs) present us with what is, simultaneously, one of the most inte resting, and one of the most difficult and tantalizing problems in quantum informatio n [1–46]. SIC-POVMs are important practically, with applications to quantum t omography and cryptography [ 4,8,12,15,20,29], and to classical signal processing [ 24,36]. However, without in any way wishing to impugn the significance of the a pplications which have so far been proposed, it appears to us that the interes t of SIC-POVMs stems less from these particular proposed uses than from rather broader, more gen- eral considerations: the sense one gets that SICs are telling us so mething deep, and hitherto unsuspected about the structure of quantum stat e space. In spite of its being the central object about which the rest of quantum mech anics rotates, and notwithstanding the efforts of numerous investigators [ 47], the geometry of quantum state space continues to be surprisingly ill-understood. T he hope which inspires our efforts is that a solution to the SIC problem will prove to b e the key, not just to SIC-POVMs narrowly conceived, but to the geometry o f state space in general. Such things are, by nature, unpredictable. However, it is not unreasonable to speculate that a better theoretical understanding of the geo metry of quantum state space might have important practical consequences: not o nly the applica- tions listed above, but perhaps other applications which have yet to be conceived. On a more foundational level one may hope that it will lead to a much imp roved understanding of the conceptual message of quantum mechanics [7,43,45,48]. Having said why we describe the problem as interesting, let us now exp lain why we describe it as tantalizing. The trouble is that, although there is an abundance of reasons for suspecting that SIC-POVMs exist in every finite dimens ion (exact and high-precision numerical examples [ 1,2,5,11,16,19,28,39,46] having now been constructed in every dimension up to 67), and in spite of the intense efforts of many people [1–46] extending over a period of more than ten years, a general existe nce proof continues to elude us. In their seminal paper on the subject , published in 2004, Renes et al[5] say “A rigorous proof of existence of SIC-POVMs in all finite dimensions seems tantalizingly close, yet remains somehow distant.” T hey could have said the same if they were writing today. The purposeofthis paperis totryto takeourunderstandingofSI C mathematics (as it might be called) a little further forward. The research we repo rt began with a chance numerical discovery made while we were working on a differen t problem. Pursuing that initial numerical hint we uncovered a rich and interest ing set of connections between SIC-POVMs in dimension dand the Lie Algebra gl( d,C). The existence of these connections came as a surprise to us. However , in retrospect it is, perhaps, not so surprising. Interest in SIC-POVMs has, to dat e, focused on the fact that an arbitrary density matrix can be expanded in terms of a SIC-POVM. However, a SIC-POVM in dimension ddoes in fact provide a basis, not just for the space of density matrices, but for the space of all d×dcomplex matrices— i.e.the Liealgebragl( d,C). Boykin et al[49] haverecentlyshownthatthere isaconnection betweentheexistenceproblemformaximalsetsofMUBs(mutuallyu nbiasedbases) and the theory of Lie algebras. Since SIC-POVMs share with MUBs th e property of being highly symmetrical structures in quantum state space it mig ht have been anticipated that there are also some interesting connections betw een SIC-POVMs and Lie algebras.2 Our main result (proved in Sections 3,4and5) is that the proposition, that a SIC-POVM exists in dimension d, is equivalent to a proposition about the adjoint representation of gl( d,C). Our hope is that transforming the problem in this way, from a question about quantum state space to a question about Lie algebras, may help to make the SIC-existence problem tractable. But even if this h ope fails to materialize we feel that this result, along with the many other result s we obtain, provides some additional insight into these structures. Inddimensional Hilbert space Hda SIC-POVM is a set of d2operatorsE1, ...,Ed2of the form Er=1 dΠr (1) where the Π rare rank-1 projectors with the property Tr(ΠrΠs) =/braceleftigg 1r=s 1 d+1r/ne}ationslash=s(2) We will refer to the Π ras SIC projectors, and we will say that {Πr:r= 1,...,d2} is a SIC set. It follows from this definition that the Ersatisfy d2/summationdisplay r=1Er=I (3) (sotheyconstitute aPOVM),andthattheyarelinearlyindependen t (sothePOVM is informationally complete). It is an open question whether SIC-POVMs exist for all values of d. However, examples have been constructed analytically in dimensions 2–15 inclus ive [1,2,11, 16,19,28,39,46], and in dimensions 19, 24, 35 and 48 [ 16,46]. Moreover, high precisionnumerical solutionshave been constructed in dimensions 2 –67inclusive [ 5, 46]. Thislendssomeplausibilitytothe speculationthat theyexistinalldime nsions. For a comprehensive account of the current state of knowledge in this regard, and many new results, see the recent study by Scott and Grassl [ 46]. All known SIC-POVMs have a group covariance property. In other words, there exists (1) a group Ghavingd2elements (2) a projective unitary representation of GonHd:i.e.a mapg→UgfromG to the set of unitaries such that Ug1Ug2∼Ug1g2for allg1,g2(where the notation “ ∼” means “equals up to a phase”) (3) a normalized vector |ψ/an}bracketri}ht(the fiducial vector) such that the SIC-projectors are given by Πg=Ug|ψ/an}bracketri}ht/an}bracketle{tψ|U† g (4) (where we label the projector by the group element g, rather than the integer ras above). Most known SIC-POVMs are covariant under the action of the Weyl- Heisenberg group (though not all—see Renes et al[5] and, for an explicit example of a non Weyl-Heisenberg SIC-POVM, Grassl [ 19]). Here the group is Zd×Zd, and the projective representation is p→Dp, wherep= (p1,p2)∈Zd×ZdandDpis the3 corresponding Weyl-Heisenberg displacement operator Dp=d−1/summationdisplay rτ(2r+p1)p2|r+p1/an}bracketri}ht/an}bracketle{tr| (5) In this expression τ=eiπ(d+1) d, the vectors |0/an}bracketri}ht,...|d−1/an}bracketri}htare an orthonormal basis, and the addition in |r+p1/an}bracketri}htis modulod. For more details see, for example, ref. [ 16]. One should not attach too much weight to the fact that all known SI C-POVMs have a group covariance property as this may only reflect the fact that group co- variant SIC-POVMs are much easier to construct. So in this paper w e will try to prove as much as we can without assuming such a property. One pot ential benefit ofthis attitude is that, by accumulatingenough facts about SIC-P OVMsin general, we may eventually get to the point where we can answer the question , whether all SIC-POVMs actually do have a group covariance property. The fact that the d2operatorsΠ rare linearly independent means that they form a basis for the complex Lie algebra gl( d,C) (the set of all operators acting on Hd). Since the Π rare Hermitian, then iΠrforms a basis also for the real Lie algebra u(d) (the set of all anti-Hermitian operators acting on Hd). So for any operator A∈gl(d,C) there is a unique set of expansion coefficients arsuch that A=d2/summationdisplay r=1arΠr (6) To find the expansion coefficients we can use the fact that d2/summationdisplay s=1Tr(ΠrΠs)/parenleftbiggd+1 dδst−1 d2/parenrightbigg =δrt (7) from which it follows ar=d+1 dTr(ΠrA)−1 dTr(A) (8) Specializing to the case A= ΠrΠswe find ΠrΠs=d+1 d d2/summationdisplay t=1TrstΠt −dδrs+1 d+1I (9) where Trst= Tr(Π rΠsΠt) (10) To a large extent this paper consists in an exploration of the proper ties of these important quantities, which we will refer to as the triple products. T hey are inti- mately related to the geometric phase, in which context they are us ually referred to as 3-vertex Bargmann invariants (see Mukunda et al[50], and references cited therein). We have, as an immediate consequence of the definition, Trst=Ttrs=Tstr=T∗ rts=T∗ tsr=T∗ srt (11) It is convenient to define Jrst=d+1 d(Trst−T∗ rst) (12) Rrst=d+1 d(Trst+T∗ rst) (13)4 SoJrstis imaginary and completely anti-symmetric; Rrstis real and completely symmetric. Both these quantities play a significant role in the theory . It follows from Eq. ( 9) that [Πr,Πs] =d2/summationdisplay t=1JrstΠt (14) So theJrstare structure constants for the Lie algebra gl( d,C). As an immediate consequence of this they satisfy the Jacobi identity: d2/summationdisplay b=1/parenleftbig JrsbJtba+JstbJrba+JtrbJsba/parenrightbig = 0 (15) for allr,s,t,a. The Jacobi identity holds for any representation of the structu re constants. In the following sections we will derive many other identit ies which are specific to this particular representation. Turning to the quantities Rrst, it follows from Eq. ( 9) that they feature in the expression for the anti-commutator {Πr,Πs}=/summationdisplay tRrstΠt−2(dδrs+1) d+1I (16) They also play an important role in the description of quantum state s pace. Let ρbe any density matrix and let pr=1 dTr(Πrρ) be the probability of obtaining outcomerin the measurement described by the POVM with elements1 dΠr. Then it follows from Eq. ( 8) thatρcan be reconstructed from the probabilities by ρ=d2/summationdisplay r=1/parenleftbigg (d+1)pr−1 d/parenrightbigg Πr (17) Suppose, now, that the prareanyset ofd2real numbers. So we do not assume that theprare even probabilities, let alone the probabilities coming from a density matrix according to the prescription pr=1 dTr(Πrρ). Then it is shown in ref. [ 34] that theprare in fact the probabilities coming from a pure state if and only if they satisfy the two conditions d2/summationdisplay r=1p2 r=2 d(d+1)(18) d2/summationdisplay r,s,t=1Rrstprpspt=2(d+7) d(d+1)2(19) Let us look at the quantities JrstandRrstin a little more detail. For each r choose a unit vector |ψr/an}bracketri}htsuch that Π r=|ψr/an}bracketri}ht/an}bracketle{tψr|. Then the Gram matrix for these vectors is of the form Grs=/an}bracketle{tψr|ψs/an}bracketri}ht=Krseiθrs(20) where the matrix θrsis anti-symmetric and Krs=/radicalbigg dδrs+1 d+1(21) Note that the SIC-POVM does not determine the angles θrsuniquely since making the replacements |ψr/an}bracketri}ht →eiφr|ψr/an}bracketri}htleaves the SIC-POVM unaltered, but changes5 the angles θrsaccording to the prescription θrs→θrs−φr+φs. This freedom to rephase the vectors |ψr/an}bracketri}htis not usually important. However, it sometimes has interesting consequences (see Section 9). It can be thought of as a kind of gauge freedom. The Gram matrix satisfies an important identity. Every SIC-POVM ha s the 2-design property [ 5,17] d2/summationdisplay r=1Πr⊗Πr=2d d+1Psym (22) wherePsymis the projector onto the symmetric subspace of Hd⊗Hd. Expressed in terms of the Gram matrix this becomes d2/summationdisplay r=1Gs1rGs2rGrt1Grt2=d d+1/parenleftbig Gs1t1Gs2t2+Gs1t2Gs2t1/parenrightbig (23) Turning to the triple products we have Trst=GrsGstGtr=KrsKstKtreiθrst(24) where θrst=θrs+θst+θtr (25) Note that the tensor θrstis completely anti-symmetric. In particular θrst= 0 if any two of the indices are the same. Note also that re-phasing the vect ors|ψr/an}bracketri}htleaves the tensors Trstandθrstunchanged. They are in that sense gauge invariant. Finally, we have the following expressions for JrstandRrst: Jrst=2i d√ d+1sinθrst (26) Rrst=2(d+1) dKrsKstKtrcosθrst (27) Like the triple products, JrstandRrstare gauge invariant. For later reference let us note that the matrix Jr, with matrix elements (Jr)st=Jrst (28) is the adjoint representative of Π rin the SIC-projector basis: adΠrΠs= [Πr,Πs] =d2/summationdisplay t=1JrstΠt (29) It can be seen that all the interesting features of the tensor Grs(respectively, the tensors Trst,JrstandRrst) are contained in the order-2 angle tensor θrs(re- spectively, the order-3 angle tensor θrst). It is also easy to see that, for any unitary U, the transformation Πr→UΠrU†(30) leaves the angle tensors invariant. This suggests that we shift our focus from indi- vidual SIC-POVMs to families of unitarily equivalent SIC-POVMs—SIC- families, as we will call them for short. We begin our investigation in Section 2by giving necessary and sufficient con- ditions for an arbitrary tensor θrs(respectively θrst) to be the rank-2 (respectively rank-3) angle tensor corresponding to a SIC-family. We also show t hat either angle tensor uniquely determines the corresponding SIC-family. Finally we describe a6 method for reconstructing the SIC-family, starting from a knowle dge of either of the two angle tensors. In Sections 3,4and5we prove the central result of this paper: namely, that the existence of a SIC-POVM in dimension dis equivalent to the existence of a certain very special set of matrices in the adjoint representation of gl(d,C). In Section3we show that, for any SIC-POVM, the adjoint matrices Jrhave the spectral decomposition Jr=Qr−QT r (31) whereQris a rankd−1 projector which has the remarkable property of being orthogonal to its own transpose: QrQT r= 0 (32) We refer to this feature of the adjoint matrices as the Q-QTproperty. In Section 3 we also show that from a knowledge of the Jmatrices it is possible to reconstruct the corresponding SIC-family. In Section 4we characterize the general class of projectors which have the property of being orthogonal to their own transpose. Then, in Section 5, we prove a converse of the result established in Section 3. The Q-QTproperty is not completely equivalent to the property of being a SIC set. However, it turns out that it is, in a certain