On the equivalence of the CH and CHSH inequalities for two three-level systems

0.28 MB·English

by  Jose L. Cereceda

On the equivalence of the CH and CHSH inequalities for two
three-level systems
∗
Jos´e L. Cereceda†
C/Alto del Le´on 8, 4A, 28038 Madrid, Spain
January 6, 2004
4
0 Abstract
0
2 In this paper we show a Clauser-Horne (CH) inequality for two three-level quantum systems or
n qutrits, alternative to the CH inequality given by Kaszlikowski et al. [Phys. Rev. A 65, 032118
a (2002)]. In contrast to this latter CH inequality, the new one is shown to be equivalent to the Clauser-
J Horne-Shimony-Holt (CHSH) inequality for two qutrits given by Collins et al. [Phys. Rev. Lett. 88,

5
040404 (2002)]. Both the CH and CHSH inequalities exhibit the strongest resistance to noise for a

nonmaximallyentangledstateforthecaseoftwovonNeumannmeasurementspersite,asfirstshownby
3
v Ac´ın et al. [Phys. Rev. A 65, 052325 (2002)]. This equivalence, however, breaks down when one takes
7 into account the less-than-perfect quantum efficiency of detectors. Indeed, for the noiseless case, the
1 thresholdquantumefficiencyabovewhichthereisnolocalandrealisticdescriptionoftheexperimentfor
1 the optimal choice of measurements is found to be (9 √33)/4 0.814 for the CH inequality, whereas
2 − ≈
it is equalto ( 3+√33)/2 0.828 for theCHSH inequality.
1 − ≈
2 Keywords: Quptrit, Bell’s inequality,no-signaling condition, noise admixture, detector inefficiency.
0
/
h
p 1 Introduction and notation
-
t
n Recently,twokinds ofBellinequalities[1]havebeen introducedfortwothree-dimensionalquantumsystems
a
(so-called qutrits). The scenario for both types of inequalities involves two parties: Alice can carry out two
u
q possiblemeasurements,A orA ,ononeofthequtrits,whereasBobisallowedtoperformthemeasurements
1 2
v: B1 or B2 on the other qutrit. Each measurement has three possible outcomes Ai,Bj = 1,2,3 (i,j = 1,2).
i Then, denoting by P(A = B +k) the probability that the measurements A and B have outcomes that
X i j i j
differ by k modulo d (in our case d=3), the Collins et al. [2] arrived at the following Bell inequality:
r
a
I = P(A =B )+P(B =A +1)+P(A =B )+P(B =A )
3 1 1 1 2 2 2 2 1
P(A =B 1) P(B =A ) P(A =B 1)
1 1 1 2 2 2
− − − − −
P(B =A 1) 2. (1)
2 1
− − ≤
TheBellinequality(1)isofthe CHSHtypebecauseitreducestothe familiarCHSHinequality[3]ford=2.
Actually, inequality (1) is a particular case of the family of Bell inequalities (CGLMP-set):
I (c ,c ,c ,c )=P(A =B +c )+P(B =A +c )+P(A =B +c )
3 1 2 3 4 1 1 1 1 2 2 2 2 3
+P(B =A +c ) P(A =B (c +c +c )) P(B =A (c +c +c ))
2 1 4 1 1 2 3 4 1 2 1 3 4
− − − −
P(A =B (c +c +c )) P(B =A (c +c +c )) 2, (2)
2 2 1 2 4 2 1 1 2 3
− − − − ≤
∗Thispaperhasbeenoriginallypublishedin: International JournalofQuantumInformation1,115-133(2003).
†Electronicmail: [email protected]
1
where
c =0, 1,
i
±
c +c +c +c =0mod3,
1 2 3 4
6
and where the sum is modulo 3 for the c ’s. Inequality (1) is obtained for c = +1 and c = c = c = 0.
i 2 1 3 4
There are 54 combinations of c ’s (with c =0, 1) fulfilling the condition c +c +c +c =0mod3.
i i 1 2 3 4
± 6
On the other hand, denoting by Pij(a ,b ) the joint probability of obtaining by Alice and Bob simulta-
i j
neously the results a and b (a ,b =1,2,3) for the pair of observables A and B , and denoting by Pi(a )
i j i j i j i
(Qj(b )) the single probability of obtaining the result a (b ) by Alice (Bob) irrespective of Bob’s (Alice’s)
j i j
outcome, the Kaszlikowskiet al. [4] arrived at the following Bell inequality:1
W = P11(2,1)+P12(2,1) P21(2,1)+P22(2,1)
3
−
+P11(1,2)+P12(1,2) P21(1,2)+P22(1,2)
−
+P11(2,2)+P12(1,1) P21(2,2)+P22(2,2)
−
P1(1) P1(2) Q2(1) Q2(2) 0. (3)
− − − − ≤
As noticed in Ref. [4], the Bell inequality (3) is the sum of two CH inequalities plus one term which bears
a resemblance to an incomplete CH inequality. (Previous derivations of Bell inequalities for two three-level
systems based on the originalClauser-Horneinequalities [5] can be found, for example, in Refs. [6] and [7].)
