Ensemble density matrix 397
whereCk(λ)=〈φk|Ψ(λ)〉. Substituting eqn. (10.2.2) into eqn. (10.2.1) yields
〈Aˆ〉=
1
Z
∑Z
λ=1
∑
k,l
Ck(λ)∗Cl(λ)〈φk|Aˆ|φl〉
=
∑
k,l
(
1
Z
∑Z
λ=1
Cl(λ)C(λ)
∗
k
)
〈φk|Aˆ|φl〉. (10.2.3)
Eqn. (10.2.3) is in the form of the trace of a matrix product. Hence,let us introduce
a matrix
ρlk=
∑Z
λ=1
C(lλ)C(λ)
∗
k (10.2.4)
and a normalized matrix ̃ρlk=ρlk/Z. The matrixρlk(or, equivalently, ̃ρlk) is known
as theensemble density matrix. Introducingρlkinto eqn. (10.2.3), we obtain
〈Aˆ〉=
1
Z
∑
k,l
ρlkAkl=
1
Z
∑
l
(ˆρAˆ)ll=
1
Z
Tr(ˆρAˆ) = Tr( ̃ρAˆ). (10.2.5)
Here,Akl=〈φk|Aˆ|φl〉and ˆρis the operator whose matrix elements in the basis areρlk.
Thus, we see that the expectation value ofAˆis expressible as a trace of the product
ofAˆwith the ensemble density matrix. According to eqn. (10.2.2), the operator ˆρcan
be written formally using the microscopic state vectors:
ρˆ=
∑Z
λ=1
|Ψ(λ)〉〈Ψ(λ)|. (10.2.6)
It is straightforward to show that this operator has the matrix elements given in eqn.
(10.2.4).
According to eqn. (10.2.6), ˆρis a Hermitian operator, so that ˆρ†= ˆρand ̃ρ†= ̃ρ.
Therefore, its eigenvectors, which satisfy the eigenvalue equation
ρ ̃|wk〉=wk|wk〉, (10.2.7)
form a complete orthonormal basis on the Hilbert space. Here, we have definedwkas
being an eigenvalue of ̃ρ. In order to see what the eigenvalues of ̃ρmean physically, let
us consider eqn. (10.2.5) for the choiceAˆ=Iˆ. Since〈Iˆ〉= 1, it follows that
1 =
1
Z
Tr(ˆρ) = Tr( ̃ρ) =
∑
k
wk. (10.2.8)
Thus, the eigenvalues of ̃ρmust sum to 1. Next, letAˆbe a projector onto an eigenstate
of ̃ρ,Aˆ=|wk〉〈wk|≡Pˆk. Then