Optimal quantum state discrimination via nested binary measurements

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by  Matteo Rosati

Optimal quantum state discrimination via nested binary measurements
Matteo Rosati,1 Giacomo De Palma,2 Andrea Mari,1 and Vittorio Giovannetti1
1NEST, Scuola Normale Superiore and Istituto Nanoscienze-CNR, I-56127 Pisa, Italy.
2QMATH, Department of Mathematical Sciences, University of Copenhagen,
Universitetsparken 5, 2100 Copenhagen, Denmark
A method to compute the optimal success probability of discrimination of N arbitrary quantum
states is presented, based on the decomposition of any N-outcome measurement into sequences of
nestedtwo-outcomeones. Inthiswaytheoptimizationofthemeasurementoperatorscanbecarried
out in successive steps, optimizing first the binary measurements at the deepest nesting level and
then moving on to those at higher levels. We obtain an analytical expression for the maximum
success probability after the first optimization step and examine its form for the specific case of
N =3,4statesofaqubit. Inthiscase,atvariancewithpreviousproposals,weareabletoprovidea
compactexpressionforthesuccessprobabilityofanysetofstates,whosenumericaloptimizationis
7
straightforward;theresultsthusobtainedhighlightsomelesser-knownfeaturesofthediscrimination
1
problem.
0
2

n I. INTRODUCTION the string of binary outcomes of the previous ones. This
a
result was already obtained in Ref. [42]. When applied
J
to state discrimination, it acquires a more operational
1
The discrimination of quantum states [1] is one of the meaning: each binary POVM can be seen as discrimi-
1
fundamentalproblemsinQuantumInformationandaba- nating between two subsets of the initial set of states,
] sic task for several applications in communication [2–6], identified by previous outcomes. Hence the sequence of
h cryptography [7–9], fundamental questions [10–13], mea- measurementsinducesasequenceofdiscriminationprob-
p
surement and control [14, 15] and algorithms [16]. Trig- abilities, so that, if the optimization problem is solved
-
t gered by the observation that non-orthogonal quantum independentlyforanysetofafixednumberofstates,the
n
statescannotbeperfectlydiscriminated,thissubjecthas result can be employed in the optimization problem for
a
u stimulatedmuchwork,bothfromatheoreticalandprac- larger sets of states.
q tical point of view: the seminal works of Helstrom [17], In the second part of the article, employing this decom-
[ Holevo [18] and Yuen et al. [19] formalized the problem, position and the two-state optimal probability [17], we

obtaining a set of conditions for the optimal measure- obtain an expression for the success probability of dis-
2
v ment operators, which in turn provide the optimal suc- criminationofanyN =3,4states,dependingonasingle
3 cess probability, then solved it for sets of states symmet- measurement operator, and solve the problem analyti-
3 ric under a unitary transformation; more recently, ac- cally for specific sets of states. Then we restrict our at-
2 knowledging that a general analytical solution is hard to tention to qubit states and obtain a compact expression
2 find, research focused on finding a solution for sets with which can be easily optimized numerically case by case,
0
more general symmetries [20–22], computing explicitly atvariancewithlesscompactresultsforN =3presented
.
1 theoptimalmeasurementsforthemostinterestingsetsof in previous works based on Bloch-space geometry [25–
0 states [23–27] and studying the implementation of such 27]. We recover the results of those works and highlight
7
measurements with available technology (see for exam- in particular some interesting lesser-known implications
1
ple [28–40] for the case of two optical coherent states, of Ref. [26].
:
v the most relevant for optical communication). Also, the Thearticleisstructuredasfollows: inSec.IIwedescribe
Xi problemofdiscriminationhasbeenidentifiedasaconvex thedecompositionintermsofnestedbinaryPOVM’sand
optimizationone,arguingthatitcanbesolvedefficiently provideaproofofitsvalidity,similartothatofRef.[42];
r
a with numerical optimization methods [41]. in Sec. III we apply it to state discrimination and ob-
In this article we attempt to solve the optimal discrimi- tain an explicit expression for the case of N = 3,4 arbi-
nationofN quantumstatesfromadifferentperspective, trary states, then discuss its optimization in some spe-
by providing a structured expression for the N-outcome cific cases; in Sec. IV we treat the case of N =3,4 qubit
Positive Operator Valued Measure (POVM) used to dis- states, computing a compact expression which can be
criminate the states. Indeed it can be shown [42, 43] optimized numerically and highlighting some results ob-
that any POVM comprising N elements is equivalent to tained in this way. Eventually in Sec. V we draw some
a collection of binary POVM’s, i.e., comprising two ele- conclusions. Detailed computations of the quantities ap-
ments,astheoneemployedinRef.[6]: dependingonthe pearing in the article are provided in the Appendices.
binaryoutcomeofthefirstmeasurement,asecondoneis
applied; its binary outcome in turn affects the choice of
the third binary measurement and so on. In this way a
sequence of nested binary POVM’s can be constructed,
where the POVM applied at a given level depends on
2
II. GENERAL DECOMPOSITION OF A
N-OUTCOME MEASUREMENT INTO NESTED
BINARY ONES
In this Section we prove that any quantum measure-
ment with an arbitrary number of outcomes can be de- Figure 1. Schematic depiction of the nested decomposition
forN =4,explicitlydiscussedinthetext. Anyfour-outcome
composed into a sequence of nested measurements with
measurement M(4) acting on a state ρ is equivalent to the
binary outcomes, where the previous results determine
concatenation of two-outcome measurements: the first-step
the choice of successive measurements. We stress that
oneB(1),withresultk =0,1,andthesecond-steponesB(2),
the same result was obtained in Ref. [42]. At variance 1 k1
which are mutually exclusive and applied only if the corre-
with the latter, our proof does not make use of the spec-
sponding first outcome k was obtained.
