On quantum capacity of erasure channel assisted by back classical communication

0.17 MB·English

by  Debbie Leung

On the Capacity of Erasure Channel Assisted by Back Classical Communication
Debbie Leung1 , Joungkeun Lim2 and Peter Shor2
∗ † ‡
1 Department of Combinatorics and Optimization, and Institute for Quantum Computing
University of Waterloo, 200 University Avenue West Waterloo, Ontario, Canada N2L 3G1
2 Department of Mathematics, Massachusetts Institute of Technology
77 Massachusetts Avenue, Cambridge, MA 02139, USA
(Dated: January 2, 2010)
We present a communication protocol for the erasure channel assisted by backward classical
communication,whichachievesasignificantlybetterratethanthebestpriorresult. Inaddition,we
proveanupperboundforthecapacityofthechannel. Theupperboundissmallerthanthecapacity
of the erasure channel when it is assisted by two-way classical communication. Thus, we prove the
separationbetweenquantumcapacitiesassistedbybackwardclassicalcommunicationandtwo-way
0 classical communication.
1
0 PACSnumbers: 03.67.Hk
2

n
In quantum information theory, a capacity Q(χ) of a portation [4] and an upper bound given by Q ( ) as
a 2 Np
J channelχisatheoreticalmaximumoftheratem/nthat
2 isachievablebysomecommunicationprotocolthatsends QB(Np)≥1−2p, if p≤2/5,
m-qubit information with n uses of the channel, where Q ( ) (1 p)/3, if p 2/5, and (1)
] n tends to infinity. The above definition of Q is defined B Np ≥ − ≥
h without auxiliary resources, and additional free classi- QB(Np)≤Q2(Np)=1−p.
p
cal communication may increase the capacity. We use
- In this Letter, we present an efficient communication
t Q, Q ,Q , and Q to denote the quantum capacities of
n 1 B 2 protocol that achieves a better lower bound of QB( p),
a quantum channel when unassisted, assisted by unlim- N
a and we prove a new upper bound of Q ( ). With this
B p
u ited forward, backward, and two-way classical communi- upper bound, we show that Q ( ) < QN( ) for all p
B p 2 p
q cation, respectively. It was proved that classical forward N N
and resolve the previously open question.
[ communication alone does not increase the quantum ca-
Preliminaries and Notations– Recall the defini-
4 pacity of any channel; in other words Q(χ) = Q1(χ) for tion of von Neumann entropy H(A) = H(ψA) =
v all channels χ [1]. In contrast, Q2 is greater than Q for tr(ψAlogψA),whereψAisthedensityoperatorforsys-
3 some channels [1]. QB is also known to be greater than −tem A. The quantum mutual information and coherent
4 Qforsomechannels[2],butithasbeenanopenquestion
information are defined as
9 whether Q (χ)=Q (χ) for all χ.
B 2
5
We study the capacities of the quantum erasure chan- I(A;B)=H(A)+H(B) H(AB), and
.
0 nel, which was first introduced in [3]. The quantum era- I(A B)=H(B) H(AB−).
1 sure channel of erasure probability p, denoted by , i −
7 Np
replaces the incoming qubit, with probability p, with an The statements in the following lemma will be used in
0
: “erasurestate” 2 orthogonaltoboth 0 and 1 ,thereby the proof of a theorem in the later section.
v | i | i | i
both erasing the qubit and informing the receiver that it
Xi hasbeenerased. Inanequivalentformulation, calledthe Lemma 1. For disjoint systems A, B, and C,
isometric extension, the channel exchanges the incoming (i) I(AB;C) I(B;C) I(A;BC).
r − ≤
a qubit with the environmental system in state 2 with (ii) I(A B) I(A BC).
probability p. It was shown in [2] that the quan|tuim ca- (iii) I(Ai C)≤+I(BiC) I(AB C).