sense, very nearly equiv alent. To be more specific: let Lrbe any set of d2Hermitian operators which constitute a basis for gl(d,C) and letCrbe the adjoint representative of Lrin this basis. Then the necessary and sufficient condition for the Crto have the spectral decomposition Cr=Qr−QT r (33) whereQris a rankd−1 projector such that QrQT r= 0 is that there exists a SIC set Π rsuch thatLr=ǫr(Πr+αI) for some fixed number α∈Rand signs ǫr=±1. In particular, the existence of an Hermitian basis for gl( d,C) having the Q-QTproperty is both necesary and sufficient for the existence of a SIC -POVM in dimensiond. In Section 6we digress briefly, and consider sl( d,C) (the Lie algebra consisting of all trace-zero d×dcomplex matrices). As we have explained, this paper is motivated by the hope that a Lie algebraic perspective will cast light o n the SIC- existence problem, rather than by an interest in Lie algebras as suc h. We focus on gl(d,C) because that is the casewherethe connection with SIC-POVMsse ems most straightforward. However a SIC-POVM also gives rise to an interes ting geometrical structure in sl( d,C), as we show in Section 6. In Section 7we derive a number of additional identities satisfied by the Jand Qmatrices. The complex projectors Qr,QT rand the real projector Qr+QT rdefine three families of subspaces. It turns out that there are some interestin g geometrical relationships between these subspaces, which we study in Section 8. Finally, in Section 9we show that, with the appropriate choice of gauge, the Gram matrix corresponding to a Weyl-Heisenberg covariant SIC-fa mily has a fea- ture analogous to the Q-QTproperty, which we call the P-PTproperty. It is an open question whether this result generalizes to other SIC-families , not covariant with respect to the Weyl-Heisenberg group.7 2.The Angle Tensors The purpose of this section is to establish the necessary and sufficie nt conditions for an arbitrary tensor θrs(respectively θrst) to be the order-2 (respectively order- 3) angle tensor for a SIC-family. We will also show that either one of t he angle tensors is enough to uniquely determine the SIC-family. Moreover, we will describe explicit procedures for reconstructing the family, starting from a knowledge of one of the angle tensors. We begin by considering the general class of POVMs (not just SIC-P OVMs) which consist of d2rank-1 elements. A POVM of this type is thus defined by a set ofd2vectors|ξ1/an}bracketri}ht,...,|ξd2/an}bracketri}htwith the property d2/summationdisplay r=1|ξr/an}bracketri}ht/an}bracketle{tξr|=I (34) Note that/summationtextd2 r=1/vextenddouble/vextenddouble|ξr/an}bracketri}ht/vextenddouble/vextenddouble2=d, so the vectors |ξr/an}bracketri}htcannot all be normalized. In the particular case of a SIC-POVM the vectors all have the same norm/vextenddouble/vextenddouble|ξr/an}bracketri}ht/vextenddouble/vextenddouble=1√ d. However in the general case they may have different norms. Given a set of such vectors consider the Gram matrix Prs=/an}bracketle{tξr|ξs/an}bracketri}ht (35) Clearly the Gram matrix cannot determine the POVM uniquely since if Uis any unitary operator then the vectors U|ξr/an}bracketri}htwill define another POVM having the same Gram matrix. However, the theorem we now prove shows that this is the only free- dom. In other words, the Gram matrix fixes the POVM up to unitary e quivalence. The theorem also provides us with a criterion for deciding whether an arbitrary d2×d2matrixPis the Gram matrix corresponding to a POVM of the specified type. As a corollary this will give us a criterion for deciding whether an arbitrary tensorθrsis specifically the order-2 angle tensor for a SIC-family. Theorem 1. LetPbe anyd2×d2Hermitian matrix. Then the following conditions are equivalent: (1)Pis a rankdprojector. (2)Psatisfies the trace identities Tr(P) = Tr(P2) = Tr(P3) = Tr(P4) =d (36) (3)Pis the Gram matrix for a set of d2vectors|ξr/an}bracketri}ht(not all normalized) such that|ξr/an}bracketri}ht/an}bracketle{tξr|is a POVM: /an}bracketle{tξr|ξs/an}bracketri}ht=Prs (37) d2/summationdisplay r=1|ξr/an}bracketri}ht/an}bracketle{tξr|=I (38) SupposePsatisfies these conditions. To construct a POVM correspondi ng toP let thedcolumn vectors ξ11 ξ12 ... ξ1d2 , ξ21 ξ22 ... ξ2d2 ,..., ξd1 ξd2 ... ξdd2 (39)8 be any orthonormal basis for the subspace onto which Pprojects. Define |ξr/an}bracketri}ht=d/summationdisplay a=1ξ∗ ar|a/an}bracketri}ht (40) where the vectors |a/an}bracketri}htare any orthonormal basis for Hd. ThenPis the Gram matrix for the vectors |ξ1/an}bracketri}ht,...,|ξd2/an}bracketri}ht. Moreover, the necessary and sufficient condition for any other set of vectors |η1/an}bracketri}ht,...,|ηd2/an}bracketri}htto have Gram matrix Pis that there exist a unitary operator Usuch that |ηr/an}bracketri}ht=U|ξr/an}bracketri}ht (41) for allr. Proof.We begin by showing that (3) = ⇒(1). Suppose |ξ1/an}bracketri}ht,...|ξd2/an}bracketri}htis any set of d2vectors such that |ξr/an}bracketri}ht/an}bracketle{tξr|is a POVM. So d2/summationdisplay r=1|ξr/an}bracketri}ht/an}bracketle{tξr|=I (42) Let Prs=/an}bracketle{tξr|ξs/an}bracketri}ht (43) be the Gram matrix. Then Pis Hermitian. Moreover, P2=Psince d2/summationdisplay t=1PrtPts=/an}bracketle{tξr| d2/summationdisplay t=1|ξt/an}bracketri}ht/an}bracketle{tξs| |ξr/an}bracketri}ht =/an}bracketle{tξr|ξs/an}bracketri}ht =Prs (44) Also Tr(P) =d2/summationdisplay r=1/an}bracketle{tξr|ξr/an}bracketri}ht=d (45) (as can be seen by taking the trace on both sides of Eq. ( 42)). SoPis a rank-d projector. We next show that (1) = ⇒(3). LetPbe a rank-dprojector, and let the d column vectors ξ11 ξ12 ... ξ1d2 , ξ21 ξ22 ... ξ2d2 ,..., ξd1 ξd2 ... ξdd2 (46) be an orthonormal basis for the subspace onto which it projects. So d2/summationdisplay r=1ξ∗ arξbr=δab (47) for alla,b, and d2/summationdisplay a=1ξarξ∗ as=Prs (48)9 for allr,s. Now let |ξ1/an}bracketri}ht,...|ξd2/an}bracketri}htbe the vectors defined by Eq. ( 40). Then it follows from Eq. ( 47) that d2/summationdisplay r=1|ξr/an}bracketri}ht/an}bracketle{tξr|=d/summationdisplay a,b=1 d2/summationdisplay r=1ξ∗ arξbr |a/an}bracketri}ht/an}bracketle{tb| =d/summationdisplay a=1|a/an}bracketri}ht/an}bracketle{ta| =I (49) implying that |ξr/an}bracketri}ht/an}bracketle{tξr|is POVM. Also, it follows from Eq. ( 48) that /an}bracketle{tξr|ξs/an}bracketri}ht=d/summationdisplay a=1ξarξ∗ as=Prs (50) implying that the |ξr/an}bracketri}hthave Gram matrix P. We next turn to condition (2). The fact that (1) = ⇒(2) is immediate. To prove the reverse implication observe that condition (2) implies Tr(P4)−2Tr(P3)+Tr(P2) = 0 (51) Letλ1,...,λ d2be the eigenvalues of P. Then Eq. ( 51) implies d2/summationdisplay r=1λ2 r(λr−1)2= 0 (52) It follows that each eigenvalue is either 0 or 1. Since Tr( P) =dwe must have d eigenvalues = 1 and the rest all = 0. So Pis a rank-dprojector. It remains to show that the POVM corresponding to a given rank- dprojector is unique up to unitary equivalence. To prove this let Pbe a rank-dprojector, let |ξr/an}bracketri}htbe the vectors defined by Eq. ( 40), and let |η1/an}bracketri}ht,...,|ηd2/an}bracketri}htbe any other set of vectors such that /an}bracketle{tηr|ηs/an}bracketri}ht=Prs (53) for allr,s. Define ηar=/an}bracketle{tηr|a/an}bracketri}ht (54) Then d2/summationdisplay r=1η∗ arηbr=/an}bracketle{ta| d2/summationdisplay r=1|ηr/an}bracketri}ht/an}bracketle{tηr| |b/an}bracketri}ht=δab (55) (because |ηr/an}bracketri}ht/an}bracketle{tηr|is a POVM) and d/summationdisplay a=1ηarη∗ as=Prs (56) (because the |ηr/an}bracketri}hthave Gram matrix P). So thedcolumn vectors η11 η12 ... η1d2 , η21 η22 ... η2d2 ,..., ηd1 ηd2 ... ηdd2 (57)10 are an orthonormal basis for the subspace onto which Pprojects. But the column vectors ξ11 ξ12 ... ξ1d2 , ξ21 ξ22 ... ξ2d2 ,..., ξd1 ξd2 ... ξdd2 (58) are also an orthonormal basis for this subspace. So there must ex ist ad×dunitary matrixUabsuch that ηar=d/summationdisplay b=1Uabξbr (59) for alla,r. Define U=d/summationdisplay a,b=1U∗ ab|a/an}bracketri}ht/an}bracketle{tb| (60) Then |ηr/an}bracketri}ht=U|ξr/an}bracketri}ht (61) for allr. /square In the case of a SIC-POVM we have |ξr/an}bracketri}ht=1√ d|ψr/an}bracketri}ht (62) where the vectors |ψr/an}bracketri}htare normalized, and Prs=1 dGrs=1 dKrseiθrs(63) whereGis the Gram matrix of the vectors |ψr/an}bracketri}htandθrsis the order-2 angle tensor. In the sequel we will distinguish these matrices by referring to Gas the Gram matrix and Pas the Gram projector. We have Corollary 2. Letθrsbe a real anti-symmetric tensor. Then the following state- ments are equivalent: (1)θrsis an order- 2angle tensor corresponding to a SIC-family. (2)θrssatisfies d2/summationdisplay t=1KrtKtsei(θrt+θts)=dKrseiθrs(64) for allr,s. (3)θrssatisfies d2/summationdisplay r,s,t=1KrsKstKtrei(θrs+θst+θtr)=d4(65) and d2/summationdisplay r,s,t,u=1KrsKstKtuKurei(θrs+θst+θtu+θur)=d5(66)11 LetΠr,Π′ rbe two different SIC-sets, and let θrs,θ′ rsbe corresponding order- 2 angle tensors. Then there exists a unitary Usuch that Π′ r=UΠrU†(67) for allrif and only if θ′ rs=θrs−φr+φs (68) for some arbitrary set of phase angles φr(in other words two SIC-sets are unitarily equivalent if and only if their order- 2angle tensors are gauge equivalent). A SIC-family can be reconstructed from its order- 2angle tensor θrsby calculating an orthonormal basis for the subspace onto which the Gram pro jector Prs=1 dKrseiθrs(69) projects, as described in Theorem 1. Remark. The sense in which we areusing the term “gaugeequivalence”is explain ed in the passage immediately following Eq. ( 21). Note that condition (2) imposes d2(d2−1)/2 independent constraints (taking account of the anti-symmetry of θrs). Condition (3), by contrast, only imposes 2 independent constraints. It is to be observed, however, that th e price we pay for the reduction in the number of equations is that Eqs. ( 65) and (65) are respectively cubic and quartic in the phases, whereas Eq. ( 64) is only quadratic. Proof.Letθrsbe an arbitrary anti-symmetric tensor, and define Prs=1 dKrseiθrs(70) The anti-symmetry of θrsmeans that Pis automatically Hermitian. So it follows from Theorem 1that a necessary and sufficient condition for Prsto be a rank- d projector, and for θrsto be the order-2 angle tensor of a SIC-family, is that d2/summationdisplay t=1KrtKtsei(θrt+θts)=dKrseiθrs(71) for allr,s. To prove the equivalence of conditions (1) and (3) note that the co nditions Tr(P) = Tr(P2) =dare an automatic consequence of Phaving the specified form. So it follows from Theorem 1thatθrsis the order-2 angle tensor of a SIC-family if and only if Eqs. ( 65) and (66) are satisfied. Now let Πr, Π′ rbe two SIC-sets and let θrs,θ′ rsbe order-2 angle tensors corre- sponding to them. Then there exist normalized vectors |ψr/an}bracketri}ht,|ψ′ r/an}bracketri}htsuch that Πr=|ψr/an}bracketri}ht/an}bracketle{tψr| Π′ r=|ψ′ r/an}bracketri}ht/an}bracketle{tψ′ r| (72) for allr, and /an}bracketle{tψr|ψs/an}bracketri}ht=Krseiθrs/an}bracketle{tψ′ r|ψ′ s/an}bracketri}ht=Krseiθ′ rs (73) for allr,s. Suppose, first of all, that there exists a unitary Usuch that Π′ r=UΠrU†(74)12 Then there exist phase angles φrsuch that |ψ′ r/an}bracketri}ht=eiφrU|ψr/an}bracketri}ht (75) for allr, which is easily seen to imply that θ′ rs=θrs−φr+φs (76) for allr,s. Soθrs,θ′ rsare gauge equivalent. Conversely, suppose there exist phase angles φrsuch that θ′ rs=θrs−φr+φs (77) Define |ψ′′ r/an}bracketri}ht=e−iφr|ψ′ r/an}bracketri}ht (78) Then /an}bracketle{tψ′′ r|ψ′′ s/an}bracketri}ht=Krseiθrs=/an}bracketle{tψr|ψs/an}bracketri}ht (79) for allr,s. So it follows from Theorem 1that there exists a unitary Usuch that |ψ′′ r/an}bracketri}ht=U|ψr/an}bracketri}ht (80) for allr. Consequently Π′ r=|ψ′′ r/an}bracketri}ht/an}bracketle{tψ′′ r|=UΠrU†(81) for allr. So Πrand Π′ rare unitarily equivalent. /square We now turn to the order-3 angle tensors. We have Theorem 3. Letθrstbe a real completely anti-symmetric tensor. Then the follow - ing conditions are equivalent: (1)θrstis the order- 3angle tensor for a SIC-family (2)For some fixed aand allr,s,t θars+θast+θatr=θrst (82) and for all r,s d2/summationdisplay t=1KrtKtseiθrst=dKrs (83) (3)For some fixed aand allr,s,t θars+θast+θatr=θrst (84) and d2/summationdisplay r,s,t=1KrsKstKtreiθrst=d4(85) d2/summationdisplay r,s,t,u=1KrsKstKtuKurei(θrst+θtur)=d5(86)13 LetΠr,Π′ rbe two different SIC-sets and let θrst,θ′ rstbe the corresponding order- 3angle tensors. Then the necessary and sufficient condition fo r there to exist a unitaryUsuch that Π′ r=UΠrU†(87) for allris thatθ′ rst=θrstfor allr,s,t(in other words two SIC-sets are unitarily equivalent if and only if their order- 3angle tensors are identical). Letθrstbe the order- 3angle tensor corresponding to a SIC-family. Then the order-2angle tensor is given by (up to gauge freedom) θrs=θars (88) for any fixed a, from which the SIC-family can be reconstructed using the me thod described in Theorem 1. Remark. Unlike the order-2tensor, the order-3angletensoris gaugeinvar iant. This means that it provides what is, in many ways, a more useful charact erization of the SIC-family. For that reason we will be almost exclusively concern ed with the order-3 tensor in the remainder of this paper. Proof.The fact that (1) = ⇒(2) is an immediate consequence of the definition of theorder-3angletensorandcondition(2)ofCorollary 2. Toprovethat(2) = ⇒(1) letθrstbe a completely anti-symmetric tensor such that condition (2) holds . Define θrs=θars (89) for allr,s. Then Eq. ( 83) implies d2/summationdisplay t=1KrtKtsei(θrt+θts)=eiθrs d2/summationdisplay t=1KrtKtseiθrst ∗ =dKrseiθrs(90) for allr,s. It follows from