1
tral decomposition of the initial measurement operators;
we present it here in a form adapted to the main pur-
pose of the article. Let us suppose we want to perform (cid:113) (cid:113)
is B(1)ρ B(1); if then the second-step outcome is
a quantum measurement with N possible outcomes: it 0 0
k = 0, the final unnormalized state of the system is
can be expressed in general as a POVM M(N) of ele- 2
(cid:113) (cid:113) (cid:113) (cid:113)
ments Ej, one for each outcome j = 0,··· ,N −1, sat- B0(2,0) B0(1)ρ B0(1) B0(2,0). Thismeansthatthenested
isfying the positivity and completeness conditions, i.e., POVM has a more explicit representation as
respectively E ≥ 0 and (cid:80)N−1E = 1, where 1 is the
ibdeenmtietaysuorpeedr.atjoTrhoisnetxhpereHsisljibo=en0rtcasjnpabcee oinfttehreprseytsetdemastao N(4) =(cid:40)Fk1,k2 =(cid:12)(cid:12)(cid:12)(cid:12)(cid:113)Bk(21),k2(cid:113)Bk(11)(cid:12)(cid:12)(cid:12)(cid:12)2(cid:41) , (1)
one-shot measurement with several possible results and k1,k2=0,1
its practical realization may often be very hard. On the
where |X|2 = X†X is the square of the absolute value
other hand we could restrict to performing only mea-
of an operator X. In the general case, let us indicate a
surements with two outcomes, as described by binary
sequence of b−a+1 bits as
POVM’s: B ≡ M(2) = {B ,B }. This may be useful
0 1
when limited technological capabilities or specific theo- (cid:40)
k ,k ,··· ,k , b≥a
reticalrequirementsconstrainthenumberofallowedout- k = a a+1 b (2)
(a,b) ∅, b<a,
comesandthecomplexityofourmeasurement. Itisthen
naturaltoaskwhetherthissmallersetofresourcesissuf-
(cid:110) (cid:111)
ficient to describe a general quantum measurement. We and define as B(u) = B(u) ,B(u) the
answer positively by showing that the more general N- k(1,u−1) k(1,u−1),0 k(1,u−1),1
binary POVM to be performed at the u-th step if the
outcome formalism can be broken up into several binary
previous u−1 measurements had a sequence of results
steps and interpreted as a sequence of nested POVM’s
k . Then we can define a nested POVM N(N) of
withtwooutcomes,tradingaone-shot,multiple-outcome (1,u−1)
order N =2uF as
measurement for a multiple-step, yes-no measurement.
(cid:110) (cid:111) (cid:110) (cid:111)
The nested POVM can be expressed in terms of con- N(N) = B(uF) ◦···◦ B(1)
dpofilteiptoerneabvlyiobuiitsnsaerlrefys,utlPotsObViseMoab’pstpaBliine(cid:126)kded=o.n{FlyBor(cid:126)kif,0ex,aBasm(cid:126)kp,p1e}cle,ififecoarcshtNricno=gm(cid:126)k4- =(cid:40)Fkk((11,,uuFF−)1=) (cid:12)(cid:12)(cid:12)(cid:12)k(cid:114)1,·B··,k(ku(u1FF,u)−F1)=·0·,·1(cid:113)Bk(11)(cid:12)(cid:12)(cid:12)(cid:12)2(cid:41), (3)
the nested POVM can be realized in two steps and writ-
tNen(4)co=m(cid:110)paBc0(t2l)y,Bas1(2t)h(cid:111)e◦c(cid:8)oBlle(1c)t(cid:9)io,nporofptherrleyecboimnaproysPedOaVsMfo’sl-: ia.lel.,ptrheevicooulslecotuitocnoomfe2suFk(−1,u1−b1i)naartyaPgOivVeMn’ssteBpk(u(u1),ua−n1)d,faolrl
lowsandshowninFig.1. Themeasurementstartsbyap- steps u = 1,··· ,u . We can certify that N(N) so con-
(cid:110) (cid:111) F
plying the first-step binary POVM B(1) = B(1) structed actually is a POVM by checking positivity and
t(cid:110)hen, d(cid:111)epending on its outcome k1, it selectks1 Bkk(211)=0=,1 cmoemnptliestetrnievsisalo,fwithsileeletmheenlattstFerk(f1o,ullFo)w:sthfreomfortmheerfarecqtutihraet-
B(2) among the two POVM’s available in the eachbinaryPOVMiscomplete,asshowninAppendixA.
k1,k2 k2=0,1 (cid:110) (cid:111) Inlightofthepreviousdiscussionwecannowstatethe
second-step collection B(2),B(2) . Eventually the cho- main theorem:
0 1
sen second-step POVM is applied, receiving an outcome
Theorem 1. Any N-outcome POVM M(N) =
k . The total outcome is a string of two bits, i.e., k ,k ,
2 1 2 {E } is equivalent to a nested POVM N(N˜),
whose value identifies one of four possible outcomes, as j j=0,···,N−1
desired. Suppose now to apply this measurement on a N˜ = 2uF, as in Eq. (3), composed exclusively of binary
state ρ of some physical system: if the first-step out- POVM’s Bk(u(1),u−1), with a total number of steps uF equal
come is k =0, the resulting unnormalized evolved state to:
1
3
1. log N, if N is a power of 2; is complete on the support of the operator preceding it
2
in the decomposition, B(u−1) .
2. (cid:100)log N(cid:101) otherwise, where (cid:100)·(cid:101) is the ceiling func- k(1,u−1)
2 Generalizing the previous discussion, at the u-th step we
tion, equal to the smallest integer following the ar-
can define the elements of the 2u−1 possible POVM’s
gument.
B(u) as
Proof. Consider the first case above, i.e., N = 2uF ≡ k(1,u−1)
N˜. We start by providing a binary representation of (cid:113) −1 (cid:113) −1
the labels j of the initial POVM M(N), i.e., we define B(u) = B(u−1) ··· B(1) (6)
E ≡ E , with j(k) = (cid:80)uF 2u−1k . In order to k(1,u) k(1,u−1) k1
prko(v1,eu)the thj(eko)rem we have to suh=o1w thatuby combining · (cid:88) E (cid:113)B(1)−1···(cid:113)B(u−1) −1.
