Table Of ContentNo-broadcasting of non-signalling boxes via operations which transform local boxes
into local ones
P. Joshi1,2, A. Grudka3, K. Horodecki2,4, M. Horodecki1,2, P. Horodecki2,5 and R. Horodecki1,2
1Faculty of Mathematics,Physics and Informatics, University of Gdan´sk, 80–952 Gdan´sk,Poland
2National Quantum Information Center of Gdan´sk, 81–824 Sopot, Poland
3Adam Mickiewicz University, 61-614 Poznan´, Poland
4Institute of Informatics, University of Gdan´sk, 80–952 Gdan´sk, Poland and
5Technical University of Gdan´sk, 80–233 Gdan´sk, Poland
We deal with families of probability distributions satisfying non-signalling condition, called non-
signallingboxesandconsiderclassofoperationsthattransformlocalboxesintolocalones(theone
thatadmitLHVmodel). Weprovethatanyoperationfromthisclasscannotbroadcastanonlocal
boxin2×2case. Weconsiderafunctioncalledanti-Robustnesswhichcannotdecreaseunderthese
operations. The proof reduces to showing that anti-Robustness would decrease after broadcasting.
1
1
0
2 I. INTRODUCTION also [20]). It was shown later, that this kind of no-
broadcastingisequivalenttothepreviousmentionedone,
v
(with general operations) in [21, 22].
o Given a quantum bipartite state and a set of mea-
N surements on its both subsystems, one ends up with a Inpresentarticle,weconsideravariantof’local’broad-
casting of non-signalling boxes. Namely we assume that
family of probability distributions obtained from these
8 input box is known, and it is processed by locality pre-
measurement on the quantum state. Such a family can
haveinterestingfeatures,e.g. canviolatesomeoftheBell serving operations. By locality preserving operations we
]
mean here the ones that transform local boxes (those
h inequalities [1]. Moreover such a family satisfies the so-
p called non-signalling condition: change of measurement with local hidden variable model) into local ones. We
- showthatanynon-localboxin2×2cannotbebroadcast
byonepartycannotchangestatisticsoftheotherparty.
t
n One can then ask after Popescu and Rohrlich [2], if any intwocopies(whichexcludesbroadcastinginarbitraryn
a copies), andproveitusingideaofmonotones, inanalogy
set of non-signalling distributions (called a box), can be
u
to entanglement theory. Namely we introduce monotone
reproduced by measurement on quantum state. The an-
q
called anti-Robustness, which can not decrease under lo-
[ swerisno,andtheproofisgivenbythefact,thatcertain
(called Popescu-Rohrlich) boxes violate CHSH inequal- cality preserving operations. For a related entanglement
1 monotone see [23]. We then show that if broadcast were
ity up to 4, while maximal violation via measurements
v possible, anti-Robustness would increase under locality
on quantum states of this inequality is due to Cirel’son’s
1 √
preserving operations, which gives desired contradiction.
8 limit 2 2 [3].
Our proof has two main parts: we first show this for
7 Since this discovery by Popescu and Rohrlich, non-
states which are mixture of PR and anti-PR boxes and
1 signalling boxes have been treated as a resource in dif-
1. ferent contexts [4]. In particular it has been shown, that thenshowthatbroadcastingofanyothernon-localboxes
impliesbroadcastingofthelattercase. Bysymmetry,we
1 theybareanalogousfeaturestothoseofentangledstates
then extend the argument to all 2 × 2 boxes. We be-
1 [5] such as non-shareability [5], monogamy of correla-
1 tions [6], offering secret key [7–9] which lead to the so gin however with analogous question in quantum case:
: canonebroadcastaquantumbipartitestatewhichisen-
v called device independent security (see [10] and refer-
tangledbymeansofoperationsthattransformseparable
i ences therein). The distillation of PR-boxes and cost of
X statesintoseparableones,andanswerinnegativetothis
non-localityhasattractedrecentlymuchattentionaswell
r [11–14], as an analogue of distillation of entanglement. question in section II. The main tools are introduced in
a section IV and IVA. The two parts of the proof of main
Another context in which non-signalling principle was
result are in sections IVB and IVC. Appendix contains
considered, are the well known no-goes of quantum the-
proofs of some needed facts.
ory: no-cloning [15] and no-broadcasting [16]. The first
states that there is no universal machine which given
unknown input produces its copies, while the second is
stronger: itstatesthatthereisnouniversalmachinethat II. NO-BROADCASTING IN QUANTUM CASE
given unknown state ρ produces a state whose subsys-
tems are in state ρ. Analogous results for non-signalling In this section we show that any entangled states can
boxes were shown in [17]. not be broadcast, which is in fact an immediate implica-
There is also a bipartite version of no-cloning and no- tion of known facts from entanglement theory.
broadcasting theorems. In case of bipartite quantum Let ρ be a state and Λ be a broadcast map which
AB
states one requires that the input state of machine is maps ρ to ρ i.e.
AB ABAA(cid:48)B(cid:48)
known, but the operations which machine uses are not
all quantum operations but local operations [18, 19] (see Λ(ρ )=ρ , (1)
AB ABA(cid:48)B(cid:48)
2
and Ω . Each of the spaces may contain (the same number
B
of) n systems. We will consider only boxes that satisfy
TrAB(ρABA(cid:48)B(cid:48))=TrA(cid:48)B(cid:48)(ρABA(cid:48)B(cid:48))=ρAB. (2) certainnon-signallingconditions. Tospecifythisweneed
to define general non-signalling condition between some
We show now that such a map does not exist if ρ is
AB partitions of systems.
entangled if it preserves the set of separable states.
