By “state” of a system, we mean a collection of all possible knowledge we
can gather about the system, which is practically achieved by studying the
distribution of outcomes of measurements made on the system. In the case of
pure states, these outcomes together are described by a ray in Hilbert space.
Consider measuring an observable
ˆ
Q with N possible eigenvalues q
i
with
corresponding eigenstates |q
i
i. If we obtain a particular result q
i
, then we can
say that the projection operator
ˆ
P
i
= |q
i
ihq
i
| has acted on the state of the
system.
If we know the state to be the pure state |ψi, then the state is as well de-
scribed by a projector |ψihψ| along this direction. The probability of outcome
q
i
is given by
P(q
i
) = hψ|
ˆ
P
i
|ψi = hψ|q
i
ihq
i
|ψi
= hq
i
|ψihψ|q
i
i
= hq
i
|ˆρ|q
i
i (5.1)
This defines the density operator ˆρ for a pure state described by a single state
vector:
ˆρ
pure
= |ψihψ|. (5.2)
In general, the system could be composed of a number of (pure) states |ψ
n
i
where n = 1, 2…d, with classical probability p
n
: 0 ≤ p
n
≤ 1,
P
n
p
n
= 1. This
mixture {p
n
, |ψ
n
i} is referred to as an ensemble of pure states with associated
probabilities. In this case, the probability of obtaining the outcome q
i
on
measuring
ˆ
Q is
P(q
i
) =
X
n
p
n
hψ
n
|
ˆ
P
i
|ψ
n
i =
X
n
p
n
hψ
n
|q
i
ihq
i
|ψ
n
i =
X
n
p
n
hq
i
|ψ
n
ihψ
n
|q
i
i
= hq
i
|
X
n
p
n
|ψ
n
ihψ
n
|
!
|q
i
i (5.3)
where in the third equality, we have moved the term hq
i
|ψ
n
i to the beginning
Mixed States, Open Systems, and the Density Operator 79
of the expression since it is a number. (This is an illustration of manipulating
expressions using the Dirac notation.)
The piece within parentheses in the middle is identified as the density
operator or the statistical operator ˆρ for that state. This operator is com-
pletely given by the initial state.
Definition 5.1. The density operator for a system consisting of a mixed
ensemble of states {p
n
, |ψ
n
i} is
ˆρ =
X
n
p
n
|ψ
n
ihψ
n
|. (5.4)
The sum over states in this expression looks like a superposition of states:
but this is an incoherent superposition, as opposed to coherent superposition
of basis states that defines a pure state. The incoherence stems from the fact
that the relative phases of the states |ψ
n
i are not available to us.
This operator uniquely prescribes the probabilities of outcomes on mea-
surements on the system. Exactly as in Equation 5.1, we can then write the
probability of obtaining the outcome q
i
as
P(q
i
) = hq
i
|ˆρ|q
i
i. (5.5)
The expectation value of
ˆ
Q in a state ˆρ is
h
ˆ
Qi
ρ
=
X
i
q
i
P(q
i
) =
X
i
q
i
hq
i
|ρ|q
i
i =
X
i,j
q
i
hq
i
|q
j
ihq
j
|ρ|q
i
i
=
X
j
hq
j
|ρ
X
i
q
i
|q
i
ihq
i
|
!
|q
j
i.
Here we have introduced the resolution of identity =
P
j
|q
j
ihq
j
| in the third
line, and then moved the term hq
i
|q
j
i to the end of the expression since it is
a number. In the last line, we identify the term in the parentheses as the
spectral representation of the operator
ˆ
Q.
Definition 5.2. The trace of an operator
ˆ
A is defined by
Tr
ˆ
A =
X
j
hj|
ˆ
A|ji,
a simple generalization of the trace of a matrix as the sum of its diagonal
elements.