It has been shown that both the CHSH inequality (1) and the CH inequality (3) give the same threshold
value of noise admixture (for which it is still not possible to build a local classical model for the predicted
probabilities)for the maximallyentangledstate [2,4](see alsoRefs. [8,9,10]). This notwithstanding,aswe
will show, inequalities (1) and (3) are not equivalent. This might seem rather surprising in view of the fact
that, for bipartite two-dimensionalsystems, the familiar CHSH andCH inequalities are equivalent provided
that the correlations cannot be used for instantaneous communication between Alice and Bob [11, 12]. So
a non-trivial question is whether the set of CH inequalities introduced in Ref. [4] does exhaust all possible
instances of CH inequalities for two three-level systems. In this paper we answer this question—a negative
one—by exhibiting a CH inequality having the same structure as inequality (3), and which is equivalent to
the CHSHinequality(1). Indeed, itwillbe arguedthattoeachofthe CHSHinequalities inthe CGLMP-set
(2), there corresponds one and only one independent CH inequality which is equivalent to it.
Central to the derivation of our results is the above-mentioned property of causal communication (also
termed “physical locality” [11]). For the experiment considered, this means that the marginal probabilities
for one party should be independent of the measurement chosen by the other party:
3 3
P(A =m,B =n)= P(A =m,B =n), (4)
i 1 i 2
n=1 n=1
X X
and
3 3
P(A =m,B =n)= P(A =m,B =n), (5)
1 j 2 j
m=1 m=1
X X
1The Bell inequality (3) is a member of theset of CH inequalities introduced in [4], namely,
P1+α1+β(2+x,1+y)+P1+α2+β(2+x,1+y) P2+α1+β(2+x,1+y)
−
+P2+α2+β(2+x,1+y)+P1+α1+β(1+x,2+y)+P1+α2+β(1+x,2+y)
P2+α1+β(1+x,2+y)+P2+α2+β(1+x,2+y)+P1+α1+β(2+x,2+y)
−
+P1+α2+β(1+x,1+y) P2+α1+β(2+x,2+y)+P2+α2+β(2+x,2+y)
−
P1+α(1+x) P1+α(2+x) Q2+β(1+y) Q2+β(2+y) 0,
− − − − ≤
where α,β =0,1; x,y =0,1,2, and where the addition is modulo 2 for α,β and modulo 3 for x,y. Inequality (3) is
obtained for α,β=0 and x,y=0.
2
foranyi,j =1,2andm,n=1,2,3. Thefulfillmentof(4)and(5)constitutesaphysicallysoundrequirement
sinceaviolationofeither(4)or(5)would,inprinciple,allowthetwopartiestocommunicatesuperluminally.
Both quantum mechanics and classical theories satisfy the requirement of causal communication (the “no-
signaling” condition), and hence the predictions by such theories do satisfy each of the constraints in (4)
and (5). In addition to this, the joint probabilities are required to satisfy the normalization condition:
3
P(A =m,B =n)=1, (6)
i j
m,n=1
X
for any i,j =1,2.
The paper is organized as follows. In Sec. 2, we provide a CH-type inequality for bipartite systems of
qutrits, alternative to the original CH inequality introduced in Ref. [4]. Using the conditions in (4)–(6), we
show that the given CH inequality is equivalent to the CHSH inequality (1). We point out that, actually, a
similar relationshipcould be establishedbetweeneachCHSH inequality in the set(2) and someappropriate
CH inequality. Conditions (4)–(6) are also used to show that inequalities (1) and (3) are not equivalent. In
Sec.3,wedescribetheoptimalsetofmeasurementsgivingthe maximalviolationofboththeCHandCHSH
inequalities, and determine the resistance to noise of such inequalities for the optimal choice of observables.
