This document discusses density operators and their use in quantum information and computing. It begins by introducing density operators and how they can be used to describe quantum systems whose state is not precisely known or composite systems. The key properties of density operators are that they must have a trace of 1 and be positive operators. The document then covers reduced density operators which describe subsystems by taking the partial trace. Finally, it discusses how the reduced density operator gives the correct measurement statistics for observations on a subsystem.

1. Quantum information and computing
lecture 3: Density operators
Jani-Petri Martikainen
Jani-Petri.Martikainen@helsinki.fi
http://www.helsinki.fi/˜jamartik
Department of Physical Sciences
University of Helsinki
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 1/39

2. Entanglement application:superdense coding
Simple example demonstrating the application of basic
quantum mechanics.
Alice is in possesion of two classical bits of information
and want’s to send these bits to Bob. However, she can
only send a single qubit. Is her task possible?
Answer: Yes!
Assume that Alice and Bob share a pair of qubits in an
entangled state
|00 + |11
|ψ = √ (7)
2
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 8/39

3. Entanglement application:superdense coding
Note: someone else might have prepapared the state
|ψ and just sending the qubits to Alice and Bob before
hand.
If Alice wishes to transmit 00, she does nothing to her
qubit.
If Alice wishes to transmit 01, she applies phase flip Z to
the her qubit.
If Alice wishes to transmit 10, she applies quantum NOT
gate X to the her qubit.
If Alice wishes to transmit 11, she applies iY to the her
qubit. 2 3
0 1
iY = 4 5 (8)
−1 0
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 9/39

4. Entanglement application:superdense coding
The states map according to:
|00 + |11
00 : |ψ → √ (9)
2
|00 − |11
01 : |ψ → √ (10)
2
|10 + |01
10 : |ψ → √ (11)
2
|01 − |10
11 : |ψ → √ (12)
2
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 10/39

5. Entanglement application:superdense coding
Alice then sends her qubit to Bob
The four states above are an orthonormal basis of the
2-qubit Hilbert space. (known as the Bell basis, Bell
states, and EPR pairs)
Orthogonal states can be distinguished by making an
appropriate quantum measurement
From the measured state Bob can then identify which of
the four alternatives Alice send him.
In some sense, this is delayed communication. The
qubits where entangled and in order for them to get
entangled they must have interacted in the past. The
channel capacity was already “waiting” as a resource in
the entangled state.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 11/39

6. Density operator
QM can also be formulated in terms of density operator
or density matrix not just state vector.
Suppose system is at state |ψi with probability pi . We
call {pi , |ψi } an ensemble of pure states
The density operator is defined through
ρ= pi |ψi ψi | (13)
i
Evolution: ρ → U ρU †
If the initial state was |ψi then the probability of the
outcome m is
† †
p(m|i) = ψi |Mm Mm |ψi = T r(Mm Mm |ψi ψi |) (14)
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 12/39

7. Density operator
For the ensemble the probability of m is
†
p(m) = p(m|i)pi = T r(Mm Mm ρ) (15)
i
After measurement result m: If initially |ψi then
m Mm |ψi
|ψi = (16)
†
ψi |Mm Mm |ψi
For the ensemble
†
m m Mm ρMm
ρm = p(i|m)|ψi ψi | = †
(17)
i T r(Mm Mm ρ)
since p(i|m) = p(m|i)pi /p(m)
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 13/39

8. Density operator
Pure state:We know the state exactly and ρ = |ψ ψ|.
Also, for pure states T r(ρ2 ) = 1
Otherwise a mixed state. For a mixed state T r(ρ2 ) < 1
Assume that or record for the result m was lost. We
could now have state ρm with probability p(m), but we
no longer know the value m. Such a system would be
described by a density operator
†
† Mm ρMm
ρ = p(m)ρm = T r(Mm Mm ρ) †
m m T r(Mm Mm ρ)
†
= Mm ρMm (18)
m
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 14/39

9. Density operator: requirements
For an operator ρ to be a density operator:
1. ρ must have a trace equal to one
2. ρ must be a positive operator
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 15/39

10. Density operator
Postulate 4: If each subsystem is described by a density
matrix ρi the joint state of the total system is
ρ1 ⊗ ρ2 ⊗ · · · ρn .
Density operators shine when, a) describing a quantum
system whose state is not known and b) describing
subsystems of a composite quantum system.
Eigenvalues and eigenvectors of the density matrix do
NOT have a special significance with regard too the
ensemble of quantum states represented by that
density matrix.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 16/39