i i ≤ i
pacities Q,Q , and Q for are given by (iv) I(A BC) I(A B) 2H(CE), where E is any sub-
1 2 Np set of B.i − i ≤
Q( p)=Q1( p)=max 0,1 2p and Proof. Subadditivity and strong subadditivity inequali-
N N { − }
Q ( )=1 p. ties [5] easily give (i),(ii),(iii),
2 p
N −
H(CDE) H(D)+H(CE),
However, until the current investigation, little has been ≤
H(AD) H(CE)+H(ADCE), and
known about Q ( ) except for two lower bounds that
B p ≤
N
followstraightforwardly from1-wayhashing[1]andtele- H(D)+H(ADE) H(AD)+H(DE),
≤
for E B and D = B/E. Adding these three inequali-
⊂
ties yields (iv).
∗[email protected]
†[email protected] We consider only near-perfect communication proto-
‡[email protected] colsthatproduce,withhighprobability, outputstatesof
2
high fidelity with the input states. The fidelity of states where resources on the left-hand side simulate those on
ρ and ρ is defined to be theright, denotesoneuseoftheerasurechannel,and
in out p
N
Qbit denotes one use of the noiseless qubit channel. We
F(ρin,ρout)≡trqρ1in/2ρoutρ1in/2. hWavitehufsreede bΦacakndclaΓssaiscaslhcoormthmanudnifcoarti|oΦni,hΦon|eanudse|ΓofihΓ|.
p
N
can prepare one ebit with probability 1 p. Hence,
From now on, we call the sender, the receiver, and the
−
environment Alice, Bob, and Eve.
1 (1 p) Φ . (4)
p AB
Communication Protocol– we derive an improved N ≥ −
lower bound for Q ( ) by providing a communication We combine equations (3) and (4) to get
B p
N
protocol. The protocol combines two subprotocols that
utilize coherent teleportation introduced in [6]. 1 Np ≥ 1−2p Qbit, if p≤1/2 and
ψCo=heare0nt+Teble1porintatsiyosnt–emGiMvenaannduannkneobwitn(qsuombiettsitmaetes 1 Np ≥ 11+−2pp Qbit, if p≥1/2.
|caliledan|EiPRp|aiirorBellstate) Φ = 1 (00 + 11 ) Hence, the rate of the first subprotocol is
| iAB √2 | i | i
bteeltewpeoerntaAtiloince[a4n].dIBnotbh,eAolircigeicnaanl tteralenpsomrtitat|ψiointoprBotoobcobly, 1−2p , if p≤1/2 and
the change of basis takes the initial state |ψiM|ΦiAB to 11+−2pp , if p≥1/2.
Second Subprotocol–Thismethodonlydiffersfromthe
1 ij XiZj ψ . (2)
2 | iMA | iB previous subprotocol in that ij will be sent using a co-
Xij herent version of superdense|coiding. More specifically,
in this case, Alice and Bob first share an ebit Φ
Reference [6] proposes a coherent variant of teleporta- | iC1C2
where C belongs to Alice and C belongs to Bob. Af-
tion in which Alice does not measure ij but in- 1 2
| iMA ter the change of basis (see equation (2)), Alice applies
stead, coherently copies ij to two ancillary sys-
| iMA control-X from M to C and control-Z from A to C ,
tems C C and transmits them coherently to Bob. 1 1
1 2 resulting in the joint state
Mathematically, Alice and Bob share the joint state
1 ij ij XiZj ψ . After receiving C C ,
B2oPbijca|niaMpApl|y aiCc1Con2trol-X|frioBm C1 to B and the1n2a 21Xij |ijiMA |ΦijiC1C2 XiZj|ψiB,
control-Z from C to B. Alice and Bob then share the
2
state 1 ij ij ψ , with ψ transmitted and sends C to Bob using the erasure channel. The
2 ij| iMA | iC1C2 | iB | i 1
and twoPebits shared between Alice and Bob. states Φij = XiZj Φ are orthogonal (they form the
| i | i
First Subprotocol– Suppose Alice and Bob already Bell basis) [5]. In case of erasure, Bob and Eve share
share an ebit, and Alice teleports ψ to Bob by at- Φ and Alice and Bob will take another ebit and
| i | ijiC1C2
temptingtousetheerasurechannelforcoherentclassical repeat the superdense coding procedure, until Bob re-
communication of each of i and j (see previous ceives the transmission (call the two-qubit system in his
| iC1 | iC2
subsection on coherent teleportation). Bob tells Alice possessionD D ). Then,Bobappliesthetransformation
1 2
whether the communication is erased or not. If so, Al- Φ ij and coherently reverts the XiZj
| ijiD1D2 → | iD1D2
ice copies and sends it again until Bob receives it. Note not only in XiZj ψ but also in all the Φ he shares
B ij
| i | i
that the transmission is coherent if it is not erased in with Eve (by acting only on his halves), so that the final
the first trial. If i and j are erased k and l times before state becomes
theyaresentsuccessfully, thestatebecomes(afterBob’s
controlled-X and Z) 12 |ijiMA|ijiD1D2 |Φi⊗EkB |ψiB.
Xij
12 |ijiA|ii⊗Ek|ji⊗El|ijiB|ψiB where k again denotes the number of erasures before the
Xij successful transmission. In this method, Alice and Bob
∼|Γi⊗AB(1Ek+1l)|ΦiA⊗B(2−1k−1l)|ψiB, alwOanycseshaagraein2, eAbliitcseantetehdesetnod.apply superdense coding
where 1k = 0 if k = 0 and 1k = 1 if k > 0 and similarly 11p timesonaverage. Thisgivestheasymptoticresource
for1 , Γ = 1 (000 + 111 ),and denotesequivalence in−equality,
l | i √2 | i | i ∼
up to a unitary transformation on E. Φ + 1 +Φ
Since the success probability of each transmission is AB 1 p Np AB
t1e−rpi,aAndlicje.tHrieensc1e−1sphteimtreasnosmniatvsera2geqtuobsietnsdtheraocuhgrhegtihse- ≥ 1 Qb−it(cid:2)+2 ΦAB +((cid:3)1−1p −1) ΦBE.
1 p Note that the above consumes more ebits than it pro-
channel. Both 1 and 1 have exp−ectation p. In asymp-
k l duces for all p; thus, we use equation (4) to supply the
totic resource inequality [6],
needed ebits, and obtain
1−2p Np+ΦAB ≥ 1Qbit+2(1−p)ΦAB+2pΓABE, (3) 1 Np ≥(1−p)2 Qbit.
3
Hence the rate of the second subprotocol is (1 p)2. statesthatasufficientamountofmutualinformation(2m
−
Rate of Communication Protocol– Applying the two for m ebits) has to be delivered to Bob. Part (ii) states
protocols selectively, the rate of the protocol is thatthemoremutualinformationislosttoEve,themore
transmissions are needed to nullify the lost information.
(1 p)2 , if p 1/2 and
− ≤ (5) Theorem 2. If the fidelity between the input and output
11+−2pp , if p≥1/2. states is at least 1 ǫn, then
−
(i) I(S ;B R) 2m 2(2√2m√ǫ +1).
thiUspsepcetrionBoisuntodpfroorveththeaCt aQpBa(cNitpy)–≤Th11e−+ppp.urpBoysethoef (ii)PPi∈i∈BE I(Sii;Bii−−11R)≥≤n−−m+4(2√2mn√ǫn+1).
definition of the capacity, for each n, there is a protocol Proof. (i) For each i , apply part (i) of lemma 1 on
Pn that uses back classical communication and Np at the systems Si,Bi 1,∈anBd R to obtain
most n times and transmits n(Q ( ) δ ) qubits from −
B p n
N −
Alice to Bob with fidelity at least 1 ǫ and probability I(B ;R) I(B ;R) I(B S ;R) I(B ;R)
at least 1 ǫ , where ǫ ,δ 0 as−n n . i − i−1 ≤ i−1 i − i−1
n n n I(S ;B R).
− → →∞ i i 1
Our strategy to show the upper bound is as follows. ≤ −
We consider any protocol that transmits m qubits with Thus,
n uses of the channel. In particular, such protocol must
be able to transmit m halves of ebits shared between Al- I(S ;B R) (I(B ;R) I(B ;R))
i i 1 i i 1
ice and a reference system R [7], without entangling Eve Xi − ≥Xi − −
and R (or else the transmission to Bob will be noisy). ∈B ∈B
=I(B ;R)=I(B(1)B(2);R)
This translates to bounds on quantum mutual informa- n
tion between Bob, Eve, and R that will be contradicted I(B(1);R)
≥
if m/n is larger than our stated upper bound.
=H(B(1))+H(R) H(B(1)R)
IfAlicetransmitsherhalvesoftheebitssharedwithR −
directlythroughthechannel,anylosstoEvecanneverbe 2(H(R) H(B(1)R)).
≥ −
recovered. Thus, Alice has to transmit quantum states
whosepotentialentanglementwithRcanbematerialized NotethatthefidelitybetweenthestateµinB(1)Rand
ornullifieddependingonBob’sbackcommunicationand Φ⊗m isatleast1−ǫn. LetD = 21tr|µ−Φ⊗m|bethetrace
Alice’s future transmissions. The finalizing or nullifying distance [5] between µ and Φ⊗m. By page 415 of [5],
process requires further uses of the channel, giving an
upper bound to the capacity. D ≤ 1−F(µ,Φ⊗m)2 ≤√2ǫn.
Toquantifytheaboveidea,denotebyS ,S , S the p
1 2 n
qubits transmitted by Alice through the chann·e·l·. Each By Fannes’ inequality [5],
S is delivered to Bob with probability 1 p or lost to
Eive with probability p. Let = iS sent−to Bob and H(B(1)R)= H(µ) H(Φ⊗m) 2Dm 2Dlog(2D)
B { | i } | − |≤ −
Eere=d{tio|SBi osebntantodEEvvee}. bWe ethdeeifinndeexEset=s of qubits deliSv- ≤2√2m√ǫn+1.
i 1 j i,j j
to be Eve’s system after the ith transmissSion≤.≤For∈BEob, (ii) Using 2, 3, and 4 to denote the use of parts
the most general procedure after each transmission is an (ii),(iii), and (iv) of lemma 1 respectively, we have
isometry followed by a measurement. By double-block
coding and by extending Theorem 10 in [8], any such 2
I(S B R) c+ I(S B R)
measurement can be approximated by a von Neumann Xi ii i−1 ≤ Xi ii n
measurement on part of Bob’s system (turning the mea- ∈E ∈E
sured qubits into classical data). Let B˜i be Bob’s quan- 3 c+I( Si BnR)=c+I(En B(1)B(2)R)
≤ i i
tum system immediately after the ith channel use, and i[
∈E
Bi be his quantum system after his measurement and 4
cisladsesilcivaelrfeededtboaBckobt,oaAndlicBe˜. =ThBus B˜iif=SBiis−l1os∪tStoi iEfvSei. ≤3 c+I(EniB(1)B(2))+2H(B(1)R)
Suppose a total of c qubitsiare mi−ea1suredi by Bob in the ≤c+I(EnRiB(1)B(2))−I(RiB(1)B(2))+2H(B(1)R)
protocol. After the final decoding operation, Bob pro- =c+I(E R B ) I(R B(1)B(2))+2H(B(1)R)
n n
duces an m-qubit system B(1) that is almost maximally i − i
entangledwiththesystemR. WedenotetherestofBob’s where the equalities use the fact that Bob’s decoding is
quantum system by B(2). isometric. I(EnR Bn) is upper bounded by n c.