InSec.4,weconsidertherealisticcaseofdetectorswithafinitequantumdetectorefficiencyη <1. Aswewill
see, for this case the equivalence between the CH and CHSH inequalities does not follow any more. For the
noiselesscase,wecalculatethecriticalquantumefficiencyneededtoruleoutalocalandrealisticdescription
of the considered experiment for both the CH and CHSH inequalities. Finally, the main conclusions are
summarized in Sec. 5.
In order to abbreviate the notation, we will henceforth use at our convenience the following shorthand
notation for the various joint probabilities:
p P11(1,1), p P11(1,2), p P11(1,3), p P11(2,1),
1 2 3 4
≡ ≡ ≡ ≡
p P11(2,2), p P11(2,3), p P11(3,1), p P11(3,2),
5 6 7 8
≡ ≡ ≡ ≡
p P11(3,3), p P12(1,1), p P12(1,2), p P12(1,3),
9 10 11 12
≡ ≡ ≡ ≡
p P12(2,1), p P12(2,2), p P12(2,3), p P12(3,1),
13 14 15 16
≡ ≡ ≡ ≡
p P12(3,2), p P12(3,3), p P21(1,1), p P21(1,2), (7)
17 18 19 20
≡ ≡ ≡ ≡
p P21(1,3), p P21(2,1), p P21(2,2), p P21(2,3),
21 22 23 24
≡ ≡ ≡ ≡
p P21(3,1), p P21(3,2), p P21(3,3), p P22(1,1),
25 26 27 28
≡ ≡ ≡ ≡
p P22(1,2), p P22(1,3), p P22(2,1), p P22(2,2),
29 30 31 32
≡ ≡ ≡ ≡
p P22(2,3), p P22(3,1), p P22(3,2), p P22(3,3).
33 34 35 36
≡ ≡ ≡ ≡
2 Alternative CH inequality
Let us consider the inequality
K = P11(1,1)+P12(1,1) P21(1,1)+P22(1,1)
3
−
+P11(2,2)+P12(2,2) P21(2,2)+P22(2,2)
−
+P11(2,1)+P12(1,2) P21(2,1)+P22(2,1)
−
P1(1) P1(2) Q2(1) Q2(2) 0. (8)
− − − − ≤
As in the caseof the CH inequality (3), the inequality (8) canbe written as the sum oftwo CH inequalities,
CH and CH , and some additional term, with CH being
1 2 1
P11(1,1)+P12(1,1) P21(1,1)+P22(1,1) P1(1) Q2(1),
− − −
3
and CH being
2
P11(2,2)+P12(2,2) P21(2,2)+P22(2,2) P1(2) Q2(2).
− − −
Note that the single probabilities appearing in (8) are the same as those appearing in (3).
In what follows we show that the CH inequality K 0 is equivalent to the CHSH inequality I 2 in
3 3
≤ ≤
(1). Moreprecisely,usingrelations(4)–(6),weshowthatthelefthandsideofinequality(8)canbeexpressed
as K =(I 2)/3. Clearly, the inequality I 2, then implies the inequality K 0. Conversely, starting
3 3 3 3
− ≤ ≤
fromthe left hand side ofinequality (1), and using relations (4)–(6), we show that I =2+3K . Therefore,
3 3
it is also the case that the inequality K 0 implies the inequality I 2. To this end, we first write both
3 3
≤ ≤
I and K as a sum of joint probabilities. From (1), it can readily be seen that
3 3
I = p +p +p +p +p +p +p +p +p +p +p +p
3 1 5 9 10 14 18 20 24 25 28 32 36
p +p +p +p +p +p +p +p +p +p +p +p , (9)
2 6 7 12 13 17 19 23 27 29 33 34
−
where we have used the(cid:0)notation in (7). On the other hand, putting the single probabilit(cid:1)ies P1(1), P1(2),
Q2(1), and Q2(2) as2
P1(1)= P12(1,1)+P12(1,2)+P12(1,3),
P1(2)= P11(2,1)+P11(2,2)+P11(2,3),
(10)
Q2(1)= P22(1,1)+P22(2,1)+P22(3,1),
Q2(2)= P22(1,2)+P22(2,2)+P22(3,2),
and susbstituting (10) into the left hand side of (8), we obtain
K =p p p +p p p p p p p . (11)
3 1 6 12 14 19 22 23 29 34 35
− − − − − − − −
Of course, due to the constraints in (4)–(6), not all the probabilities p ,p ,...,p are independent. Now,
1 2 36
inorderto compareexpressions(9)and(11),we haveto knowthe relationsthatcanbe establishedbetween
such probabilities. The normalization (6) plus no-signaling conditions (4)–(5) constitute a linear system
of 16 equations and 36 unknowns p ,p ,...,p . It can be shown [13] that such a system determines 12
1 2 36
probabilities at most among p ,p ,...,p . So, for example, we can solve the system of equations (4)–(6)
1 2 36
with respect to the set of variables p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p to find, in particular,
3 4 8 11 15 16 21 22 26 30 31 35
{ }
that
1
p = 1+p p 2p p +2p +p +p p +2p +p p 2p
22 1 2 5 6 7 9 10 12 13 14 17 18
3 − − − − − −
(cid:0)p +p p 2p 2p p 2p p +p +2p p +p , (12)
19 20 23 24 25 27 28 29 32 33 34 36
− − − − − − − −
(cid:1)
and
1
p = 1+p +2p +p p p 2p 2p p p +p +2p +p
35 1 2 5 6 7 9 10 12 13 14 17 18
3 − − − − − −
(cid:0)p 2p p +p +p +2p +p p 2p p p 2p . (13)
19 20 23 24 25 27 28 29 32 33 34 36
− − − − − − − −
(cid:1)
Thus, inserting (12) and (13) into (11) gives K =(I 2)/3, with I being the expression in (9).
