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Eavesdropping in Quantum Cryptography - BB84 vs. Six States Presented by Igor Tselniker 236823 - Seminar in Quantum Information processing
Eavesdropping in Quantum Cryptography - BB84 vs. Six States
Presented by Igor Tselniker
236823 - Seminar in Quantum Information processing
Talk Outline• The Six-State Quantum Cryptosystem• Eavesdropping (single qubit attack)• Eve’s attacks and it’s analysis (under symmetry
assumption)• Introduction to Classical and Quantum information
theory• Calculation of Mutual Information and it’s
maximization for Eve ???????????????????• Comparison between four state (BB84) and six state
schemes
References[1] Dagmar Bruss, “Optimal eavesdropping in quantum cryptography with six states” Phys. Rev. vol. 81, 3018 (1998), quant-ph/9805019[2] C. A. Fuchs, N. Gisin, R. B. Griffiths, C.-S. Niu, and A. Peres, Phys. Rev. A 56, 1163 (1997).[3] C. A. Fuchs and A. Peres, Phys. Rev. A 53, 2038 (1996).
The Six State Scheme• Alice wishes to generate a shared secret key with Bob
using a quantum channel and an authenticated classical channel.
• Alice selects randomly (with equal probability 1/3) the basis and then she sends the appropriate qubit in the base she chose:
0 1 0 10 0 0 0
2 20 1 0 1
1 1 1 12 2
z x y i
z x y i
i
i
+ += = + = = + =
− −= = − = = − =
Alice Bob
Quantum channel
Classical channel
• Such a scenario is a straightforward extension of the traditional protocol (BB84). The main purpose of this presentation is to point out that this generalized scheme is principally more secure than the original BB84.
• This is due to the fact that in this scenario the optimal strategy an eavesdropper, Eve, can design (in order to gather information) by performing some unitary transformation on the transmitted qubit, gives her less information for a fixed disturbance of Bob’s (received) qubit.
• As Alice increases the set of inputs, it is more difficult for Eve to learn something in transit – prove???.
The Six State Scheme (Cont.)
The Six State Scheme (Cont.)• After Bob receives all the qubits, Alice
announces on the classical channel which bases were used.
• Now Bob measures in Alice’s basis. The sent qubit and measured qubit should agree. These values will be used to form the key.
Alice Bob
Quantum channel
Classical channel
X(0) X(1) Z(0) Y(1) Z(0) Z(1) Z(0) Y(0) X(1) Z(1) Z(0) Y(1)
Disadvantages of six-state scheme comparing to BB4 (four-state)
• In this presentation we assume that Bob has a quantum memory, so he performs his measurements AFTER Alice reveals for him the bases she used. In practical, Bob measures first and THEN Alicepublish the bases she used, so only 1/3 of qubits are kept and 2/3 are discarded, which is less than in four-state scheme – only half of the received qubitsare discarded.
• The practical implementation of six-state scheme is much involved and complicated than four-state.
• Proves of security of six state might be more complicated (due to more inputs).
Eavesdropping• In addition to Alice and Bob, there is Eve:
• Eve is not very nice and she wants the key. In an attempt to learn about the key, she may listen to the classical channel and do quantum operations on the channel and some qubits at her lab. Quantum operations are unitary.
∑⎯⎯→⎯j
BobEvejiU
AliceEvejEi eve
,0...00
No Cloning of Qubits• Unitarity of quantum operations means that
qubits can’t be copied exactly (no-cloning):
++⎯→⎯+
⎯→⎯
+EE
EEU
U
0
0000 0Proof:
The left side of above is normalized, and unitary operations preserve length, so the right side is normalized. Inner product is preserved so
inner product of the left sides and right sides are equal:?
0 0
0
1 10 0 0 0 0 022
2 1
E E E E E E
E E
+ +
+
+ = + = = + + =
⇒ = >
Any attempt to learn qubits, disturbs them, so Eve causes Errors!