To this end consider first cloning of known bipartite Definition 1 Consideraboxofsomenumberofsystems
state by LOCC operations which is a smaller class of op- mdividedintwosets: C1,...,Ci andD1,...,Dm−i. Abox
erations then the separability preserving ones. Such a on these systems given by probability table P(c,d|x,y) is
problem was considered in [24] and entanglement mea- non-signalling w.r.t. to C1,...,Ci|D1,...,Dm−i cut if the
suredefinedbyqualityofcloningwassuggested. Indeed, following conditions are satisfied:
by LOCC one can clone any separable state, so if the (cid:88) (cid:88)
∀ P(c,d|x,y)= P(c,d|x,y(cid:48)), (5)
quality of cloning is not perfect the state must be en- c,x,y,y(cid:48)
tangled. At that time it was not known whether any d d
(cid:88) (cid:88)
entangled state can be cloned. Note that, if we knew an ∀ P(c,d|x,y)= P(c,d|x(cid:48),y). (6)
d,x,x(cid:48),y
entanglement measure E, that for any entangled state c c
E(ρ ⊗ρ )>E(ρ ), (3) In what follows we will consider only
AB A(cid:48)B(cid:48) AB
boxes that satisfy non-signalling condition in
this would imply impossibility of cloning entangled state A ,...,A |B ,...,B cut and also that do not-signal
1 n 1 n
byLOCC.Indeed,bycloningwewouldincreasethemea- in all A B |A ,...,A ,...A ,B ,...,B ,...B cuts for
i i 1 i n 1 i n
sure E, which is impossible by LOCC. Consider now
i = 1,..,n where i denotes the lack of A term in the
i
broadcasting by LOCC. Again, if we knew entanglement
sequence [4, 28].
measure E which satisfies even more, namely
By locally realistic box we mean the following ones:
E(ρ )>E(ρ ), (4) Definition 2 Locally realistic box of 2n systems
ABA(cid:48)B(cid:48) AB
A ,...,A ,B ,...,B is defined as
1 n 1 n
for any entangled state ρ and any state ρ be-
ing a broadcast copy of ρAB , i.e. Tr ρ ABA(cid:48)=B(cid:48)ρ (cid:88)p(λ)P(a|x)(λ) ⊗P(b|y)(λ) , (7)
AB AB ABA(cid:48)B(cid:48) AB A1,...,An B1,...,Bn
and TrA(cid:48)B(cid:48)ρABA(cid:48)B(cid:48) = ρAB, then broadcasting of known λ
entangled state by LOCC would be impossible. Indeed,
for some probability distribution p(λ), where we as-
like in cloning case, by broadcasting, we would increase
sume that boxes P(a|x)(λ) do not signal in
the measure E, which is impossible by LOCC. A1,...,An
A |A ,...,A ,...A cut for all i = 1,...,n, and analo-
Such a measure is actually known. Namely, in [25] it i 1 i n
was shown that entanglement of formation satisfies this gously for P(b|y)(λ) . The set of all such boxes
equation,andasacorollary,itwasobtainedthatcloning we denote as LR B1.,...A,Blnl boxes that are non-signalling
ns
of arbitrary entangled state by LOCC is impossible (and in A ,...,A |B ,...,B cut, and are non-signalling in
1 n 1 n
also broadcasting, as we now see). A B |A ,...,A ,...A ,B ,...,B ,...B cut for i=1,...,n,
i i 1 i n 1 i n
One can strengthen the result by referring to a later
but do not satisfy the condition (7), are called non-
analogous result by Marco Piani [26], who showed that
LR .
ns
relativeentropyofentanglementsatisfiesequation(4)too
Havingdefinedrelevantclassesofboxes, wecandefine
for any entangled state. Now, since relative entropy of
relevant class of operations. We consider a family L of
entanglement does not increase on arbitrary operations
operations Λ on a box shared between Alice and Bob,
whichpreservethesetofseparablestates[27](callednon-
which preserve locality, as defined below.
entangling operations), we obtain
Definition 3 AnoperationΛiscalledlocalitypreserving
Corrolary 1 Arbitrary entangled state cannot be broad-
if it satisfies the following conditions:
cast by non-entangling operations.
(i) validity i.e. transforms boxes into boxes.
Finally, let us mention, that broadcasting by means of
(ii) linearity i.e. for each mixture X =pP +(1−p)Q,
local operations was also considered, and it was shown
there is Λ(X)=pΛ(P)+(1−p)Λ(Q)
in [18] that a state can be broadcast by means of local
(iii) locality preserving i.e. transforms boxes from
operations only when the state is classical (i.e. it is a
LR into boxes from LR .
ns ns
state of two classical registers).
(iv) non-signalling i.e. it preserves non-
signalling in A ,...,A |B ,...,B cut and in all
1 n 1 n
A B |A ,...,A ,...A ,B ,...,B ,...B cuts for i=1,..,n.
i i 1 i n 1 i n
III. STATEMENT OF THE BROADCASTING
The set of all such operations we denote as L.
PROBLEM IN BOX SCENARIO
Theproblemofbroadcastingisthenif there exists Λ∈
L which makes a broadcast of a box i.e.