The expectation value of the observable we are measuring is thus given by
h
ˆ
Qi
ρ
= Tr(ρ
ˆ
Q). (5.6)
(The trace here is apparently taken in the {|q
i
i} basis, but trace is basis-
independent, as you will prove.)
80 Introduction to Quantum Physics and Information Processing
This averaging of the physical property Q is twofold: first the quantum
average hQi
n
= Tr(ρ
n
Q) over each of the (pure) states comprising the mixture,
and the usual statistical average over the whole ensemble with each state
average weighted by the probability p
n
of its occurrence. We can make this
explicit by writing
h
ˆ
Qi
ρ
= h
ˆ
Qi =
X
n
p
n
Tr(ρ
n
ˆ
Q). (5.7)
Exercise 5.1. Show that for vectors |φ
i
i and |φ
j
i, Tr(|φ
i
ihφ
j
|) = hφ
j
|φ
i
i.
Exercise 5.2. Show that the trace of an operator is independent of the basis
chosen to evaluate it.
Exercise 5.3. Show that trace as an operation is linear, i.e., Tr(A + B) = TrA +
TrB and Tr(λA) = λTrA.
Exercise 5.4. Show that the trace of products of operators is invariant under
cyclic permutations of the operators. i.e., Tr(AB) = Tr(BA), Tr(ABC) =
Tr(BCA) = Tr(CAB). etc.
Exercise 5.5. Show that the Pauli matrices are traceless.
The matrix representation of the density operator, called the density ma-
trix of the system, is useful for computations. In the computational basis {|ii},
we can represent the density operator (Equation 5.2) as a matrix:
|ψi =
X
i
c
i
|ii (5.8)
=⇒ ρ
pure
=
X
i,j
c
i
c
∗
j
|iihj|. (5.9)
For a mixed state, in this basis we can represent the density matrix as
ρ
mixed
=
X
i,j
ρ
ij
|iihj|. (5.10)
If a system consists of equal mixtures of all possible computational basis states
it is said to be maximally mixed. In n dimensions, such a state is represented
by a multiple of the identity matrix:
ρ
max
=
1
n
n×n
.
Mixed States, Open Systems, and the Density Operator 81
Example 5.1.1. The density matrix for the unpolarized electron beam dis-
cussed above is the maximally mixed state
ρ
m
=
1
2
|0ih0| +
1
2
|1ih1| =
1
2
“
1 0
0 1.
#
=
1
2
.
In contrast, the density matrix for the pure state |↑
x
i =
1
√
2
(|0i + |1i) is
ρ
p
=
1
√
2
(|0i + |1i)
1
√
2
(h0| + h1|)
=
1
2
(|0ih0| + |0ih1| + |1ih0| + |1ih1|)
=
1
2
“
1 1
1 1
#
.
Note the difference, though when beams in either state are passed through an
SG
z
machine, we get ↑ and ↓ outputs with equal probability! However, when
passed through an SG
x
machine, the mixed state gives the same result, while
the pure state |↑
x
i gives only an ↑ beam with probability 1.
Example 5.1.2. We’ll see how the usual results regarding experimental mea-
surements follow using density matrices for pure states. Consider the state
|↑
x
i of the above example. The probability of obtaining +1 on measuring σ
z
in this state is
P
+
= h0|ρ|0i = [ 1 0 ]
“
1
0
#
=
1
2
.
The matrix σ
z
ρ =
1
2
“
1 0
0 −1
#”
1 1
1 1
#
=
“
1 1
−1 −1
#
, and
the expectation value of σ
z
is the sum of its diagonal elements = 0.
Example 5.1.3. A mixed state need not necessarily be composed of orthogonal
states. For example, one could have a mixture containing 20% of the state
|0iand 80% of the state |↑
x
i, whose density matrix would be given by
ρ =
1
5
|0ih0| +
4
5
|↑
x
ih↑
x
| =
1
5
“
1 0
0 0
#
+
2
5
“
1 1
1 1
#
=
1
5
“
3 2
2 2
#
Leave a Reply