Difficulty of distinguishing product states locally

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by  Sarah Croke

Difficulty of distinguishing product states locally
Sarah Croke1,∗ and Stephen M. Barnett1
1School of Physics and Astronomy, University of Glasgow, Glasgow G12 8QQ, UK
Non-localitywithoutentanglementisarathercounter-intuitivephenomenoninwhichinformation
may be encoded entirely in product (unentangled) states of composite quantum systems in such a
way that local measurement of the subsystems is not enough for optimal decoding. For simple
examplesofpureproductstates,thegapinperformanceisknowntoberathersmallwhenarbitrary
local strategies are allowed. Here we restrict to local strategies readily achievable with current
technology; those requiring neither a quantum memory nor joint operations. We show that, even
for measurements on pure product states there can be a large gap between such strategies and
theoretically optimal performance. Thus even in the absence of entanglement physically realizable
local strategies can be far from optimal for extracting quantum information.
7
1
0 A composite quantum system is more than the sum of twoexamplesofsuchsets,oneofwhichmaybeperfectly
2 itsparts;itcanhavepropertiesthatcannotbeexplained discriminated with only two rounds of classical commu-
as a result of properties of its constituent particles. This nication, while the other, the so-called domino states,
n
a maybethoughtofasaresultofthesuperpositionprinci- requires quantum communication for perfect discrimina-
J ple, and has far-reaching consequences, giving quantum tion. We show that any one-way scheme succeeds with

0 theory a far richer structure than classical probability probabilityatmost(cid:39)85%and(cid:39)84%respectively,asig-
1 theory. Indeed entangled states, characterized by corre- nificantdeficitcomparedtothegloballyoptimalschemes.

lations between particles rather than local properties of Physically, current experimental capabilities are such
]
h constituent particles, famously exhibit correlations that that sequential measurement (i.e. one-way) strategies
p cannot be reproduced by any local classical theory [1]. may be readily implemented in a variety of physical
-
In addition, there exist properties of composite systems systems. Additional rounds of measurement and classi-
t
n
that are not accessible through only local measurement cal communication are more technologically challenging;
a
of the subsystems: for example, a measurement which they not only introduce additional errors and inefficien-
u
q distinguishesbetweenthej =0andj =1subspacesofa cies, but also require some sort of quantum memory in
[ two spin-half system cannot be performed through only which to store the local systems in a many round proto-

local measurements on the spins. col. Ourexamplesdemonstratethatevenforthesimplest
1
v Incontrasttoclassicalprobabilitytheorytherefore,lo- cases of discriminating pure, orthonormal, unentangled
9 cal measurements together with post-processing of mea- states, quantum memories or joint control can be neces-
7 surement results is not enough to perform all measure- sary for optimal, or even close-to-optimal performance.
5
mentsallowedbyquantumtheory. Infact,duetotheex- We begin with a simple example, first considered by
2
istence of incompatible observables, even classical com- Groisman and Vaidman [9], which demonstrates the
0
. munication between subsystems (allowing the choice of asymmetry of classical communication as a resource for
1
measurementononesystemtodependontheoutcomeof performing joint measurements, and serves to illustrate
0
measurement on the other) gives the measuring parties some features of optimal measurement strategies we will
7
1 new capabilities. In general, each of one-way classical need later. Consider the product basis of two-qubit
: communication, two-way classical communication, and states:
v
quantum communication (the ability to send quantum
i |ψ (cid:105) = |0(cid:105) ⊗|0(cid:105) , |ψ (cid:105) = |1(cid:105) ⊗|0+1(cid:105) ,
X states) is more powerful than the last [2–27]. Remark- 00 A B 10 A B (1)
|ψ (cid:105) = |0(cid:105) ⊗|1(cid:105) , |ψ (cid:105) = |1(cid:105) ⊗|0−1(cid:105) .
