There is another sense in which density operators are a useful way to
describe nature. In general, it is impossible to isolate the system of interest
from some parts of its environment. We then have to regard our system as a
subsystem of a larger system: “system + environment”. If the large system in a
pure quantum state consists of subsystems, then the state of any subsystem is
essentially described by a density operator. The way to get there is to perform
a reduction of the density matrix of the larger system, by a procedure called
the partial trace over all subsystems except the one of interest.
5.3.1 Reduced density operator
Consider a composite of two systems A and B, described by a pure state
density operator ρ
AB
.
FIGURE 5.5: Illustrating a bipartite composite system.
For the purposes of this book, we will only concentrate on systems consist-
Mixed States, Open Systems, and the Density Operator 93
ing of two subsystems, the so-called bipartite systems (Figure 5.5). We can
perform a partial trace over the system B alone to obtain the state of system
A. If the set {|k
B
i} forms a basis for system B then
ρ
A
= Tr
B
ρ
AB
=
X
k
hk
B
|ρ
AB
|k
B
i. (5.34)
Trace operation is linear, and if we demand that partial trace is also linear in
its inputs, we can compute partial traces in practice.
Definition 5.4. If subsystems A and B are given by Hilbert spaces spanned
by the bases {|i
A
i} and {|j
B
i} respectively, we define the partial trace of ρ
AB
with respect to subsystem A as
Tr
A
ρ
AB
=
X
i
hi
A
|ρ
AB
|i
A
i (5.35)
which will be an operator on the Hilbert space of subsystem B alone.
Example 5.3.1. Consider a simple example where the system state can be
written in separable form:
ρ
AB
= σ
A
1
⊗ σ
B
2
.
Then quite trivially,
ρ
A
= Tr
B
(σ
A
1
⊗ σ
B
2
) = σ
A
1
Tr
B
σ
B
2
= σ
A
1
.
Example 5.3.2. A less trivial case where the two subsystems are entangled,
so that the state of the system is a Bell state:
|ψ
AB
i =
1
√
2
|0
A
i|0
B
i + |1
A
i|1
B
i
.
=⇒ ρ
AB
= |ψ
AB
ihψ
AB
|
=
1
2
(|00ih00| + |00ih11| + |11ih00| + |11ih11|) .
To obtain ρ
A
by a partial trace over B, we sandwich each term between the
basis states of B and add up:
ρ
A
= Tr
B
ρ
AB
= h0
B
|ρ
AB
|0
B
i + h1
B
|ρ
AB
|1
B
i
94 Introduction to Quantum Physics and Information Processing
We illustrate the calculation of this by first evaluating the contribution by the
first term in ρ
AB
:
h0
B
|(|00ih00|) |0
B
i + h1
B
|(|00ih00|) |1
B
i
= |0ih0
B
|0ih0|h0|0
B
i + |0ih1
B
|0ih0|h0|1
B
i
= |0ih0| + 0.
Evaluating the other terms similarly, we find that
ρ
A
=
1
2
(|0ih0| + |1ih1|) =
1
2
.
Thus the subsystem A is in a maximally mixed state! Similarly,
ρ
B
= Tr
A
ρ
AB
=
1
2
.
This result is a hallmark of entanglement: though the composite system is
in a well-defined state, i.e., its density operator contains maximal informa-
tion about all measurement outcomes in the state, we can say nothing about
measurement outcomes on either of the component subsystems: they are in
maximally mixed states.
Exercise 5.9. Calculate the density matrices for both subsystems for the other
three Bell states.
Exercise 5.10. Consider a 2-qubit system AB with the density matrix ρ =
1
2
|β
00
ihβ
00
| +
1
2
|10ih10|. Compute the reduced density matrices ρ
A
and
ρ
B
.
The fact that the reduced density matrices for entangled systems represent
mixed states is generic, and can be used to characterize entanglement. As we
have already seen, the reduced density matrices for separable systems will
always be pure.
Box 5.1: POVM from Projective Measurements on a Composite System
POVM measurements on quantum systems can be realized as projective
measurements on an extended “system+ancilla” Hilbert space. Let’s consider
a system A that is not interacting with the independent ancilla B. The com-
bined AB system is in a product state that can be represented by the density
operator
ρ
AB
= ρ
A
⊗ ρ
B
.