11. Density operator
3 1
For example, ρ = 4 |0 0| + 4 |1 1| probability of being in
|1 1/4 and probability of being in |0 3/4 (???) Not
necessarily!
suppose 1/2 prob. for both
|a = 3/4|0 + 1/4|1 (19)
and
|b = 3/4|0 − 1/4|1 (20)
The density matrix
ρ = 1/2|a a| + 1/2|b b| = 3/4|0 0| + 1/4|1 1|
Different ensembles can give rise to the same density
matrix
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 17/39

12. Density operator
What class of emsembles gives rise to the same
density matrix?
˜
Vectors (not necessarily normalized) |ψi generate the
˜ ˜
operator ρ ≡ |ψi ψi |... connection to usual ensemble
˜i = √pi |ψi .
picture of density operators : |ψ
˜ ˜
Answer:The sets |ψi and |φi generate the same
density matrix if and only if
˜
|ψi = ˜
uij |φj , (21)
j
˜
where uij is a unitary matrix. (Pad which ever set |ψi or
˜
|φi is shorter with additional vectors having probability
0 so that both sets have the same number of elements.)
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 18/39

13. Density operator
˜
Proof: suppose |ψi = ˜
uij |φj . Then
j
˜ ˜
|ψi ψi | = ˜ ˜
uij u∗ |φj φk | (22)
ik
i ijk
= u∗ uij ˜ ˜
|φj φk | = ˜ ˜
δjk |φj φk |
ki
jk i jk
= ˜ ˜
|φj φj | (23)
j
so they generate the same operator. Conversely, suppose
˜ ˜ ˜ ˜
A = i |ψi ψi | = j |φj φj |.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 19/39

14. Density operator
Let A = k λk |k k| be a decomposition of A into
˜
orthonormal states |k with λk > 0. We wish to relate |ψi to
√
states |k ˜ ˜
˜ = λk |k and similarly |φi to states |k . Let |ψ be
˜
any vector orthonormal to space spanned by |ki , so
˜
ψ|k = 0 for all k . Therefore
0 = ψ|A|ψ = ˜ ˜
ψ|ψi ψi |ψ = ˜
| ψ|ψi |2 (24)
i i
˜ ˜
and thus ψ|ψi = 0 for all i. It follows that each |ψi is a
˜
linear combination |ψi = k cik |k .˜
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 20/39

15. Density operator
Since A = ˜ ˜
|k k| = ˜ ˜
|ψi ψi | we see that
k i
˜ ˜
|k k| = cik c∗ |k ˜
˜ l|. (25)
il
k kl i
Since operators |k ˜ are linearly independent
˜ l|
i cik c∗ = δkl . This ensures that we may append extra
il
columns to c to obtain a unitary matrix v such that
˜ ˜
|ψi = k vik |k where we have appended zero vectors to
˜
the list |k . Similarly we can find w, such that
˜ ˜ ˜ ˜
|φi = k wik |k . Thus |ψi = j uij |φi where u = vw† is
unitary.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 21/39

16. Reduced density operator
Describe subsystems by using a reduced density operator
Suppose the system is composed of A and B , then the
reduced density operator for system A is
ρA ≡ T rB (ρAB ). (26)
Above the partial trace is defined by
T rB (|a1 a2 | ⊗ |b1 b2 |) ≡ |a1 a2 |T r(|b1 b2 |) (27)
where |ai and |bi are any vectors in the respective
state spaces and T r(|b1 b2 |) = b2 |b1 as usual. Partial
trace must also be linear in its input.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 22/39

17. Reduced density operator
Reduced density matrix gives the correct measurement statistics (this justifies its use
physically)
√
Take the Bell state (|00 + |11 )/ 2 (pure state):
|00 00| + |11 00| + |00 11| + |11 11|
ρ= (28)
2
Trace out the second qubit:
T r2 (|00 00|) + T r2 (|11 00|) + T r2 (|00 11|) + T r2 (|11 11|)
ρ1 = T r2 (ρ) =
2
|0 0| 0|0 + |1 0| 1|0 + |0 1| 0|1 + |1 1| 1|1 |0 0| + |1 1|
= = = I/2
2 2
This state is a mixed state even though the composite state was pure! Hallmark of
entanglement.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 23/39