01 A B 11 A B
r ably, this is true even for measurements on unentangled
a
states,aphenomenonknownas“non-localitywithouten- where{|0(cid:105),|1(cid:105)}formanorthonormalbasis,byconvention
tanglement” [2–5]. In practical terms however, for mea- the eigenbasis of the Pauli operator σz, and |0±1(cid:105) =
surements on product states, the known bounds on the √1 (|0(cid:105)±|1(cid:105)) are the eigenstates of σx. Clearly, the
2
gaps in performance are really rather small [4, 5]. states are perfectly distinguishable given one-way com-
Inthispaperweintroduceanewpieceinthepuzzleof munication from Alice to Bob, but not the other way
local detection of quantum information, and show that around: if Alice can send a message to Bob, she simply
two-way classical communication can significantly im- needs to measure in the {|0(cid:105),|1(cid:105)} basis, and send the re-
provethedistinguishabilityofpure,orthonormalproduct sult of her measurement to Bob. Given result “0”, Bob
states. Such sets of states have two important features: measures in the {|0(cid:105),|1(cid:105)} basis, while given result “1” he
they can be prepared in separated laboratories without measures in the {|0±1(cid:105)} basis. On the other hand, if
eitherclassicalorquantumcommunicationbetweenlabs, BobcansendmessagestoAlicebutnotvice-versa, there
and they are perfectly distinguishable through a joint isclearlynoprocedurethatallowsAliceandBobtoper-
measurement. We derive optimal one-way strategies for fectlydistinguishthestates: Bob’slocalstatesareeigen-
2
states of incompatible observables σ and σ , and any states. Consider the following orthonormal product ba-
z x
measurement giving information about one must neces- sis of a 2⊗4 level system (introduced also in [17]):
sarily disturb the other, thus destroying the orthogonal-
ity of at least one pair of states.
It is instructive to consider what is the best Alice and |ψ (cid:105) = |0(cid:105) |0(cid:105) , |ψ (cid:105) = |0+1(cid:105) |2(cid:105) ,
00 A B 02 A B
Bob can do if they are limited to communication only |ψ (cid:105) = |0(cid:105) |1(cid:105) , |ψ (cid:105) = |0+1(cid:105) |3(cid:105) ,
01 A B 03 A B
from Bob to Alice. Note that Alice’s system is essen- |ψ (cid:105) = |1(cid:105) |0+1(cid:105) , |ψ (cid:105) = |0−1(cid:105) |2+3(cid:105) ,
10 A B 12 A B
tially classical in that regardless of the information ob- |ψ (cid:105) = |1(cid:105) |0−1(cid:105) , |ψ (cid:105) = |0−1(cid:105) |2−3(cid:105) .
11 A B 13 A B
tained from Bob, the only sensible measurement she can (3)
make is in the z-basis, which allows her to determine These states are perfectly distinguishable given just two
perfectlytheindexiinthelabelling{|ψij(cid:105)},butgivesno rounds of classical communication, while any one-way
information about the index j. The role of Bob’s mea- scheme succeeds with probability at most cos2 π, as we
8
surement therefore is to provide the information which now show.
allows Alice to distinguish between the states within the
pair {|ψ (cid:105),|ψ (cid:105)}, for each possible outcome i of Alice’s We first note that Bob can learn in which subspace
i0 i1
measurement. It follows that Bob’s measurement must {|0(cid:105),|1(cid:105)}or{|2(cid:105),|3(cid:105)}hisstatelieswithoutdisturbingany
ruleoutaswellaspossibleonestatefromeachpair,leav- ofthestates,viaavonNeumannmeasurementwithpro-
ing Alice with two remaining allowed states, which are jectors {|0(cid:105)(cid:104)0|+|1(cid:105)(cid:104)1|,|2(cid:105)(cid:104)2|+|3(cid:105)(cid:104)3|}, thus learning to
perfectly distinguishable through Alice’s measurement. which of the two subsets S0 = {|ψ00(cid:105),|ψ01(cid:105),|ψ10(cid:105),|ψ11(cid:105)}
There are four subsets of two states with the property or S1 = {|ψ02(cid:105),|ψ03(cid:105),|ψ12(cid:105),|ψ13(cid:105)} the shared state be-
that the states are perfectly distinguishable on Alice’s longs. Each subset is equivalent (up to local unitaries)
side: tothesimplersetdiscussedabove. Thisobservationsim-
plifies the analysis of optimal schemes.
S ={|ψ (cid:105),|ψ (cid:105)}, S ={|ψ (cid:105),|ψ (cid:105)},
00 00 10 01 00 11
There are three cases of interest: i) Two-way classical
S ={|ψ (cid:105),|ψ (cid:105)}, S ={|ψ (cid:105),|ψ (cid:105)}.
