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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