ByboxX wemeanafamilyofprobabilitydistributions
that have support on Cartesian product of spaces Ω × Λ(B) =B(2), (8)
A ABA(cid:48)B(cid:48)
3
where B(2) is arbitrary box with 4 inputs and 4 out- The set of boxes having two inputs and two outputs,
puts whose marginals are box B again. More generally, forms an 8 dimensional polytope with 24 vertices [4]. 16
one can ask if there exists Λ which produces n broadcast of these are deterministic boxes while rest 8 boxes are
copies. equivalent to PR-boxes, defined like this:
(cid:26)
1/2 if a⊕b=xy⊕rx⊕sy⊕t
IV. NO-BROADCASTING THEOREM IN BOX B = (10)
rst 0 else.
SCENARIO
The polytope geometry is very compact since every PR-
Ourapproachislikeinentanglementtheory(seeinthis boxisdirectlyconnectedto8deterministicboxes. These
context [29]). We pick up a monotone and show that it 8 deterministic boxes span the set of LR 2×2 boxes
ns
couldbesmallerafterbroadcastingwhichisnotpossible. that satisfy all the CHSH inequalities [4] −2 ≤ β ≤ 2
rst
Our monotone will be anti-Robustness of a box, defined with:
as follows:
β ≡ (−1)t(cid:104)00(cid:105)+(−1)s+t(cid:104)01(cid:105)+(−1)r+t(cid:104)10(cid:105)
rst
Definition 4 Let, A and B be any two NS boxes and A +(−1)r+s+t+1(cid:104)11(cid:105), (11)
be a non-LR . Anti-Robustness of A denoted as R¯(A)
ns
is defined as, where (cid:104)ij(cid:105)=P(a=b|ij)−P(a(cid:54)=b|ij).
We describe now 4 twirling operations τ , and show
rs
R¯(A)=max{q|qA+(1−q)X ∈LR }, (9) that they preserve corresponding CHSH quantities. The
ns
X twirling τ is introduced in [29].
00
where X is arbitrary box.
Definition 5 A twirling operation τ is defined by flip-
rs
The name anti-Robustness comes from the fact that if
ping randomly 3 bits δ,γ,θ and applying the following
a given q is anti-Robustness of some box A, then 1−q
transformation to a 2×2 box P(a,b|x,y):
is minimal weight with which one needs to admix some
box X to make A local, i.e. 1−q reports how ’robust’ x → x⊕δ
is A against admixing of some other boxes in terms of
y → y⊕γ
non-locality.
a → a⊕γx⊕δγ⊕θ⊕sγ
Observation 1 R¯ is non decreasing under locality pre- b → b⊕δy⊕θ⊕rδ
serving operations. (12)
Proof.- Let us fix arbitrary box A. Let Λ be linear op-
eration taking LR boxes into LR boxes. Let also q
ns ns 0
bethevalueofR¯(A). ThenthereexistsboxX suchthat Wethenmakethefollowingobservation, whichiseasy
q A+(1−q )X ∈LR . LetusapplyΛtoq A+(1−q )X to check:
0 0 ns 0 0
by linearity of Λ it reads q Λ(A)+(1−q )Λ(X), and by
0 0 Observation 3 Twirling τ maps all 2×2 boxes into
its locality, this box is LR, hence q is a candidate for rs
value R¯(Λ(A)), but by definition th0e latter can be at linepBrst+(1−p)Brst¯; τrs(Brst)=Brst andτrs(Brst¯)=
most higher, hence proving R¯(A)≤R¯(Λ(A)). Brst¯where t¯denotes binary negation of t.
Wearereadytoshowthattwirlingpreservesappropri-
Inthefollowing,wewillneedalsoatechnicalproperty
ateCHSHquantity,whichisformulatedinlemmabelow:
of anti-Robustness, that can be viewed as connectivity:
if it is attained at q, it could be attained at all p<q: Lemma 1 The CHSH quantities β satisfy:
rst
Observation 2 If there exists L = qA+(1−q)X with βrst(P)=βrst(τrs(P)). (13)
q > p, then there exists also a box X(cid:48) such that L =
pA+(1−p)X(cid:48).
Proof.- It is straightforward to check that β (P) =
Proof.- The proof is straightforward with X(cid:48) = (q − rst
p)/(1−p)A+(1−q)/(1−p)X. (cid:104)2(Brst −Brst¯)|P(cid:105), where (cid:104).|.(cid:105) denotes Euclidean scalar
product and hence there is
(cid:88)
β (τ (P))=(cid:104)β |τ (P)(cid:105)= q (cid:104)β |π P(cid:105). (14)
rst rs rst rs i rst i
A. Extremal non-local boxes, twirlings and CHSH
i
quantities
Here we use the fact that each twirling is mixture of
permutations. We have then
We will use numerously the operation of twirling of
a box [29], which maps all boxes into smaller subset of (cid:88)
β (τ (P))= q (cid:104)π β |P(cid:105)=(cid:104)τ (β )|P(cid:105), (15)
boxes. In what follows we will consider 4 such twirlings rst rs i i rst rs rst
and show that they preserve corresponding CHSH quan- i
tities. To this end we need also recall the geometry of which ends the proof, since τ (B ) = B and
rs rst rst
the set of 2×2 boxes. τrs(Brst¯)=Brst¯by observation 3.