Attacks (examples)Measure/Resend
• The simplest attack Eve can think about is to chose one of the bases with equal probability, then measure the Alice’squbit (and so she destroys the qubit) and resend the new qubit (according to what she’s got in her measure) to Bob
• Let’s calculate Bob’s error rate in this case:
Pr( ) Pr( , ) Pr( , )Pr( | ) Pr( ) Pr( | ) Pr( )
1 1 1 2 112 2
: #
Bob obtains the right bitEve chose the right baseEve chose the wrong base
k kk k k
where k of bases
= + == ⋅ + ⋅
− + −= ⋅ + ⋅ =
⋅=
○○○ ○
Attacks (examples)Measure/Resend (Cont.)
In our (six state) case:• With probability 1/3 Eve guesses the right basis and
so Bob receives the qubit without any error!• With probability 2/3 Eve guesses the wrong basis,
but still Bob can obtain in his measure, with probability ½, the right bit.
• All in all, Bob obtains the right bit with probability 2/3, or, in other words, his error rate in this case is 1/3 (which is higher than BB84 four-state scheme –advantage or disadvantage???)!!!
Attacks (examples) (Cont.)
CNOT Attack
0 0 0 0
1 0 1 1
1 10 0 0 1 12 2
z z z z EE
z z z zE E
z z z z zE E E E E
→
→
⎡ ⎤ ⎡ ⎤+ → + = + + + − −⎣ ⎦ ⎣ ⎦
Hence, in the X basis, Bob’s outcome becomes random, and with probability ½ he decides on wrong bit !!
•In this attack Bob and Alice can discover Eve by comparing QBER in different bases!!!
•In the Z basis:
So Bob receives his qubit with no error!•In the X basis:
Eavesdropping (Cont.)• The most general unitary transformation Eve
can design is of the form:
( )1
0 0 1
1 1 0
0
UA E B E B E
UA E B E B E
A B
C D
unitarity A D B C
Ψ ⎯⎯→ +
Ψ ⎯⎯→ +
⇒ ⇒ + =
• The first qubit is the one sent to Bob and acted on by Eve. Eve’s initial state is , and
refer to her un-normalized states after the interaction.
EΨ
, , ,E E E E
A B C D
Eavesdropping (Cont.)• Eve wants to learn the key as much as she can, while
keeping Bob’s error rate as low as she can, so she won’t be discovered!
• Let’s calculate the error rate for Bob, assuming the only noise that may occur in channel is due to Eve’s persistence:
• Under assumption that Alice sends her qubits with equal probability ½ and Eve disturbance for ‘0’ is the same as for ‘1’ we get (symmetry assumption):
( ) ( ) ( )( ) ( ) ( ) ( )( ) ( )
'0 ', '1' '1', '0 '
'1' '0 ' | '1' '0 ' '1' | '0 '
'1' '0 '
P Bob is wrong P Bob Alice P Bob Alice
P Alice P Bob Alice P Alice P Bob Alice
P Alice D D P Alice B B
= = = + = = =
= = = + = = = =
= = + =
( ) 12
dd
P Bob is wrong D D B B d==
⎡ ⎤⎢ ⎥= +⎢ ⎥⎣ ⎦
Eavesdropping (Cont.)• Let’s write Eve’s transformation for other states
transmitted by Alice as well.• For the , states:
( ) ( ) ( ) ( )
( ) ( )
( ) ( )
1 12 2
12
1 12 2
1 1 12
0 01 11
02
0
12
UA E EA
B BE
B E B E
E EB B
EU
A E B E B E
UA E E B B
B E B E
E E
AE
B
E
B
A
C D
C D
A B
A B
A B A B
A B
C D C D
C D CB D
E F
A
⎡ ⎤+ Ψ = ⎡ + ⎤ Ψ ⎯⎯→ + =⎣ ⎦ ⎣ ⎦
⎡ ⎤= + =⎣ ⎦
⎡ ⎤ ⎡ ⎤=
+
+ − − + + + −
+ + + − −⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⇒ + Ψ ⎯⎯→ + + −
⎡ ⎤ ⎡ ⎤− Ψ = ⎡ − ⎤ Ψ ⎯⎯→ + − + − +⎣ ⎦ ⎢ ⎥ ⎢ ⎥⎣ ⎦
+
+ + − + + − −
+ −+ +
−+⎦
−−⎣
⇒ − Ψ UE B E B E
G H⎯⎯→ − + +
+ −
Eavesdropping (Cont.)