3 3 3
−
Alternatively,wecansolvethe systemofequations(4)–(6)withrespecttothe setofvariables p ,p ,p ,
2 4 9
{
p ,p ,p ,p ,p ,p ,p ,p ,p . In the Appendix we write down the resulting expressions for the
11 13 18 20 24 25 28 32 36
}
2Wenotethat,sincethejointprobabilitiesPij(ai,bj)satisfytheconditionsinEqs.(4)–(5),thesingleprobabilities
Pi(ai)andQj(bj)canactuallybeexpressedintermsofPij(ai,bj)intwoequivalentforms. So,forexample,P1(1)in
(10) can be put alternatively as, P1(1)=P11(1,1)+P11(1,2)+P11(1,3), Q2(1) as Q2(1)=P12(1,1)+P12(2,1)+
P12(3,1), etc. Of course, the result that K3 = (I3 2)/3 or I3 = 2+3K3 can be obtained by using either one of
−
thetwo equivalentforms foreach Pi(ai) and Qj(bj). Theadvantageof usingthechoice in (10) isthat it leads toan
expression for K3 involvingonly ten joint probabilities.
4
probabilitiesp ,p ,p ,p ,p ,p ,p ,p ,p ,andp (see(A.1)–(A.10)intheAppendix). Substituting
2 9 13 18 20 24 25 28 32 36
now (A.1)–(A.10) into (9) we obtain I =2+3K , with K being the expression in (11).
3 3 3
Summing up, from the conditions of normalization and causal communication we have derived the rela-
tionsK =(I 2)/3andI =2+3K ,hence itfollowstheequivalenceoftheBellinequalities: K 0and
3 3 3 3 3
− ≤
I 2. We would like to emphasize that the no-signaling condition is satisfied by both quantum mechanics
3
≤
and local realistic theories, so the relation K = (I 2)/3 (or I = 2+3K ) is certainly fulfilled by the
3 3 3 3
−
probabilities predicted by such theories.
In the same way, one could equally show that each of the CHSH inequalities in the CGLMP-set (2) is
associated with some appropriate CH inequality. So, for example, consider the inequality in (2) for which
c =+1 and c =c =c =0:
4 1 2 3
′
I = P(A =B )+P(B =A )+P(A =B )+P(B =A +1)
3 1 1 1 2 2 2 2 1
P(A =B 1) P(B =A 1) P(A =B 1)
1 1 1 2 2 2
− − − − − −
P(B =A ) 2. (14)
2 1
− ≤
′
In terms of joint probabilities, I can be written in the form
3
′
I = p +p +p +p +p +p +p +p +p +p +p +p
3 1 5 9 11 15 16 19 23 27 28 32 36
p +p +p +p +p +p +p +p +p +p +p +p . (15)
2 6 7 10 14 18 21 22 26 29 33 34
−
(cid:0) (cid:1)
Consider now the CH-type inequality
K′ = P11(1,1) P12(1,1)+P21(1,1)+P22(1,1)
3 −
+P11(2,2) P12(2,2)+P21(2,2)+P22(2,2)
−
+P11(2,1) P12(2,1)+P21(1,2)+P22(2,1)
−
P2(1) P2(2) Q1(1) Q1(2) 0. (16)
− − − − ≤
We mention, incidentally, that inequality (16) contains the following two CH inequalities:
CH =P11(1,1) P12(1,1)+P21(1,1)+P22(1,1) P2(1) Q1(1),
1
− − −
and
CH =P11(2,2) P12(2,2)+P21(2,2)+P22(2,2) P2(2) Q1(2).
