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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