• For the , states:i+ i−
( )
( ) ( ) ( ) ( )
( ) ( )
1 12 2
12
1 12
1 1 0
1
0 0 1
2
1 122
0
Ui A E EA
i iB BE E
Ui i iA E B E B E
Ui iA E
B EB E B E
i i i iE EB
B E
i i i iE
E BA
EB BB
A B
A i i B
A
i i
i B
C D
i i C i D
C i D C i D
C
A i B
A i B
i
I J
i
⎡ ⎤+ Ψ = ⎡ + ⎤ Ψ ⎯⎯→ + =⎣ ⎦ ⎣ ⎦
⎡ ⎤= + + =⎣ ⎦
⎡ ⎤ ⎡ ⎤= + + + − −
+
− + + − +
⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⇒ + Ψ ⎯⎯→ + + −
− Ψ = ⎡ − ⎤ Ψ ⎯⎯→ + −⎣ ⎦
+
+ + − + + −− + + −
− +
−
+ +
( ) ( )12i B
E EU
i i iA E B E B E
A i Bi D C i D
K L
⎡ ⎤ ⎡ ⎤+ − +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⇒ − Ψ ⎯⎯→ − + +
− −+
Assumptions on attacks• Assume that Eve is much smarter, so• Assume, as well, Eve is clever enough to treat all
bases in the same way (i.e., with same disturbance for Bob) – otherwise Alice and Bob could find out about her existence by comparing error rates in different bases. This assumption results some constrains which the scalar products of Eve’s states have to fulfill. Let’s derive these constrains differently for four states and six states accordingly!
103
d≤ ≤
Calculations of Bob’s fidelity
0( )
14
1
14 1
1 2 2 Re4
1unitarity
A A A B A C A D
B A B B B C B DE E
C A C B C C C D
D A D B D C D D
d A B A C A D
B A d B C B D
C A C B d C D
D A D B D C d
A B A C B DA D C C DB=
⎡ + + + + ⎤⎢ ⎥
+ + + +⎢ ⎥= =⎢ ⎥+ + + +⎢ ⎥
⎢ ⎥+ + +⎣ ⎦⎡ − + + + +⎤⎢ ⎥
+ + + +⎢ ⎥= =⎢ ⎥+ + − + +⎢ ⎥
⎢ ⎥+ + +⎣ ⎦⎡ ⎤⎧ ⎫⎪ ⎪⎢ ⎥= + + + + + =⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎭⎩⎣ ⎦
=
+
}{1 Re2
A B A C B D C D⎡ ⎤+ + + +⎣ ⎦
Calculations of Bob’s fidelity (Cont.)
( )
}{
}{
0( )
14
1 2 2 Re4
1 1 Re2
1 1 R2
'e
unitarity
A A A B A C A D
B A B B B C B DG G
C A C B C C C D
D A D B D C D D
A B A C B D C D
A B A C B D C D
E E A B A C B D C D symmetry at
A D B C
Ev s te
=
⎡ − + − + ⎤⎢ ⎥− + − + +⎢ ⎥
= =⎢ ⎥− + − +⎢ ⎥⎢ ⎥− + − +⎣ ⎦
⎡ ⎤⎧ ⎫⎪ ⎪⎢ ⎥= + − + − + − =⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎭⎩⎣ ⎦
⎡ ⎤= + − + + −⎣ ⎦
⎡ ⎤= + + + + ≡⎣
+
⎦
}{ }{
}{
Re 01
1 1 R
e
2
12
e
R
a
E E G G A
ck
G G
C B
A B AA C B D C D B
D
C D+ =
≡
⎡ ⎤≡ = + − + + − ⇒⎣ ⎦
⇒ ⎡ ⎤= = + +⎣ ⎦
Calculations of Bob’s fidelity (Cont.)