4
B. No-broadcasting for mixtures of PR and and according to its values choose inputs x and y to first
anti-PR box (second) box. Then compare the outputs a and b. If the
outputs satisfy PR condition a⊕b=xy, the value of C
1
Here we show, for a subclass of non-LRns boxes that (C2) is set to 4. If they does not satisfy this condition,
they cannot be broadcast. These boxes are a family of i.e. satisfy anti-PR correlations a⊕b=xy⊕1, then the
convex combinations of B ≡ B (PR-box) and B ≡ valueissetto−4. Itisstraightforwardtocheckthatthe
000 001
B˜ (anti-PR) boxes averagevalueofthisrandomvariableonaboxequalsthe
CHSH quantity β of this box.
000
B =αB+(1−α)B˜, (16) We will be interested now in joint probability distri-
α
butions of the variables (C ,C ) applied to boxes L, Bˆ
1 2 α
where α ∈ [1,3), such that when α = 1 ⇒ B = B ( and X. Note that this transformation of mapping the
4 α
PR box) and when α = 3 ⇒ B = K (say) which is an box to a probability distribution is linear. With a box
4 α
LR box. One can express K in terms of B and B˜ as we associate a corresponding probability distribution of
ns α
(C ,C ):
follows: 1 2
K =p B +(1−p )B˜, (17)
α α α
L→{p(cid:48) ,p(cid:48) ,p(cid:48) ,p(cid:48) }, (21)
11 12 21 22
where p = 3 . We show that for B ∈ {αB +
(1−α)B˜α(cid:107)α ∈ [41α,3]} broadcasting is possibαle only when with p(cid:48)ij = P(C1 = (−1)i+14,C2 = (−1)j+14), similarly
4 for a broadcast copy:
B = K. And this is known fact that LR boxes can
α ns
be broadcast.
Bˆ →{p˜ ,p˜ ,p˜ ,p˜ }, (22)
We can pass to the main result of this section: α 11 12 21 22
Theorem 1 For any broadcast copy Bˆ of B there is and the X:
α α
R¯(B )>R¯(Bˆ ). (18) X →{p(cid:48)1(cid:48)1,p(cid:48)1(cid:48)2,p(cid:48)2(cid:48)1,p(cid:48)2(cid:48)2}. (23)
α α
If there exists L and X such that (20) holds, then there
Proof.-SupposewecanbroadcastB andconsiderBˆ also exists the one which is permutationally invariant
α α w.r.t. to copies, because Bˆ is such. Hence without loss
be the broadcast of it. Let Λ be the operation that α
of generality, we can assume that distribution L is per-
achieves broadcast. Since it transforms LR boxes into
ns mutationally invariant, therefore we have p(cid:48) =p(cid:48) and
LRns boxes,ifappliedtoK itwilltransformitintosome 12 21
p˜ = p˜ .
local box L. Thus using (17) we would obtain: 12 21
Now, by assumption, the box L is LR . Hence, if we
ns
L=pαBˆα+(1−pα)X. (19) perform operation defined by the random variable C1 on
first copy, given we observe value 4, the second copy is
We will show however in lemma 2 below, that for any also an LR box: it is a mixture of induced product
ns
broadcast copy Bˆ , any LR box L and any box X the boxes on the second system given input i,j and output
α ns
above equality does not hold. Now, if there does not a,b on the first system. Analogous property holds if we
exist an LR box L satisfying the above equality, by ask how looks like the second system given the C had
ns 1
definition of anti-Robustness p = R¯(B ) can not be value −4. Thus the CHSH inequality of conditional box
α α
anti-RobustnessofBˆ . Thelattercannotbealsohigher onsecondsystemmustholdinbothcases. Recallthatthe
α
than pα or otherwise (19) would be satisfied, because we average value of C2 on a box equals the CHSH quantity
have the observation 2 which ends the proof. β000 of this box, thus if the measured copy results 4 for
C , the CHSH inequality −2 ≤ β ≤ 2 for the second
1 000
Corrolary 2 The boxes B for α∈[1,3) are not broad- copy is
α 4
castable.
Proof.- The proof follows from the above theorem and −2≤ 4p(cid:48)11+(−4)p(cid:48)12 ≤2, (24)
monotonicity of anti-Robustness (observation 1). p(cid:48) +p(cid:48)
11 12
We can proceed now with the proof of crucial lemma
mentioned in the proof of the theorem above: and if the outcome is −4 then CHSH inequality is
Lemma 2 For any broadcast copy Bˆα with α ∈ [1,34), −2≤ 4p(cid:48)21+(−4)p(cid:48)22 ≤2. (25)
any LR box L and any box X the equality p(cid:48) +p(cid:48)
ns 21 22
L=p Bˆ +(1−p )X (20) Usingthefactthatp(cid:48) =p(cid:48) ,p˜ =p˜ andnormalization
α α α 12 21 12 21
condition,werewriteaboveinequalitiesinsimplifiedform
does not hold. as follows
Proof.-WedefinetworandomvariablesC andC each
1 2
inthesameway: pickuprandomlyindependently2bits, 0≤6p(cid:48) −2p(cid:48) , (26)
11 12
5
2p(cid:48) −6p(cid:48) ≤0, (27) P2
11 12 1.0
0.8
0≤2p(cid:48) +10p(cid:48) −2, (28)
11 12 0.6
0.4 9 3
6p(cid:48)11+14p(cid:48)12−6≤0. (29) 16,16
0.2
These inequalities give an LR polytope.
ns 0.00.0 0.2 0.4 0.6 0.8 1.0P1
FIG.2: Asetofpairsofparameters(p(cid:48) ,p(cid:48) )ofboxL,satis-
1.0P2 fyinglocalityconditions(26)-(29)(shad11edr1e2gion)andtheset
of the parameters (p˜ ,p˜ ) of box Bˆ scaled by p (dashed
11 21 α α
0.8
line).