10 01 10 11 01 11
communication: Bob measures first, and tells Alice to
ThusAliceandBob’sstrategy,ifclassicalcommunication which of the subsets S or S the state belongs. Each
0 1
is allowed from Bob to Alice only, is for Bob’s measure- subset is perfectly distinguishable with one-way commu-
menttooptimallyassignthestatetooneofthesesubsets, nication from Alice to Bob, so just one more round of
while Alice’s measurement discriminates between the re- communication is needed for perfect discrimination. ii)
maining two states within a subset. One-way communication from Alice to Bob: within each
If the states are equiprobable the measurement on subset, the shared states are perfectly distinguishable
Bob’s system maximising the probability of correctly with one-way measurement from Alice to Bob. How-
identifying the state is in fact quite well-known, as these ever, Alice doesn’t know in which subset the state lies.
states arise in the BB84 protocol for quantum key dis- ClearlyAlice’smeasurementmustsimultaneouslydistin-
tribution [28]. The index i corresponds to the sender’s guish, as well as possible, between the states {|0(cid:105),|1(cid:105)}
choice of basis, while j denotes the bit value. Naturally and{|0+1(cid:105),|0−1(cid:105)}. Butthisissimplythesameproblem
enoughduetothesymmetryoftheset,theoptimalmea- aswehaveseenpreviously,andanoptimalmeasurement
surementisinabasisintermediatetothex-basisandthe for Alice is again in the Breidbart basis eqn. (2). Alice
z-basis: the so-called Breidbart basis [29]: communicates the result of measurement to Bob; given
π π outcome 0, Bob’s measurement discriminates perfectly
|φ0(cid:105)=cos 8|0(cid:105)+sin 8|1(cid:105), between the states {|0(cid:105),|1(cid:105)} and {|2(cid:105),|3(cid:105)}, given out-
π π come1heinsteadmeasuresinthebasis{|0±1(cid:105),|2±3(cid:105)}.
|φ (cid:105)=sin |0(cid:105)−cos |1(cid:105). (2)
1 8 8 The combination of Alice’s and Bob’s measurement re-
sults leads to a unique guess as to the state, which is
Outcome 0 leads Bob to guess that the state belongs to
correct with probability cos2 π. iii) One-way communi-
subset S while outcome 1 leads to a guess of S . In 8
00 11 cationfromBobtoAlice: BobknowsinwhichsubsetS ,
fact the optimal measurement is degenerate: measure- 0
S the state lies. In each case as we have noted, the re-
mentofeitherofthespinobservables √12(σz±σx)results su1lting set is equivalent to our simple example discussed
in an optimal strategy. This succeeds with probability
above, for which the optimal one-way strategy succeeds
cos2 π8 = 12(1 + √12) (cid:39) 0.85, a significant deficit com- with probability cos2 π. Thus any one-way scheme, re-
pared to the unit probability of success achievable when 8
gardlessofthedirectionofcommunication,succeedswith
one-waycommunicationisallowedfromAlicetoBob[9]. probability at most cos2 π (cid:39) 85%, while two rounds of
Using this simple set as a building block we can con- 8
classicalcommunicationaresufficienttodiscriminatethe
structourfirstexampleforwhichtwo-wayclassicalcom-
states perfectly.
municationprovidesasignificantadvantageoverone-way
classical communication for discriminating unentangled For our second example, we turn to the domino states
3
[3],anorthonormalbasisofa3⊗3levelsystemgivenby: Proving that this is indeed optimal is less straight-
forward than the previous cases. Our strategy is to
|ψ (cid:105) = |0(cid:105)|0−1(cid:105), |ψ (cid:105) = |1+2(cid:105)|0(cid:105),
00 10 place an upper bound on the probability of success of
|ψ (cid:105) = |0(cid:105)|0+1(cid:105), |ψ (cid:105) = |1(cid:105)|1(cid:105),
01 11 any sequential strategy, by considering a simpler, re-
|ψ (cid:105) = |0−1(cid:105)|2(cid:105), |ψ (cid:105) = |0+1(cid:105)|2(cid:105),
02 12 (4) lated discrimination problem, and then show that this
|ψ (cid:105) = |1−2(cid:105)|0(cid:105),
20 bound is achievable. Thus, to bound the probability of
|ψ (cid:105) = |2(cid:105)|1−2(cid:105),
21 success in identifying the state, rather than trying to
|ψ (cid:105) = |2(cid:105)|1+2(cid:105),
22 discriminate between all nine states, let us suppose in-
stead that we simply try to determine to which of the
A useful graphical representation of these states is given
threesubsets{|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)},{|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)},or
in Fig. 1. This example is of a different flavour to the 00 01 02 10 11 12
{|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)} the state belongs; that is, with our
20 21 22
choice of labeling |ψ (cid:105), we try to determine the index j,
jk
without worrying about k. This is equivalent to discrim-
inating between the equiprobable mixed states
1(cid:88)
ρ = |ψ (cid:105)(cid:104)ψ |. (6)
j 3 jk jk
k
This problem of subset discrimination is strictly easier
than our original problem: any measurement to discrim-
inatebetweenallninedominostatesmayalsobeusedfor
subset discrimination, and performs at least as well for
thistask. Thusthesuccessprobabilityoftheoptimalse-
quentialmeasurementdiscriminatingthestates{ρ }isat
i
least as high as the probability of success of any sequen-
tial measurement discriminating all nine domino states:
Pseq({ψ })≤Pseq({ρ }).
corr i corr i
FIG. 1. Graphical representation of the domino states.