( ) }{
}{ }{0( )
14
1 2 2 Im 2Re4
1 1 Im Re2
unitarity
A A i A B A C i A D
i B A B B i B C B DI I
C A i C B C C i C D
i D A D B i D C D D
A B C D A C B D
A B C D A C B
A D
D
B C=
⎡ − + + + ⎤⎢ ⎥
+ + − +⎢ ⎥= =⎢ ⎥− + + +⎢ ⎥
⎢ ⎥− − − +⎣ ⎦⎡ ⎤⎧ ⎫
⎪ ⎪⎢ ⎥= + − − + − =⎨ ⎬⎢ ⎥⎪ ⎪⎢ ⎥⎭⎩⎣ ⎦
⎡ ⎤= + − + −
+
⎣ ⎦
Calculations of Bob’s fidelity (Cont.)
{ }
{ } { }
}{ }{
0(
14
1 2 2Im 2 Re4
1 1 Im Re2
1 1 Im e2
'R
unitarity
A A i A B A C i A D
i B A B B i B C B DK K
C A i C B C C i C D
i D A D B i D C D D
A B C D A C B D
A B C D A C B D
I I A B C D A C B D s
A D B C
Eve s
=
⎡ + + − + ⎤⎢ ⎥− + − − +⎢ ⎥
= =⎢ ⎥+ + − +⎢ ⎥⎢ ⎥− + +⎣ ⎦
⎡ ⎤⎧ ⎫⎪ ⎪⎢ ⎥= + − + + + − =⎨ ⎬⎢ ⎥⎪ ⎪⎩ ⎭⎣ ⎦
⎡ ⎤= − − + −⎣ ⎦
⎡ ⎤= + − + − ≡⎣
+
⎦
{ } { } }{
}{1 1 Re
1 1 Im Re m 02
2
I
I I K K A C
y
B
mmetry attack
K A BK A B C D
D
C D C DA B
≡
⎡ ⎤= − − + −
⎡ ⎤⇒ = = + −
⎦ − =⎣
⎣
⇒
⎦
Constrains on the scalar products of Eve’s states (four-state scheme)
}{
( )
Eve is clever enough to treat all bases in the s
Summary on constrains on the scalar products of Eve s
0 , 1 , , :
ame wa1 1 Re 12
:tates:
:
y
1 0
for chosen states
A A C C
E E G G A C B D d
unitarity
s
A D B
ymmet
C
ry
+ −
= ≡
≡ ≡
⎡
+
⎤≡ = = + + −⎣ ⎦
=
⇒
( ) }{( )( ) { }
Re 0
1
2
3/ :
4 Re 1 2base treatment fidelit
A B C D
A A C C d
A C B D dy
+ =
= =⎧ −
+ = −⎪⎨⎪⎩
Constrains on the scalar products of Eve’s states (four-state scheme)
}{
( )
0 , 1 , , :
1 1 R
Eve is clever enough to treat all bases in the same
Summary on constrains on the scalar products of Eve states:
e
a
2
y
1
1
w
0:
i ifor chosen states
A A C C
I I K K A C B D d
unitarity
symmetr
A D B C
+ −
= ≡
≡ ≡
⎡ ⎤≡ = = + − ⎦
=
−
⇒
+
⎣
( ) { }( )( ) { }
: 2
3/ :
4
Im 0
1
Re 1 2
y
base treatment fideli
A B C D
A A C C d
A Ct
B Dy
d
− =
= = −
− = −
⎧⎪⎨⎪⎩
Constrains on the scalar products of Eve’s states (four-state scheme)
}{
}{Eve is clever enough to treat all bases in the same w
, , , :
1 1 Re2
Summary on constrains on the scalar products o
ay
f Eve states
1 1 Re 12
:
i ifor chosen states
E E G G A C B D
I I K K A C B D d
unitarity
+ − + −
⎡ ⎤= = + + ≡⎣ ⎦
≡ ≡
⎡ ⎤≡ = = + − −⎣ ⎦
⇒
( ){ } { }( )
( ) { }( ) { }
0
Re 0 Im 0
0
Re 0
R
: 1
:2
3/ :
e 24 1
A D B C
A B C D and A B C D
A B D Csymmetry
base treatment fidelitB D
A dy
C
⎧⎪⎨
+ =
+ = − =
⇒ + =
=
= −
⎪⎩⎧⎪⎨⎪⎩
Constrains on the scalar products of Eve’sstates (extension to six-state scheme)
}{
}{
Eve is clever enough to treat all bases in the same way
Eve is clever enough to treat all bases in the s
1 1 Re2
1 1 Re 12
ame way
Summary on constrains on the s
A A C C
E E G G A C B D
I I K K A C B D d
= ≡
≡ ≡
⎡ ⎤= = + + ≡⎣ ⎦
≡ ≡
⎡ ⎤= = + − −⎣ ⎦
⇒
( ){ } { }( )
( )( ) { }( ) { }
0
Re 0 Im 0
0
1
Re 0
: 1
:2
3
/ : 4
5
calar products of Eve states:
Re 1 2
A D B C
A B C D and A B C D
unitarity
symmetry
base treatment fide
A B D C
A A C C d
B D
A C
t
d
li y
⎧⎪
+ =
+ = − =
⇒ + =
= = −
⎨⎪⎩
⎧⎪
=
=
⎨
−
⎪
⎪⎪⎩•As one may notice, in case of six state scheme there are more constrains
on Eve!!!