Corollary 2thatθrsis the order-2 and θrstthe order-3 angle tensor of a SIC-family. The equivalence of conditions (1) and (3) is proved similarly. It remains to show that two SIC-sets are unitarily equivalent if and o nly if their order-3 angle tensors are identical. To see this let Πr=|ψr/an}bracketri}ht/an}bracketle{tψr|and Π′ r= |ψ′ r/an}bracketri}ht/an}bracketle{tψ′ r|be two different SIC-sets having the same order-3 angle tensor θrst. Let θrs(respectively θ′ rs) be the order-2 angle tensor corresponding to the vectors |ψr/an}bracketri}ht (respectively |ψ′ r/an}bracketri}ht). Choose some fixed index a. We have θ′ ar+θ′ sa+θ′ rs=θar+θsa+θrs (91) for allr,s. Consequently θ′ rs=θrs+φr−φs (92) for allr,s, where φr=θar−θ′ ar (93) Soθ′ rsandθrsare gauge equivalent. It follows from Corollary 2that Πrand Π′ rare unitarily equivalent. Conversely, suppose that Πrand Π′ rare unitarily equivalent, and letθrs,θ′ rsbe order-2 angle tensors corresponding to them. It follows from Corollary 2thatθrsandθ′ rsare gauge equivalent. It is then immediate that the order-3 angle tensors are identical. /square14 Finally, let us note that when expressed in terms of the triple produc ts Eq. (83) reads d2/summationdisplay t=1Trst=dK2 rs (94) while Eq. ( 85) reads d2/summationdisplay r,s,t=1Trst=d4(95) For Eq. ( 86) we have to work a little harder. We have d2/summationdisplay r,s,t,u=11 K2 rtTrstTtur=d5(96) from which it follows d5=d2/summationdisplay r,s,t,u=1/parenleftbig −dδrt+d+1/parenrightbig TrstTtur = (d+1)d2/summationdisplay r,s,t,u=1TrstTtur−dd2/summationdisplay r,s,u=1K2 rsK2 ru = (d+1)d2/summationdisplay r,s,t,u=1TrstTtur−d5(97) Consequently d2/summationdisplay s,u=1Tr/parenleftbig TsTu/parenrightbig =d2/summationdisplay r,s,t,u=1TrstTtur=2d5 d+1(98) This equation be alternatively written d2/summationdisplay r,s=1Tr/parenleftbig TrTs/parenrightbig =2d5 d+1(99) whereTris the matrix with matrix elements ( Tr)uv=Truv. When they are written like this, in terms of the triple products, the f act that Eq. (94) implies Eqs. ( 95) and (98) becomes almost obvious. The reverse implica- tion, by contrast, is rather less obvious. 3.Spectral Decompositions LetTr,Jr,Rrbe thed2×d2matrices whose matrix elements are (Tr)st=Trst (Jr)st=Jrst (Rr)st=Rrst(100) whereJrst,RrstarethequantitiesdefinedbyEqs.( 12)and(13). SoJristheadjoint representation matrix of Π r. In this section we derive the spectral decompositions of these matrices. To avoid confusion we will use the notation |ψ/an}bracketri}htto denote a ket in ddimensional Hilbert space Hd, and/bardblψ/an}bracketri}ht/an}bracketri}htto denote a ket in d2dimensional Hilbert15 spaceHd2. In terms of this notation the spectral decompositions will turn ou t to be: Tr=d d+1Qr+2d d+1/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (101) Jr=Qr−QT r (102) Rr=Qr+QT r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (103) In these expressions the vector /bardbler/an}bracketri}ht/an}bracketri}htis normalized, and its components in the stan- dard basis are all real. Qris a rankd−1 projector such that Qr/bardbler/an}bracketri}ht/an}bracketri}ht=QT r/bardbler/an}bracketri}ht/an}bracketri}ht= 0 (104) and which has, in addition, the remarkable property of being orthog onal to its own transpose (also a rank d−1 projector): QrQT r= 0 (105) Explicit expressions for /bardbler/an}bracketri}ht/an}bracketri}htandQrwill be given below. It will be convenient to define the rank 2( d−1) projector ¯Rr=Qr+QT r (106) We have ¯Rr=J2 r (107) and Rr=¯Rr+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (108) SinceQris Hermitian we have QT r=Q∗ r (109) whereQ∗ ris the matrix whose elements are the complex conjugates of the cor re- sponding elements of Qr. So¯Rris twice the real part of Qrand−iJris twice the imaginary part. In Section 5we will show that Eq. ( 102) is essentially definitive of a SIC-POVM. To be more specific, let Lrbe any set of d2Hermitian matrices which constitute a basis for gl( d,C), and letCrbe the adjoint representative of Lrin that basis. Then we will show that Crhas the spectral decomposition Cr=Qr−QT r (110) whereQris a rankd−1 projector which is orthogonal to its own transpose if and only if the Lrare a family of SIC projectors up to multiplication by a sign and shifting by a multiple of the identity. Having stated our results let us now turn to the task of proving the m. We begin byderivingthespectraldecompositionof Tr. Multiplyingboth sidesoftheequation ΠrΠs=d+1 dd2/summationdisplay t=1TrstΠt−K2 rsI (111) by Πrwe find ΠrΠs=d+1 dd2/summationdisplay t=1TrstΠrΠt−K2 rsΠr16 =(d+1)2 d2d2/summationdisplay t=1(Tr)2 stΠt−d+1 dd2/summationdisplay t=1TrstK2 rtI−K2 rsΠr(112) We have d2/summationdisplay t=1TrstK2 rt=1 d+1d2/summationdisplay t=1Trst(dδrt+1) =1 d+1 dTrsr+d2/summationdisplay t=1Trst =2d d+1Tsrr =2d d+1K2 rs (113) Consequently ΠrΠs=d+1 dd2/summationdisplay t=1/parenleftbiggd+1 d(Tr)2 st−K2 rsK2 rt/parenrightbigg Πt−K2 rsI (114) Comparing with Eq. ( 111) we deduce (Tr)2 rs=d d+1Trst+d d+1K2 rsK2 rt (115) Now define /bardbler/an}bracketri}ht/an}bracketri}ht=/radicalbigg d+1 2dd2/summationdisplay s=1K2 rs/bardbls/an}bracketri}ht/an}bracketri}ht (116) where the basis kets /bardbls/an}bracketri}ht/an}bracketri}htare given by (in column vector form) /bardbl1/an}bracketri}ht/an}bracketri}ht= 1 0 ... 0 ,/bardbl2/an}bracketri}ht/an}bracketri}ht= 0 1 ... 0 ,...,/bardbld2/an}bracketri}ht/an}bracketri}ht= 0 0 ... 1 (117) It is easily verified that /bardbler/an}bracketri}ht/an}bracketri}htis normalized. Eq. ( 115) then becomes T2 r=d d+1Tr+2d2 (d+1)2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (118) Using Eq. ( 113) we find /an}bracketle{t/an}bracketle{ts/bardblTr/bardbler/an}bracketri}ht/an}bracketri}ht=/radicalbigg d+1 2dd2/summationdisplay t=1TrstK2 rt =/radicalbigg 2d d+1K2 rs =2d d+1/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht (119) So/bardbler/an}bracketri}ht/an}bracketri}htis an eigenvector of Trwith eigenvalue2d d+1.17 Also define Qr=d+1 dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (120) So in terms of the order-3 angle tensor the matrix elements of Qrare Qrst=d+1 dKrsKrt/parenleftbig Ksteiθrst−KrsKrt/parenrightbig (121) Qris Hermitian (because Tris Hermitian). Moreover Q2 r=(d+1)2 d2T2 r−4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl=Qr (122) SoQris a projection operator. Since Tr(Tr) =/summationdisplay uTruu=d2/summationdisplay u=1K2 ru=d (123) we have Tr(Qr) =d−1 (124) We have thus proved that the spectral decomposition of Tris Tr=d d+1Qr+2d d+1/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (125) whereQris a rankd−1 projector, as claimed. We next prove that QT r/bardbler/an}bracketri}ht/an}bracketri}ht= 0. The fact that the components of /bardbler/an}bracketri}ht/an}bracketri}htin the standard basis are all real means /an}bracketle{t/an}bracketle{ts/bardblTT r/bardbler/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{ter/bardblTr/bardbls/an}bracketri}ht/an}bracketri}ht=2d d+1/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht (126) So/bardbler/an}bracketri}ht/an}bracketri}htis an eigenvector of TT ras well asTr, again with the eigenvalue2d d+1. In view of Eq. ( 120) it follows that QT r/bardbler/an}bracketri}ht/an}bracketri}ht= 0. Turning to the problem of showing that Qris orthogonal to its own transpose. We have QrQT r=/parenleftbiggd+1 dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg/parenleftbiggd+1 dTT r−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg =(d+1)2 d2TrTT r−4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (127) It follows from Eq. ( 24) that /an}bracketle{t/an}bracketle{ts/bardblTrTT r/bardblt/an}bracketri}ht/an}bracketri}ht=d2/summationdisplay u=1TrsuTrtu =GrsGrtd2/summationdisplay u=1GsuGtuGurGur (128) In view of Eq. ( 23) (i.e.the fact that every SIC-POVM is a 2-design) this implies /an}bracketle{t/an}bracketle{ts/bardblTrTT r/bardblt/an}bracketri}ht/an}bracketri}ht=2d d+1|Grs|2|Grt|2 =2d d+1K2 rsK2 rt18 =4d2 (d+1)2/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardblt/an}bracketri}ht/an}bracketri}ht (129) So TrTT r=4d2 (d+1)2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (130) and consequently QrQT r= 0 (131) Eqs. (102) and (103) are immediate consequences of the results already proved and the definitions of Jr,Rr. We definedthe Jmatricestobe theadjointrepresentativesofthe SIC-projecto rs, considered as a basis for the Lie algebra gl( d,C), and that is certainly a most important fact about them. However, the results of this section s how that, along with the vectors /bardbler/an}bracketri}ht/an}bracketri}ht, they actually determine the whole structure. Specifically, we have Qr=1 2/parenleftbig Jr+J2 r/parenrightbig (132) Rr=J2 r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (133) Tr=d 2(d+1)/parenleftig Jr+J2 r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightig (134) Moreover, if we know the Tmatrices then we know the order-3 angle tensor, which in view of Theorem 3means we can reconstruct the SIC-projectors. Since the vectors/bardbler/an}bracketri}ht/an}bracketri}htare given, once and for all, this means that the problem of proving th e existenceofa SIC-POVMin dimension dis equivalent to the problem ofprovingthe existence of a certain remarkable structure in the adjoint repres entation of gl( d,C) (as we will see in more detail in Section 5). In the Introduction webegan with the concept ofa SIC-POVM,and then defined theJmatrices in terms of it. However, one could, if one wished, go in the op posite direction, and take the Lie algebraic structure to be primary, with t he SIC-POVM being the secondary, derivative entity. 4.TheQ-QTProperty The next five sections are devoted to a study of the Jmatrices which, as we will see, have numerous interesting properties. We begin our investiga tion by trying to get some additional insight into what we will call the Q-QTproperty: namely, the fact that the Jmatrices have the spectral decomposition Jr=Qr−QT r (135) whereQrisarankd−1projectorwhichisorthogonaltoits owntranspose. We wish to characterize the general class of matrices which are of this typ e. The following theorem provides one such characterization. Theorem 4. LetAbe a Hermitian matrix. Then the following statements are equivalent:19 (1)Ahas the spectral decomposition A=P−PT(136) wherePis a projector which is orthogonal to its own transpose. (2)Ais pure imaginary and A2is a projector. Proof.To show that (1) = ⇒(2) observe that the fact that Pis Hermitian means PT=P∗(137) whereP∗is the matrix whose elements are the complex conjugates of the cor re- sponding elements of P. So Eq. ( 136) implies that the components of Aare pure imaginary. Since PPT= 0 it also implies that A2is a projector. To show that (2) = ⇒(1) observe that the fact that A2is a projector means that the eigenvalues of A=±1 or 0. So A=P−P′(138) whereP,P′are orthogonal projectors. Since Ais pure imaginary we must have PT−(P′)T=AT=A∗=−A=P′−P (139) PTand (P′)Tare also orthogonal projectors. So if PT|ψ/an}bracketri}ht=|ψ/an}bracketri}ht, and|ψ/an}bracketri}htis nor- malized, we must have 1 =/an}bracketle{tψ|PT|ψ/an}bracketri}ht =/angbracketleftbig ψ/vextendsingle/vextendsingle/parenleftbig PT−(P′)T/parenrightbig/vextendsingle/vextendsingleψ/angbracketrightbig =/an}bracketle{tψ|P′|ψ/an}bracketri}ht−/an}bracketle{tψ|P|ψ/an}bracketri}ht (140) Since 0≤ /an}bracketle{tψ|P′|ψ/an}bracketri}ht ≤1 (141) 0≤ /an}bracketle{tψ|P|ψ/an}bracketri}ht ≤1 (142) we must have /an}bracketle{tψ|P′|ψ/an}bracketri}ht= 1, implying P′|ψ/an}bracketri}ht=|ψ/an}bracketri}ht. Similarly P′|ψ/an}bracketri}ht=|ψ/an}bracketri}htimplies PT|ψ/an}bracketri}ht=|ψ/an}bracketri}ht. So P′=PT(143) /square We also have the following statement, inspired in part by Ref. [ 51], Theorem 5. The necessary and sufficient condition for a matrix Pto be a projector which is orthogonal to its own transpose is that P=SDST(144) whereSis an any real orthogonal matrix and Dhas the block-diagonal form D= σ ... 0 0...0 ............ 0... σ 0...0 0...0 0...0 ............ 0...0 0...0 (145)20 with σ=1 2/parenleftbigg 1−i i1/parenrightbigg (146) In other words Dhasncopies ofσon the diagonal, where n= rank(P), and0 everywhere else. Proof.Sufficiency is an immediate consequence of the fact that σis a rank 1 pro- jector such that σσT= 0. To prove necessity let dbe the dimension of the space and nthe rank of P. It will be convenient to define |1/an}bracketri}ht= 1 0 ... 0 ,|2/an}bracketri}ht= 0 1 ... 0 , ... |d/an}bracketri}ht= 0 0 ... 