0.6
0.4 C. General Case - No-broadcasting for all 2×2
non-LR boxes
ns
0.2
In this section we show no-broadcasting for all 2×2
0.00.0 0.2 0.4 0.6 0.8 1.0P1 non-LR boxes. To this end we will need the following
ns
FIG.1: Asetofpairsofparameters(p(cid:48) ,p(cid:48) )ofboxL(shaded crucial lemma, proved in Appendix.
11 12
region), satisfying locality conditions (26)-(29).
Lemma 3 For any r,s,t in {0,1}, and any box P sat-
AB
isfying β (P )≥2 there is
Let us consider constraints on Bˆ , i) normalisation con- rst AB
α
dition ii) symmetry of broadcast i.e. p˜ = p˜ iii) if we R¯(P )=R¯(τ (P )), (32)
12 21 AB rs AB
traceoutonecopyofthebroadcastcopythenthesecond
copy has to be Bα and hence where τrs is twirling given in def 5.
We are ready to state our main result:
p˜ +p˜ =α, (30)
11 21 Theorem 2 Anynon-LR boxin2×2cannotbebroad-
ns
cast.
as α is probability of obtaining 4 (i.e. value 4 of C ) on
1 Proof.-Wewillshowfirstthatanyboxwithβ (P)>
B box. We can rewrite eq.(20) as, 000
α 2 is not broadcastable. Suppose by contradiction that
it is broadcastable, i.e. there is transformation Λ that
L−p Bˆ =(1−p )X. (31)
α α α makes from P a box which is its broadcast P .
AB ABA(cid:48)B(cid:48)
We will use now monotonicity of anti-Robustness under
Thankstolinearityofthemapgiving(C ,C )fora box,
1 2 linear operations that transform LR boxes into LR
ns ns
the same relation holds for the related probability distri-
boxes(Observation1). Frommonotonicityandtheabove
butions{p(cid:48) }, {p˜ }and{p(cid:48)(cid:48)}. SinceX isabox, thedis-
ij ij ij Lemma 3 we get,
tribution {p(cid:48)(cid:48)} should always have positive coefficients.
ij
We check if for any Bα it becomes negative. So, now R¯(τrsPAB)=R¯(PAB)≤R¯(PABA(cid:48)B(cid:48))= (33)
we have a complete set of LRns boxes mapped on the R¯(cid:104)(τAB ⊗τA(cid:48)B(cid:48))(P )(cid:105). (34)
shaded region as shown in fig.1 let us denote it S1. The rs rs ABA(cid:48)B(cid:48)
image under mapping to distribution {p˜ } of one copy
ij
ofBˆ isnothingbutastraightlinegivenbytheeq. (30). But this contradicts equation (18). This reduction ar-
α
gument proves no-broadcasting of boxes P satisfying
We draw this straight line scaled by the factor p , and
α
β (P) > 2. The whole set of 2×2 non-LR boxes
denoteresultingsetofpointsasS . Interestingly,forany 000 ns
2
α ∈ [3,1], we get the same line. Changing value of α, can be written compactly as
4
simply shifts points on the line.
1
We observe from fig. 2, that only the intersection point (cid:91)
{P :β (P)>2}, (35)
rst
of these two sets S and S will give non-negative value
1 2 r,s,t=0
of both coordinates of (1−p )X. The intersection point
α
is ( 9 , 3 ) and the sum of coordinates is α = 3. This hence we need to have proof for 7 other values of string
16 16 4
intersectionpointpreciselycorrespondstothecasewhen rst. We prove that if boxes with β > 2 are non-
000
B =K which we know that it can be broadcast. Hence broadcastable then so are those with β > 2 for
α r(cid:48)s(cid:48)t(cid:48)
for any α ∈ (3,1) the eq. (31) does not hold with both r(cid:48)s(cid:48)t(cid:48) (cid:54)= 000. This is because by definition of B there
4 rst
positive X. This ends proof of the lemma 2. islocaloperationwhichmapsB intoB andB
000 r(cid:48)s(cid:48),t(cid:48) 001
6
into Br(cid:48)s(cid:48)t¯(cid:48). Hence if boxes αBr(cid:48)s(cid:48)t(cid:48)+(1−α)Br(cid:48)s(cid:48)t¯(cid:48) were thenR¯(P)=maxXqX(P). Nowforβ(P)>2,β(X)≤2,
broadcastable for α∈[1,3), then the corresponding box any q >qX there is Y ∈/ LR. Thus qX(P)≤qX for any
4 0 0
αB + (1 − α)B = B would be broadcastable, X.
000 001 α
which is disproved in section IVB. Thus we have no- However, we have a lemma 4 that if β(A) = 2, then
broadcasting on a line between Br(cid:48)s(cid:48)t(cid:48) and Br(cid:48)s(cid:48)t¯(cid:48) with A∈LR(seesectionbelow). Henceforq =q0X,β(Y)=2,
β > 2. To prove this for all β > 2 boxes, we and therefore Y ∈ LR. This implies that for any X,
r(cid:48)s(cid:48)t(cid:48) r(cid:48)s(cid:48)t(cid:48)
note, that reduction argument as shown above applies, q (P) = qX. Thus for β(P) ≥ 2 we can equivalently
X 0
with r = r(cid:48) s = s(cid:48) t = t(cid:48) in lemma 3. This proves the write definition of anti-Robustness as
theorem.
