Quantum Key Distribution with Qubit Pairs

0.06 MB·English

by  Tabish Qureshi

Quantum Key Distribution with Qubit Pairs
Tabish Qureshi∗ and Mohd. Asad Siddiqui†
Centre for Theoretical Physics, Jamia Millia Islamia, New Delhi, India.
We propose a new Quantum Key Distribution method in which Alice sends pairs of qubits to
Bob, each in oneof four possible states. Bob uses onequbitto generatea secure keyand theother
to generate an auxiliary key. For each pair he randomly decides which qubit to use for which key.
The auxiliary key has to be added to Bob’s secure key in order to match Alice’s secure key. This
schemeprovides an additional layer of security over thestandard BB84 protocol.
PACSnumbers: 03.67.Dd
3
1 Quantum cryptography [1] is a field which has ma- Alice and Bob keep only the data from those mea-
•
0 tured since the first protocol, BB84, given by Bennett surements for which their bases are the same, dis-
2 andBrassardin1984[2]. Theideaistogenerateasecret carding all the rest.

n keybetweentworemoteparties,traditionallycalledAlice
This data is interpreted as a binary sequence ac-
a and Bob using a quantum channel. The secret key may
•
J then be used in sending encryptedmessages throughthe cording to the coding scheme 0 x = 0, 1 x = 1,
8 so-called Vernam Cipher [3] or one-time pad. Vernam + z = 1, z = 0. | i | i
| i |−i
1 cipher has been provedto be verysecure,under the con-
Alice announcesthe resultsofasmallsubsetofher

dition that one shared key is to be used only once. •
] measurements. Bob checks if he has identical re-
h Inprinciple, BB84is provento be hundredpercentse- sults. Any discrepency here indicates a possible
p cure [4]. However, when implemented in real life with
evesdropping attempt.
- non-ideal sources and detectors, several attacks have
t
n beensuccessfullydemonstratedagainstcommercialQKD If there is no discrepancy, the rest of the binary
a •
systems [5, 6]. sequence is treated as the new key, and is identical
u
InthetraditionalBB84scheme,Alicesendsastreamof for both Alice and Bob.
q
[ single qubits to Bob which eventually leads to the gen-
eration of a secure key. Here we propose a new QKD Inour new scheme,Alice sends pairsofqubits to Bob,
1 randomly chosen to be in one of the following states,
scheme in which Alice sends a stream pairs of qubits to
v
Bob. We will show that this scheme introduces an addi-
4 1 1 1 2,
1 tional feature over the standard BB84 protocols, which | i | i
0 should make it more difficult to break. |0i1|0i2,
5 To set the ball rolling, we describe the BB84 QKD 1
1. protocol [2]. √2(|+i1|+i2+|−i1|−i2)
0 1
3 Alice sends single qubits to Bob randomly in one (+ 1 2+ 1 + 2), (1)
• √2 | i |−i |−i | i
1 ofthefollowingstates: 0 , 1 , + and ,where
iv: |±i=(|0i±|1i)/√2. | i | i | i |−i washseorceia|t±ediiw=ith(|e1aicih±st|0aitie)/s√he2.senFdosraArelicaes,ftohlleowkes:y bits
X
Bob measures the incoming qubit’s state by ran-
ar • domly choosing a measurement of either the x- 1 1 1 2 1,
| i | i →
component of the qubit or the z-component, with
0 1 0 2 0,
equal probability. Let us say, 0 , 1 are eigen- | i | i →
| i | i 1
statesofthe z-componentofthe qubit, and are (+ 1 + 2+ 1 2) 0,
the eigenstates of the x-component. |±i √2 | i | i |−i |−i →
1
(+ 1 2+ 1 + 2) 1. (2)
BobpubliclytellsAlicewhichbasesheusedforeach √2 | i |−i |−i | i →
•
qubit he received (but, of course not the result of
In addition, she calls the first two states as z-basis, and
his measurement).
the other two as x-basis.
Bob uses one qubit for his secure key and one for the
Alice publicly tells Bob which basis she sent each
• auxiliary key. For generating the secure key, he ran-
qubit in.
domly measures the x-component or the z-component
of the qubit. For the auxiliary key he measures only
the x-component. For his secure key measurement re-
∗Electronicaddress: [email protected] sults, he uses the followingconventionfor key bit values:
†Electronicaddress: [email protected] 1 1, 0 0, + 1, 0. For his auxiliary
| i → | i → | i → |−i →
2
Basis x z x x z x z z x z z x z x x his non-matching secure key bit will make it identical to
Alice key bit 0 1 1 0 1 0 0 0 1 1 1 0 0 0 1 Alice’s bit.
An evesdropper trying to figure out the shared key in
Bob secure state - 1 - - 1 - 0 0 - 1 1 - 0 + -
between, will have his task made difficult in more ways
Bob secure key bit 0 1 0 0 1 0 0 0 0 1 1 0 0 1 0
than it is in BB84. For correctly figuring out the auxil-
Bob auxiliary state - + + - - - + - + - + - - + +
iary key, the evesdropper has to know which of the two
Bob auxiliary bit 0 0 1 0 0 0 0 0 1 0 0 0 0 1 1
qubits was used for the auxiliary key for each and every
Bob secure + auxil 0 1 1 0 1 0 0 0 1 1 1 0 0 0 1 pair,animpossibletask. Ifweassumethattheevesdrop-
per is able to correctly guess which of the two qubits is
TABLE I: Result of a typical sequence of 15 qubits sent by
usedfortheauxiliarykey,hecansimplymakeanx-basis
Alice toBob, wherethemeasurement basis ofAlice and Bob
measurement on those qubits, and generate the correct
matches. When the auxiliary key is added to Bob’s secure
auxiliary key without Bob’s knowledge. However, even
key,it matches with Alice’s key.
inthisnearimpossiblecircumstance,therestofthecom-
munication still remains as the standard BB84.
key measurement results, he uses the following conven- In conclusion,we have introduceda new QKD scheme
tionforkeybitvalues: + 1, 0,ifhemeasured in which Alice sends pairs of qubits to Bob in four pos-
x-component for the sec|urie→key;|−+i→ 0, 0, if he sible states. Bob randomly chooses one of the pair for
measured z-component for the se|cuire→key.|−i→ his secure key and one for the auxiliary key. The auxil-
AliceandBobannouncetheirbasesforsecurekeypub- iary key has to be added to Bob’s secure key in order to
licly, and discard those qubits for which their bases do get the correct shared key. An evesdropper job is made
not match. Various measurement results for rest of the more difficult as he has to correctly predict which parti-
caseswhere the basesagree,will be correlatedin the fol- cle Bob is going to use for secure key and which one for
lowingfashion. If Alice sends z-basis,the securekey bits auxiliary key, for every pair used in the communication.
ofAlice andBobwill be identical. If Alice sends x-basis, Eveniftheevesdroppermanagestoacheivethisnearim-
the secure key bits of Alice and Bob will be identical if possible feat, he is still left with the job of cracking the
theauxiliarykeybitis0;the securekeybitsofAliceand BB84 security. We believe this scheme should be easily
Bob will be different if the auxiliary key bit is 1. implementable in practice.
Thus,forasequenceofqubitpairssent,AliceandBob’s
securekeybitswillbeidenticalifthecorrespondingaux-
iliary bit is 0. Alice and Bob’s secure key bits will not Acknowledgments
matchforcaseswherethe auxiliarybitis1. So,Bobjust
needs to add his final auxiliary key to his secure key bit Asad Siddiqui thanks the University Grants Commis-
by bit, modulo 2. This way, his auxiliary bit 1 added to sion for financial support.
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