2
− − −
Putting the single probabilities P2(1), P2(2), Q1(1), and Q1(2) as
P2(1)= P21(1,1)+P21(1,2)+P21(1,3),
P2(2)= P22(2,1)+P22(2,2)+P22(2,3),
(17)
Q1(1)= P11(1,1)+P11(2,1)+P11(3,1),
Q1(2)= P21(1,2)+P21(2,2)+P21(3,2),
and substituting (17) into the left hand side of (16), we obtain
′
K =p p p p p p p p +p p . (18)
3 5− 7− 10− 13− 14− 20− 21− 26 28− 33
′ ′
InordertorelatetheexpressionforK in(18)withthequantityI in(15),wesolvethesystemofequations
3 3
(4)–(6) with respect to the set of variables p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p . This gives, in
3 4 8 12 13 17 20 24 25 30 31 35
{ }
particular,
1
p = 1 2p p +p +2p p +p p +p p 2p 2p p
13 1 2 5 6 7 9 10 11 14 15 16 18
3 − − − − − − − −
+(cid:0)p p +2p +p p 2p +p p 2p p +2p +p , (19)
19 21 22 23 26 27 28 29 32 33 34 36
− − − − − −
5 (cid:1)
and
1
p = 1+p +2p +p p p 2p p 2p p +p +p +2p
20 1 2 5 6 7 9 10 11 14 15 16 18
3 − − − − − −
(cid:0)2p p p 2p p +p +p +2p +p p p 2p . (20)
19 21 22 23 26 27 28 29 32 33 34 36
− − − − − − − −
′ ′ (cid:1)
Thus, using (19) and (20), in (18), we get K = (I 2)/3. On the other hand, solving with respect to
3 3 −
the variables p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p , and replacing the resulting probabilities p ,
1 6 8 11 15 16 19 23 27 30 32 34 1
{ }
′ ′ ′
p , p , p , p , p , p , p , p , and p in (15), we would find that I = 2+3K , with K being the
6 11 15 16 19 23 27 32 34 3 3 3
′ ′
expression in (18). Therefore, the inequality I 2 implies the inequality K 0, and conversely, the
3 ≤ 3 ≤
′ ′
inequality K 0 implies the inequality I 2.
3 ≤ 3 ≤
ItisimportanttonotethatagivenCHSHinequalitycanonlyberelatedtooneindependentCHinequality.
To see this, suppose insteadthat the CHSH inequality I 2 is relatedto two independent CH inequalities,
3
≤
′ ′
K 0 and K 0, through the respective relations I = 2+3K and I = 2+3K . Then it trivially
3 ≤ 3 ≤ ′ 3 3 3 3 ′
followsfromsuchrelationsthatK =K ,andhencetheinitialsuppositionthatK andK areindependent
3 3 3 3
cannot be true. Analogously, it follows that a given CH inequality can only be related to one independent
CHSH inequality. It should be noticed, however,that there is not a one-to-one correspondence between the
set of CHSH inequalities in (2) and the set of CH inequalities since, for example, as we argue in the next
paragraph,the CH inequality (3) is not equivalent to any of the inequalities in (2).