R¯(P)=maxqX ≡max{q :qβ(P)+(1−q)β(X)=2},
0
X X
(40)
V. CONCLUSIONS
but we know by lemma 1 that twirl of a box has same
valueofCHSHasthatoftheboxforthesameCHSHi.e.
We have proven result that locality preserving opera- β(P)=β(τP) [29] Hence,
tionsdonotbroadcastnon-localboxesin2×2. Itisintu-
itive in a sens that non-locality is a resource, and it can R¯(P)=max{q :qβ(τP)+(1−q)β(X)=2}. (41)
not be brought into for free, which broadcast would do. X
We developed idea of monotones in boxes paradigm, in-
But according to (40) this is nothing but the definition
troducing anti-Robustness (or equivalently Robustness), ofanti-RobustnessofτP i.e. R¯(τP). AndhenceR¯(P)=
aquantityinterestingonitsown. Theproofusescounter- R¯(τP) for β(P)>2. For β(P)=2 we have β(τ(P))=2
intuitive property of this monotone: it does not change
by lemma 1. Hence by lemma 4 we have that both P
under irreversible operation of twirling, resembling the
and τ(P) are local. It is easy to see, that for local boxes
fact that CHSH value is preserved under twirling. In
anti-Robustness is 1, hence the desired weak inequality.
this proof we have used heavily some properties of 2×2
boxes. It would be interesting to show the same for ar-
bitrary nonlocal box, which is an open question. We
B. locality of β (X)=2 hyperplane
considerhereexactbroadcasting. Itwouldbeinteresting rst
to prove its non-exact version as well.
Themainresultofthissectionisthelemmabelow. We
first show the proof of this lemma, and then the proof of
VI. APPENDIX theorem (3) which is crucial to this proof.
Lemma 4 For any r,s,t ∈ {0,1} and any box X,
In this section we prove some results including proof
β (X)=2 implies X ∈LR .
rst ns
of lemma 3.
Proof Let us fix r,s,t. By theorem (3) there is X ∈
conv{x ,x˜(rst),...,x˜(rst)}wherex˜(rst)arepointsfromthe
0 1 n i
half plain defined by β (x) = 2 which belongs to ray
A. Proof of the lemma 3 rst
starting at x = B and passing through x which is
0 rst i
the i-th of 23 (apart from x ) extremal point of the set
WefirstprovethatR¯(P)=R¯(τ P)withβ (P)>2. 0
rs rst of non-signalling boxes. In other words x˜(rst) =p B +
We fix values r,s,t and omit them in the following proof i i rst
as thanks to lemma 1 it goes the same way for all these (1−p )x such that β (x˜(rst)) = 2. Thus X = p x +
i i rst i 0 0
indices. (cid:80)n p x˜(rst). Now, since β (X) = 2 there is X =
i=1 i i (rst)
To this end consider an arbitrary box X (cid:54)= P and (cid:80) p x˜(rst) i.e. theweightp ofx iszerointhemixture.
Y=qP+(1-q)X. Then, i i i 0 0
But it is easy to check that all x˜(rst) are local, hence X
i
β(Y)=qβ(P)+(1−q)β(X). (36) mustbelocalitself. Toseethiswecheckthatforallr(cid:48)s(cid:48)t(cid:48)
there is
To make Y local, we need clearly β(X) ≤ 2. Let qX be
0
solution of −2≤β (x˜(rst))≤2, (42)
r(cid:48)s(cid:48)t(cid:48) i
2=q0Xβ(P)+(1−q0X)β(X). (37) i.e. that x˜(rst) belongs to the LR in 2×2. To this end
i ns
Let us observe that we first compute from the assumption β (x˜(rst)) = 2
rst i
the probability p and check for all values r(cid:48)s(cid:48)t(cid:48) (cid:54)= r,s,t
R¯(P)=maxmax{q|qP +(1−q)X ∈LR}, (38) the value of β i of x˜(rst). The last check is easy if we
X q r(cid:48)s(cid:48)t(cid:48) i
observe that β (B ) ∈ {−4,0,4} and β (L) ∈
r(cid:48)s(cid:48)t(cid:48) rst r(cid:48)s(cid:48)t(cid:48)
and denote {−2,2},whereLstandsforanylocallyrealisticextremal
box. This holds because both nonlocal boxes B and
rst
q (P):=max{q|qP +(1−q)X ∈LR}, (39)
X locally realistic extremal ones L can be represented (not
q
7
uniquely) as vectors v of 1s and −1s (4 of them in to- Since β(x) ≥ 2, we have x ∈ H which proves x ∈LHS.
i +
tal each corresponding to one pair of x and y), where Take now the converse: x ∈ LHS. This means that
1 denotes maximal correlations of a distribution and −1 x = (cid:80)n p x , hence x = p x +(1−p )(cid:80)n p /(1−
i=0 i i 0 0 0 i=1 i
denotes maximal anticorrelations. Each value of β p )x which means x ∈ cone(x ,conv{x ,...,x }), and
r(cid:48)s(cid:48)t(cid:48) 0 i 0 1 n
can be represented as an Euclidean scalar product of v hence x ∈ C, which taking into account x ∈ H , gives
i +
with again vector of 4 1s and −1s depending on sign of x∈RHS which proves the thesis.