The measure of information (classical)
• The entropy of a random variable is defined as:
• The conditional entropy is defined as:
• The mutual information is defined as:
2( ) logb bb
H B p p= −∑
| 2 |( | ) loga b a b aa b
H B A p p p= −∑ ∑
( ; ) ( ) ( | )I A B H B H B A= −
The measure of information (quantum)• We talk about Eve’s accessible information, so we need to
quantify it. Since we talk about quantum states, we need to use Quantum Information Theory notations.
• The entropy of a quantum state (pure or mixed) is defined as (also called von Neumann entropy):
• But how do we calculate ?
( ) ( )2logS trρ ρ ρ= −
2log ρ
The measure of information (Cont.)• The function of any square matrix is defined as follows:
( ) ( )
( ) [ ]( )
1
1
iiii
if A Q BQthenf A Q f B Q
when B is a diagonal matrix
and f B f B
−
−
=
=⎡ ⎤⎣ ⎦
The measure of information (Cont.)• One of the main Quantum Information properties is a Holevo
bound:
• Actually Eve is interested in maximum information access in order to learn the key as much as she can while introducing minimum disturbance on Bob’s qubit (so she won't be discovered).
• So she will try to maximize her mutual information as much as she can and try to keep Bob’s disturbance as low as she can.
• Therefore , calculating the Holevu bound for Eve and Alice will give us the upper bound of Eve and Alice possible mutual information.
• Maximizing it over d – is our main goal of this presentation!!!
( ) ( ) ( ); B Ba a a a
a aI A B S p p Sρ ρ≤ ∑ −∑
Calculations of Mutual Information • The mutual information between Alice and Bob is
given by:
• By construction, Bob’s disturbance d is the same no matter which state was sent by Alice.
( ) ( )( ) ( )
2 2
2 2
2 2
1 1 1 1( ) log log 12 2 2 2
( | ) log 1 log 1
( ; ) 1 log 1 log 1
H B
H B A d d d d
I A B d d d d
= − − =
= − + − −⎡ ⎤⎣ ⎦⇒ = + ⋅ + − ⋅ −
Calculations of Mutual Information (Cont.)
• Let’s try to calculate the Holevo bound for Bob.• The calculation is for each qubit and qubit.• Suppose that Alice sent a qubit in Z basis.• After Alice reveals for him which base she chose, Bob’s
density matrix is in case Alice sent 0 and
in case Alice sent 1.
1 0 0 1 01 1 10 0 1 0 12 2 2
B Ba a
a
d dp
d dρ ρ
−⎡ ⎤ ⎡ ⎤ ⎡ ⎤⇒ = ∑ = + =⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦
0
1 00
Ba
dd
ρ =
−⎡ ⎤= ⎢ ⎥⎣ ⎦
1
00 1
Ba
dd
ρ =
⎡ ⎤= ⎢ ⎥−⎣ ⎦
Calculations of Mutual Information (Cont.)
• Thus Holevo bound for Alice and Bob is:
• In other words the mutual information between Alice and Bob reaches its Holevo bound and thus it is maximized!
• We calculated this bound under assumption that Alicechose base Z. But it is obvious that it is doesn’t matter which base we chose except the fact that calculating it in other bases is little bit more complicated (we need to diagonalize some matrices).
( ) ( )
( ) ( )2 2
1 0 1 0 01 1 10 1 0 0 12 2 2
1 log 1 log 1 ( ; )
B Ba a a a
a a
d dS p p S S S S
d d
d d d d I A B
ρ ρ⎛ ⎞ ⎛ − ⎞ ⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎡ ⎤
∑ −∑ = − − =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎢ ⎥−⎣ ⎦ ⎣ ⎦ ⎣ ⎦⎝ ⎠ ⎝ ⎠ ⎝ ⎠+ ⋅ + − ⋅ − =
Calculations of Mutual Information (Cont.)
• It was shown (in [3]) that it is sufficient for Eve to use only two qubits in order to extract the maximal information. (Intuitively – there are only four possible states at Eve’safter an interaction: )
• Let’s expand and where the complex coefficients have to satisfy
• And similarly for • We are free to chose as one of the four basis vectors,
e.g., and can fulfill the third constraint by setting, without loss of generality.
, , ,E E E E
A B C D
B00B d=
11D d=
3
0i
iA iα
== ∑
3 2
01i
idα
=∑ = −
B and D
3
0i
iC iγ
== ∑
Calculations of Mutual Information (Cont.)
• From the additional three constraints we get:
• Now we can try to calculate the Holevo bound between Alice and Eve:
( ) { }
( )( ) *
3*
0
* *0
*3
3 3
0
0
3 0
Re Re5 2
1
2
0
1
0
i ii
A C d
A B D C d
A D B C d d
d
α γ
α γ γ
γ γ
α
α α
=
+ =
⎧ ⎫= = −⎨ ⎬⎭⎩
+ = + = ⇒ =
+ = ⇒ = −
−
∑
The Holevo bound between Alice and Eve2 * * *
0 0 1 0 2 0 32* * *
1 0 1 1 2 1 30 2* * *
2
2 * *1 2
2 *1 1 1
0 2 1 2 2 32* * *
3 0 3 1 3 2 3
* * *3 3 3 0 3
3 0
3 0
* * *3 0 0
2 11 2*
2 2 1 2 22* *
1 20 0
Ea
Ea
d
dα α α α α α α
α α α α α α αρ
α α α α α α α
α α α α α α α
α α α α α
α α
α α
α α α α
γ γ
γ γ γ γ γρ
γ γ γ γ γ
γ γ α
=
=
⎡ ⎤+⎢ ⎥
⎡ ⎤− −⎢ ⎥⎢ ⎥− −⎢ ⎥=⎢ ⎥− −⎢ ⎥⎢ ⎥− − +⎣ ⎦
⎢ ⎥= ⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦
0 112
E E E Ea a a a
apρ ρ ρ ρ= =⎡ ⎤= ∑ = +⎣ ⎦
The Holevo bound between Alice and Eve (Cont.)
• So Holevo bound for Alice and Eve is:
• In order to calculate the expression written above we need to diagonalize three 4x4 parametric matrices, which is a real suicide!!!
• For some reason (which I couldn’t find anywhere) the author of the paper derive the mutual information between Alice and Eve in this form:
( ) ( ) ( ) ( )0 1 0 1212
1 12
E E Ea a a a
Ea a
E Ea a
a aS p S S Sp Sρ ρ ρ ρρ ρ= ===
⎛ ⎞−∑ = − −⎜ ⎟⎝
⎡ ⎤∑ +⎣ ⎦ ⎠
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
2 2 2 2 2 2 2 20 0 2 0 0 1 1 2 1 1
2 2 2 2 2 2 2 22 2 2 2 2 3 3 2 3 3
2 2 2 2 2 2 2 20 2 0 1 2 1 2 2
2 2 2 2 2 2 2 20 2 0 1 2 1 2 2 2 3 2 3
2 3 2 3log log log l
log log log l
( ; ) 1
12 log log
log lo
o
og
g
g
I A E
d
d d
d
d
d
d
d
α γ α γ α γ α
α α α α α α α
γ
α γ α γ α γ
γ γ γ γ γ
α
α
γ
γ
γ γ
= +
⎫+⎪⎪+ − ⎪⎬
− + + + + − + + − ⎪
− + + − + + + +
+ + + + +
+
+
+ +
⎭
+⎧⎪⎪⎪⎨⎪⎪ ⎪⎪ ⎪⎩
Maximization of • The task is to maximize along with the constraints.• The method of Lagrange multipliers leads to a set of
equations which cannot be simultaneously fulfilled unless . This means that the best
solution for Eve is to use states such that , which one would have expected.