We now show that the CH inequality (3) and the CHSH inequality (1) are not equivalent. To see this,
we first write W in the equivalent form
3
W =p p p p p p p p +p p . (21)
3 2 6 10 12 16 20 22 23 31 35
− − − − − − − −
Then,solvingthe systemofequations(4)–(6)withrespecttothe setofvariables p ,p ,p ,p ,p ,p ,p ,
3 4 8 11 15 16 21
{
p ,p ,p ,p ,p , and substituting the resulting probabilities p , p , p , and p into (21), we would
22 26 30 31 35 16 22 31 35
}
obtain
1
W = I 2 p +p +p p +p p p +p . (22)
3 3 1 2 13 14 19 20 28 29
3 − − − − −
On the other hand, solving wit(cid:0)h respe(cid:1)ct to the variables
p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,p ,
1 5 9 11 13 18 19 24 26 28 32 36
{ }
and replacing the resulting probabilities p , p , p , p , p , p , p , p , p , and p in (9), we would
1 5 9 13 18 19 24 28 32 36
obtain
I =2+3 p p p p +p p +p p p p , (23)
3 4 6 7 12 14 23 25 29 34 35
− − − − − − − −
which can be written equivale(cid:0)ntly as (cid:1)
I =2+3W +3 p p p +p p +p +p p , (24)
3 3 1 2 13 14 19 20 28 29
− − − −
(cid:0) (cid:1)
with W being the expression in (21), and where we have made use of the relations
3
p p p =p p p ,
25 4 7 1 19 22
− − − −
and
p +p p p =p p ,
10 16 31 34 28 13
− − −
[cf. (5)] in passing from (23) to (24). Of course, the expression for I in (24) can also be obtained directly
3
from (22). From this latter equation we can deduce that the inequality I 2 would imply the inequality
3
≤
W 0 provided that, for I 2, the following inequality:
3 3
≤ ≤
p +p +p +p p +p +p +p , (25)
2 13 19 29 1 14 20 28
≤
6
is satisfied. On the other hand, from (24), it can be seen that the inequality W 0 would imply the
3
≤
inequality I 2 provided that, for W 0, the following inequality:
3 3
≤ ≤
p +p +p +p p +p +p +p , (26)
1 14 20 28 2 13 19 29
≤
is satisfied. Obviously,the conditions in(25) and(26) aremutually exclusive exceptfor the particularevent
in which
p +p +p +p =p +p +p +p .
1 14 20 28 2 13 19 29
This means that it is in general not possible for the inequality W 0 to imply the inequality I 2, and
3 3
≤ ≤
simultaneously for the inequality I 2 to imply the inequality W 0. In other words, such inequalities
3 3
≤ ≤
are not equivalent. Similarly, it could equally be shown that the CH inequality (3) is not equivalent to any
of the CHSH inequalities in (2). Furthermore, it seems likely that this conclusion also applies to any one of
the 36 CH-type inequalities given in Ref. [4].
Weendthissectionbynotingthattherequirementofcausalcommunicationdoesnotbyitselfpreventthe
sum of probabilities in I [cf. (9)] from reaching its maximum value, I =4. Indeed, there exist probability
3 3
distributions p ,p ,...,p satisfying all the constraints in (4)–(6), and which give I = 4. An example
1 2 36 3
{ }
of such a distribution is:
1
p =p =p =p =p =p =p =p =p =p =p =p = ,
1 5 9 10 14 18 20 24 25 28 32 36
3
with allother probabilitieszero. This example is the generalizationto twothree-levelsystemsof the finding
by Popescu and Rohrlich [14] that, for d=2, relativistic causality does not constrain the maximum CHSH
sum of correlations to 2√2, but instead it allows for probability distributions giving the maximum level of
violation. Arguably, this conclusion generalizes to any dimension d.
3 Optimal set of measurements
Let us consider a Bell experiment for which the source produces pairs of qutrits in the entangled state:
1
ψ =cosθ 2 2 + sinθ(1 1 + 3 3 ), (27)
A B A B A B
| i | i | i √2 | i | i | i | i
where 1 , 2 , 3 denotes an orthonormal basis in the state space of qutrit A (B). The
A(B) A(B) A(B)
{| i | i | i }
maximally entangled state is obtained for cosθ = 1/√3 and sinθ = 2/3. We now describe the set of
measurements giving the maximal quantum violation of both the CH and CHSH inequalities [2, 4, 15] (see
p
alsoRefs.[8,9, 10]). Firstly,for eachofthe emitted pairsofqutrits, Alice (Bob)applies aunitary operation
Ua (Ub), a,b=1,2, on qutrit A (B), with Ua and Ub given by