1 (147) In terms of these basis vectors we have P=d/summationdisplay r,s=1Prs|r/an}bracketri}ht/an}bracketle{ts| (148) Now let |a1/an}bracketri}ht,...,|an/an}bracketri}htbe an orthonormal basis for the subspace onto which P projects, and let |a∗ r/an}bracketri}htbe the column vector which is obtained from |ar/an}bracketri}htby tak- ing the complex conjugate of each of its components. Taking comple x conjugates on each side of the equation P|ar/an}bracketri}ht=|ar/an}bracketri}ht (149) gives P∗|a∗ r/an}bracketri}ht=|a∗ r/an}bracketri}ht (150) So|a∗ 1/an}bracketri}ht,...,|a∗ n/an}bracketri}htis an orthonormal basis for the subspace onto which PT=P∗ projects. Since PTis orthogonal to Pwe conclude that /an}bracketle{tar|a∗ s/an}bracketri}ht= 0 (151) for allr,s. Next define vectors |b1/an}bracketri}ht,...,|b2n/an}bracketri}htby |b2r−1/an}bracketri}ht=1√ 2/parenleftbig |a∗ r/an}bracketri}ht−|ar/an}bracketri}ht/parenrightbig (152) |b2r/an}bracketri}ht=i√ 2/parenleftbig |a∗ r/an}bracketri}ht+|ar/an}bracketri}ht/parenrightbig (153) By construction these vectors are orthonormal and real. So we c an extend them to an orthonormal basis for the full space by adding a further d−2nvectors |b2n+1/an}bracketri}ht,...,|bd/an}bracketri}ht, which can also be chosen to be real. We have P=n/summationdisplay r=1|ar/an}bracketri}ht/an}bracketle{tar| =1 2n/summationdisplay r=1/parenleftig |b2r−1/an}bracketri}ht/an}bracketle{tb2r−1|−i|b2r−1/an}bracketri}ht/an}bracketle{tb2r|+i|b2r/an}bracketri}ht/an}bracketle{tb2r−1|+|b2r/an}bracketri}ht/an}bracketle{tb2r|/parenrightig (154)21 So if we define S=d/summationdisplay r=1|br/an}bracketri}ht/an}bracketle{tr| (155) thenSis a real orthogonal matrix such that P=SDST(156) where D=1 2n/summationdisplay r=1/parenleftig |2r−1/an}bracketri}ht/an}bracketle{t2r−1|−i|2r−1/an}bracketri}ht/an}bracketle{t2r|+i|2r/an}bracketri}ht/an}bracketle{t2r−1|+|2r/an}bracketri}ht/an}bracketle{t2r|/parenrightig (157) is the matrix defined by Eq. ( 145). /square This result implies the following alternative characterization of the cla ss of ma- trices to which the Jmatrices belong Corollary 6. LetAbe a Hermitian matrix. Then the following statements are equivalent: (1)Ahas the spectral decomposition A=P−PT(158) wherePis a projector which is orthogonal to its own transpose. (2)There exists a real orthogonal matrix Ssuch that A=SDST(159) whereDhas the block diagonal form D= σy...0 0...0 ............ 0... σ y0...0 0...0 0...0 ............ 0...0 0...0 (160) σybeing the Pauli matrix σy=/parenleftbigg0−i i0/parenrightbigg (161) In other words Dhasncopies ofσyon the diagonal, where n=1 2rank(A), and0everywhere else (note that a matrix of this type must have eve n rank). Proof.Immediate consequence of Theorem 5. /square 5.Lie Algebraic Formulation of the Existence Problem This section is the core of the paper. We show that the problem of pr oving the existence of a SIC-POVM in dimension dis equivalent to the problem of proving the existence of an Hermitian basis for gl( d,C) all of whose elements have the Q-QT property. We hope that this new way of thinking will help make the SIC -existence problem more amenable to solution. The result we prove is the following:22 Theorem 7. LetLrbe a set ofd2Hermitian matrices forming a basis for gl(d,C). LetCrstbe the structure constants relative to this basis, so that [Lr,Ls] =d2/summationdisplay t=1CrstLt (162) and letCrbe the matrix with matrix elements (Cr)st=Crst. Then the following statements are equivalent (1)EachCrhas the spectral decomposition Cr=Pr−PT r (163) wherePris a rankd−1projector which is orthogonal to its own transpose. (2)There exists a SIC-set Πr, a set of signs ǫr=±1and a real constant α/ne}ationslash=−1 dsuch that Lr=ǫr(Πr+αI) (164) Remark. The restriction to values of α/ne}ationslash=−1 dis needed to ensure that the matrices Lrare linearly independent, and therefore constitute a basis for gl( d,C) (otherwise they would all have trace = 0). The Q-QTproperty continues to hold even if α does =−1 d. It will be seen that it is not only SIC-sets which have the Q-QTproperty, but also any set of operators obtained from a SIC-set by shifting by a c onstant and multiplying by an r-dependent sign. Sothe Q-QTpropertyis not strictly equivalent to the property of being a SIC-set. However, it could be said that t he properties are almost equivalent. In particular, the existence of an Hermitian b asis for gl(d,C) having the Q-QTproperty implies the existence of a SIC-POVM in dimension d, and conversely. Proof that (2) =⇒(1).Taking the trace on both sides of [Πr,Πs] =d2/summationdisplay t=1JrstΠt (165) we deduce that d2/summationdisplay t=1Jrst= 0 (166) Then from the definition of Lrin terms of Π rwe find Crst=ǫrǫsǫtJrst (167) Consequently Cr=Pr−PT r (168) where Pr=ǫrSQrS (169) Sbeing the symmetric orthogonal matrix S= ǫ10...0 0ǫ2...0 ......... 0 0... ǫ d2 (170) The claim is now immediate.23 Proof that (1) =⇒(2).Forthis we need to workharder. Since the proofis rather lengthy we will break it into a number of lemmas. We first collect a few ele mentary facts which will be needed in the sequel: Lemma 8. LetLrbe any Hermitian basis for gl(d,C), and letCrstandCrbe the structure constants and adjoint representatives as defi ned in the statement of Theorem 7. Letlr= Tr(Lr). Then (1)Thelrare not all zero. (2)TheCrstare pure imaginary and antisymmetric in the first pair of indi ces. (3)TheCrstare completely antisymmetric if and only if the Crare Hermitian. (4)In every case d2/summationdisplay t=1Crstlt= 0 (171) for allr,s. (5)In the special case that the Crare Hermitian d2/summationdisplay r=1lrLr=κI (172) where κ=1 d d2/summationdisplay r=1l2 r >0 (173) Proof.To prove (1) observe that if the lrwere all zero it would mean that the identity was not in the span of the Lr—contrary to the assumption that they form a basis. To prove(2) observethat taking Hermitian conjugates on both sid es of Eq. ( 162) gives −[Lr,Ls] =d2/summationdisplay t=1C∗ rstLt (174) from which it follows that C∗ rst=−Crst. The fact that Csrt=−Crstis an imme- diate consequence of the definition. (3) is now immediate. (4) is proved in the same way as Eq. ( 166). To prove (5) observe that if the Crare Hermitian it follows from (2) and (3) that d2/summationdisplay r=1lrCrst= 0 (175) for alls,t. Consequently the matrix d2/summationdisplay r=1lrLr (176)24 commutes with everything. But the only matrices for which that is tr ue are multi- ples of the identity. It follows that d2/summationdisplay r=1lrLr=κI (177) for some real κ. Taking the trace on both sides of this equation we deduce d2/summationdisplay r=1l2 r=dκ (178) The fact that κ>0 is a consequence of this and statement (1). /square We next observe that if the Crhave theQ-QTproperty they must, in particular, be Hermitian. It turns out that that is, by itself, already a very str ong constraint. Before stating the result it may be helpful if we explain the essential idea on which it depends. Although we have not done so before, and will not d o so again, it will be convenient to make use of the covariant/contravariantinde x notation which is often used to describe the structure constants. Define the me tric tensor Mrs= Tr(LrLs) (179) and letMrsbe its inverse. So d2/summationdisplay t=1MrtMts=Mr s=/braceleftigg 1r=s 0r/ne}ationslash=s(180) We can use these tensors to raise and lower indices (we use the Hilber t-Schmidt inner product for this purpose because the fact that gl( d,C) is not semi-simple means that its Killing form is degenerate [ 52–55]). In particular, the matrices Lr=d2/summationdisplay t=1MrsLs (181) are the basis dual to the Lr: Tr(LrLs) =Mr s (182) Suppose we now define structure constants ˜Crstby [Lr,Ls] =d2/summationdisplay t=1˜CrstLt(183) (so in terms of the Crstwe have ˜Ct rs=Crst). It follows from the relation ˜Crst= Tr/parenleftbig [Lr,Ls]Lt/parenrightbig = Tr/parenleftbig Lr[Ls,Lt]/parenrightbig (184) that the ˜Crstare completely antisymmetric for any choice of the Lr. If we now require that the matrices Crbe Hermitian it means that, not only the ˜Crst, but also theCrstmust be completely antisymmetric. Since the two quantities are related by ˜Crst=d2/summationdisplay u=1CrsuMut (185)25 this is a very strong requirement. It means that the Lrmust, in a certain sense, be close to orthonormal (relative to the Hilbert-Schmidt inner prod uct). More precisely, it means we have the following lemma: Lemma 9. LetLr,CrstandCrbe defined as in the statement of Theorem 7, and letlr= Tr(Lr). Then the Crare Hermitian if and only if Tr(LrLs) =βδrs+γlrls (186) whereβ,γare real constants such that β >0andγ <1 d. If this condition is satisfied we also have d2/summationdisplay r=1lrLr=β 1−dγI (187) d2/summationdisplay r=1l2 r=dβ 1−dγ(188) Proof.To prove sufficiency observe that, in view of Eq. ( 185), the condition means ˜Crst=βCrst+γltd2/summationdisplay u=1Crsulu (189) In view of Lemma 8, and the fact that β/ne}ationslash= 0, this implies Crst=1 β˜Crst (190) Since the ˜Crstare completely antisymmetric we conclude that the Crstmust be also. It follows that the Crare Hermitian. To prove necessity let ˜Cr(respectively M) be the matrix whose matrix elements are˜Crst(respectively Mst). Then Eq. ( 185) can be written ˜Cr=CrM (191) Taking the transpose (or, equivalently, the Hermitian conjugate) on both sides of this equation we find ˜Cr=MCr (192) implying [M,Cr] = 0 (193) for allr. Since the Lrare a basis for gl( d,C) we deduce [M,adA] = 0 (194) for allA∈gl(d,C). Eq. (186) is a straightforward consequence of this, the fact that gl(d,C) has the direct sum decomposition CI⊕sl(d,C), the fact that sl( d,C) is simple, and Schur’s lemma [ 52–55]. However, for the benefit of the reader who is not so familiar with the theory of Lie algebras we will give the argument in a little more detail.26 Given arbitrary A=/summationtextd2 r=1arLr, let/bardblA/an}bracketri}ht/an}bracketri}htdenote the column vector /bardblA/an}bracketri}ht/an}bracketri}ht= a1 a2 ... ad2 (195) So /bardblLr/an}bracketri}ht/an}bracketri}ht= 1 0 ... 0 /bardblL2/an}bracketri}ht/an}bracketri}ht= 0 1 ... 0 /bardblLd2/an}bracketri}ht/an}bracketri}ht= 0 0 ... 1 (196) In view of Lemma 8we then have /bardblI/an}bracketri}ht/an}bracketri}ht=1 κd2/summationdisplay r=1lr/bardblLr/an}bracketri}ht/an}bracketri}ht (197) Since Tr(A) =d2/summationdisplay r=1arlr=κ/an}bracketle{t/an}bracketle{tI/bardblA/an}bracketri}ht/an}bracketri}ht (198) we have that A∈sl(d,C) if and only if /an}bracketle{t/an}bracketle{tI/bardblA/an}bracketri}ht/an}bracketri}ht= 0. Now observe that it follows from Lemma 8and the definition of Mthat M/bardblI/an}bracketri}ht/an}bracketri}ht=κ/bardblI/an}bracketri}ht/an}bracketri}ht (199) IfMis a multiple of the identity we have Mrs=κδrsand the lemma is proved. OtherwiseMhas at least one more eigenvalue, βsay. Let Ebe the corresponding eigenspace. Since Eis orthogonal to /bardblI/an}bracketri}ht/an}bracketri}htit follows from Eq. ( 198) thatE⊆sl(d,C). SinceMcommutes with every adjoint representation matrix we have adAE⊆E (200) for allA∈sl(d,C). SoEis an ideal of sl( d,C). However sl( d,C) is a simple Lie algebra, meaning it has no proper ideals [ 52–55]. So we must have E= sl(d,C). It follows that if we define ˜Lr=Lr−lr dI (201) then M/bardblLr/an}bracketri}ht/an}bracketri}ht=lr dM/bardblI/an}bracketri}ht/an}bracketri}ht+M/bardbl˜Lr/an}bracketri}ht/an}bracketri}ht (202) =κlr d/bardblI/an}bracketri}ht/an}bracketri}ht+β/bardbl˜Lr/an}bracketri}ht/an}bracketri}ht (203) =d2/summationdisplay s=1(βδrs+γlrls)/bardblLs/an}bracketri}ht/an}bracketri}ht (204) whereγ=1 d/parenleftig 1−β κ/parenrightig . Eqs. (186), (187) and (188) are now immediate (in view of Lemma8).27 It remains to establish the bounds on β,γ. LetA=/summationtextd2 r=1arLrbe any non-zero element of sl( d,C). Then/summationtextd2 r=1arlr= 0, so in view of Eq. ( 186) we have 0<Tr(A2) =βd2/summationdisplay r=1a2 r (205) It follows that β >0. Also, using Lemma 8once more, we find lr=1 κd2/summationdisplay s=1lsTr(LrLs) =βlr κ+γlr κd2/summationdisplay s=1l2 s =lr/parenleftbiggβ κ+dγ/parenrightbigg (206) Since thelrcannot all be zero this implies β κ= 1−dγ (207) Sinceβ κ>0 we deduce that γ <1 d. /square Eq. (186) only depends on the Crbeing Hermitian. If we make the assumption that theCrhave theQ-QTproperty we get a stronger statement: Corollary 10. LetLr,CrstandCrbe as defined in the statement of Theorem 7. Suppose that the Crhave the spectral decomposition Cr=Pr−PT r (208) wherePris a rankd−1projector which is orthogonal to its own transpose. Then (1)For allr Tr(Lr) =ǫ′ rl (209) (2)For allr,s Tr(LrLs) =d d+1δrs+ǫ′ rǫ′ s d/parenleftbigg l2−1 d+1/parenrightbigg (210) (3) d2/summationdisplay r=1ǫ′ rLr=dlI (211) for some real constant l>0and signsǫ′ r=±1. Proof.The proof relies on the fact that the Killing form for gl( d,C) is related to the Hilbert-Schmidt inner product by [ 55] Tr(adAadB) = 2dTr(AB)−2Tr(A)Tr(B) (212) Specializing to the case A=B=Lrand making use of the Q-QTproperty we find d−1 =dTr(L2 r)−l2 r (213) Using Lemma 9we deduce l2 r=dβ−d+1 1−dγ(214)28 It follows that lr=ǫ′ rl (215) for some real constant l≥0 and signs ǫ′ r=±1. The fact that the Lrare a basis for gl(d,C) means the lrcannot all be zero. So we must in fact have l>0. Using this result in Eq. ( 188) we find β+d2l2γ=dl2(216) while Eq. ( 214) implies dβ+dl2γ=d−1+l2(217) This gives us a pair of simultaneous equations in βandγ. Solving them we obtain β=d d+1(218) γ=1 dl2/parenleftbigg l2−1 d+1/parenrightbigg (219) Substituting these expressions into Eqs. ( 186) and (187) we deduce Eqs. ( 210) and (211). /square The next lemma shows that each Lris a linear combination of a rank-1projector and the identity: Lemma 11. LetLbe any Hermitian matrix ∈gl(d,C)which is not a multiple of the identity. Then rank(ad L)≥2(d−1) (220) The lower bound is achieved if and only if Lis of the form L=ηI+ξP (221) wherePis a rank- 1projector and η,ξare any pair of real numbers. The eigenvalues ofadLare then ±ξ(each with multiplicity d−1) and0(with multiplicity d2−2d+2). Proof.Letλ1≥λ2≥ ··· ≥λdbe the eigenvalues of Larranged in decreasing order, and let |b1/an}bracketri}ht,|b2/an}bracketri}ht,...,|bd/an}bracketri}htbe the correspondingeigenvectors. We mayassume, without loss of generality, that the |br/an}bracketri}htare orthonormal. We have adL/parenleftbig |br/an}bracketri}ht/an}bracketle{tbs|/parenrightbig =/bracketleftbig L,|br/an}bracketri}ht/an}bracketle{tbs|/bracketrightbig = (λr−λs)|br/an}bracketri}ht/an}bracketle{tbs| (222) So the eigenvalues of ad Lareλr−λs. SinceLis not a multiple of the identity we must haveλr/ne}ationslash=λr+1for somerin the range 1 ≤r≤d−1. We then have that λs−λt/ne}ationslash= 0 if either s≤r<tort≤r<s. There are 2 r(d−r) such pairs s,t. So rank(ad L)≥2r(d−r)≥2(d−1) (223) Suppose, now that the lower bound is achieved. Then r(d−r) =d−1, implying