the elements of the initial N-outcome POVM M(N) one k(1,u),k(u+1,uF) k1 k(1,u−1)
can always define a set of binary POVM’s B(u) , for k(u+1,uF)
k(1,u−1)
all k ,··· ,k = 0,1 and u = 1,··· ,u , such that: i) These elements are positive by construction and they
1 u−1 F
their nested composition is a POVM of the form N(N), satisfy the weak completeness relation B(u) +
Eq. (3); ii) the elements of the latter are equal to the B(u) = 1 , which is sufficient tokd(1e,ufi−n1e),0the
elements of M(N). k(1,u−1),1 k(1,u−1)
Firstofallweconstructthebinaryelementsateachstep POVM N(N), as discussed in Appendix A. Hence we are
u, by taking the sum of all the elements E left to show that, when combining the binary elements
with a fixed value of the first u bits, then rke(1n,ou)r,mk(ua+li1z,uinFg) Eq.(6)asinEq.(3),theelementsFk(1,uF) soconstructed
it by all previous binary elements, as in a Square Root are equal to the E . Indeed let us evaluate Eq. (6)
Measurement [3, 44]. For example define the elements of for u = u , i.e., akt(1,tuhFe) last step, noting that the sum
F
the first-step POVM B(1) as contains only one term:
Bk(11) = (cid:88) Ek1,k(2,uF), (4) B(uF) =(cid:114)B(uF−1) −1···(cid:113)B(1)−1
k(2,uF) k(1,uF) k(1,uF−1) k1
for each value of the outcome k = 0,1. Being a sum of (cid:113) −1 (cid:114) −1
positiveoperators,theelementss1odefinedarethemselves · Ek(1,uF) Bk(11) ··· Bk(u(1F,u−F1−)1) . (7)
positive; moreover their sum equals the sum of all the
elements of M(N), implying that they are complete. At Let us then successively invert the outer square roots on
the second step define the elements of the two possible the left-hand side of the equation exactly u −1 times,
F
POVM’s B(2) as to obtain the relation
k1
Bk(2(1),2) =(cid:113)Bk(11)−1k((cid:88)3,uF)Ek(1,2),k(3,uF)(cid:113)Bk(11)−1, (5) Ek(1,uF) =(cid:12)(cid:12)(cid:12)(cid:12)(cid:114)Bk(u(1F,u)F)···(cid:113)Bk(11)(cid:12)(cid:12)(cid:12)(cid:12)2 ≡Fk(1,uF), (8)
where the inverse of an operator is to be computed only which demonstrates that we can recover the initial
on its support, while it is equal to 0 on its kernel, i.e., POVM with the procedure outlined above.
its pseudo-inverse. Also in this case the defined ele- ThiscompletestheproofwhenN isanexactpowerof2.
ments are positive by construction, but they are not If this is not the case, it means that log N is not an in-
2
complete. Indeed it is easy to show, employing the def- tegeranditsufficestoconsiderthenesteddecomposition
ipnriotijoecnto(r4)o,nthtahteBsuk(21p),p0o+rtBok(f21),t1h=e p1rke1v.iouHseroeut1cko1mies othpe- fN˜or=the2unFe.xtLheitghuesrtihnetengetrr,ivii.ael.l,yseetxpuaFnd=t(cid:100)hleogi2nNiti(cid:101)al+N1-,
erator, B(1), which may have a non-trivial kernel, so outcome POVM to a N˜-outcome one as
k1
that in general it holds 1 ≤ 1. This problem may (cid:110) (cid:111)
k1 M(N˜) =M(N)∪ E =0,∀j(k) >N −1 , (9)
be overcome easily by redefining the POVM elements k(1,uF)
as B˜(2) = B(2) ⊕(1−1 )/2, i.e., trivially expand-
k(1,2) k(1,2) k1 by adding N˜ −N null elements. The nested decompo-
ing the support of those already defined in (5), so that
B˜(2) + B˜(2) = 1 ⊕ (1−1 ) = 1. This operation sition N(N˜) equivalent to M(N˜) can be computed again
k1,0 k1,1 k1 k1 by Eqs. (3,6) and it comprises N˜ −N null elements too.
is trivial because, in the construction (3) of the nested
If we isolate these elements from the rest we obtain a
POVM, the operators Bk(2(1),2) always act after the oper- decomposition
ator B(1), so that the value of the former outside the
supportk1of the latter is completely irrelevant. In other N(N˜) =N(N)∪(cid:110)F =0,∀j(k) >N −1(cid:111),(10)
k(1,uF)
words, completeness of the binary POVM’s is not neces-
sary for the definition of N(N) as a proper POVM; it is whereN(N)canbeinterpretedasanestedrepresentation
sufficienttoaskforweak completeness,i.e.,thatB(u) of the initial POVM M(N).
k(1,u−1)
4
III. AN APPLICATION: OPTIMAL QUANTUM of remaining measurements can be seen as a nested
STATE DISCRIMINATION POVM of order N/2:
(cid:110) (cid:111) (cid:110) (cid:111)
sitIionnthtoistSheecptiroonblwemeaopfpolypttihmeaplrsetvaitoeudsisPcOriVmMinadteioconm. Lpeot- Nk(1N/2) = Bk(u1,Fk)(2,uF−1) k2,···,kuF−1=0,1◦···◦ Bk(21) .
ussupposewearegivenonecopyofaquantumstate,rep- Hence we can easily rewrite the probability of discrim-
resented by a positive and trace-one operator ρ , chosen inating the set of states S(N) with the POVM N(N),
j
atrandomfromasetS(N) ={ρ˜ =p ρ } ofN Eq. (13), as the probability of discriminating the set
j j j j=0,···,N−1
states weighted with probability p , so that (cid:80)N−1p = S(N/2) with the POVM N(N/2) if the first measurement
j j=0 j k1 k1
1; we have to perform a measurement to decide which had an outcome k1, averaged over all values of k1 =0,1:
state was sent. If the states are not orthogonal, i.e., (cid:16) (cid:17) (cid:88)
ρjρk (cid:54)= 0 for some values of j,k, and we are constrained PSucc S(N),N(N) = pSucc(k1)
togiveaconclusiveanswer, thereexistsnomeasurement k1,k(2,uF)
that can succeed with unit probability. The average suc- (cid:34)(cid:12)(cid:114) (cid:113) (cid:12)2 (cid:35)
cweistshparoNba-obuiltictyomoef dPiOscVriMminMat(iNng),tahsedseetfinoefdstianteSsecS.(NII), ·Tr (cid:12)(cid:12)(cid:12) Bk(u1,Fk)(2,uF)··· Bk(21),k2(cid:12)(cid:12)(cid:12) τ˜k1,k(2,uF)
can be computed as =(cid:88)p (k )P (cid:16)S(N/2),N(N/2)(cid:17).(15)
Succ 1 Succ k1 k1
(cid:16) (cid:17) N(cid:88)−1 k1
P S(N),M(N) = Tr[E ρ˜ ], (11)
Succ j j This expression suggests a recursive optimization: if
j=0 the optimal discrimination problem is solved for any set
where each measurement outcome E is associated with S(N/2) of a fixed size N/2, possibly restricting to a spe-
j
thedetectionoftherespectiveweightedstateρ˜ . Weare cific Hilbert space, e.g., qubits or continuous-variable
j
particularlyinterestedintheoptimalsuccessprobability, states, thentheresultcanbepluggedinto(15)toobtain
obtained by optimizing over all measurements: anexpressionforthediscriminationprobabilityofasetof
doublesize,whichdependsonasinglecoupleofmeasure-
P (S(N))= maxP (cid:16)S(N),M(N)(cid:17). (12) ment operators, i.e., the first-step binary POVM B(1).