(cid:104)ij(cid:105) in definition (11) where the number of −1 is always We can now prove the following lemma, which enables
odd. The numbers {−4,−2,0,2,4} follows from the fact us to state the main question of this section:
that for each B v has always odd number of −1s, and
rst i
Lemma 7 H has one point of intersection with each of
each L has always even number of them.
the segments [x ,x ], denoted as x˜ .
0 i i
proofWehaveL ={x:(1−α)x +αx =x},H ={z :
i 0 i
β(z)=2}. WewanttoprovethatH∩L={x˜ }. Tothis
1. geometrical theorem i
end we observe that x∈H∩L implies β(x)=2. Taking
this into account and β(x ) = 4 as well as β(x ) ≤ 2 we
Following Bengtsson and Z˙yczkowski [30], by a cone 0 i
have by lemma 5 that there exists unique α∈[0,1] such
with apex x and some body such that x (cid:54)∈ body as a
0 0 that x = (1−α)x +αx , call it x˜ i.e. x˜ ∈ [x ,x ] as
base we mean the set of points obtained by the following 0 i i i 0 i
we claimed.
operation: taking rays (half lines) that connect x and
0 To prove theorem 3, we first show the following inclu-
each point of the body.
sion:
Thus we consider operation cone which makes
cone from the body defined in the following way: Lemma 8 H ∩B ⊇conv({x ,x˜ ,...,x˜ }) .
+ 0 1 n
cone(x0,conv{x1,...,xn}) where x0 does not belong to Proof
conv{x1,...,xn} and x1,...,xn are extremal points of the TakexfromRHS.Firstweprovethatx∈H+. Thisis
body. In our case, the apex will be any of the maxi- easy since x=γ x +(cid:80)n γ x˜ , by linearity of function
mally non-local boxes Brst, and the body will be convex β we have β(x)0=0γ0β(xi0=)1+i(cid:80)ini=1γi2 since for each x˜i
combinationofother23extremalpointsofthesetofnon- we have β(x˜ ) = 2. Thus, following β(x ) > 2 we have
i 0
signallingboxes. Werecallthatβrst(Brst)=4. Equality also β(x)>2 i.e. x∈H+.
βrst(X)=2definesahyperplaneH(rst). BythesetofX We prove now that x ∈ B. To this end note that
swaetifisfxyirn,gs aβnrdst(tXa)nd≥o2mwiteimt,eaasntHhe+(rpsrt)o.ofIngowehsatthefosllaomwes b(cid:80)yni=d1efiγniαitii)oxn0+x˜i(cid:80)=ni=α1iγxi0(1+−(α1i−)xiα∈i)xcio,nhve{nxc0e,xx1,=...,(xγn0}+.
way for all indices. To prove the converse inclusion: H+ ∩ B ⊆
The main thesis of this section is the following conv({x0,x˜1,...,x˜n}) we need the following lemma:
Theorem 3 H ∩B =conv({x ,x˜ ,...,x˜ }) where x˜ is Lemma 9 Equivalent definition of a cone C is the set of
+ 0 1 n i
(cid:80)
apointfromH (thehalfplaindefinedbyβ(x)=2)which all points satisfying x = x + α r with r = x −x
0 i i i i i 0
belongs to ray starting at x and passing through x . and α are non-negative coefficients.
0 i i
In what follows we use numerously the following Proof
lemma: We have the following chain of equivalences. x=x +
0
(cid:80) (cid:80)
α r . This is if and only if x = x + α (x −x ),
βL(exm)m=aλβ5(zI)f+x(=1−azλ+)β(1(x−0a)),xt0hean∈eiRthaerndλl=ineaarorfuβn(czt)io=n wtohiαich(cid:80)i iisiαiffi/(cid:80)αxiiα+ix(i1+−(1α−)x(cid:80)0 iwαhii)cxh0waen0daitmheisdiistoieqpuiriovvael.e0nt
β(x0). This lemma gives the following
Proof
By linearity of β we have β(x)=aβ(z)+(1−a)β(x ) Corrolary 3 Equivalent definition of C is the set of all
0 (cid:80)
but such a combination is unique in real numbers, hence points satisfying x=x0+ iγir˜i where r˜i =x˜i−x0 and
either a=λ or β(z)=β(x0), which ends the proof. γi are non-negative coefficients.
In what follows, we will have β(x )(cid:54)=β(z), but we do Proof
0
not state it each time. Armed with this lemma, we can We know that x˜i = λixi+(1−λi)x0 where λi ∈ R+
observe the following property: hencer˜i =λi(xi−x0)=λiri,whichendstheproof,since
r˜ are just scaled r and the proof goes with similar lines
i i
Lemma 6 H ∩B = H ∩C where H is a half space to that of lemma 9.
+ + +
defined by β(x)≥2, B is body spanned by {x ,x ,...,x } To complete the proof of theorem 3 we now proceed
0 1 n
distinct points, and C is a cone obtained by operation with the proof of the converse inclusion: H ∩ B ⊆
+
cone(x ,conv{x ,...,x }). conv({x ,x˜ ,...,x˜ }). Thanks to lemma 6, we may as-
0 1 n 0 1 n
Proof sume that x ∈ H ∩ C. Now, thanks to lemma 5, if
+
If x ∈RHS, then β(x) ≥ 2 and there exists y ∈ we take the ray with beginning x crossing H in point
0
conv{x ,...,x } such that x = αx +(1−α)y for α ∈ x˜ , that passes through x then if x ∈ H , there is x ∈
1 n 0 0 +
R . By lemma 5, this means that x ∈ [x ,y] because [x ,x˜ ]. Thisisbecauseβ(x )>2andβ(x )≥β(x)>2
≥0 0 0 0 0 0
β(x )=4, β(y)<2andβ(x)≥2andwehaveβ(x)≤4. while β(x˜ )=2.
