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) =/braceleftigg
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)/parenleftig
Jr+J2
r+4/bardbler/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{ter/bardbl/parenrightig
(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/parenleftig
|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|/parenrightig
(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/parenleftig
|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|/parenrightig
(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=/braceleftigg
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/parenleftig
1−β
κ/parenrightig
. 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=/braceleftigg
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=/braceleftigg
L′
r r≤m
2(d−2)
d2I−d−4
dL′
rr>m(242)
Consequently
˜T1rr=/braceleftigg
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=/braceleftigg
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=/radicaligg
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=/braceleftigg
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/parenleftigg/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/parenrightigg
(282)
and
Tr/parenleftbig
QrQT
s/parenrightbig
= Tr/parenleftigg/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/parenrightigg
(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/parenleftigg/vextenddouble/vextenddoubleP|ψ/an}bracketri}ht/vextenddouble/vextenddouble
/vextenddouble/vextenddouble|ψ/an}bracketri}ht/vextenddouble/vextenddouble/parenrightigg
(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/parenleftig
1+Re/parenleftbig
eiφ/an}bracketle{tur|us/an}bracketri}ht/parenrightbig/parenrightig
(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/parenleftig
1√d+1/parenrightig
.
(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/parenleftig
1√d+1/parenrightig
.
(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/parenleftig/radicalig
1
d+1/parenrightig
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/radicalig
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=/parenleftig
/an}bracketle{t/an}bracketle{tt/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightig∗
(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=/parenleftig
/an}bracketle{t/an}bracketle{tfrs/bardblfrs/an}bracketri}ht/an}bracketri}ht/parenrightig∗
= 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/parenleftig
/bardbles/an}bracketri}ht/an}bracketri}ht−/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig
(333)
/bardblf∗
rs/an}bracketri}ht/an}bracketri}ht=−/bardblfsr/an}bracketri}ht/an}bracketri}ht−i/radicalbigg
2
d/parenleftig
/bardbles/an}bracketri}ht/an}bracketri}ht−/bardbler/an}bracketri}ht/an}bracketri}ht/parenrightig
(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=/radicaligg
d
2(d+1)/parenleftig
/bardblr/an}bracketri}ht/an}bracketri}ht+/bardblv0/an}bracketri}ht/an}bracketri}ht/parenrightig
(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=/radicaligg
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/parenleftig
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/parenrightig
+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/parenleftig
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/parenrightig
+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/parenleftig
K2
raTrsb+K2
rbTras+K2
rsTrab+K2
raK2
rb/parenrightig
(360)
/an}bracketle{t/an}bracketle{ter/bardblTsTr/bardblb/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd
2(d+1)/parenrightbigg3
2/parenleftig
2K2
rsK2
rb+K2
rb+Trsb/parenrightig
(361)
/an}bracketle{t/an}bracketle{ta/bardblTrTs/bardbler/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd
2(d+1)/parenrightbigg3
2/parenleftig
2K2
rsK2
ra+K2
ra+Tras/parenrightig
(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/parenleftig
GraGasGsbGbr
+K2
raTrsb+K2
rbTras+K2
raK2
rb/parenrightig
=d2
(d+1)2/parenleftig
(d+1)TrasTrsb
+K2
raTrsb+K2
rbTras+K2
raK2
rb/parenrightig
(364)
/an}bracketle{t/an}bracketle{ter/bardblTT
sTr/bardblb/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd
2(d+1)/parenrightbigg3
2/parenleftig
K2
rsK2
rb+K2
rb+2Trsb/parenrightig
(365)
/an}bracketle{t/an}bracketle{ta/bardblTrTT
s/bardbler/an}bracketri}ht/an}bracketri}ht= 2/parenleftbiggd
2(d+1)/parenrightbigg3
2/parenleftig
K2
rsK2
ra+K2
ra+2Tras/parenrightig
(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/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(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/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(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)/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(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)/parenleftig
/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/parenrightig
=¯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/parenleftig
TsarTsrb−K2
saK2
srTsrb
−K2
srK2
sbTsar+K2
saK4
srK2
sb/parenrightig
(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/parenleftigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg
+Trba/parenrightigg
(403)
d2/summationdisplay
s=1K2
saK2
srTsrb=d
d+1/parenleftigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+2d+1
d(d+1)/parenrightigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg
+1
d+1Trba/parenrightigg
(404)
d2/summationdisplay
s=1K2
srK2
sbTsar=d
d+1/parenleftigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+2d+1
d(d+1)/parenrightigg
+1
d+1Trba/parenrightigg
(405)
d2/summationdisplay
s=1K2
saK4
srK2
sb=d
(d+1)/parenleftigg
d+2
d+1/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg
+d
(d+1)3δab+d+2
(d+1)3/parenrightigg
(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/parenleftigg
Trba−/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{ta/bardbl¯er/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg/parenleftigg/radicalbigg
d−1
d+1/an}bracketle{t/an}bracketle{t¯er/bardblb/an}bracketri}ht/an}bracketri}ht+1
d/parenrightigg/parenrightigg
(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)/parenleftig
/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/parenrightig
=d2/summationdisplay
s=1
(s/negationslash=r)/parenleftig
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/parenrightig
= 2/parenleftig
¯Rr−(d−1)/bardbl¯er/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{t¯er/bardbl
−1
d+1/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig/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)/parenleftig
/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/parenrightig
=¯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)/parenleftig
I−/bardblv0/an}bracketri}ht/an}bracketri}ht/an}bracketle{t/an}bracketle{tv0/bardbl/parenrightig
(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. But if it turns
out to be incorrect, the amount of effort which has been expended on this problem
over a period of more than ten years, so far without fruit, sugges ts to us that the
solution will depend on finding some other structural feature of a S IC, which is not
obvious, and which has hitherto escaped attention.
11.Acknowledgements
The authors thank I. Bengtsson for discussions. Two authors, D MA and CAF,
were supported in part by the U. S. Office of Naval Research (Gran t No. N00014-
09-1-0247). Research at Perimeter Institute is supported by th e Government of
Canada through Industry Canada and by the Province of Ontario t hrough the
Ministry of Research & Innovation.
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