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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)