Succ Succ
M(N) However, if a general solution for the discrimination of a
smallersetofstatesisnotavailable,theproblemremains
(cid:16) (cid:17)
Following Sec. II, we can always decompose the discrim- hard, since when optimizing P S(N/2),N(N/2) one
ination measurement into a sequence of nested binary Succ k1 k1
stillhastotakeintoaccountthedependenceofthestates
ones, writing the success probability as
of S(N/2) on the first-step POVM, which is itself subject
P (cid:16)S(N),N(N)(cid:17)= (cid:88) Tr(cid:104)F ρ˜ (cid:105) to opk1timization afterwards, thus making the states arbi-
Succ k(1,uF) k(1,uF) trary.
k(1,uF) Fortunately the first step of the recursion has a well-
=k((cid:88)1,uF)Tr(cid:34)(cid:12)(cid:12)(cid:12)(cid:12)(cid:114)Bk(u(1F,u)F)···(cid:113)Bk(11)(cid:12)(cid:12)(cid:12)(cid:12)2ρ˜k(1,uF)(cid:35), (13) known soluPtiSouncc[(1S7](:2))=(1+||ρ˜0−ρ˜1||1)/2, (16)
where we have introduced the binary representation where ||·|| = Tr[|·|] is the trace norm of the argument.
1
k forthelabelsj ofthestatesandmeasurementop- Then by plugging this expression into the optimization
(1,uF)
erators and employed the definition (3) for the elements of Eq. (15) for N =4 states we can write:
of the nested POVM. This decomposition is interesting
because it establishes a relation between the discrimina- P (cid:16)S(4)(cid:17)=max(cid:88)p (k )1+||τ˜k1,0−τ˜k1,1||1
tion probability of a given set of states and that of its Succ B(1) Succ 1 2
k1
subsets of smaller size. Let us indeed suppose that the (cid:32)
first measurement is successful, i.e., that an outcome k1 =max(cid:88)1 Tr(cid:104)B(1)(ρ˜ +ρ˜ )(cid:105)
othcecufirrsstifbointewoafstphreessetnatt(cid:104).esTρhkis1,kh(a2,pupFe)nws(cid:105)iwthitthhpatrovbaalbuielitoyf (cid:12)B(cid:12)(cid:113)(1) k1 2 k1 (cid:113)k1,0 (cid:12)(cid:12)k1,(cid:33)1
pSucc(k1)=(cid:80)k(2,uF)Tr Bk(11)ρ˜k1,k(2,uF) . Inthiscasethe +(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) Bk(11)(ρ˜k1,0−ρ˜k1,1) Bk(11)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) .(17)
possible weighted states after the measurement are 1
(cid:113) (cid:113) We can write the latter equation more compactly by in-
τ˜ = B(1)ρ˜ B(1)/p (k ), (14) troducing the function
k1,k(2,uF) k1 k1,k(2,uF) k1 Succ 1
(cid:104) (cid:12)(cid:112) (cid:112) (cid:12)
forming a set of size N/2: S(N/2) = FQ(A,B,C)=Tr QA+(cid:12)(cid:12) QB Q(cid:12)(cid:12)
(cid:110)τ˜k1,k(2,uF)(cid:111)k2,···,kuF=0,1. Moreover thek1collection +(cid:12)(cid:12)(cid:12)(cid:112)1−QC(cid:112)1−Q(cid:12)(cid:12)(cid:12)(cid:105), (18)
5
where Q is a positive and less-than-one operator, while positive and negative support of A; ii) B and C have
the arguments A,B,C are hermitian operators, and its a definite sign; iii) A, B and C all commute with each
maximum over Q, i.e., other.
F(A,B,C)= max F (A,B,C). (19) Remark 1. In the first case of Proposition 1, i.e., that
Q
1≥Q≥0
the operators B and C have support respectively on the
Setting B(1) =Q and B(1) =1−Q, we obtain: positive and negative support of A, the expression (22)
0 1 can be simplified as
(cid:16) (cid:17) p +p (cid:16) (cid:17)
P S(4) = 1,0 1,1 +F A(4),B(4),C(4) , (20)
Succ 2 F(A,B,C)=Tr[A+]+||B||1+||C||1. (23)
with A(4) = (ρ˜ +ρ˜ −ρ˜ −ρ˜ )/2, B(4) = (ρ˜ − Remark 2. The optimal success probability is invariant
0,0 0,1 1,0 1,1 0,0
ρ˜ )/2 and C(4) = (ρ˜ −ρ˜ )/2. Similarly for N = 3 underexchangeofthestates, i.e., underrelabellingofthe
0,1 1,0 1,1
states we have: indices k1,k2 in our case N =3,4. Hence it can happen
that the conditions listed in Proposition 1 are valid only
(cid:16) (cid:17) (cid:16) (cid:17)
P S(3) =p +F A(3),B(3),C(3) , (21) for A,B,C given by a specific ordering of the states.
Succ 1,0
with A(3) = (ρ˜ +ρ˜ )/2−ρ˜ , B(3) = B(4) as before The previous remark implies that, when checking
0,0 0,1 1,0
whether a set of states satisfies the conditions of Propo-
and C(3) = 0. Thus the optimal discrimination problem
sition 1 or not, one has to consider all possible sets of
of N = 3,4 states has been reduced to the evaluation
A,B,C obtainable by different orderings of the states,
of the function F, which requires an optimization over a
not only the conventional one employed in Eqs. (20,21).
single operator Q.
Alternatively, one can apply this symmetry under ex-
As already discussed, if the problem of Eq. (20) were to
change of the states to obtain recursive relations for
be solved exactly for any set of states S(4), then the re-
F(A,B,C), e.g., for N = 3 and by exchanging (0,0) ↔
sult could be plugged into Eq. (15), obtaining an expres-
(1,0), it holds
sion for the optimal discrimination probability of N =8
states dependent only on the first binary POVM. Unfor-
F(−3B−A,B−A,0)
tunately a solution of Eqs. (20,21) can be found only in F(A,B,0)= +Tr[A+B]; (24)
2
some specific cases, listed below and discussed in detail
in Appendix B. In the following we employ the positive
then Proposition 1 holds on the right-hand side of (24)
part of an operator X, defined as X+ =(X+|X|)/2. when B(cid:48) = B −A has a definite sign, but the latter is
simplyB(cid:48) =(ρ˜ −ρ˜ )/2,anexpressionoftheoperator
Proposition 1. The value of the function F(A,B,C) of 1,0 0,1
B for the new ordering of the states.