thatr= 1 ord−1. Also we must have λs=λs+1for alls/ne}ationslash=r. So either L=λ2I+(λ1−λ2)|b1/an}bracketri}ht/an}bracketle{tb1| (224) or L=λd−1I−(λd−1−λd)|bd/an}bracketri}ht/an}bracketle{tbd| (225) Either way Land the spectrum of ad Lare as described. /square The final ingredient needed to complete the proof is29 Lemma 12. LetLr,CrstandCrbe as defined in the statement of Theorem 7. Suppose that the Crhave the spectral decomposition Cr=Pr−PT r (226) wherePris a rankd−1projector which is orthogonal to its own transpose. Let l, ǫ′ rbe as in the statement of Corollary 10. Then there is a fixed sign ǫ=±1such that Πr=ǫǫ′ rLr−ǫl−1 dI (227) is a rank- 1projector for all r. Proof.Define L′ r=ǫ′ rLr−l−1 dI (228) Then it follows from Corollary 10that Tr(L′ r) = 1 (229) for allr, Tr(L′ rL′ s) =dδrs+1 d+1(230) for allr,s, and d2/summationdisplay r=1L′ r=dI (231) It is also easily seen that if we define C′ rst=ǫ′ rǫ′ sǫ′ tCrstthen [L′ r,L′ s] =d2/summationdisplay t=1C′ rstL′ t (232) and C′ r=P′ r−P′ rT(233) whereP′ ris a rank-1 projector which is orthogonal to its own transpose (se e the first part of the proof of Theorem 7). In particular rank/parenleftbig adL′r/parenrightbig = 2(d−1) (234) and the eigenvalues of ad L′rall equal to ±1 or 0. So, taking account of the fact that Tr(L′ r) = 1, we can use Lemma 11to deduce that there is a family of rank-1 projectors Π′ rand signsξr=±1 such that L′ r=ξrΠ′ r+1−ξr dI (235) Ifξr= +1 (respectively −1) for allrthen Eq. ( 227) holds with Π r= Π′ randǫ= +1 (respectively −1). Also, if d= 2 thenL′ ris a rank-1 projector irrespective of the value ofξr, so Eq. ( 227) holds with Π r=L′ randǫ= +1. The problem therefore reduces to showing that if d >2 it cannot happen that ξr= +1 for some values ofrand−1 for others. We will do this by assuming the contrary and deducing a contradiction. Letmbe the number of values of rfor whichξr= +1. We are assuming that mis in the range 1 ≤m≤d2−1. We may also assume, without loss of generality,30 that the labelling is such that ξr= +1 for the first mvalues ofr, and−1 for the rest. So L′ r=/braceleftigg Π′ r ifr≤m 2 dI−Π′ rifr>m(236) Now define ˜Trst= Tr/parenleftbig L′ rL′ sL′ t/parenrightbig (237) Eqs. (230) and (231) mean that the same argument which led to Eq. ( 9) can be used to deduce L′ rL′ s=d+1 d d2/summationdisplay t=1˜TrstL′ t −K2 rsI (238) SinceL′ 1is a projector it follows that L′ 1L′ s=/parenleftbig L′ 1/parenrightbig2L′ s=d+1 d d2/summationdisplay t=1˜T1stL′ 1L′ t −K2 1sL′ 1 (239) By essentially the same argument which led to Eq. ( 118) we can use this to infer /parenleftbig˜T′ 1/parenrightbig2=d d+1˜T1+2d2 (d+1)2/bardble1/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{te1/bardbl (240) where˜T′ 1is the matrix with matrix elements ˜T′ 1rsand/bardble1/an}bracketri}ht/an}bracketri}htis the vector defined by Eq.(116). Asbefore /bardble1/an}bracketri}ht/an}bracketri}htisaneigenvectorof ˜T′ 1witheigenvalue2d d+1. Consequently the matrix ˜Q1=d+1 d˜T′ 1−2/bardble1/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{te1/bardbl (241) is a projector. But that means Tr( ˜Q1) must be an integer. We now use this to derive a contradiction. It follows from Eq. ( 236) that (L′ r)2=/braceleftigg L′ r r≤m 2(d−2) d2I−d−4 dL′ rr>m(242) Consequently ˜T1rr=/braceleftigg K2 1r r≤m 2(d−2) d2−d−4 dK2 1rr>m(243) and so Tr(˜Q1) =d+1 dd2/summationdisplay r=1˜T1rr−2 =d+1−4d2+2m(d−2) d3(244) So if Tr( ˜Q1) is an integer/parenleftbig 4d2+2n(d−2)/parenrightbig /d3must also be an integer. But the fact that 1 ≤m<d2, together with the fact that d>2 means 4 d<4d2+2m(d−2) d3<2 (245)31 Ifd= 3 or 4 there are no integers in this interval, which gives us a contrad iction straight away. If, on the other hand, d≥5 there is the possibility 4d2+2m(d−2) d3= 1 (246) implying m=d2(d−4) 2(d−2)(247) This equationhasthe solution d= 6,m= 9(this is in fact the only integersolution, ascanbe seenfrom ananalysisofthe possible primefactorizationso fthe numerator and denominator on the right hand side). To eliminate this possibility de fine L′′ r=2 dI−L′ d2+1−r (248) for allr. It is easily verified that Tr(L′′ rL′′ s) =dδrs+1 d+1(249) d2/summationdisplay r=1L′′ r=dI (250) and L′′ r=/braceleftigg Πr r≤d2−m 2 dI−Πrr>d2−m(251) So we can go through the same argument as before to deduce d2−m=d2(d−4) 2(d−2)(252) Eqs. (247) and (252) have no joint solutions at all with d/ne}ationslash= 0, integer or otherwise. /square To complete the proof of Theorem 7observe that Eqs. ( 210) and (227) imply Tr(ΠrΠs) =dδrs+1 d+1(253) So the Π rare a SIC-set. Moreover Lr=ǫr(Πr+αI) (254) whereǫr=ǫǫ′ randα= (ǫl−1)/d. 6.The Algebra sl(d,C) The motivation for this paper is the hope that a Lie algebraic perspec tive may cast some light on the SIC-existence problem, and on the mathemat ics of SIC- POVMs generally. We have focused on gl( d,C) as that is the case where the con- nection with Lie algebras seems most straightforward. However, it may be worth mentioning that a SIC-POVM also gives rise to an interesting geometr ical structure in sl(d,C) (the Lie algebra consisting of all trace-zero d×dcomplex matrices).32 Let Πrbe a SIC-set and define Br=/radicaligg d+1 2(d2−1)/parenleftbigg Πr−1 dI/parenrightbigg (255) SoBr∈sl(d,C). Let /an}bracketle{tA,A′/an}bracketri}ht= Tr(ad AadA′) = 2dTr(AA′) (256) be the Killing form [ 55] on sl(d,C). Then /an}bracketle{tBr,Bs/an}bracketri}ht=/braceleftigg 1 r=s −1 d2−1r/ne}ationslash=s(257) So theBrform a regular simplex in sl( d,C). Since sl( d,C) isd2−1 dimensional theBrare an overcomplete set. However, the fact that d2/summationdisplay r=1Br= 0 (258) means that for each A∈sl(d,C) there is a unique set of numbers arsuch that A=d2/summationdisplay r=1arBr (259) and d2/summationdisplay r=1ar= 0 (260) Thearcan be calculated using ar=d2−1 d2/an}bracketle{tA,Br/an}bracketri}ht (261) Similarly, given any linear transformation M: sl(d,C)→sl(d,C), there is a unique set of numbers Mrssuch that MBr=d2/summationdisplay s=1MrsBs (262) and d2/summationdisplay s=1Mrs=d2/summationdisplay s=1Msr= 0 (263) for allr. TheMrscan be calculated using Mrs=d2−1 d2/an}bracketle{tBs,MBr/an}bracketri}ht (264) In short, the Brretain many analogous properties of, and can be used in much the same way as, a basis. It could be said that they form a simplicial basis.33 7.Further Identities In the preceding pages we have seen that there are five different f amilies of ma- trices naturally associated with a SIC-POVM: namely, the projecto rsQrtogether with the matrices Jr=Qr−QT r (265) ¯Rr=Qr+QT r (266) Rr=Qr+QT r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (267) Tr=d d+1Qr+2d d+1/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (268) (see Section 3). As we noted previously, it is possible to define everything in terms of the adjoint representation matrices Jrand the rank-1 projectors /bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl: Qr=1 2Jr(Jr+I) (269) ¯Rr=J2 r (270) Rr=J2 r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (271) Tr=d 2(d+1)Jr(Jr+I)+2d d+1/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl (272) In that sense the structure constants of the Lie algebra, supple mented with the vectors/bardbler/an}bracketri}ht/an}bracketri}ht, determine everything else. In the next section we will show that there are some interesting geo metrical relationships between the hyperplanes onto which Qr,QT rand¯Rrproject. In this section, as a preliminary to that investigation, we prove a number of identities satisfied by the Q,Jand¯Rmatries. We start by computing their Hilbert-Schmidt inner products: Theorem 13. For allr,s Tr/parenleftbig QrQs/parenrightbig =d3δrs+d2−d−1 (d+1)2(273) Tr/parenleftbig QrQT s/parenrightbig =d2(1−δrs) (d+1)2(274) Tr/parenleftbig JrJs/parenrightbig =2(d2δrs−1) d+1(275) Tr/parenleftbig¯Rr¯Rs/parenrightbig =2(d−1)(d2δrs+2d+1) (d+1)2(276) Tr/parenleftbig Jr¯Rs/parenrightbig = 0 (277) Proof.Let us first calculate some auxiliary quantities. It follows from the de finition ofTr, andthe factthat the matrix P=1 dGdefined byEq.( 63) isarankdprojector, that Tr(TrTs) =d2/summationdisplay u,v=1TruvTsvu34 =d2/summationdisplay u,v=1K2 uvGruGusGsvGvr =d d+1d2/summationdisplay u=1K2 ruK2 su+d4 d+1d2/summationdisplay u,v=1PruPusPsvPvr =d2(dδrs+d+2) (d+1)3+d4 d+1/vextendsingle/vextendsinglePrs/vextendsingle/vextendsingle2 =d2(dδrs+d+2) (d+1)3+d2 d+1K2 rs =d2/parenleftbig d(d+2)δrs+2d+3/parenrightbig (d+1)3(278) Also Tr/parenleftbig TrTT s/parenrightbig =d2/summationdisplay u,v=1TruvTsuv =d2/summationdisplay u=1GruGsu d2/summationdisplay v=1GuvGuvGvrGvs =2d d+1d2/summationdisplay u=1GruGsuGurGus =2d2 (d+1)2/parenleftbig 1+K2 rs/parenrightbig =2d2(dδrs+d+2) (d+1)3(279) where we made two applications of Eq. ( 23) (i.e.the fact that every SIC-POVM is a 2-design). Finally, it is a straightforward consequence of the defi nitions ofTr,TT r and/bardbler/an}bracketri}ht/an}bracketri}htthat /an}bracketle{t/an}bracketle{ter/bardblTs/bardbler/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{ter/bardblTT s/bardbler/an}bracketri}ht/an}bracketri}ht =d+1 2dd2/summationdisplay u,v=1TsuvK2 ruK2 rv =1 2d(d+1) d2Tsrr+dd2/summationdisplay v=1Tsrv+dd2/summationdisplay u=1Tsur+d2/summationdisplay u,v=1Tsuv =d 2(d+1)/parenleftbig 3K2 rs+1/parenrightbig =d(3dδrs+d+4) 2(d+1)2(280)35 and /an}bracketle{t/an}bracketle{ter/bardbles/an}bracketri}ht/an}bracketri}ht=d+1 2dd2/summationdisplay u=1K2 ruK2 su =dδrs+d+2 2(d+1)(281) Using these results in the expressions Tr/parenleftbig QrQs/parenrightbig = Tr/parenleftigg/parenleftbiggd+1 dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg/parenleftbiggd+1 dTs−2/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardbl/parenrightbigg/parenrightigg (282) and Tr/parenleftbig QrQT s/parenrightbig = Tr/parenleftigg/parenleftbiggd+1 dTr−2/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightbigg/parenleftbiggd+1 dTT s−2/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardbl/parenrightbigg/parenrightigg (283) the first two statements follow. The remaining statements are imme diate conse- quences of these and the fact that Jr=Qr−QT r (284) ¯Rr=Qr+QT r (285) /square Now define /bardblv0/an}bracketri}ht/an}bracketri}ht=1 dd2/summationdisplay r=1/bardblr/an}bracketri}ht/an}bracketri}ht (286) where/bardblr/an}bracketri}ht/an}bracketri}htis the basis defined in Eq. ( 117). The following result shows (among other things) that the subspaces onto which the Qr(respectively QT r,Rr) project span the orthogonal complement of /bardblv0/an}bracketri}ht/an}bracketri}ht. Theorem 14. For allr Qr/bardblv0/an}bracketri}ht/an}bracketri}ht=QT r/bardblv0/an}bracketri}ht/an}bracketri}ht=Jr/bardblv0/an}bracketri}ht/an}bracketri}ht=Rr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (287) Moreover d2/summationdisplay r=1Qr=d2/summationdisplay r=1QT r=d2 d+1/parenleftbig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig (288) d2/summationdisplay r=1Jr= 0 (289) d2/summationdisplay r=1¯Rr=2d2 d+1/parenleftbig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig (290) Proof.Some of this is a straightforward consequence of the fact that Jris the adjoint representative of Π r. Since d2/summationdisplay s=1Πs=dI (291)36 we must have d2/summationdisplay s,t=1JrstΠt=d2/summationdisplay s=1adΠrΠs= 0 (292) In view of the antisymmetry of the Jrstit follows that d2/summationdisplay r=1Jr= 0 (293) and Jr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (294) Using the relations Qr=1 2Jr(Jr+I) (295) QT r=1 2Jr(Jr−I) (296) ¯Rr=J2 r (297) we deduce Qr/bardblv0/an}bracketri}ht/an}bracketri}ht=QT r/bardblv0/an}bracketri}ht/an}bracketri}ht=¯Rr/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (298) It remains to prove Eqs. ( 288) and (290). It follows from Eq. ( 120) that d2/summationdisplay r=1Qrst=d+1 dd2/summationdisplay r=1Trst−2d2/summationdisplay r=1/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardblt/an}bracketri}ht/an}bracketri}ht = (d+1)K2 st−d+1 dd2/summationdisplay r=1K2 rsK2 rt =d2δst−1 d+1(299) from which it follows d2/summationdisplay r=1Qr=d2/summationdisplay r=1QT r=d2 d+1/parenleftbig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightbig (300) Eq. (290) follows from this and the fact that Rr=Qr+QT r. /square 8.Geometrical Considerations In this section we show that there are some interesting geometrica l relationships between the subspaces onto which the operators Qr,QT rand¯Rrproject. The original motivation for this work was an observation concerning the subspaces onto which the ¯Rrproject. ¯Rris a real matrix, and so it defines a 2( d−2) subspace inRd2, which we will denote Rr. We noticed that for each pair of distinct indices randsthe intersection Rr∩Rsis a 1-dimensional line. This led us to speculate that a set of hyperplanes parallel to the Rrmight be the edges of an interesting polytope. We continue to think that this could be the case. Unfortu nately we have not been able to prove it. However, it appears to us that the result s we obtained37 while trying to prove it have an interest which is independent of the tr uth of the motivating speculation. We will begin with some terminology. Let Pbe any projector (on either RN orCN), letPbe the subspace onto which Pprojects, and let |ψ/an}bracketri}htbe any non-zero vector. Then we define the angle between |ψ/an}bracketri}htandPin the usual way, to be θ= cos−1/parenleftigg/vextenddouble/vextenddoubleP|ψ/an}bracketri}ht/vextenddouble/vextenddouble /vextenddouble/vextenddouble|ψ/an}bracketri}ht/vextenddouble/vextenddouble/parenrightigg (301) (soθis the smallest angle between |ψ/an}bracketri}htand any of the vectors in P). Suppose, now, that P′is another projector, and let P′be the subspace onto whichP′projects. We