• Now we have only two parameters, for ,• and can write
( ; )I A E( ; )I A E
0 0 3 30 0andα γ α γ= = = =0A B C D= =
1 1andα γ ( ; )I A E
( ) ( ) ( ) ( )( ) ( ) ( ) ( )
( ) ( )
2 2 2 2 2 2
2 2
2 21 1 2 1 1 2 2 2 2 2
2 2 2 2 2
2 2 2 21 2 1 2 2 2
2 2 2 21
1 1 2 1 1
2 21 2 1 2 2 2
2 2 2 21 2 1 1 22 1 2 1
1
1 1
( ; )
112 log log
log llog log
log 1 lo112 log
og
log 1 log 1 g 1
2 2
d d d
A
d
d
I E
α γ α γ α γ α γ
α γ α γ
α α γ γ γ γ
γ γ γ γ
α α
α α α α
α γ
=
⎧ ⎫+ −⎪ ⎪= + =⎨ ⎬− + + − + +⎪ ⎪⎭⎩
+
++ −= +
− + + −
+
+ − − − −−
−
− −
− −
−
( ) ( )2 2 21 2 1 1log 2 2d α γ
⎧ ⎫⎪ ⎪⎨ ⎬
− − −⎪ ⎪⎭⎩
Maximization of (Cont.)• It is straightforward to write down the system of equations
which has to be fulfilled in order to maximize .• If we chose then
where• By checking the higher derivatives, one confirms that this
is a maximum, which is, due to concavity, the absolutemaximum. Using the last restriction allows us to find the “best” relative phase between and thus leads to the solution for the highest mutualinformation that Eve can extract from measuring her twoqubits:
( ; )I A E
( ; )I A E
( ) ( )22 1( ; ) 1 1I A E d h α= − − ⋅2 2
1 11 dα γ= − −
( ) ( ) ( )2 2 2log 1 log 1h x x x x x− − − −
A and C{ }Re 1 2C A d= −
[ ]( )
[ ] ( )
2( ; ) 1
1 11 2 32
I A E f h K f
where K f f ff
= − ⋅
⎛ ⎞= + −⎜ ⎟
⎝ ⎠
Comparison between four state (BB84) and six state schemes
• It has been found [4] that for four-state scheme the maximal mutual information between Alice and Eve is given by form:
• And we found that for six-state scheme it is:
( ) ( )
( ) ( ) ( ) ( ) ( )
84
2 2
1; 2 121 log 1 1 log 1
BBI A E f f
where z z z z z
φ
φ
⎡ ⎤= −⎣ ⎦
+ + + − −
[ ]( )
[ ] ( )
2( ; ) 1
1 11 2 32
I A E f h K f
where K f f ff
= − ⋅
⎛ ⎞= + −⎜ ⎟
⎝ ⎠
• Plotting the two curves on the same graph we get:
Comparison between four state (BB84) and six state schemes(Cont.)
Conclusions• It is worth mentioning that the optimal unitary
transformation, which leads to maximum mutual information between Alice and Eve that we found herein, disturbs all Bloch vectors in the same way, not only the six states used by Alice, and allows Eve to gain the same information in all possible bases.
• In other words: the optimal eavesdropping action for six states is shown to be a universal transformation. Thismeans that using a bigger number of states cannot increase security.
• The gain in security described in this presentation isdue to the fact that the three bases are spanning the full Bloch sphere, as opposed to the case of BB84, where only atwo-dimensional plane is spanned.