A B A B
1 eiαa e2iαa 1 eiβb e2iβb
1 1
Ua = 1 λeiαa µe2iαa , Ub = 1 µeiβb λe2iβb , (28)
A √3  B √3 
1 µeiαa λe2iαa 1 λeiβb µe2iβb
   
where λ = exp(2πi/3), µ = λ∗ = exp(4πi/3), and where α (β ) is the phase defining Ua (Ub). For each
a b A B
run of the experiment, Alice (Bob) has the freedom to choose the transformation U1 or U2 (U1 or U2) to
A A B B
be appliedonqutritA(B). The unitaryoperationsin(28)canbe realizedby meansofanunbiasedsix-port
beam splitter [4]. A detailed description of such devices can be found in Ref. [6]. Finally, once Ua and Ub
A B
have been applied on the respective qutrit, Alice (Bob) measures the state of the transformed qutrit A (B)
in the initial basis 1 , 2 , 3 . Thus the joint probability distribution of outcomes predicted
A(B) A(B) A(B)
{| i | i | i }
by quantum mechanics for the initial state (27) is the following:
7
1
Pab(1,1)=Pab(2,2)=Pab(3,3)= 1+sin2θcos2φ +√2sin2θcosφ ,
ψ ψ ψ 9 ab ab
1h 1 i
Pab(1,2)=Pab(2,3)=Pab(3,1)= 1 sin2θ cos2φ +√3sin2φ
ψ ψ ψ 9 − 2 ab ab
h 1 (cid:0) (cid:1)
sin2θ cosφ √3sinφ , (29)
ab ab
− √2 −
1 1 (cid:0) (cid:1)i
Pab(1,3)=Pab(2,1)=Pab(3,2)= 1 sin2θ cos2φ √3sin2φ
ψ ψ ψ 9 − 2 ab− ab
h 1 (cid:0) (cid:1)
sin2θ cosφ +√3sinφ ,
ab ab
− √2
(cid:0) (cid:1)i
where φ =α +β . Using the probabilities (29) in (9), we obtain
ab a b
√3
I (ψ )= sin2θ √3cos2φ +sin2φ +√3cos2φ sin2φ
3 11 11 12 12
| i 6 −
(cid:0) √3cos2φ sin2φ +√3cos2φ +sin2φ
21 21 22 22
− −
1
+ sin2θ √3cosφ sinφ +√3cosφ +sinφ (cid:1)
11 11 12 12
√6 −
√3(cid:0)cosφ +sinφ +√3cosφ sinφ . (30)
21 21 22 22
− −
(cid:1)
Ofcourse,fromtheresultsoftheprecedingsection,thequantumpredictionforK willbegivenbyK (ψ )=
3 3
| i
(I (ψ ) 2)/3,asonecancheckdirectlybyusingthejointprobabilities(29)ineither(8)or(11),andputting
3
| i −
the single probabilities equal to 1.
3
Consider now the following values of the phases
π π π
α =α + , β = α + , β = α , (31)
2 1 1 1 2 1
3 − 6 − − 6
where α , β , and β are given in terms of the variable phase α . For the settings in (31), expression (30)
2 1 2 1
reduces to
2
I (ψ )=2sin2θ+2 sin2θ, (32)
3
| i 3
r
which is independent of α . The corresponding expression for K is
1 3
2 2
K (ψ )= sin2θ cos2θ . (33)
3
| i 3 3 −
r
(cid:16) (cid:17)
In Fig. 1, the functions in (32) and (33) have been plotted for 0 θ π. The maximum values are
≤ ≤
Imax(ψ )=1+ 11/3 2.915,
3 | i ≈
p
and
Kmax(ψ )=( 11/3 1)/3 0.305,
3 | i − ≈
and they are attained for an angle θ 60.74◦.pExplicitly, the state leading to the maximal violation is
max
≈
11 √33 11+√33
ψ = − 2 2 + (1 1 + 3 3 ). (34)
mv A B A B A B
| i s 22 | i | i s 44 | i | i | i | i
Previous numerical work [15] shows the optimality of the chosen set of measurements, the values Imax(ψ )
3 | i
and Kmax(ψ ) indeed being the maximum ones predicted by quantum mechanics for the case in which two
3 | i
von Neumann measurements are performed by each of the parties. In particular, these values are slightly
larger than those obtained for the maximally entangled state, namely,
8
Fig. 1. I3(θ) and K3(θ) as predicted by quantum mechanics for pairs of qutrits in the state
(27), where I3(θ) and K3(θ) are evaluated in the case of condition (31). Such functions are
related to each other by I3=2+3K3.
I (ψ )=(12+8√3)/9 2.873,
3 me
| i ≈
K (ψ )=(8√3 6)/27 0.291.
3 me
| i − ≈
Note that the inequalities are not violated by the states in (27) for which θ = nπ/2 (n = 0, 1, 2,...).
± ±
When n is even such states correspondto product states, and when n is odd they correspondto maximally
entangled states of the form (1/√2)(1 1 + 3 3 ). The latter state describes two entangled qutrits
A B A B
| i | i | i | i
eachofthemliving inatwo-dimensionalstate space. Suchastate doesnotexploitthe fulldimensionalityof
thequtritsspace,andhenceitcannotviolatetheinequalities. Ontheotherhand,thestatesin(27)that,for
the measurements considered, yield a violationof either the CHSH inequality (I 2) or the CH inequality
3
≤
(K 0) are those for which arctan 3/8<θ <π/2 (mod π).