Eq. (19) is
Remark 3. In all the cases listed in Proposition 1, with
F(A,B,C)=Tr[(A+|B|−|C|) ]+||C|| , (22)
+ 1 the conventional ordering of the states of Eqs. (20,21),
when at least one of the following conditions holds: i) the optimal success probabilities for the discrimination of
the operators B and C have support respectively on the N =3,4 states become
P (cid:16)S(3)(cid:17)=p +Tr(cid:20)(ρ˜0,0+ρ˜0,1−2ρ˜1,0+|ρ˜0,0−ρ˜0,1|)+(cid:21), (25)
Succ 1,0 2
PSucc(cid:16)S(4)(cid:17)= p1,0+2 p1,1 +Tr(cid:20)(ρ˜0,0+ρ˜0,1−ρ˜1,0−ρ˜1,1+|2ρ˜0,0−ρ˜0,1|−|ρ˜1,0−ρ˜1,1|)+(cid:21)+(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)ρ˜1,0−2 ρ˜1,1(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) .(26)
1
IV. A NUMERICAL EXAMPLE: THE N =3,4 known that the density matrices ρ of this system can be
QUBIT CASE representedasarealvector(cid:126)v insideathree-dimensional
ρ
unit sphere (the Bloch sphere), i.e. ρ = (1 +(cid:126)v ·(cid:126)σ)/2
2 ρ
where 1 is the identity operator and (cid:126)σ = (σ ,σ ,σ ) is
2 1 2 3
InthisSectionweanalyzethediscriminationprobabil- the vector of Pauli matrices
ityobtainedwiththenestedPOVMdecompositioninthe
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
case of N = 3,4 qubit states. Indeed, since Eqs. (20,21) 0 1 0 −i 1 0
σ = , σ = , σ = .
seem not to be solvable analytically for generic sets of 1 1 0 2 i 0 3 0 −1
states, itisinterestingtotackletheproblembychoosing
thesimplestpossibleHilbertspaceforthemeasuredsys- In particular, pure states are situated on the sphere’s
tem, i.e., the qubit space H of dimension two. It is well surface, i.e., v = |(cid:126)v | = 1 for ρ = |ψ(cid:105)(cid:104)ψ|, while the
2 ρ ρ
6
completely mixed state 1 /2 is at the origin. More gen- in the Bloch space, i.e.
2
erally, any hermitian operator X on the qubit space can
X =c 1 +(cid:126)r ·(cid:126)σ, (27)
be expressed in terms of four real coefficients: a scalar X 2 X
c , which represents the normalization coefficient of the
X the trace of the operator being determined by Tr[X] =
operator,andavector(cid:126)r ,whichrepresentstheoperator
X 2c , while its eigenvalues by λ(±) =c ±r with r =
X X X X X
|(cid:126)r |.
X
Employing the representation described above we can
hence rewrite the function F (A,B,C) as (see Ap-
Q
pendix C for details)
(cid:32) (cid:114)
(cid:12) (cid:16) (cid:17)(cid:16) (cid:17)
F (A,B,C)(cid:12) =2 c c +(cid:126)r ·(cid:126)r + (c c +(cid:126)r ·(cid:126)r )2+ (r )2−(c )2 (c )2−(r )2
Q (cid:12) Q A Q A Q B Q B B B Q Q
H2
(cid:114) (cid:33)
(cid:16) (cid:17)(cid:16) (cid:17)
+ ((1−c )c −(cid:126)r ·(cid:126)r )2+ (r )2−(c )2 (1−c )2−(r )2 , (28)
Q C Q C B B Q Q
when B and C do not have a definite sign, or
(cid:32) (cid:33)
(cid:12)
F (A,B,C)(cid:12) =2 c c +(cid:126)r ·(cid:126)r +c c +(cid:126)r ·(cid:126)r +(1−c )c −(cid:126)r ·(cid:126)r
Q (cid:12) Q A Q A Q |B| Q |B| Q |C| Q |C|
H2
(cid:0) (cid:1)
=2 c c +(cid:126)r ·(cid:126)r +c , (29)
Q A+|B|−|C| Q A+|B|−|C| |C|
whenbothBandC haveadefinitesign,withc =±c , its argument is positive and zero otherwise.
|B| B
(cid:126)r =±(cid:126)r respectivelyforB ≥0andB ≤0andsimilar As for the other case, in which Eq. (28) is valid, un-
|B| B
definitions for |C|. fortunately the function cannot be completely optimized
For each set of N =3,4 qubit states to discriminate, the analytically. Nevertheless its numerical optimization is
operatorsA,B,C,i.e.,theircoefficientscand(cid:126)r,arefixed straightforward and thus, together with Eqs. (20, 21), it
andtheoptimizationof Eqs.(28, 29)is tobecarriedout provides a convenient method to obtain the optimal suc-
only over Q, i.e., on its coefficients c and (cid:126)r under the cess probability of discrimination and the optimal mea-
Q Q
constraints surement operators for N = 3,4 qubit states. As an
example let us consider N = 3 equiprobable pure qubit
1≥c ≥0, r ≤min[c ,1−c ], (30)
Q Q Q Q states situated on the (x,y) plane of the Bloch sphere,
that ensure the positivity of Q and the fact that it must i.e., a combination of 1 , σ and σ ; this is a simple
2 1 2
besmallerthanone. InparticularforN =3statesC =0 choice for the sake of clarity, but we stress that no addi-
and one can show that c +r = λ(+) = 1 is optimal. tional optimization difficulties are met when considering
Q Q Q
Moreover the optimal(cid:126)r lies on the plane of(cid:126)r and(cid:126)r , non-equiprobable and mixed states. Let us fix the first
Q A B
sothatitcanbedefinedintermsofitsnormr andasin- statetobeonthexaxis, i.e.,(cid:126)r =(1,0,0), withoutloss
Q ρ1
gle angle φ as(cid:126)r ·(cid:126)r =r r cosφ . Then in this case of generality. Then we can study the optimal success
Q Q A Q A Q
itisonlyrequiredtooptimizetwoparameters,namelyc probability by varying the angles of the other two vec-
Q
and φ . For N = 4 instead there are no further simpli- tors with respect to the first one: (cid:126)r =(cosφ ,sinφ ,0)
Q ρ2 2 2
fications and one has to optimize four parameters, with and (cid:126)r = (cosφ ,sinφ ,0). If the states are also sym-
ρ3 3 3
constraints. metrically distributed at constant angles along the cir-
Let us now consider the case in which Eq. (29) is valid, cumference, i.e., φ = φ = 2π/3, the result is well-
2 3
(cid:16) (cid:17)
i.e., B and C have a definite sign. It is one of the situa- known [17]: P S(3) = 2/3, which is the maxi-
Succ sym
tions considered in Proposition 1, thus the optimization
mum success probability of discrimination for any three
of (29) must match the expression (22). This fact is al-
equiprobablequbitstates(becausenootherconfiguration
ready quite clear if we express Eq. (29) directly in terms
can achieve a lower average state-overlap than this one).