18. Reduced density operator
Why the partial trace?
Partial trace turns out to be a unique operation which
gives rise to the correct description of observable
quantities for subsystems of composite systems.
Suppose M is an observable on A and we have a
measuring device capable of realizing the
˜
measurement of M ...let M be the corresponding
measurement on the composite system AB
If the system is in state |m |ψ with |ψ some arbitrary
state on B and |m an eigenstate of M , the measuring
device must give the result m with a probability of 1.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 24/39

19. Reduced density operator
Therefore, if Pm is the projector onto the m eigenspace
of the observable M , then the corresponding projector
˜
on M must be Pm ⊗ IB ...we have
˜
M= mPm ⊗ IB = M ⊗ IB (29)
m
Then let us show that the partial trace procedure gives
the correct measurement statistics for observations on
a subsystem.
Physical consistency requires that any prescription
associating a ’state’ ρA to system A, must have the
property that measurement averages be the same as
for the whole system i.e.
˜
T r(M ρA ) = T r(M ρAB ) = T r((M ⊗ IB )ρAB ) (30)
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 25/39

20. Reduced density operator
This equation is certainly satisfied if ρA = T rB (ρAB ).
In fact, partial trace turns out to be a unique function
having this property.
Let f (·) be any map of density operators on AB to
density operators on A such that
T r(M f (ρAB )) = T r((M ⊗ IB )ρAB ) (31)
Let Mi be orthonormal basis of operators for the space
of Hermitian operators with respect to Hilbert-Schmidt
inner product (see text book page 76) (X, Y ) = T r(XY ),
then we can expand f (ρAB ) in this basis...
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 26/39

21. Reduced density operator
...
f (ρAB ) = Mi T r(Mi f (ρAB ))
i
= Mi T r((M ⊗ IB )ρAB ) (32)
i
Therefore, f is uniquely determined by Eq. (30)
Moreover, the partial trace satisfies Eq. (30), so it is the
unique function having this property.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 27/39

22. Schmidt decomposition
Suppose |ψ is a pure state of the composite system
AB . Then there exists orthormal states |iA and |iB
such that
|ψ = λi |iA |iB , (33)
i
where λi ≥ 0 are known as Schmidt coefficients and
satisfy i λ2 = 1
i
Consequence: the reduced density matrices
ρA = λ2 |iA iA |
i (34)
i
ρB = λ2 |iB iB |
i (35)
i
have the same eigenvalues.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 28/39

23. Schmidt decomposition and Purification
Schmidt number is the number of non-zero λi and in
some sense it quantifies the amount of entanglement
between systems A and B .
a state of the composite system is a product state (and
thus not entangled) if and only if it has a Schmidt
number 1.
Purification: suppose we are given a state ρA of a
quantum system A. We can introduce another system
R and a pure state |AR there, such that
ρA = T rR (|AR AR|)
This mathematical procedure is known as purification
and enables us to associate pure states with mixed
states.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 29/39

24. EPR and Bell inequality
What is actually the difference between quantum
mechanics and the classical world? What makes
quantum mechanics non-classical?
In QM an unobserved particle does not possses
properties that exist independent of observation. For
example, a qubit does not possess definite properties of
’spin in the z-direction’, and ’spin in the x-direction’ each
of which can be revealed by performing the appropriate
measurement.
In the ’EPR-paper’ Einstein, Podolsky and Rosen
proposed a thought experiment which they believed
demonstrated the incompleteness of QM as a theory of
Nature.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 30/39

25. EPR and Bell inequality
Attempt to find ’elements of reality’ which were not
included in QM. Introduced what they claimed was a
sufficient condition for a physical property to be an
element of reality...namely that it is possible to predict
with certainty the value of that property, immediately
before measurement.
Consider, an entangled pair of qubits (spin singlet)
belonging to Alice and Bob
|01 − |10
√ (36)
2
Measure spin along − axis i.e. − · − for both spins
→
v → →
v σ
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 31/39

26. EPR and Bell inequality
No matter what we choose for − the two
→v
measurements are always opposite to one another.
That, if first qubit measurement yields +1 the second
will give −1 and vice versa.
Suppose |a and |b are the eigenstates of − · − then
→ →
v σ
|0 = α|a + β|b , |1 = γ|a + δ|b (37)
and
|01 − |10 |ab − |ba
√ = (αδ − βγ) √ (38)
2 2
But αδ − βγ is a determinant of a unitary matrix and
thus just equal to a phase factor eiθ
As if the 2nd qubit knows the result of the measurement
on the 1st
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 32/39