This set further has the rather nice property that Bob’s
systemisessentiallyclassical;hisstateisalwaysdiagonal
previous one, as the states are globally perfectly distin-
inthe{|0(cid:105),|1(cid:105),|2(cid:105)}basis,thusregardlessofthemeasure-
guishable, but two-way classical communication, even in
mentonA,theonlysensiblemeasurementonB isinthis
the limit of infinite rounds of measurement and commu-
basis. The role of Alice’s measurement then, as in our
nication,isnotenoughtoperfectlydistinguishthestates
firstexample,issimplytoinformBobhowtointerprethis
[3]. The known upper bound on the probability of er-
measurement result. By inspection of the states ρ (eq.
ror of any scheme with two-way communication however j
6)weseethatifBobobtainsoutcome0,Alice’smeasure-
is so small as to be negligible for all practical purposes:
1.9×10−8[4]. Wefind,bycontrast,thatthebestone-way ment must discriminate between the orthonormal states
{|0(cid:105),|1+2(cid:105),|1−2(cid:105)},toprovideBobwiththeinformation
strategy has an error of more than 16%.
required to identify one of the states {ρ ,ρ ,ρ } respec-
We begin by simply stating the optimal sequential 0 1 2
tively. Similar observations for Bob’s outcomes 1 and 2
measurement,whichissomewhatintuitive,andgivesome
reveal that the role of Alice’s measurement is to distin-
of the technical details later. We assume the states are
guishsimultaneouslyaswellaspossiblethestateswithin
equiprobable, and we suppose that A is measured first,
the three bases {|0(cid:105),|1±2(cid:105)}, {|0(cid:105),|1(cid:105),|2(cid:105)}, {|0±1(cid:105),|2(cid:105)}.
whichduetothesymmetryofthestateswecandowith-
There are 27 subsets containing exactly one state from
out any loss of generality. There are 8 subsets of the set
each basis: Alice’s job is to optimally discriminate be-
of domino states which are perfectly distinguishable on
tween these subsets. This sounds like rather a daunting
system B alone, these are:
task, however it turns out, as we discuss below, that the
S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, optimal strategy is one that only ever identifies those
0 00 01 02 4 10 11 02
S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, subsets {S ,···,S } given in eqn. (5).
1 00 01 12 5 10 11 12 0 7
S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, Thus Alice’s measurement giving the optimal one-way
2 10 21 22 6 20 11 02
S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}, S = {|ψ (cid:105),|ψ (cid:105),|ψ (cid:105)}. strategyforthissimplerproblemispreciselythatconjec-
3 20 21 22 7 20 11 12
(5) tured to be optimal for discriminating all nine domino
The optimal sequential strategy assigns the state to one states. This succeeds with probability 83.6%, which
of these subsets in the first step, and discriminates per- as argued above is therefore an upper bound on the
fectly the states within the subset in the second step. success probability for discriminating the domino states
Thissucceedswithprobability(cid:39)83.6%,asweshowlater. withonlyone-wayclassicalcommunication. Further,this
4
(cid:80)
boundisachievableviatheconjecturedsequentialscheme Denoting Γ = p ρ π , an alternative, equivalent con-
i i i i
given above, which we therefore conclude is an optimal dition, which is sometimes easier to use in practice, is
one-way strategy. In the remainder of the paper we dis- obtained by summing over i in eqn. 9, giving
cuss some of the technical details of the derivation of
(Γ−p ρ )π =0. (10)
Alice’s optimal measurement. We give more details and j j j
an alternative proof elsewhere [30]. Ourstrategytofindtheoptimalprobabilityofsuccess
Alice wishes to optimally assign the state to one of forAlice’smeasurementistousetheHelstromconditions
the subsets {Si}. She performs a measurement on her constructivelytofindΓ= 18(cid:80)iσiπi suchthattheopera-
system, and upon obtaining outcome j takes this to in- tors Γ−1σ are rank-two positive semi-definite for all j.