oftheinitialoperators; furthermoreitcanalsobeshown
For more general states the results are shown in Fig. 2
by direct analytical optimization that in this case
where we plot the optimal success probability (21), com-
(cid:12)
F(A,B,C)(cid:12)(cid:12) =|cA+|B|−|C||+cA+|B|−|C|+2cC (31) puted by numerical optimization of (28) over cQ, (cid:126)rQ, as
H2 a function of the third angle φ3 and for several choices
and the optimal value of Q is r = 0 and c = of φ = 0,π/6,π/2,2π/3,π. It can be seen that, for all
Q Q 2
θ(c ), withθ(·)thestepfunction, valued1when values of φ , there is a range of values of φ for which
A+|B|−|C| 2 3
7
(cid:16) (cid:17)
Figure 2. Plot of P S(3) , the optimal success prob-
Succ
ability (21) for a set of three equiprobable pure qubit
Figure3. Plotofthevectorsofthethreestates(cid:126)r =(1,0,0)
states, identified by the Bloch vectors (cid:126)r = (1,0,0), (cid:126)r = ρ1
(cosφ ,sinφ ,0)and(cid:126)r =(cosφ ,sinφρ,10),asafunctioρn2 of (black fixed on x axis), (cid:126)rρ3 = (cosφ3,sinφ3,0) (black) and
φ for2severa2lvaluesofρφ3 =0,π/36,π/2,32π/3,π (respectively (cid:126)rρ2 = (cosφ2,sinφ2,0) (red/dark gray at the top and or-
3 2 ange/light gray at the bottom) on the (x,y) Bloch plane,
from yellow/light-gray to black). The results are obtained
as well as the triangles formed by them (same color codes
by numerical optimization of Eq. (28) over c , (cid:126)r . Observe
Q Q as (cid:126)r ). The labels a,b,c refer respectively to values of
that,forallvaluesofφ ,thereisarangeofvaluesofφ where ρ2
(cid:16) (cid:17) 2 3 φ3 = π/15,3π/5,7π/5, also highlighted in Fig. 2, while two
PSucc S(3) attainsthemaximumallowedfornon-orthogonal values of φ2 =2π/3,π/6 (respectively red/dark gray and or-
states, i.e., the same as for symmetric states. Outside of this ange/light gray figures) are considered. By comparison with
(cid:16) (cid:17)
range the quantity decreases, reaching a constant minimum Fig. 2 it is evident that P S(3) is maximum when the
Succ
forφ ≤φ (seetextandFig.3foranexplanation). Thecases
3 2 triangle formed by the states contains the origin (c), while it
φ =π/6,2π/3 are explicitly depicted in Fig. 3 for three val-
2 is lower otherwise. In particular, when φ ≤ φ (a,b top, a
ues of φ identified by the labelled dotted lines. 3 2
3 bottom)thelargestsideofthetriangleformedbythestatesis
always(cid:126)r −(cid:126)r and this determines completely the optimal
ρ2 ρ1
success probability.
P (cid:0)S(3)(cid:1) is equal to the maximum value of 2/3, even
Succ
though the states are not symmetrically distributed on
the circumference. In other words there is a wide class Indeed in this case the polytope formed by the states
of states that can be discriminated with performance as is already maximal in the Bloch sphere and R = 1/3,
good as if they were symmetric. Outside of this range, as for the symmetric set. On the other hand, if the
whosewidthdependsonφ ,thevalueofP (cid:0)S(3)(cid:1)de- polytopedoesnotcontaintheorigin,theoptimalsuccess
2 Succ
creasesanditreachesaconstantminimumwhenφ ≤φ . probability is strictly lower than the maximum one, i.e.,
3 2 (cid:16) (cid:17) (cid:16) (cid:17)
These peculiarities can be explained by referring to P S(3) < P S(3) , as in Fig. 3a,b. Indeed
Succ (cid:43)0 Succ sym
Refs. [25–27]. In particular Ref. [26] states that the
in this second case the polytope formed by the states
optimal success probability of discrimination of a set
is not maximal and can be expanded until its largest
Se(qN) ={ρj/N}j=0,···,N−1 of N equiprobable qubit states side matches a diameter of the circumference, so that
can be found by: i) considering the geometric figure de- R < 1/3. As for the region φ ≤ φ where P (cid:16)S(3)(cid:17)
termined by the weighted states in the Bloch space, i.e., 3 2 Succ (cid:43)0
their polytope of vertices {(cid:126)r /N}; ii) finding the poly- is minimum and constant (as in Fig. 3a,b top and a bot-
tope similar to the latter anρjd that is also maximal in tom), it can be explained by observing that the largest
the Bloch sphere; iii) computing the ratio R between side of the triangle determined by the states, which in
the original and the maximal polytope. Then the op- turn determines R, is always the one that connects (cid:126)rρ1
(cid:16) (cid:17) and(cid:126)r , independentlyof(cid:126)r . Since(cid:126)r −(cid:126)r isconstant
timal success probability is PSucc Se(qN) = N1 +R. In for coρn2stant φ , R is constaρn3t too in tρh2is caρs1e.
