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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 |
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