3
≤
If the initial state (27) is mixed with some amount of uncolored noise, the state becomes
p
11
ρ=λ ψ ψ +(1 λ) , (35)
| ih | − 9
where 0 λ 1. Quantum mechanics now predicts the probabilities:
≤ ≤
1 λ
Pij(a ,b )=λPij(a ,b )+ − ,
ρ i j ψ i j 9
1 λ
Pi(a )=λPi(a )+ − ,
ρ i ψ i 3
1 λ
Qj(b )=λQj(b )+ − .
ρ j ψ j 3
Since Pi(a )=Qj(b )= 1, then the same holds true for the new single probabilities, Pi(a )=Qj(b )= 1.
ψ i ψ j 3 ρ i ρ j 3
Correspondingly, I and K change to I (ρ) = λI (ψ ) and K (ρ) = λK (ψ ) 2(1−λ), respectively.
3 3 3 3 | i 3 3 | i − 3
Therefore, for the case in which I (ψ )>2, the inequality I 2 will be violated by quantum mechanics if
3 3
| i ≤
and only if
2
λ> , (36)
I (ψ )
3
| i
and, similarly, for the case in which K (ψ ) > 0, the inequality K 0 will be violated by quantum
3 3
| i ≤
mechanics if and only if
2
λ> . (37)
2+3K (ψ )
3
| i
9
Note that, as expected, the conditions in (36) and (37) are exactly the same since I (ψ ) = 2+3K (ψ ).
3 3
| i | i
So there exists a critical value λ = 2/I (ψ ) above which a local realistic description of the experiment is
3
| i
not possible. The optimal, minimum value of λ is obtained when I (ψ ) is maximum, i.e.
3
| i
λ =2/Imax(ψ )=(√33 3)/4 0.686,
min 3 | i − ≈
and this optimal value being achieved for the state ψ . Put it another way, the maximum amount of
mv
| i
uncolorednoisethat canbe addedto the two-qutritsystemwhile still getting aviolationofBell’s inequality
is:
1 λ =(7 √33)/4 0.314.
min
− − ≈
We conclude this section by noting that one could equally measure the strength of the inequality I 2
3
≤
or K 0 by mixing the initial state with some kind of noise other than uncolored noise. For example,
3
≤
one could consider the possibility of mixing the initial entangled state with the closest separable one, or to
mix it with the tensor product state of the reduced density matrices. Remarkably, it turns out [15] that
′ ′′
the optimal values of λ, λ and λ , provided by these alternative measures of nonlocality for the state
min min
ψ coincide, and they are equal to the optimal value obtained when ψ is mixed with some amount of
mv mv
| i ′ ′′ | i
(uncolored) noise, i.e. λ =λ =λ .
min min min
4 Finite detector efficiency
Now we consider the case in which eachof the detectors in our Bell experiment is endowedwith a quantum
efficiency η (0 η 1), where η is meant to be the probability that a detector “cliks” when a particle
≤ ≤
(qutrit) impinges on it. (We note that, quite generally, η may also account for all possible losses of the
particles on their way from the source to the detectors.) Quantum mechanics then predicts the modified
probabilities:
Pij(a ,b )=η2Pij(a ,b ),
η i j ψ i j
Pi(a )=ηPi(a ),
η i ψ i
Qj(b )=ηQj(b ),
η j ψ j
so the new single probabilities are given by Pi(a )=Qj(b )= η. Correspondingly, the quantum prediction
η i η j 3
for I and K becomes
3 3
Iη(ψ )=η2I (ψ ), (38)
3 | i 3 | i
4
Kη(ψ )=η2K (ψ )+ η(η 1), (39)
3 | i 3 | i 3 −
where I (ψ ) and K (ψ ) represent the quantum prediction of I and K evaluated for η = 1. Thus, for
3 3 3 3
| i | i
the case in which I (ψ ) > 2, it follows from (38) that the inequality I 2 will be violated by quantum
3 3
| i ≤
mechanics if and only if
2
η > . (40)
sI3(ψ )
| i
Similarly, for the case in which K (ψ )>0, it follows from (39) that the inequality K 0 will be violated
3 3
| i ≤
by quantum mechanics if and only if
4 4
η > = . (41)
4+3K (ψ ) 2+I (ψ )
3 3
| i | i
From (40) and (41), we can see that the condition for the violation of the inequality I 2 differs from the
3
≤
condition for the violation of the inequality K 0. This is a consequence of the fact that, actually, the
3
≤
10