2
light of this observation, we can explain the results of
Fig. 2 by plotting the states in the Bloch sphere and
studying the polygon formed by their vertices, as done
V. CONCLUSIONS
in Fig. 3 for two values of φ =π/6,2π/3 used in Fig. 2
2
and φ = π/15,3π/5,7π/5, corresponding to the dot-
3 In this article we proposed a method to compute the
ted lines labelled a,b,c in Fig. 2. If the polytope de-
optimaldiscriminationprobabilityandoptimalmeasure-
termined by the qubits, usually a triangle, contains the
ment operators of an arbitrary set of N states. We
origin, then the optimal success probability is still max-
(cid:16) (cid:17) (cid:16) (cid:17) showed how to decompose any multiple-outcome mea-
imum, i.e., PSucc S⊇(30) ≡ PSucc Ss(y3)m , as in Fig. 3c. surement into several binary-outcome steps, which could
8
be of interest also in other contexts. For the discrimina- of states or investigating different kinds of measurement
tionproblemthisdecompositionintroducesaconnection decompositions.
betweenthesuccessprobabilitiesofsetsofdifferentsizes,
possiblysimplifyingtheoptimizationprocedure,butdoes
not allow to reach a general analytical solution. Never-
theless it proves to be a useful tool for quickly determin-
VI. ACKNOWLEDGMENTS
ing the optimal discrimination probability of N = 3,4
qubit states, requiring just a simple numerical optimiza-
tion. Indeed with this method we were able to highlight GdP acknowledges financial support from the Eu-
someinterestingproperties,explicitlyverifyingthevalid- ropean Research Council (ERC Grant Agreement no
ityandgeometricinsightofsomepreviousresults[25–27]. 337603), the Danish Council for Independent Research
Future lines of work could focus on simplifying the opti- (SapereAude)andVILLUMFONDENviatheQMATH
mization for higher-dimensional systems and larger sets Centre of Excellence (Grant No. 10059).
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9
macher, M. Westmoreland, and W. K. Wootters, ibid. start by summing over the last bit k = 0,1, coupling
uF
54, 1869 (1996). elements that differ only for its value, i.e., F
andF . Thesearemadeofthesamesekq(u1,eunFc−e1)o,0f
k(1,uF−1),1
operators apart from the most interior ones, B(uF)
Appendix A: Completeness of the nested POVM k(1,uF−1),0
andB(uF) ,whichareinsteadtwodifferentelements
In this appendix we show that the nested POVM de- oftheks(a1m,uFe−b1)in,1aryPOVMB(uF) ,thussatisfyacom-
finedin(3)iscomplete,i.e.,thesumofitselementsisthe k(1,uF−1)
pleteness relation and their sum can be simplified. The
identity on the whole Hilbert space of the system. This
same procedure is then applied recursively on previous
can be shown by employing the completeness of each bi-
bits as follows:
nary POVM’s B(u) at each step u. Indeed we can
k(1,u−1)
(cid:88) (cid:88) (cid:16) (cid:17)
F = F +F
k(1,uF) k(1,uF−1),0 k(1,uF−1),1
k1,···,kuF k1,···,kuF−1
= (cid:88) (cid:113)B(1)···(cid:114)B(uF−1) (cid:16)B(uF) +B(uF) (cid:17)(cid:114)B(uF−1) ···(cid:113)B(1)
k1 k(1,uF−1) k(1,uF−1),0 k(1,uF−1),1 k(1,uF−1) k1
k1,···,kuF−1
(cid:113) (cid:114) (cid:114) (cid:113)
= (cid:88) B(1)··· B(uF−2) B(uF−1) B(uF−2) ··· B(1)
k1 k(1,uF−2) k(1,uF−1) k(1,uF−2) k1
k1,···,kuF−1
= (cid:88) (cid:113)B(1)···(cid:114)B(uF−2) (cid:16)B(uF−1) +B(uF−1) (cid:17)(cid:114)B(uF−2) ···(cid:113)B(1)
k1 k(1,uF−2) k(1,uF−2),0 k(1,uF−2),1 k(1,uF−2) k1
k1,···,kuF−2
=···=(cid:88)(cid:113)B(1)(cid:16)B(2) +B(2) (cid:17)(cid:113)B(1) =B(1)+B(1) =1. (A1)
k1 k1,0 k1,1 k1 0 1
k1
We note that the previous result does not change tive in all its arguments, i.e., for any set of operators
if instead of employing complete binary POVM’s, we {A ,B ,C } it holds:
j j j j=1,···,n
relax to weak completeness, as defined in Sec. II,
(cid:88) (cid:88) (cid:88) (cid:88)
i.e., that each measurement B(u) is complete on F( Aj, Bj, Cj)≤ F(Aj,Bj,Cj). (B1)
k(1,u−1)
the support of the operator that preceeds it in the j j j j
nested decomposition, B(u−1) . In this case it still This follows from the subadditivity of the trace norm.
k(1,u−1)
holds (cid:113)B(u−1) (cid:16)B(u) +B(u) (cid:17)(cid:113)B(u−1) = We can now state some lemmas that help demonstrate
k(1,u−1) k(1,u−1),0 k(1,u−1),1 k(1,u−1) Proposition 1. Throughout the Appendix, the notation
B(u−1) and the equalities in (A1) are unchanged. 1 represents the projector on the support of the opera-
k(1,u−1) toXr X.
Lemma1. LetussupposethatAispositivesemidefinite,
Appendix B: Detailed study of F B has support inside the support of A and that C and A
have orthogonal supports. Then F(A,B,C) = Tr[A]+
In this appendix we study the function FQ(A,B,C) ||B||1+||C||1.
appearing in Eqs. (18), and discuss its optimization in
Proof. Consider the set of operators
the cases mentioned in Sec. III. The optimization of
{A =Aδ ,B =Bδ ,C =Cδ } and ap-
F (A,B,C) is difficult because of the competing inter- j j,1 j j,2 j j,3 j=1,2,3
Q
ply the subadditivity property (B1):
ests of the three terms composing it. Indeed each single
term of Eq. (18) is maximized by a different operator Q:
F(A,B,C)≤F(A,0,0)+F(0,B,0)+F(0,0,C)
the first one is maximum when Q is the projector on the
=Tr[A]+||B|| +||C|| . (B2)
positive support of A; the second one is maximum when 1 1
Q is the identity on the whole Hilbert space of the sys- Thelatterinequalitycanbesaturatedunderthehypothe-
tem;thethirdoneismaximumwhenQiszero. Hencewe ses of this lemma, by taking Q=1 .
