50
3 Introducing Quantum Key Distribution
where the r.h.s. entropy is computed on the following state:
˜
ρ R A ET = (E R A ⊗ id ET ) Tr B
˜
ρ AB E T
=
1
4
(E R A ⊗ id ET ) Tr B
t
|φ
t
AB E φ
t
AB E | ⊗ |tt | T
=:
1
4
t
ρ
t
R A E ⊗ |tt | T ,
(3.39)
where the quantum map
E R A (σ) =
1
a=0
|aa | |a|σ|a
represents the measurement performed by Alice for key generation, i.e. a projection
onto the Z basis. Being the state in Eq. (3.39) a c.q. state, its entropy simplifies to:
H (R A |E T ) ˜
ρ =
1
4
t
H (R A |E) ρ t .
(3.40)
The last part of the proof shows that H (R A |E) ρ t is actually independent of t and
equal to conditional entropy of the original state H (R A |E) ρ . This is clear if the state
ρ
t
R A E is made explicit. From Eq. (3.39) we have that:
ρ
t
R A E = (E R A ⊗ id ET ) Tr B
|φ
t
AB E φ
t
AB E |
,
(3.41)
where |φ
t
AB E is the purification of one of the four states in (3.34) prepared by
Eve according to the random variable T . For definiteness, let’s fix that state to be
(X ⊗ X ) ρ AB (X ⊗ X ), although an analogous reasoning holds for any other state in
Eq. (3.34). By writing ρ AB in its spectral decomposition:
ρ AB =
λ
λ|λλ|,
(3.42)
we can immediately explicit |φ
t
AB E as follows:
|φ
t
AB E =
λ
√
λ|λ
t
AB ⊗ |e λ E ,
(3.43)
where the eigenstates of the operator (X ⊗ X ) ρ AB (X ⊗ X ) read: |λ
t
= (X ⊗
X )|λ. By substituting (3.43) into (3.41) and by making explicit the map E R A we
obtain the following chain of equalities: