
Mixed States, Open Systems, and the Density Operator 87
Proof. We can see this by invoking the convexity property (Equation 5.16).
Suppose our pure state density matrix admits such a decomposition
ˆρ
pure
= λ|ψ
1
ihψ
1
| + (1 − λ)|ψ
2
ihψ
2
|
= λρ
1
+ (1 − λ)ρ
2
.
Now since the state is pure, there exists some vector |ui such that
ˆρ
pure
= |uihu|.
Consider an orthogonal vector |vi : hu|vi = 0.
=⇒ hv|ˆρ
pure
|vi = hv|uihu|vi = 0.
=⇒ λhv| ˆρ
1
|vi + (1 − λ)hv| ˆρ
2
|vi = 0.
Since λ and (1−λ) are positive, this equation can only be satisfied if hv| ˆρ
1
|vi =
0 = hv| ˆρ
2
|vi. This means ˆρ
1
and ˆρ
2
are orthogonal to |vi. But |vi can be any
vector orthogonal to |ui. So we must have