Reply to Fahmi and Golshani's comment on "Quantum key distribution in the Holevo limit"

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by  Adan Cabello

Reply to Fahmi and Golshani’s comment on “Quantum key distribution in the Holevo
limit”
∗
Ad´an Cabello
Departamento de F´ısica Aplicada II, Universidad de Sevilla, E-41012 Sevilla, Spain
(Dated: February 2, 2008)
AsFahmiandGolshanicorrectlypointout,aprotocolintroducedinA.Cabello,Phys. Rev. Lett.
85,5635 (2000),toshowthataquantumkeydistributionprotocolcanhaveefficiencyone(i.e.,can
achieve the Holevo limit), does indeed not have efficiency one. The corrected protocol, introduced
in A. Cabello, Recent. Res. Devel. Physics 2, 249 (2001), is reproduced here.
PACSnumbers: 03.67.Dd,03.67.Hk,03.65.Ud
8
0
0 As Fahmi andGolshanicorrectlypoint outin the pre- η0 ,“01”ifthe stateis η1 ,“10”ifthestateis η2 ,and
2 | i | i | i
cedingComment[1],aprotocolintroducedin[2]toshow “11” if the state is η3 ). Since the four states (1a)–(1d)
| i
n that a quantum key distribution protocol can have ef- areorthogonal,Bobcanunambiguouslydiscriminatebe-
a
ficiency one (i.e., can achieve the Holevo limit), where tween them and identify which is the one sent by Alice.
J
efficiencyisdefinedasthenumberofsecretbitspertrans- Ascanbeeasilychecked,therevisedprotocoldoesnot
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1 mittedbitplusqubit,doesindeednothaveefficiencyone. only satisfy Mor’s requirements to prevent cloning when
This error was already corrected in [3, 4]. For complete- Eve has a one-by-one access to the qubits (namely, that

] ness’ sake, the corrected protocol introduced in [3, 4], the reduced states of the first subsystem must be non-
h
with efficiency one, is reproduced here. orthogonal and non-identical, and the reduced states of
p
- Supposethatthequantumchanneliscomposedoftwo the second subsystem must be non-orthogonal [5]) for
nt qubits (1 and2)preparedwith equalprobabilitiesinone any two states chosen from (1a)–(1d), but is also secure
a of four orthogonalpure states: against the double C-NOT eavesdropping strategy pro-
u
posed by Fahmi and Golshani in [1].
q 1
[ |η0i= √3(|00i+|01i+|10i), (1a)

1 1
v |η1i= √3(|00i−|01i+|11i), (1b)
8 1 ∗ Electronic address: [email protected]
0 η2 = (00 10 11 ), (1c) [1] A. Fahmi and M. Golshani, preceding Comment, Phys.
5 | i √3 | i−| i−| i
Rev.Lett. 100, 018901 (2008).
.2 η3 = 1 (01 10 + 11 ). (1d) [2] A. Cabello, Phys. Rev.Lett. 85, 5635 (2000).
1 | i √3 | i−| i | i [3] A. Cabello, Recent. Res. Devel.Physics 2, 249 (2001).
0 [4] A. Cabello, in F´ısica Cu´antica y Realidad. Quantum
8 Alice sends both qubits to Bob. Eve cannot access PhysicsandReality,editedbyC.MataixandA.Rivadulla
0 qubit 2 while she still holds qubit 1. Each pair of qubits (Editorial Complutense, Madrid, 2002), p.333.
v: encodes2bitsofthekey(forinstance,“00”ifthestateis [5] T. Mor, Phys. Rev.Lett. 80, 3137 (1998).
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