Entanglement
A 2-qubits pure state
|
ψ
⟩
A
B
∈
H
A
⊗
H
B
{\displaystyle |\psi \rangle _{AB}\in H_{A}\otimes H_{B}}
can be written (using Schmidt decomposition) as
|
ψ
⟩
A
B
=
∑
j
λ
j
|
j
⟩
A
|
j
⟩
B
{\textstyle |\psi \rangle _{AB}=\sum _{j}\lambda _{j}|j\rangle _{A}|j\rangle _{B}}
, where
{
|
j
⟩
A
}
,
{
|
j
⟩
B
}
{\displaystyle \{|j\rangle _{A}\},\{|j\rangle _{B}\}}
are the bases of
H
A
,
H
B
{\displaystyle H_{A},H_{B}}
respectively, and
∑
j
λ
j
2
=
1
,
λ
j
≥
0
{\textstyle \sum _{j}\lambda _{j}^{2}=1,\lambda _{j}\geq 0}
. Its density matrix is
ρ
A
B
=
∑
i
,
j
λ
i
λ
j
|
i
⟩
A
⟨
j
|
A
⊗
|
i
⟩
B
⟨
j
|
B
{\textstyle \rho ^{AB}=\sum _{i,j}\lambda _{i}\lambda _{j}|i\rangle _{A}\langle j|_{A}\otimes |i\rangle _{B}\langle j|_{B}}
. The degree in which it is entangled is related to the purity of the states of its subsystems,
ρ
A
=
tr
B
(
ρ
A
B
)
=
∑
j
λ
j
2
|
j
⟩
A
⟨
j
|
A
{\textstyle \rho ^{A}=\operatorname {tr} _{B}(\rho _{AB})=\sum _{j}\lambda _{j}^{2}|j\rangle _{A}\langle j|_{A}}
, and similarly for
ρ
B
{\displaystyle \rho ^{B}}
(see partial trace). If this initial state is separable (i.e. there's only a single
λ
j
≠
0
{\displaystyle \lambda _{j}\neq 0}
), then
ρ
A
,
ρ
B
{\displaystyle \rho ^{A},\rho ^{B}}
are both pure. Otherwise, this state is entangled and
ρ
A
,
ρ
B
{\displaystyle \rho ^{A},\rho ^{B}}
are both mixed. For example, if
|
ψ
⟩
A
B
=
|
Φ
+
⟩
=
1
2
(
|
0
⟩
A
⊗
|
0
⟩
B
+
|
1
⟩
A
⊗
|
1
⟩
B
)
{\textstyle |\psi \rangle _{AB}=|\Phi ^{+}\rangle ={\frac {1}{\sqrt {2}}}(|0\rangle _{A}\otimes |0\rangle _{B}+|1\rangle _{A}\otimes |1\rangle _{B})}
which is a maximally entangled state, then
ρ
A
,
ρ
B
{\displaystyle \rho ^{A},\rho ^{B}}
are both completely mixed.
For 2-qubits (pure or mixed) states, the Schmidt number (number of Schmidt coefficients) is at most 2. Using this and Peres–Horodecki criterion (for 2-qubits), a state is entangled if its partial transpose has at least one negative eigenvalue. Using the Schmidt coefficients from above, the negative eigenvalue is
−
λ
0
λ
1
{\displaystyle -\lambda _{0}\lambda _{1}}
. The negativity
N
=
−
λ
0
λ
1
{\displaystyle {\mathcal {N}}=-\lambda _{0}\lambda _{1}}
of this eigenvalue is also used as a measure of entanglement – the state is more entangled as this eigenvalue is more negative (up to
−
1
2
{\textstyle -{\frac {1}{2}}}
for Bell states). For the state of subsystem
A
{\displaystyle A}
(similarly for
B
{\displaystyle B}
), it holds that:
ρ
A
=
tr
B
(
|
ψ
⟩
A
B
⟨
ψ
|
A
B
)
=
λ
0
2
|
0
⟩
A
⟨
0
|
A
+
λ
1
2
|
1
⟩
A
⟨
1
|
A
{\displaystyle \rho ^{A}=\operatorname {tr} _{B}(|\psi \rangle _{AB}\langle \psi |_{AB})=\lambda _{0}^{2}|0\rangle _{A}\langle 0|_{A}+\lambda _{1}^{2}|1\rangle _{A}\langle 1|_{A}}
And the purity is
γ
=
λ
0
4
+
λ
1
4
=
(
λ
0
2
+
λ
1
2
)
2
−
2
(
λ
0
λ
1
)
2
=
1
−
2
N
2
{\displaystyle \gamma =\lambda _{0}^{4}+\lambda _{1}^{4}=(\lambda _{0}^{2}+\lambda _{1}^{2})^{2}-2(\lambda _{0}\lambda _{1})^{2}=1-2{\mathcal {N}}^{2}}
.
One can see that the more entangled the composite state is (i.e. more negative), the less pure the subsystem state.