will say that P′is uniformly inclined to Pif every vector in P′makes the same angle θwithP. Ifθ= 0 this means that P′⊆P, while ifθ=π 2 it means P′⊥P. Suppose, on the other hand, that 0 < θ <π 2. Let|u′ 1/an}bracketri}ht,...,|u′ n/an}bracketri}ht be any orthonormal basis for P′, and define |ur/an}bracketri}ht= secθP|u′ r/an}bracketri}ht. Then/an}bracketle{tur|ur/an}bracketri}ht= 1 for allr. Moreover, if P,P′are complex projectors, /an}bracketle{tu′ r+eiφu′ s|P|u′ r+eiφu′ s/an}bracketri}ht= 2cos2θ/parenleftig 1+Re/parenleftbig eiφ/an}bracketle{tur|us/an}bracketri}ht/parenrightbig/parenrightig (302) for allφandr/ne}ationslash=s. On the other hand it follows from the assumption that P′is uniformly inclined to Pthat /an}bracketle{tu′ r+eiφu′ s|P|u′ r+eiφu′ s/an}bracketri}ht= 2cos2θ (303) for allφandr/ne}ationslash=s. Consequently /an}bracketle{tur|us/an}bracketri}ht=δrs (304) for allr,s. It is easily seen that the same is true if P,P′are real projectors. Suppose we now make the further assumption that dim( P′) = dim( P) =n. Then |u1/an}bracketri}ht,...,|un/an}bracketri}htis an orthonormal basis for P, and we can write P=n/summationdisplay r=1|ur/an}bracketri}ht/an}bracketle{tur| (305) P′=n/summationdisplay r=1|u′ r/an}bracketri}ht/an}bracketle{tu′ r| (306) Observe that /an}bracketle{tu′ r|us/an}bracketri}ht=/an}bracketle{tu′ r|P|us/an}bracketri}ht= cosθ/an}bracketle{tur|us/an}bracketri}ht= cosθδrs (307) for allr,s. Consequently P′|ur/an}bracketri}ht= cosθ|ur/an}bracketri}ht (308) for allr. It follows that /vextenddouble/vextenddoubleP′|ψ/an}bracketri}ht/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay r=1cosθ/an}bracketle{tur|ψ/an}bracketri}ht|u′ r/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble|ψ/an}bracketri}ht/vextenddouble/vextenddouble (309) for all|ψ/an}bracketri}ht ∈P. SoPis uniformly inclined to P′at the same angle θ. It follows from Eqs. ( 305) and (306) that PP′P= cos2θP (310) P′PP′= cos2θP′(311) Eq. (310), or equivalently Eq. ( 311), is not only necessary but also sufficient for the subspaces to be uniformly inclined. In fact, let P,P′be any two subspaces38 which have the same dimension n, but which are not assumed at the outset to be uniformly inclined, and let P,P′be the corresponding projectors. Suppose PP′P= cos2θP (312) for someθin the range 0 ≤θ≤π 2. It is immediate that P=P′ifθ= 0, and P⊥P′ifθ=π 2. Either way, the subspaces are uniformly inclined. Suppose, on the other hand, that 0 <θ<π 2. Let|u′ 1/an}bracketri}ht,...,|u′ n/an}bracketri}htbe any orthonormal basis for P′, and define |ur/an}bracketri}ht= secθP|u′ r/an}bracketri}ht. Eq. (305) then implies P= sec2θn/summationdisplay r=1P|u′ r/an}bracketri}ht/an}bracketle{tu′ r|P=n/summationdisplay r=1|ur/an}bracketri}ht/an}bracketle{tur| (313) Given any |ψ/an}bracketri}ht ∈Pwe have |ψ/an}bracketri}ht=P|ψ/an}bracketri}ht=n/summationdisplay r=1/an}bracketle{tur|ψ/an}bracketri}ht|ur/an}bracketri}ht (314) Since dim( P) =nit follows that the |ur/an}bracketri}htare linearly independent. In particular |ur/an}bracketri}ht=P|ur/an}bracketri}ht=n/summationdisplay s=1/an}bracketle{tus|ur/an}bracketri}ht|us/an}bracketri}ht (315) Since the |ur/an}bracketri}htare linearly independent this means /an}bracketle{tus|ur/an}bracketri}ht=δrs (316) So the|ur/an}bracketri}htare an orthonormal basis for P. It follows, that if |ψ′/an}bracketri}htis any vector in P′, then /vextenddouble/vextenddoubleP|ψ′/an}bracketri}ht/vextenddouble/vextenddouble=/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay r=1/an}bracketle{tu′ r|ψ′/an}bracketri}htP|u′ r/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddoublen/summationdisplay r=1/an}bracketle{tu′ r|ψ′/an}bracketri}ht|ur/an}bracketri}ht/vextenddouble/vextenddouble/vextenddouble/vextenddouble/vextenddouble= cosθ/vextenddouble/vextenddouble|ψ′/an}bracketri}ht/vextenddouble/vextenddouble(317) implying that P′is uniformly inclined to Pat angleθ. It will be convenient to summarise all this in the form of a lemma: Lemma 15. LetP,P′be any two subspaces, real or complex, having the same dimensionn. LetP,P′be the corresponding projectors. Then the following state- ments are equivalent: (a)P′is uniformly inclined to Pat angleθ. (b)Pis uniformly inclined to P′at angleθ. (c) PP′P= cos2θP (318) (d) P′PP′= cos2θP′(319) Suppose these conditions are satisfied for some θin the range 0< θ <π 2, and let|u1/an}bracketri}ht,...|un/an}bracketri}htbe any orthonormal basis for P. Then there exists an orthonormal basis|u′ 1/an}bracketri}ht,...,|u′ n/an}bracketri}htforP′such that P′|ur/an}bracketri}ht= cosθ|u′ r/an}bracketri}ht (320) P|u′ r/an}bracketri}ht= cosθ|ur/an}bracketri}ht (321) We are now in a position to state the main results of this section. Let Qr (respectively ¯Qr) be the subspace onto which Qr(respectively QT r) projects. We then have39 Theorem 16. For each pair of distinct indices r,sthe subspaces Qr,¯Qrhave the orthogonal decomposition Qr=Q0 rs⊕Qrs (322) ¯Qr=¯Q0 rs⊕¯Qrs (323) where Q0 rs⊥Qrs dim(Q0 rs) = 1 dim( Qrs) =d−2 ¯Q0 rs⊥¯Qrs dim(¯Q0 rs) = 1 dim( ¯Qrs) =d−2 We have (a)Relation of the subspaces QrandQs: (1)Q0 rs⊥QsrandQrs⊥Q0 sr. (2)Q0 rsandQ0 srare inclined at angle cos−1/parenleftbig1 d+1/parenrightbig . (3)QrsandQsrare uniformly inclined at angle cos−1/parenleftig 1√d+1/parenrightig . (b)Relation of the subspaces ¯Qrand¯Qs: (1)¯Q0 rs⊥¯Qsrand¯Qrs⊥¯Q0 sr. (2)¯Q0 rsand¯Q0 srare inclined at angle cos−1/parenleftbig1 d+1/parenrightbig . (3)¯Qrsand¯Qsrare uniformly inclined at angle cos−1/parenleftig 1√d+1/parenrightig . (c)Relation of the subspaces Qrand¯Qs: (1)Q0 rs⊥¯Qsr,Qrs⊥¯Q0 srandQrs⊥¯Qsr. (2)Q0 rsand¯Q0 srare inclined at angle cos−1/parenleftbigd d+1/parenrightbig . The relations between these subspaces are, perhaps, easier to a ssimilate if pre- sented pictorially. In the following diagrams the line joining each pair of subspaces is labelled with the cosine of the angle between them. In particular a 0 o n the line joining two subspaces indicates that they are orthogonal. Q0 rs Qrs Q0 sr Qsr0 01 d+11√d+1 0❅ ❅ ❅ ❅ ❅ ❅❅0¯Q0 rs¯Qrs ¯Q0 sr¯Qsr0 01 d+11√d+1 0❅ ❅ ❅ ❅ ❅ ❅❅0 Q0 rs Qrs ¯Q0 sr¯Qsr0 0d d+10 0❅ ❅ ❅ ❅ ❅ ❅❅040 Wewillprovethistheorembelow. Beforedoingso,however,letusst atetheother mainresult ofthis section. Let Rrbe the subspace ontowhichthe ¯Rrproject. Since ¯Rris a real matrix we regard Rras a subspace of Rd2. We have Theorem 17. For each pair of distinct indices r,sthe subspace Rrhas the decom- position Rr=R0 rs⊕R1 rs⊕Rrs (324) whereR0 rs,R1 rs,Rrsare pairwise orthogonal and dim(R0 rs) = 1 dim( R1 rs) = 1 dim( Rrs) = 2d−4 (325) We have (1)R0 rs=R0 sr. (2)R1 rs⊥RsrandRrs⊥R1 sr. (3)R1 rsandR1 srare inclined at angle cos−1/parenleftbigd−1 d+1/parenrightbig . (4)RrsandRsrare uniformly inclined at angle cos−1/parenleftig/radicalig 1 d+1/parenrightig In particular, the subspaces ¯Rrand¯Rsintersect in a line. In diagrammatic form the relations between these subspaces are R0 rs=R0 srR1 rs Rrs R1 sr Rsr0 0d−1 d+1/radicalig 1 d+1 0❅ ❅ ❅ ❅ ❅ ❅❅0✟✟✟✟✟0 ❍❍❍❍❍00 0 where, as before, each line is labelled with the cosine of the angle betw een the two subspaces it connects. Proof of Theorem 16.Let/bardbl1/an}bracketri}ht/an}bracketri}ht,...,/bardbld2/an}bracketri}ht/an}bracketri}htbethestandardbasisfor Hd2, asdefined by Eq. (117). For each pair of distinct indices r,sdefine /bardblfrs/an}bracketri}ht/an}bracketri}ht=i√ d+1Qr/bardbls/an}bracketri}ht/an}bracketri}ht (326) /bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=−i√ d+1QT r/bardbls/an}bracketri}ht/an}bracketri}ht (327) The significance of these vectors is that /bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl(respectively /bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl) will turn out to be the projectoronto the 1-dimensionalsubspace Q0 rs(respectively ¯Q0 rs).41 Note that the fact that Qris Hermitian means QT r=Q∗ r (328) (whereQ∗ ris the matrix whose elements are the complex conjugates of the cor re- sponding elements of Qr). Consequently /an}bracketle{t/an}bracketle{tt/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=/parenleftig /an}bracketle{t/an}bracketle{tt/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightig∗ (329) for allr,s,t. It is easily seen that /bardblfrs/an}bracketri}ht/an}bracketri}ht,/bardblf∗ rs/an}bracketri}ht/an}bracketri}htare normalized. In fact, it follows from Eqs. (116) and (120) that /an}bracketle{t/an}bracketle{tfrs/bardblfrs/an}bracketri}ht/an}bracketri}ht= (d+1)/an}bracketle{t/an}bracketle{ts/bardblQr/bardbls/an}bracketri}ht/an}bracketri}ht =(d+1)2 dTrss−2(d+1)/an}bracketle{t/an}bracketle{ts/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbls/an}bracketri}ht/an}bracketri}ht =(d+1)2 d/parenleftbig K2 rs−K4 rs/parenrightbig = 1 (330) for allr/ne}ationslash=s. In view of Eq. ( 329) we then have /an}bracketle{t/an}bracketle{tf∗ rs/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=/parenleftig /an}bracketle{t/an}bracketle{tfrs/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightig∗ = 1 (331) for allr/ne}ationslash=s. The fact that QrQT r= 0 means we also have /an}bracketle{t/an}bracketle{tfrs/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht= 0 (332) for allr/ne}ationslash=s. Note that, although we required that r/ne}ationslash=sin the definitions of /bardblfrs/an}bracketri}ht/an}bracketri}ht,/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht, the definitions continue to make sense when r=s. However, the vectors are then zero (as can be seen by setting r=sin Eq. (121)). The vectors /bardblfrs/an}bracketri}ht/an}bracketri}ht,/bardblf∗ rs/an}bracketri}ht/an}bracketri}htsatisfy a number of identities, which it will be conve- nient to collect in a lemma: Lemma 18. For allr/ne}ationslash=s /bardblfrs/an}bracketri}ht/an}bracketri}ht=−/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht+i/radicalbigg 2 d/parenleftig /bardbles/an}bracketri}ht/an}bracketri}ht−/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig (333) /bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=−/bardblfsr/an}bracketri}ht/an}bracketri}ht−i/radicalbigg 2 d/parenleftig /bardbles/an}bracketri}ht/an}bracketri}ht−/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig (334) (where/bardbler/an}bracketri}ht/an}bracketri}htis the vector defined by Eq. ( 116)) Qr/bardblfrs/an}bracketri}ht/an}bracketri}ht=/bardblfrs/an}bracketri}ht/an}bracketri}ht QT r/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht (335) QT r/bardblfrs/an}bracketri}ht/an}bracketri}ht= 0 Qr/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht= 0 (336) Qs/bardblfrs/an}bracketri}ht/an}bracketri}ht=−1 d+1/bardblfsr/an}bracketri}ht/an}bracketri}ht QT s/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=−1 d+1/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht(337) QT s/bardblfrs/an}bracketri}ht/an}bracketri}ht=−d d+1/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht Qs/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht=−d d+1/bardblfsr/an}bracketri}ht/an}bracketri}ht(338) /an}bracketle{t/an}bracketle{tfrs/bardblfsr/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{tf∗ rs/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht=−1 d+1(339)42 /an}bracketle{t/an}bracketle{tfrs/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{tf∗ rs/bardblfsr/an}bracketri}ht/an}bracketri}ht=−d d+1(340) Proof.It follows from Eqs. ( 116) and (120) that /an}bracketle{t/an}bracketle{tt/bardblfrs/an}bracketri}ht/an}bracketri}ht+/an}bracketle{t/an}bracketle{tt/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht=i√ d+1/parenleftbig Qrts−Qsrt/parenrightbig =i√ d+1/parenleftbiggd+1 d/parenleftbig Trts−Tsrt/parenrightbig −2/parenleftbig /an}bracketle{t/an}bracketle{tt/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbls/an}bracketri}ht/an}bracketri}ht−/an}bracketle{t/an}bracketle{tr/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardblt/an}bracketri}ht/an}bracketri}ht/parenrightbigg =i/radicalbigg 2 d/parenleftbig /an}bracketle{t/an}bracketle{tt/bardbles/an}bracketri}ht/an}bracketri}ht−/an}bracketle{t/an}bracketle{tt/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightbig (341) where we used the fact that Trts=Tsrtin the third step, and the fact that /an}bracketle{t/an}bracketle{tt/bardbles/an}bracketri}ht/an}bracketri}htis real in the last. This establishes Eq. ( 333). Eq. (334) is obtained by taking complex conjugates on both sides, and using the fact that the vectors /bardbles/an}bracketri}ht/an}bracketri}htare real. Eqs. (335) and (336) are immediate consequences of the definitions, and the fact thatQrQT r= 0. Turning to the proof of Eqs. ( 337) and (338), it follows from Eqs. (119) and (120) that Qs/bardbles/an}bracketri}ht/an}bracketri}ht= 0 (342) Using this and the fact that Qs/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht= 0 in Eq. ( 333) we find Qs/bardblfrs/an}bracketri}ht/an}bracketri}ht=−i/radicalbigg 2 dQs/bardbler/an}bracketri}ht/an}bracketri}ht (343) Since /bardbler/an}bracketri}ht/an}bracketri}ht=/radicaligg d 2(d+1)/parenleftig /bardblr/an}bracketri}ht/an}bracketri}ht+/bardblv0/an}bracketri}ht/an}bracketri}ht/parenrightig (344) and taking account of the fact that Qs/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (see Eq. ( 287)) we deduce Qs/bardblfrs/an}bracketri}ht/an}bracketri}ht=−i/radicalbigg 1 d+1Qs/bardblr/an}bracketri}ht/an}bracketri}ht=−1 d+1/bardblfsr/an}bracketri}ht/an}bracketri}ht (345) Taking complex conjugates on both sides of this equation we deduce the second identity in Eq. ( 337). In the same way, acting on both sides of Eq. ( 333) withQT swe find QT s/bardblfrs/an}bracketri}ht/an}bracketri}ht=−/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht−i/radicalbigg 2 dQT s/bardbler/an}bracketri}ht/an}bracketri}ht =−/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht−i/radicalbigg 1 d+1QT s/bardblr/an}bracketri}ht/an}bracketri}ht =−d d+1/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht (346) Taking complex conjugates on both sides of this equation we deduce the second identity in Eq. ( 338). Turning to the last group of identities we have /an}bracketle{t/an}bracketle{tfrs/bardblfsr/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{tfrs/bardblQr/bardblfsr/an}bracketri}ht/an}bracketri}ht=−1 d+1/an}bracketle{t/an}bracketle{tfrs/bardblfrs/an}bracketri}ht/an}bracketri}ht=−1 d+1(347) and /an}bracketle{t/an}bracketle{tfrs/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{tfrs/bardblQr/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht=−d d+1/an}bracketle{t/an}bracketle{tfrs/bardblfrs/an}bracketri}ht/an}bracketri}ht=−d d+1(348)43 The other two identities are obtained by taking complex conjugates on both sides of the two just derived. /square This lemma provides a substantial part of what we need to prove the theorem. The remaining part is provided by Lemma 19. For allr/ne}ationslash=s QrQsQr=1 d+1Qr−d (d+1)2/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (349) QrQT sQr=d2 (d+1)2/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (350) Proof.It follows from Eq. ( 120) that QrQsQr=d+1 dQrTsQr−2Qr/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardblQr (351) QrQT sQr=d+1 dQrTT sQr−2Qr/bardbles/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tes/bardblQr (352) In view of Eqs. ( 344), (287) and the definition of /bardblfrs/an}bracketri}ht/an}bracketri}htwe have Qr/bardbles/an}bracketri}ht/an}bracketri}ht=/radicaligg d 2(d+1)Qr/bardbls/an}bracketri}ht/an}bracketri}ht=−i√ d√ 2(d+1)/bardblfrs/an}bracketri}ht/an}bracketri}ht (353) Substituting this expression into Eqs. ( 351) and (352) we obtain QrQsQr=d+1 dQrTsQr−d (d+1)2/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (354) QrQT sQr=d+1 dQrTT sQr−d (d+1)2/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (355) The problem therefore reduces to showing QrTsQr=d (d+1)2Qr (356) QrTT sQr=d2 (d+1)2/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (357) Using Eq. ( 120) we find /an}bracketle{t/an}bracketle{ta/bardblQrTsQr/bardblb/an}bracketri}ht/an}bracketri}ht=(d+1)2 d2/an}bracketle{t/an}bracketle{ta/bardblTrTsTr/bardblb/an}bracketri}ht/an}bracketri}ht −1 2/parenleftbigg2(d+1) d/parenrightbigg3 2/parenleftig K2 ra/an}bracketle{t/an}bracketle{ter/bardblTsTr/bardblb/an}bracketri}ht/an}bracketri}ht+K2 rb/an}bracketle{t/an}bracketle{ta/bardblTrTs/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig +2(d+1) dK2 raK2 rb/an}bracketle{t/an}bracketle{ter/bardblTs/bardbler/an}bracketri}ht/an}bracketri}ht (358) /an}bracketle{t/an}bracketle{ta/bardblQrTT sQr/bardblb/an}bracketri}ht/an}bracketri}ht=(d+1)2 d2/an}bracketle{t/an}bracketle{ta/bardblTrTT sTr/bardblb/an}bracketri}ht/an}bracketri}ht −1 2/parenleftbigg2(d+1) d/parenrightbigg3 2/parenleftig K2 ra/an}bracketle{t/an}bracketle{ter/bardblTT sTr/bardblb/an}bracketri}ht/an}bracketri}ht+K2 rb/an}bracketle{t/an}bracketle{ta/bardblTrTT s/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig +2(d+1) dK2 raK2 rb/an}bracketle{t/an}bracketle{ter/bardblTT s/bardbler/an}bracketri}ht/an}bracketri}ht (359)44 Using the definitions of Tr,/bardbler/an}bracketri}ht/an}bracketri}htand Eq. ( 23) (the 2-design property) we find, after some algebra, /an}bracketle{t/an}bracketle{ta/bardblTrTsTr/bardblb/an}bracketri}ht/an}bracketri}ht=d2 (d+1)2/parenleftig K2 raTrsb+K2 rbTras+K2 rsTrab+K2 raK2 rb/parenrightig (360) /an}bracketle{t/an}bracketle{ter/bardblTsTr/bardblb/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd 2(d+1)/parenrightbigg3 2/parenleftig 2K2 rsK2 rb+K2 rb+Trsb/parenrightig (361) /an}bracketle{t/an}bracketle{ta/bardblTrTs/bardbler/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd 2(d+1)/parenrightbigg3 2/parenleftig 2K2 rsK2 ra+K2 ra+Tras/parenrightig (362) /an}bracketle{t/an}bracketle{ter/bardblTs/bardbler/an}bracketri}ht/an}bracketri}ht=d 2(d+1)/parenleftbig 3K2 rs+1/parenrightbig (363) and /an}bracketle{t/an}bracketle{ta/bardblTrTT sTr/bardblb/an}bracketri}ht/an}bracketri}ht=d2 (d+1)2/parenleftig GraGasGsbGbr +K2 raTrsb+K2 rbTras+K2 raK2 rb/parenrightig =d2 (d+1)2/parenleftig (d+1)TrasTrsb +K2 raTrsb+K2 rbTras+K2 raK2 rb/parenrightig (364) /an}bracketle{t/an}bracketle{ter/bardblTT sTr/bardblb/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd 2(d+1)/parenrightbigg3 2/parenleftig K2 rsK2 rb+K2 rb+2Trsb/parenrightig (365) /an}bracketle{t/an}bracketle{ta/bardblTrTT s/bardbler/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd 2(d+1)/parenrightbigg3 2/parenleftig K2 rsK2 ra+K2 ra+2Tras/parenrightig (366) /an}bracketle{t/an}bracketle{ter/bardblTT s/bardbler/an}bracketri}ht/an}bracketri}ht=d 2(d+1)/parenleftbig 3K2 rs+1/parenrightbig (367) where in deriving Eq. ( 364) we used the fact that GraGasGsbGbr= (d+1)TrasTrsb (in view of the fact that r/ne}ationslash=s). Substituting these expressions into Eqs. ( 358) and (359) we deduce Eqs. ( 356) and (357). /square Now define the rank d−1 projectors Qrs=Qr−/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl (368) QT rs=QT r−/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl (369) andletQ0 rs,Qrs,¯Q0 rsand¯Qrsbe, respectively, the subspacesontowhich /bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl, Qrs,/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardblandQ∗ rsproject. It is immediate that we have the orthogonal decompositions Qr=Q0 rs⊕Qrs (370) ¯Qr=¯Q0 rs⊕¯Qrs (371) Using Lemma 18we find Qsr/bardblfrs/an}bracketri}ht/an}bracketri}ht=Qrs/bardblfsr/an}bracketri}ht/an}bracketri}ht= 0 (372)45 implying that Q0 rs⊥QsrandQrs⊥Q0 sr, and /vextendsingle/vextendsingle/an}bracketle{t/an}bracketle{tfrs/bardblfsr/an}bracketri}ht/an}bracketri}ht/vextendsingle/vextendsingle=1 d+1(373) implying that Q0 rsandQ0 srare inclined at angle cos−1/parenleftbig1 d+1/parenrightbig . Using Lemma 18 together with Lemma 19we find QrsQsrQrs=QrsQsQrs =QrQsQr−/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardblQsQr−QrQs/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl +/an}bracketle{t/an}bracketle{tfrs/bardblQs/bardblfrs/an}bracketri}ht/an}bracketri}ht/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl =1 d+1Qr−1 d+1/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl =1 d+1Qrs (374) which in view of Lemma 15implies that QrsandQsrare uniformly inclined at angle cos−1/parenleftbig1√d+1/parenrightbig . This proves part (a) of the theorem. Parts (b) and (c) are prov ed similarly. Proof of Theorem 17.Define /bardblgrs/an}bracketri}ht/an}bracketri}ht=1√ 2/parenleftbig /bardblf∗ rs/an}bracketri}ht/an}bracketri}ht+/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightbig (375) /bardbl¯grs/an}bracketri}ht/an}bracketri}ht=i√ 2/parenleftbig /bardblf∗ rs/an}bracketri}ht/an}bracketri}ht−/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightbig (376) By construction the components of /bardblgrs/an}bracketri}ht/an}bracketri}ht,/bardbl¯grs/an}bracketri}ht/an}bracketri}htin the standard basis are real, so we can regard them as ∈Rd2. They are orthonormal: /an}bracketle{t/an}bracketle{tgrs/bardblgrs/an}bracketri}ht/an}bracketri}ht=/an}bracketle{t/an}bracketle{t¯grs/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 1 and /an}bracketle{t/an}bracketle{tgrs/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 0 (377) It is also readily verified, using Lemma 18, that ¯Rr/bardblgrs/an}bracketri}ht/an}bracketri}ht=/bardblgrs/an}bracketri}ht/an}bracketri}ht (378) ¯Rr/bardbl¯grs/an}bracketri}ht/an}bracketri}ht=/bardbl¯grs/an}bracketri}ht/an}bracketri}ht (379) So Rrs=¯Rr−/bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl−/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardbl (380) is a rank 2 d−4 projector. If we define R0 rs,R1 rsandRrsto be, respectively, the subspaces onto which /bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl,/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardblandRrsproject we have the orthogonal decomposition Rr=R0 rs⊕R1 rs⊕Rrs (381) It follows from Eqs. ( 333) and (334) that /bardblgrs/an}bracketri}ht/an}bracketri}ht=−/bardblgsr/an}bracketri}ht/an}bracketri}ht (382) implying that R0 rs=R0 srfor allr/ne}ationslash=s. It is also easily verified, using Lemma 18, that/vextendsingle/vextendsingle/an}bracketle{t/an}bracketle{t¯grs/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht/vextendsingle/vextendsingle=d−1 d+1(383) from which it follows that R1 rsandR1 srare inclined at angle cos−1/parenleftbigd−1 d+1/parenrightbig . We next observe that Rrs=Qrs+QT rs (384)46 Using Lemma 18once again we deduce Rrs/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht=Rsr/bardbl¯grs/an}bracketri}ht/an}bracketri}ht= 0 (385) from which it follows that R1 rs⊥RsrandRrs⊥R1 sr. Finally, we know from Theorem 16thatQT rsQsr=QrsQT sr= 0. Consequently RrsRsrRrs=QrsQsrQrs+QT rsQT srQT rs =d d+1Qrs+d d+1QT rs =1 d+1Rrs (386) In view of Lemma 15it follows that RrsandRsrare uniformly inclined at angle cos−1/parenleftbig1√d+1/parenrightbig . Further Identities. We conclude this section with another set of identities in- volving the vectors /bardblfrs/an}bracketri}ht/an}bracketri}ht,/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht,/bardblgrs/an}bracketri}ht/an}bracketri}htand/bardbl¯grs/an}bracketri}ht/an}bracketri}ht. Define /bardbl¯er/an}bracketri}ht/an}bracketri}ht=/radicalbigg 2d d−1/bardbler/an}bracketri}ht/an}bracketri}ht−/radicalbigg d+1 d−1/bardblv0/an}bracketri}ht/an}bracketri}ht (387) where/bardblv0/an}bracketri}ht/an}bracketri}htis the vector defined by Eq. ( 286). It is readily verified that /an}bracketle{t/an}bracketle{t¯er/bardbl¯er/an}bracketri}ht/an}bracketri}ht= 0 and /an}bracketle{t/an}bracketle{t¯er/bardblv0/an}bracketri}ht/an}bracketri}ht= 0 (388) So/bardbl¯er/an}bracketri}ht/an}bracketri}ht,/bardblv0/an}bracketri}ht/an}bracketri}htis an orthonormal basis for the 2-dimensional subspace spanned b y /bardbler/an}bracketri}ht/an}bracketri}ht,/bardblv0/an}bracketri}ht/an}bracketri}ht. Note that Qr/bardbl¯er/an}bracketri}ht/an}bracketri}ht=QT r/bardbl¯er/an}bracketri}ht/an}bracketri}ht=¯Rr/bardbl¯er/an}bracketri}ht/an}bracketri}ht= 0 (389) We then have Theorem 20. For allr 1 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl=Qr (390) 1 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl=QT r (391) 2 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl=¯Rr (392) 2 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardbl¯grs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯grs/bardbl=¯Rr (393) and 1 d−1d2/summationdisplay s=1 (s/negationslash=r)/bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardbl=QT r+/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+1 d2−1/parenleftig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig (394)47 1 d−1d2/summationdisplay s=1 (s/negationslash=r)/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl=Qr+/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+1 d2−1/parenleftig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig (395) 2 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblgsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgsr/bardbl=¯Rr (396) 2 d−3d2/summationdisplay s=1 (s/negationslash=r)/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯gsr/bardbl=¯Rr+4(d−1) d−3/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+4 (d+1)(d−3)/parenleftig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig (397) Proof.It follows from the definition of /bardblfrs/an}bracketri}ht/an}bracketri}htthat 1 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl=d2/summationdisplay s=1 (s/negationslash=r)Qr/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardblQr =Qr d2/summationdisplay s=1/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardbl Qr =Qr (398) where in the second step we used the fact that Qr/bardblr/an}bracketri}ht/an}bracketri}ht= 0 (as can be seen by setting r=sin Eq. (121)). Eq. (391) is obtained by taking the complex conjugate on both sides. We also have 1 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl=−d2/summationdisplay s=1 (s/negationslash=r)Qr/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardblQT r =−Qr d2/summationdisplay s=1/bardbls/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ts/bardbl QT r =−QrQT r = 0 (399) Taking the complex conjugate on both sides we find 1 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl= 0 (400) Consequently 2 d+1d2/summationdisplay s=1 (s/negationslash=r)/bardblgrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgrs/bardbl=1 d+1d2/summationdisplay s=1 (s/negationslash=r)/parenleftig /bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl+/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl +/bardblfrs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ rs/bardbl+/bardblf∗ rs/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfrs/bardbl/parenrightig =¯Rr (401)48 Eq. (393) is proved similarly. To prove the second group of identities we have to work a little harde r. Using Eqs. (116) and (120) we find 1 d−1d2/summationdisplay s=1 (s/negationslash=r)/an}bracketle{t/an}bracketle{ta/bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardblb/an}bracketri}ht/an}bracketri}ht=d+1 d−1d2/summationdisplay s=1/an}bracketle{t/an}bracketle{ta/bardblQs/bardblr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tr/bardblQs/bardblb/an}bracketri}ht/an}bracketri}ht =(d+1)3 d2(d−1)d2/summationdisplay s=1/parenleftig TsarTsrb−K2 saK2 srTsrb −K2 srK2 sbTsar+K2 saK4 srK2 sb/parenrightig (402) (where we used the