0 0
8
Hence to prove that x∈conv(x ,x˜ ,...,x˜ ), it is suffi- Acknowledgments
0 1 n
cient to show that x˜ is spanned by {x˜ ,...,x˜ }. We will
0 1 n
show it in what follows. Namely, by corollary (3), there
(cid:80)
is x˜ = γ x˜ where γ ∈ R . By linearity of β there
0 i(cid:80)i i i(cid:80) +
is β(x˜ ) = γ β(x˜ ) = γ 2. Since x˜ ∈ H, there is WethankT.SzarekandD.Reebforinterestingdiscus-
0 i i i i i (cid:80) 0
also β(x˜ ) = 2. Hence there is γ = 1, which taking sions. This research is partially funded by QESSENCE
0 i i
into account non-negativity of γ shows that γ forms a grant and Foundation of Polish Science from Interna-
i i
convexcombinationofx˜ whichweaimedtoprove. This tional PhD Project: ”Physics of future quantum-based
i
ends the proof that H ∩B = conv{x˜ ,...,x˜ }, and fol- informationtechnologies”. K.H.acknowledgesgrantBW
+ 1 n
lowing lemma 8, ends the proof of theorem 3. nr 538-5300-0637-1.
[1] J.S.Bell,Physics(LongIslandCity,N.Y.)1,195(1964). B. Schumacher, Phys. Rev. Lett. 53, 2818 (1996).
[2] S.PopescuandD.Rohrlich,Found.Phys.24,379(1994). [17] H. Barnum, J. Barrett, M. Leifer, and A. Wilce (2006),
[3] B. S. Cirel’son, Lett. Math. Phys. 4, 93 (1980). arXiv:quant-ph/0611295.
[4] J.Barrett,N.Linden,S.Massar,S.Pironio,S.Popescu, [18] M. Piani, P. Horodecki, and R. Horodecki, Phys. Rev.
and D. Roberts, PRA 71, 022101 (2005), arXiv:quant- Lett. 100, 090502 (2008), arXiv:0707.0848.
ph/0404097. [19] M. Horodecki, A. S. De, and U. Sen, PRA 70, 052326
[5] L. Masanes, A. Acin, and N. Gisin, Phys. Rev. A 73, (2004), arXiv:quant-ph/0403169.
012112 (2006), arXiv:quant-ph/0508016. [20] M. Piani, M. Christandl, C. E. Mora, and P. Horodecki,
[6] M. Paw(cid:32)lowski and C. Brukner, Phys. Rev. Lett. 102, Phys. Rev. Lett. 102, 250503 (2009), arXiv:0901.1280.
030403 (2009), arXiv:0810.1175. [21] S. Luo, Lett. Math. Phys. 92, 143 (2010).
[7] A. K. Ekert, Phys. Rev. Lett. 67, 661 (1991). [22] S. Luo and W. Sun, PRA 82, 012338 (2010).
[8] J. Barrett, L. Hardy, and A. Kent, Phys. Rev. Lett. 95, [23] G. Vidal and R. Tarrach, Phys. Rev. A 59, 141 (1999),
010503 (2005). quant-ph/9806094.
[9] L. Masanes, R. Renner, M. Christandl, A. Winter, and [24] M.Horodecki,A.Sen(De),andU.Sen,Phys.Rev.A70,
J. Barrett (2006), arXiv:quant-ph/0606049. 052326 (2004), quant-ph/0403169.
[10] E. Hanggi, Ph.D. thesis, ETH, Zurich (2010). [25] D. Yang, M. Horodecki, R. Horodecki, and B. Synak-
[11] N. Brunner and P. Skrzypczyk, Phys. Rev. Lett. 102, Radtke, Phys. Rev. Lett. 95, 190501 (2005), quant-
160403 (2009), arXiv:0901.4070. ph/0506138.
[12] J. Allcock, N. Brunner, N. Linden, S. Popescu, [26] M. Piani, Phys. Rev. Lett. 103, 160504 (2009),
P. Skrzypczyk, and T. Vertesi, Phys. Rev. A 80, 062107 arXiv:0904.2705.
(2009), arXiv:0908.1496. [27] F.G.BrandaoandM.B.Plenio(2007),arXiv:0710.5827.
[13] M. Forster, Phys. Rev. A 83, 062114 (2011), [28] J. Barrett (2005), arXiv:quant-ph/0508211.
arXiv:1105.1357. [29] A. J. Short, Phys. Rev. Lett. 102, 180502 (2009),
[14] N.Brunner,D.Cavalcanti,A.Salles,andP.Skrzypczyk, arXiv:0809.2622v1.
Phys. Rev. Lett. 106, 020402 (2011), arXiv:1009.4207. [30] I. Bengtsson and K. Z˙yczkowski, Geometry of Quan-
[15] W. K. Wootters and W. H. Zurek, Nature 299, 802 tum States. An Introduction to Quantum Entanglement
(1982). (Cambridge University Press, 2006).
[16] H. Barnum, C. M. Caves, C. A. Fuchs, R. Jozsa, and