A projective measurement on this state is the action of projection operators
Mixed States, Open Systems, and the Density Operator 95
ˆ
m
on this state. The probability of outcome m is then
P(m) = Tr
h
ˆ
m
(ρ
A
⊗ ρ
B
)
i
= Tr
A
h
Tr
B
ˆ
m
ρ
A
⊗ ρ
B
i
= Tr
A
(
ˆ
E
m
ρ
A
) (5.36)
where the
ˆ
E
m
s are operators on the system A. We can identify the matrix
elements of these operators by expressing the above equation in components:
using orthonormal bases {|ii} for the system A and {|µi} for the ancilla B,
Tr
B
ˆ
m
ρ
A
⊗ ρ
B
=
X
ijµν
(
ˆ
m
)
jνiµ
(ρ
A
)
ij
(ρ
B
)
µν
=
X
ij
(E
m
)
ji
(ρ
A
)
ij
=⇒ (E
m
)
ji
=
X
µν
(
ˆ
m
)
jνiµ
(ρ
B
)
µν
.
It is easy to see that the
ˆ
E
m
s defined this way are complete. Suppose ρ
B
is
diagonal in the basis {|µi}:
ρ
B
=
X
µ
p
µ
|µihµ|,
X
m
E
m
=
X
µ
p
µ
hµ|
X
m
ˆ
m
|µi = .
5.3.2 Schmidt decomposition
Another useful way of dealing with composite systems, the Schmidt de-
composition is about expressing the state of a bipartite system in terms of
orthonormal states of the two subsystems.
Theorem 5.2. If {|u
A
i
i} and {|v
B
j
i} are orthonormal sets of vectors in the
Hilbert spaces of subsystems A and B, respectively, the state of the combined
system can be expressed as
|ψ
AB
i =
X
i
λ
i
|u
A
i
i|v
B
i
i. (5.37)
The constants λ
i
are called Schmidt coefficients, and are non-negative real
numbers satisfying
P
i
λ
2
i
= 1. The number of terms in the expansion is known
96 Introduction to Quantum Physics and Information Processing
as the Schmidt number. While such an expansion may not in general be
unique, the Schmidt number is unique for a given state.
Proof. The Schmidt decomposition theorem (5.37) can be proved by simple
results in linear algebra.
Consider a general pure state in the computational basis {|i
A
i|j
B
i}:
|ψi =
X
ij
C
ij
|ii|ji.
Now the matrix C of complex numbers is a square matrix, and therefore (from
results in linear algebra) has a singular value decomposition (SVD) of the form
C = UDV where D is a diagonal matrix and U and V are unitaries. So we
can write
|ψi =
X
ij
X
k
U
ik
D
kk
V
kj
|ii|ji.
By defining D
kk
= λ
k
,
P
i
U
ik
|ii = |u
k
i,
P
j
V
kj
|ji = |v
k
i we get the form of
Equation 5.37 for |ψi.
In terms of density matrices,
ρ
AB
=
X
i
λ
2
i
|u
A
i
ihu
A
i
| ⊗ |v
B
i
ihv
B
i
|. (5.38)
If we perform partial traces on this, we will get
ρ
A
=
X
i
λ
2
i
|u
i
ihu
i
|, ρ
B
=
X
i
λ
2
i
|v
i
ihv
i
|, (5.39)
There are some important take-home points to note here:
• Both the reduced density matrices have the same eigenvalues.
• ρ could have zero eigenvalues and those terms are not present in the
expansion above. So the sets {|u
A
i
i} and {|v
B
j
i} are not bases for H
A
and H
B
, but can be extended to bases by including eigenvectors for the
zero eigenvalues.
If the composite system is in a product state, then there is obviously only
one term in the Schmidt decomposition. Thus the Schmidt number for product
states is always 1. Therefore an entangled state has Schmidt number > 1. This
is one of the first ways of quantifying entanglement.
Example 5.3.3. Let’s find the Schmidt form of some simple states:
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