A
can solve the problem exactly only if the three operators
exhibit specific properties. Lemma 2. Let us suppose that A is negative semidef-
We start by observing that the function is subaddi- inite, C has support inside the support of A and that
10
B and A have orthogonal supports. Then F(A,B,C) = inequalities(B7,B8)andhence(B9)aresaturatedinboth
||B|| +||C|| . the second and third cases of Proposition 1, though for
1 1
different reasons:
Proof. Consider the same set of operators of Lemma 1
• If B and C have a definite sign, then it holds B =
and apply again the subadditivity property (B2), then
B or B = B , so that Eq. (B7) is saturated and
use the fact that A≤0: + −
analogously (B8);
• If A, B, C all commute with each other, then
F(A,B,C)≤F(A,0,0)+F(0,B,0)+F(0,0,C) (B3)
Eqs. (B7,B8) are saturated by any operator Q
=||B||1+||C||1. which commutes with both B and C. Eventually
the choice Q=1 necessary to saturate
Thelatterinequalitycanbesaturatedunderthehypothe- Eq. (B10) satisfie(sA+th|Bis|−la|Ctt|)e+r condition in the case
ses of this lemma, by taking Q=1B. considered.
Finally we note that the previous case of commuting op-
Hence we can prove the first case of Proposition 1: let erators, as well as further results, can also be derived by
A = A+ ⊕(−A−) be the decomposition of A in terms applying the symmetry property of the optimal success
of its positive and negative parts, with A± ≥ 0, and probabilities(20,21)toobtainrecursiveformulas,asdis-
suppose that B, C have support respectively inside the cussed after Remark 2, but still a full solution cannot be
support of A+, A−. Then consider the set of operators found in this way.
(cid:8)A =(−1)j+1A ,B =Bδ ,C =Cδ (cid:9)
j (−1)j+1 j j,1 j j,2 j=1,2
and apply the subadditivity property (B1), together
with Lemmas 1, 2: Appendix C: Computation of FQ in the qubit case
F(A,B,C)≤F(A ,B,0)+F(−A ,0,C) InthisSectionwederivetheresults(28, 29)explicitly.
+ −
=Tr[A ]+||B|| +||C|| , (B4) Asapreliminaryrecallthat,foranythreevectors(cid:126)a,(cid:126)b,(cid:126)c∈
+ 1 1
R3 and the Pauli matrices (cid:126)σ it holds:
whichissaturatedbyameasurementoperatorQ=1 . (cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
A+ ((cid:126)a·(cid:126)σ) (cid:126)b·(cid:126)σ = (cid:126)a·(cid:126)b +i (cid:126)a×(cid:126)b ·(cid:126)σ, (C1)
This expression is equivalent to that given in (22) under
the current hypotheses, indeed in this case it holds (cid:16) (cid:16) (cid:17)(cid:17) (cid:16) (cid:17)
(cid:126)a× (cid:126)b×(cid:126)c =((cid:126)a·(cid:126)c)(cid:126)b− (cid:126)a·(cid:126)b (cid:126)c. (C2)
(A+|B|−|C|) =((A +|B|)⊕(−A −|C|))
+ + − + Moreover, given a positive operator√Q on H2, the coeffi-
=A +|B|, (B5) cients c√ , (cid:126)r√ of its square root Q can be expressed
+ Q Q
in terms of its coefficients c , (cid:126)r as:
Q Q
so that (B4) becomes  √ √
(cid:40)c =(cid:0)c√ (cid:1)2+(cid:0)r√ (cid:1)2 c√ = cQ+rQ+ cQ−rQ
F(A,B,C)=Tr[A +|B|]+||C|| Q Q Q ↔ Q √ 2√ (C3)
+ 1 rQ =2c√Qr√Q r√ = cQ+rQ− cQ−rQ.
=Tr[(A +|B|−|C|) ]+||C|| .(B6) Q 2
+ + 1
with(cid:126)r (cid:107)(cid:126)r√ . In order to evaluate F (A,B,C) we can
Q Q Q
Asfor thesecond andthirdcasesofProposition 1, let us computeitsfirsttwoterms,whilethethirdoneissimilar
first note that to the second one. Let us start with the product QA: it
is a generic operator with coefficients
(cid:12)(cid:12)(cid:112) (cid:112) (cid:12)(cid:12) (cid:12)(cid:12)(cid:112) (cid:112) (cid:12)(cid:12) (cid:12)(cid:12)(cid:112) (cid:112) (cid:12)(cid:12)
(cid:12)(cid:12) QB Q(cid:12)(cid:12) ≤(cid:12)(cid:12) QB Q(cid:12)(cid:12) +(cid:12)(cid:12) QB Q(cid:12)(cid:12)
(cid:12)(cid:12) (cid:12)(cid:12)1 (cid:12)(cid:12) + (cid:12)(cid:12)1 (cid:12)(cid:12) − (cid:12)(cid:12)1 cQA =cQcA+(cid:126)rQ·(cid:126)rA (C4)
=Tr[Q(B++B−)]=Tr[Q|B|], (B7) (cid:126)rQA =cQ(cid:126)rA+cA(cid:126)rQ+i((cid:126)rQ×(cid:126)rA), (C5)
where B are the positive and negative parts of B as computed by applying Eq. (C2). Thus the first term of
±
defined above for A, and analogously FQ is simply Tr[QA√]=2√cQA.
Asfortheproduct QB Q,itsfirstcoefficientissimple:
√ √
(cid:12)(cid:12)(cid:12)(cid:12)(cid:112)1−QC(cid:112)1−Q(cid:12)(cid:12)(cid:12)(cid:12) ≤Tr[(1−Q)|C|]. (B8) c√QB√Q = Tr[ QB Q]/2 = cQB, easily obtained by
(cid:12)(cid:12) (cid:12)(cid:12) relabelling Eq. (C4). The vector of coefficients instead is
1
(cid:0) (cid:1)
(cid:126)r√ √ =c√ (cid:126)r√ +c√ (cid:126)r√ +i (cid:126)r√ ×(cid:126)r√ (C6)
We then have QB Q QB Q Q QB QB Q
(cid:0) (cid:1) (cid:16)(cid:0) (cid:1)2 (cid:0) (cid:1)2(cid:17)
=c (cid:126)r +2 (cid:126)r√ ·(cid:126)r (cid:126)r√ + c√ − r√ (cid:126)r ,
F (A,B,C)≤Tr[Q(A+|B|−|C|)]+||C|| (B9) B Q Q B Q Q Q B
Q 1 √
≤Tr[(A+|B|−|C|) ]+||C|| .(B10) wherewehavefirstcomputedtheproductbetween QB
+ 1 √
and Q, then substituted the expression for the former
The inequality (B10), is saturated by taking Q equal to by relabelling once again the product QA and employed
the projector onto the support of (A+|B|−|C|) . The (C3).
+