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=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