27. EPR and Bell inequality
Since Alice can predict the measurement result when
Bob’s qubit is measured along − , that is an ’element of
→v
reality’ and should be included in a complete physical
theory.
Standard QM does not include any fundamental
element to represent the value of − · − for all unit
→ →
v σ
vectors − .
→
v
EPR hoped for a return to a more classical view:
system can be ascribed properties which exist
independently of measurements performed.
Nature experimentally invalidates EPR, while agrees with
QM
Key: Bell’s inequality
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 33/39

28. EPR and Bell inequality
Start with a common sense analysis (a’la EPR) and
then proceed with QM analysis...find a difference...let
nature decide.
Perform a measurement outlined in the figure: Charlie
prepares two particles and sends one to Alice and one
to Bob.
Alice can choose to measure two different things which
are physical properties labelled by PQ and PR . She
doesn’t know in advance what measurement she will
conduct...decides by flipping a coin...measurement
outcome is (for simplicity) ±1
Alice’s particle has a value Q for the property PQ . This
value is assumed to be an objective property of Alice’s
particle, which is merely revealed by the measurement.
(similarly R is the value revealed by the measurement
of PR ) ˜
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/ kvanttilaskenta/ – p. 34/39

29. EPR and Bell inequality
ALICE BOB
Q=+−1 S=+−1
R=+−1 1 particle 1 particle R=+−1
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 35/39

30. EPR and Bell inequality
Likewise Bob is capable of measuring values S and T
(±1) of properties PS and PT ..also he makes the
chooses the measurement randomly
Timing is such that Alice and Bob measure at the same
time (or at least in causally disconnected manner)...so
Alice’s measurement cannot disturb Bob’s
measurement
Look at QS + RS + RT − QT = (Q + R)S + (R − Q)T
since R, Q = ±1 it follows that either (Q + R)S = 0 or
(R − Q)T = 0
In either case QS + RS + RT − QT = ±2
suppose next that p(q, r, s, t) is the prob. that before
measurements the system is in a state with Q = q ,
R = r, S = s, and T = t.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 36/39

31. EPR and Bell inequality
These probabilities may depend on what Charlie does.
Let E(·) denote the mean value of a quantity...we have
E(QS + RS + RT − QT ) = p(q, r, s, t)(qs + rs + rt − qt)
qrst
≤ p(q, r, s, t) × 2
qrst
= 2 (39)
and
X
E (QS + RS + RT − QT ) = p(q, r, s, t)qs
qrst
X X
+ p(q, r, s, t)rs + p(q, r, s, t)rt
qrst qrst
X
− p(q, r, s, t)qt
qrst
= E(QS) + E(RS) + E(RT ) − E(QT ) (40)
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 37/39

32. EPR and Bell inequality
Combine and you have an example of a Bell inequality
E(QS) + E(RS) + E(RT ) − E(QT ) ≤ 2 (41)
By repeating the measurements and comparing their
outcomes Alice and Bob can determine the left hand
side
Now lets put QM back in...Let Charlie prepare a
quantum system of two qubits
|01 − |10
√ (42)
2
Alice and Bob perform measurements Q = Z1 , √
√
S = (−Z2 − X2 )/ 2, R = X1 , and T = (Z2 − X2 )/ 2
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 38/39

33. EPR and Bell inequality
Simple calculation (show this...) shows that
√ √ √
QS = 1/ 2, RS = 1/ 2, RT = 1/ 2, and
√
QT = −1/ 2 and thus
√
QS + RS + RT − QT = 2 2 (43)
Violates the inequality and is consistent with
experiments! ...some asumptions going into the
derivation of the Bell inequality must have been wrong
However, if Alice and Bob choose (for example),
Q = Z1 , S = −Z2 , R = X1 , and T = −X2 then
E(QS) + E(RS) + E(RT ) − E(QT ) = 2 (44)
and there is no violation.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 39/39

34. EPR and Bell inequality
Questionable assumptions:
1. Physical properties PQ , PR ,PS , and PT have definite
values which exist independent of observation
(assumption of realism)
2. Alice’s measurement does not influence Bob’s
measurement (locality)
World is not locally realistic!
entanglement can be a fundamentally new resource
which goes beyond classical resources...How to exploit
it, is the key question of quantum information and
computing.
Department of Physical Sciences, University of Helsinki http://theory.physics.helsinki.fi/˜kvanttilaskenta/ – p. 40/39