i −|E|−
In the following theorem, I(S ;B R) is the amount of I(R B(1)B(2)) is lower bounded as
i i i
mutual information carried by each transmission S . For
i
2 4
the rest of the paper, information theoretical quantities I(R B(1)B(2)) I(R B(1)) I(R B(1)T) 2H(T)
are evaluated on the states that are held at the corre- i ≥ i ≥ i −
=m 2H(B(1)R)
sponding stages of the protocol. Part (i) of the theorem
−
4
where T purifies B(1)R. Putting together the two previ- kn]=0, which is a contradiction. Hence,
ous sets of inequalities,
Xi I(SiiBi−1R)≤n−|E|−m+4(2√2m√ǫn+1). QB(Np)≤ 11−+pp. (6)
∈E
Hence,
Discussion–Wesummarizethepreviousandournew
I(S ;B R)= (H(S )+I(S B R))
Xi∈E i i−1 Xi∈E i ii i−1 rdeestuelrtmsiinneFdigaurerae1o.f QThBe(ligp)h,tegrivreengiboynitshtehperpevreiovuiosulsowuner-
N
(1+I(S B R)) and upper bounds in equation (1). The darker region is
i i 1
≤Xi i − the new undetermined area of QB( p) due to our lower
∈E N
and upper bounds in equations (5) and (6), which are
n m+4(2√2m√ǫn+1). significantly improved over previous results. Since our
≤ −
upper bound of Q ( ) is strictly less than Q ( ), we
B p 2 p
Since Alice cannot predict whether Bob or Eve will N N
prove the separation between Q and Q answering the
B 2
receive the next transmission and a certain fraction of
long-standing question raised in [2].
the transmission are lost to Eve, the same fraction of
mutual information has to be lost to Eve. Combined 1
withthetheorem, theargumentgivesanupperboundof 0.9
Q ( ). To prove this rigorously, consider the following 0.8
B p
randNom variable. 0.7
Q ( N)0.6
Xi =(cid:26) p−2(I12−(Sp)i;IB(iS−i1;RB)i−1R) iiff SSii iiss dloesltivteoreEdvteo Bob B p000...345
Then X 1 and E(X ) = 0. Note that the X ’s may 0.2
i i i
not be| in|d≤ependent variables. Let Y = i X and 0.1
i j=1 j 0
Y0 = 0. Then Y0,Y1, ,Yn is a martinPgale [9] with 0 0.1 0.2 0.3 0.4 0p.5 0.6 0.7 0.8 0.9 1
···
Y Y 1. If the fidelity between the input and
i+1 i
| − | ≤
output states is at least 1−ǫn, then from theorem 2 FIG. 1: Undetermined area of QB(Np)
Yn = p2Xi I(Si;Bi−1R)− (1−2p)Xi I(Si;Bi−1R)
∈B ∈E
≥ (1+2p)m− (1−2p)n−(2−p)(2√2m√ǫn+1). We thank Andrzej Grudka and Michal Horodecki
for pointing out an important mistake in an earlier
Assume by contradiction that QB(Np)> 11−+pp. Then, for manuscript, and for suggesting a solution that also sub-
sufficiently large n, mn ≥ 11−+pp +4k for some k > 0. The ssutapnptoiarltleydsbimyptlhifieeWst.hMe.pKroeocfk. TFohuisnrdeasteiaornchCwenatseprafortriaElxly-
above expression for Y , which holds with probability at
n treme Quantum Information Theory. P.W.S. and J.L.
least1 ǫ ,willexceedkn. Thereforelim Pr[ Y
kn]=1−. n n→∞ | n|≥ would like to thank the National Science Foundation for
support through grant CCF-0431787. J.L. thanks SLSF
However, Azuma’sinequality[9]appliedtomartingale
k2 for support. D.L. thanks NSERC, CRC, CFI, ORF, MI-
Yi gives Pr[Yn kn] e− 2 n, and limn Pr[ Yn TACS, ARO, and CIFAR for support.
| | ≥ ≤ →∞ | | ≥
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