8 j
dicate that the state belonged to the subset Sj. The Thus eqn 8 is satisfied, and choosing πj to be a weighted
mostgeneralmeasurementshecanperformonhersystem projector onto the zero eigenvalue eigenstate of Γ− 1σ
8 j
is described by a probability operator measure (POM) ensures that eqn 10 is also satisfied. If we can choose
[31], also known as a positive operator-valued measure the weights such that the resulting operators π sum to
j
(POVM) [32], that is a set of positive operators {πj} the identity, we have succeeded in finding an optimal
that form a resolution of the identity measurement. This task is facilitated by noting that the
states have rather a lot of symmetry. It is clear that the
(cid:88)
π ≥0, π =I. (7)
j j set {σ } is invariant under the action of the two unitary
k
j operators U and V. Hence we search for a measurement
{π } with the same symmetry properties. It follows that
Theprobabilityofobtainingoutcomej inameasurement i
we are searching for an operator Γ= 1(cid:80) σ π which is
on a system prepared in state ρ is given by the Born 8 i i i
invariant under U and V: Γ = UΓU† = VΓV†, which
rule: P(j|ρ)=Tr(ρπ ). Thus, the probability that Alice
j
further implies that Γ is of the form:
chooses a subset containing the state prepared is given
by:
Γ=p(|0(cid:105)(cid:104)0|+|2(cid:105)(cid:104)2|)+q|1(cid:105)(cid:104)1|
(cid:88)1 (cid:88)
P = Tr (|ψ (cid:105)(cid:104)ψ |π ) Finally, using the symmetry again, we need only check
corr 9 AB ij ij k
ij k||ψij(cid:105)∈Sk the condition (8) for σ0 and σ4, the rest follow by con-
8(cid:88)1 struction. ImposingthateachofΓ−18σ0andΓ−18σ4have
= 3 8TrA(σkπk) one zero eigenvalue allows us to solve for p and q, giving
k p(cid:39)0.110, q (cid:39)0.093, and P = 8Tr(Γ)= 8(2p+q)(cid:39)
corr 3 3
where in the last line we have defined σ = 0.836. We note without proof that we can indeed form a
k
(cid:16) (cid:17)
Tr 1(cid:80) |ψ (cid:105)(cid:104)ψ | . σ isthusthedensityop- resolution of the identity from the resulting operators
B 3 ij||ψij(cid:105)∈Sk ij ij k {π }, which therefore define an optimal measurement.
erator obtained by taking an equal mixture of the states i
Finally we recall that the simplified problem of discrim-
in the subset S , traced over Bob’s system. It follows
k inating the three mixed states {ρ ,ρ ,ρ } given above
from the above that {π } is the optimal measurement 0 1 2
k leads to a problem of discriminating between 27 subsets
to discriminate the equiprobable states {σ }. Explicitly,
k on Alice’s side. Performing the same analysis as above
the states σ are given by:
k it is readily verified that the measurement just derived,
2 1 that optimally discriminates the 8 subsets {S ,···,S }
σ = |0(cid:105)(cid:104)0|+ |0−1(cid:105)(cid:104)0−1|, 0 7
0 3 3 remains optimal. It is tedious but straight-forward to
σ =Uσ U†, σ =VUσ U†V†, σ =Vσ V†, check that the Helstrom condition 8 is satisfied for the
1 0 2 0 3 0
1 1 1 remaining states, which are therefore never identified.
σ = |0−1(cid:105)(cid:104)0−1|+ |1(cid:105)(cid:104)1|+ |1+2(cid:105)(cid:104)1+2|,
4 3 3 3 We thus find that for these cases, it is two-way clas-
σ =Uσ U†, σ =UVσ V†U†, σ =Vσ V†, sical communication that significantly boosts the distin-
5 4 6 4 7 4
guishability of the states and that quantum communica-
where we have introduced the unitary operators: tionprovidesonlyasmalladditionalboostinonecase. It
is known of course that two-way communication is more
U =−|0(cid:105)(cid:104)0|+|1(cid:105)(cid:104)1|+|2(cid:105)(cid:104)2|,
powerfulthanone-waycommunication; howeverexplicit,
V =|0(cid:105)(cid:104)2|+|1(cid:105)(cid:104)1|+|2(cid:105)(cid:104)0|.
quantitative gaps for the problem of discriminating or-
Anymeasurement{π }discriminatingoptimallybetween thogonalstateshavebeenshownintheliteratureonlyfor
j
thestates{ρ }withpriors{p }satisfiestheso-calledHel- measurements on entangled states (see e.g. [16], [21]). It
j j
strom conditions, which are known to be both necessary issurprisingthatevenasingleroundoftwo-wayclassical
and sufficient [33–36]: communication can provide such a significant improve-
ment in measurements on pure product states.
(cid:88)
pjρjπj −pkρk ≥0, (8) The optimal sequential measurement is a well-
j motivated indicator of achievable experimental perfor-
π (p ρ −p ρ )π =0. (9) mance in local measurement schemes: many rounds of
i i i j j j
5
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