fact that Qs/bardbls/an}bracketri}ht/an}bracketri}ht= 0 in the first step). After some algebra we find d2/summationdisplay s=1TsarTsrb=d d+1/parenleftigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg +Trba/parenrightigg (403) d2/summationdisplay s=1K2 saK2 srTsrb=d d+1/parenleftigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+2d+1 d(d+1)/parenrightigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg +1 d+1Trba/parenrightigg (404) d2/summationdisplay s=1K2 srK2 sbTsar=d d+1/parenleftigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+2d+1 d(d+1)/parenrightigg +1 d+1Trba/parenrightigg (405) d2/summationdisplay s=1K2 saK4 srK2 sb=d (d+1)/parenleftigg d+2 d+1/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg +d (d+1)3δab+d+2 (d+1)3/parenrightigg (406) wherewe usedEq. ( 23) to derivethe firstexpression. Substituting these expressions into Eq. ( 402) and using /an}bracketle{t/an}bracketle{ta/bardblQT r/bardblb/an}bracketri}ht/an}bracketri}ht=d+1 d/parenleftigg Trba−/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg/parenleftigg/radicalbigg d−1 d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1 d/parenrightigg/parenrightigg (407) we deduce Eq. ( 394). Taking complex conjugates on both sides we obtain Eq. ( 395). Eq. (396) is an immediate consequence of Eq. ( 392) and the fact that /bardblgsr/an}bracketri}ht/an}bracketri}ht= −/bardblgrs/an}bracketri}ht/an}bracketri}htfor allr,s.49 To prove Eq. ( 397) observe that it follows from Eqs. ( 394)–(396) that d2/summationdisplay s=1 (s/negationslash=r)/parenleftig /bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl+/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardbl/parenrightig =d2/summationdisplay s=1 (s/negationslash=r)/parenleftig 2/bardblgsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tgsr/bardbl−/bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardbl−/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl/parenrightig = 2/parenleftig ¯Rr−(d−1)/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl −1 d+1/parenleftig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig/parenrightbigg (408) Hence 2 d−3d2/summationdisplay s=1 (s/negationslash=r)/bardbl¯gsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯gsr/bardbl=1 d−3d2/summationdisplay s=1 (s/negationslash=r)/parenleftig /bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tfsr/bardbl+/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl −/bardblfsr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl−/bardblf∗ sr/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tf∗ sr/bardbl/parenrightig =¯Rr+4(d−1) d−3/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl+4 (d+1)(d−3)/parenleftig I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig (409) /square 9.TheP-PTProperty In the preceding sections the Q-QTproperty has played a prominent role. In this section we show that in the particular case ofa Weyl-Heisenberg covariantSIC- POVM, and with the appropriate choice of gauge, the Gram project or (defined in Eq. (63)) has an analogous property, which we call the P-PTproperty. Specifically one has PPT=PTP=/bardblh/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{th/bardbl (410) where/bardblh/an}bracketri}ht/an}bracketri}htis a normalized vector whose components in the standard basis are a ll real. In odd dimensions the components of /bardblh/an}bracketri}ht/an}bracketri}htin the standard basis can be simply expressed in terms of the Wigner function of the fiducial vector. I t could be said thattheprojectors PandPTarealmostorthogonal(bycontrastwiththeprojectors QrandQT rwhich are completely orthogonal). More precisely Phas the spectral decomposition P=¯P+/bardblh/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{th/bardbl (411) where¯Pis a rank (d−1) projector with the property ¯P¯PT= 0 (412) This means that the matrix JP=P−PT(413) is a pure imaginary Hermitian matrix with the property that J2 Pis a real rank 2d−2 projector ( c.f.the discussion in Section 4). Although we are mainly interested in the P-PTproperty as it applies to SIC- POVMs, itshould benoted that itactuallyholdsforanyWeyl-Heisenbe rgcovariant POVM (with the appropriate choice of gauge). So we will prove the ab ove propo- sitions for this more general case.50 Let us begin by fixing notation. Let |0/an}bracketri}ht,...,|d−1/an}bracketri}htbe an orthonormal basis for d-dimensional Hilbert space and let XandZbe the operators whose action on the |r/an}bracketri}htis X|a/an}bracketri}ht=|a+1/an}bracketri}ht (414) Z|a/an}bracketri}ht=ωa|a/an}bracketri}ht (415) whereω=e2πi dand the addition of indices in the first equation is modd. We then define the Weyl-Heisenberg displacement operators by (adopting t he convention used in, for example, ref. [ 16]) Dp=τp1p2Xp1Zp2(416) wherepis the vector ( p1,p2) (p1,p2being integers) and τ=e(d+1)πi d. Generally speaking the decision to insert the phase τp1p2is a matter of convention, and many authors define it differently, or else omit altogether. However, for the purposes of this section it is essential, as a different choice of phase at this stage would lead to a different gauge in the class of POVMs to be defined below, and the Gra m projector would then typically not have the P-PTproperty. Note thatτ2=τd2=ωin every dimension. If the dimension is odd we can write τ=ωd+1 2. Soτis adthroot of unity. However, if the dimension is even τd=−1. This has the consequence that Dp+du= (−1)u1p2+u2p1Dp (417) Soin even dimension p=q(modd) does notnecessarilyimply Dp=Dq(although the operators are, of course, equal if p=q(mod 2d)) In every dimension (even or odd) we have D† p=D−p (418) for allp (Dp)n=Dnp (419) for allp,nand DpDq=τ/angbracketleftp,q/angbracketrightDp+q (420) for allp,q. In the last expression /an}bracketle{tp,q/an}bracketri}htis the symplectic form /an}bracketle{tp,q/an}bracketri}ht=p2q1−p1q2 (421) Now let|ψ/an}bracketri}htbe any normalized vector (not necessarily a SIC-fiducial vector), and define |ψp/an}bracketri}ht=Dp|ψ/an}bracketri}ht (422) Let L=/summationdisplay p∈Z2 d|ψp/an}bracketri}ht/an}bracketle{tψp| (423) It is easily seen that/bracketleftbig Dp,L/bracketrightbig = 0 (424) for allp.51 We now appeal to the fact that there is no non-trivial subspace of Hdwhich the displacement operators leave invariant. To see this assume the contrary. Then there would exist non-zero vectors |φ/an}bracketri}ht,|χ/an}bracketri}htsuch that /an}bracketle{tφ|Dp|χ/an}bracketri}ht= 0 (425) for allp. Writing the left-hand side out in full this gives d−1/summationdisplay a=0ωp2a/an}bracketle{tφ|a+p1/an}bracketri}ht/an}bracketle{ta|χ/an}bracketri}ht= 0 (426) for allp1,p2. Taking the discrete Fourier transform with respect to p2, we have /an}bracketle{tφ|a+p1/an}bracketri}ht/an}bracketle{ta|χ/an}bracketri}ht= 0 (427) for alla,p1, implying that either |φ/an}bracketri}ht= 0 or|χ/an}bracketri}ht= 0—contrary to assumption. We can therefore use Schur’s lemma [ 55] to deduce that L=kI (428) for some constant k. Taking the trace on both sides of this equation we infer thatk=d. We conclude that1 d|ψp/an}bracketri}ht/an}bracketle{tψp|is a POVM. We refer to POVMs of this general class as Weyl-Heisenberg covariant POVMs. We refer to th e vector |ψ/an}bracketri}ht which generates the POVM as the fiducial vector (with no implication t hat it is necessarily a SIC-fiducial). Now consider the Gram projector P=/summationdisplay p,q∈Z2 dPp,q/bardblp/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tq/bardbl (429) where Pp,q=1 d/an}bracketle{tψp|ψq/an}bracketri}ht (430) and where we label the matrix elements of Pand the standard basis kets with the vectorsp,qrather than with the single integer indices r,sas in the rest of this paper. We know from Theorem 1thatPis a rankdprojector. In view of Eqs. ( 418) and (420) we have /an}bracketle{t/an}bracketle{tp/bardblP/bardblq/an}bracketri}ht/an}bracketri}ht=Pp,q =1 dτ−/angbracketleftp,q/angbracketright/an}bracketle{tψ|Dq−p|ψ/an}bracketri}ht =1 dd−1/summationdisplay a=0τp1p2+q1q2ωaq2−(q1+a)p2/an}bracketle{tψ|a+q1−p1/an}bracketri}ht/an}bracketle{ta|ψ/an}bracketri}ht(431) Hence /an}bracketle{t/an}bracketle{tp/bardblPPT/bardblq/an}bracketri}ht/an}bracketri}ht=/summationdisplay u∈Zd/an}bracketle{t/an}bracketle{tp/bardblP/bardblu/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tq/bardblP/bardblu/an}bracketri}ht/an}bracketri}ht =1 d2d−1/summationdisplay a,b,u1,u2=0τp1p2+q1q2ωu2(u1+a+b)−(u1+a)p2−(u1+b)q2 ×/an}bracketle{tψ|a+u1−p1/an}bracketri}ht/an}bracketle{tψ|b+u1−q1/an}bracketri}ht/an}bracketle{ta|ψ/an}bracketri}ht/an}bracketle{tb|ψ/an}bracketri}ht52 =1 dd−1/summationdisplay a,b=0τp1p2+q1q2ωp2b+q2a/an}bracketle{tψ|−b−p1/an}bracketri}ht/an}bracketle{tb|ψ/an}bracketri}ht/an}bracketle{tψ|−a−q1/an}bracketri}ht/an}bracketle{ta|ψ/an}bracketri}ht =/an}bracketle{t/an}bracketle{tp/bardblh/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{th/bardblq/an}bracketri}ht/an}bracketri}ht (432) where/bardblh/an}bracketri}ht/an}bracketri}htis the vector with components /an}bracketle{t/an}bracketle{tp/bardblh/an}bracketri}ht/an}bracketri}ht=1√ dd−1/summationdisplay a=0τp1p2ωp2a/an}bracketle{tψ|−a−p1/an}bracketri}ht/an}bracketle{ta|ψ/an}bracketri}ht (433) It is easily verified that /bardblh/an}bracketri}ht/an}bracketri}htis normalized, and that /an}bracketle{t/an}bracketle{tp/bardblh/an}bracketri}ht/an}bracketri}htis real. Finally, suppose that the dimension is odd. Then the Wigner function o f the state|ψ/an}bracketri}htis [56,57] W(p) =1 d/an}bracketle{tψ|DpUPD† p|ψ/an}bracketri}ht=1 d/an}bracketle{tψ|D2pUP|ψ/an}bracketri}ht (434) whereUPistheparityoperator,whoseactiononthestandardbasisis UP|a/an}bracketri}ht=|−a/an}bracketri}ht. It is straightforward to show /an}bracketle{t/an}bracketle{tp/bardblh/an}bracketri}ht/an}bracketri}ht=√ dW(−2−1p) (435) where 2−1= (d+1)/2 is the multiplicative inverse of 2 considered as an element of Zd:i.e.the unique integer 0 ≤m<dsuch that 2 m= 1 (modd). 10.Conclusion A curious fact about SIC-POVMs is that, although they are charac terized by their being highly symmetric structures, they do not wear this prop erty on their sleeve (so to speak). If one casually inspects the components of a SIC-fiducial, without knowing in advance that that is what they are, there does n ot seem to be anything special about them at all. Indeed, so far from there being any obvious pattern to the components, they seem, to a casual inspection, lik e a completely random collection of numbers. Moreover, this is just as true of an e xact fiducial as it is of a numerical one (see, for instance, the tabulations in Scott a nd Grassl [ 46]). It is only when one looks at them through the right pair of spectacles , and takes the trouble to calculate the overlaps Tr(Π rΠs), that the symmetry becomes apparent. The situation is a little reminiscent of a hologram, which only takes on th e aspect of a meaningful image when it is viewed in the right way. If one wanted to s ummarize the content of this paper in a nutshell it could be said that we have ex hibited some other pairs of spectacles—other ways of looking at a SIC—which cau se its inner secrets (or at any rate some of its inner secrets) to become manif est. Rather than focusing on the SIC-vectors |ψr/an}bracketri}ht, as is usually done, we havefocused on the angle tensors θrsandθrst, and on the T,JandRmatrices defined in terms of them. This is an important change of emphasis because, ra ther than being tied to any particular SIC, these quantities characterize ent ire families of unitarily equivalent SICs. Like the components of a SIC-fiducial, the angle tensors appear, to a casual inspection, like a random collection of numbers. However, if one examines the spectra of the T,JandRmatrices one realizes that, underlying the appearance of randomness, there is a high degree of order. I f one then goes on to examine the geometrical relationships between the subspaces o nto which the Q, QTand¯Rmatrices project, as we did in Section 8, one finds yet more instances of structure and order. To our minds what is particularly interestin g about all of53 this is that none of it is obviously suggested by the defining property of a SIC, that Tr(ΠrΠs) = 1/(d+1) forr/ne}ationslash=s. In the course of this paper we have several times expressed the h ope that the Lie algebraic perspective on a SIC will lead to a solution to the existenc e problem. Of course, that is only a hope, and it may not be fulfilled. However, we feel on rather safer ground when we suggest that the solution is likely to co me, if not from this investigation, then from one which is like it to the extent that it fo cuses on a feature of a SIC which is not immediately apparent to the eye. Specializing to the case of a Weyl-Heisenberg covariant SIC, a fiducia l vector|ψ/an}bracketri}ht is a solution to the equations /vextendsingle/vextendsingle/an}bracketle{tψ|Dp|ψ/an}bracketri}ht/vextendsingle/vextendsingle2=dδp,0+1 d+1(436) Allowing for the arbitrariness of the overall phase of |ψ/an}bracketri}ht, and taking |ψ/an}bracketri}htto be nor- malized, this gives us d2−1 conditions on only 2 d−2 independent real parameters. The equations are thus over-determined, and very highly over-de termined when d is large. Nevertheless, they have turned out to be soluble in every c ase which has been investigated to date. It seems likely that progress will depend on finding the structural feature which is responsible for this remarkable fact. The motivation for this paper is the belief that it may be structural features of the Lie algebra gl(d,C) which are responsible. That suggestion may or may not be correct. 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