Lecture notes in QI

L1 Kinematics: Qubits, Bloch sphere, quantum ensembles

Chapter: 1.1-1.3, 2.1.2 (pp. 29-31)
L2 Kinematics: Composite systems, Schmidt’s theorem, reduced density
operator and partial trace
Chapter: 2.1.1 (pp. 26-28), 2.1.2 (pp. 32-36)
L3 Dynamics: Completely positive maps, Stinespring dilation
Chapter: 2.3, 2.4.4
L4 Dynamics: Master equations (Lindblad), measurements, BB84
Chapter: 1.4, 2.3, 2.4.1, 2.4.2
L5 Entanglement: Separable and inseparable states, PPT, entanglement
measures
Chapter: 2.1.2 (pp. 36-38)
L6 Entanglement as a resource: Quantum teleportation
Multipartite entanglement: Residual entanglement, GHZ and W states
Chapter: 2.2.3
L7 Quantum non-locality: Bell’s inequalities
Signalling: No-cloning, quantum copier
Chapter: 2.2.2, 2.2.4
L8 Entropy: Shannon and von Neumann entropy, relative entropy, entan-
glement of formation
Chapter: 3.2
L9 Distance measures: Trace distance, fidelity, Uhlmann’s theorem
Chapter: 3.3

Reading instructions refer to S. Stenholm and K.A. Suominen, Quantum

Approach to Informatics (Wiley)
Alternative Reading:
- M. A. Nielsen and I. L. Chuang, Quantum computation and quantum
information (Cambridge University Press)
- J. Preskill, Lecture notes in quantum computation (Caltech)
http://www.theory.caltech.edu/people/preskill/ph229/
Lecture 1

Unit of quantum information: qubit

• The classical bit encodes information in one of two distinguishable
values, e.g., heads or tails of a coin. The explicit physical realization
of the bit does not matter. These values are therefore denoted by the
generic symbols 0 and 1.
A string of n bits can encode 2n symbols.

• The quantum-mechanical bit (qubit) encodes two distinguishable val-

ues in quantum-mechanical superpositions.
Examples: Polarization states |Hi and |V i of a photon; spin states |↑i
and |↓i of a spin- 12 particle; ground and excited states |gi and |ei of an
atom/ion/molecule.
Just as for the classical bit, the exact nature of the qubit does not
matter so we introduce the generic notation |0i and |1i that defines the
computational basis of the qubit state space.
Several qubits can be combined to form states in a 2n -dimensional
Hilbert space, in which all 2n computational basis states can be super-
posed.

• The qubit is an analog encoding (state preparation) but a digitial de-

coding (read out).
The qubit state is geometrically represented by the Bloch sphere:
a unit sphere with polar angles θ, ϕ defining the arbitrary qubit su-
perposition cos(θ/2)|0i + eiϕ sin(θ/2)|1i. Each single qubit state is in
one-to-one correspondence with a point on Bloch sphere. Distinguish-
able read-outs are given by antipodal points on the Bloch sphere.
Example: Spin- 21 states can be associated with directions in ordinary
3D-space that provides an explicit interpretation of the Bloch sphere.
Read out can be carried out in, e.g., a Stern-Gerlach setting: Different
pairs of antipodal points correspond to different orientations of the
SG-magnet.
Quantum ensembles: The density operator

• Conceptually, a mixed quantum state can be thought of as a source

that produces sometimes the state ψ1 , sometimes the state ψ2 , and
so on. The probabilities (relative frequencies) for these to happen are
p1 , p2 , . . .
• For such a source, what would be the average value of an observable
O? First, recall that if the system is described by the state |ψk i, then
the average value of O is given by the expectation value: hOiψk =
hψk |O|ψk i. It thus follows:
K
X
hOi{ψk ,pk } = p1 hψ1 |O|ψ1 i + p2 hψ2 |O|ψ2 i + . . . = pk hψk |O|ψk i,
k=1

where K is the number of different output states. Note that |ψk i need
not be mutually orthogonal, which means that K 6= dim H in general.
To proceed, we need the following identity:
hψ|O|ψi = Tr (|ψihψ|O) .
Proof: Trace is independent
 of
 basis. Choose one of the basis states to
be |ψi. We find: Tr |ψihψ|O = hψ|ψihψ|O|ψi + 0 = hψ|O|ψi.

Tr

By using the above identity and linearity of trace, we find:

K K
!
X   X
hOi{ψk ,pk } = pk Tr |ψk ihψk |O = Tr pk |ψk ihψk |O .
k=1 k=1

The sum-over-k factor is independent of the observable, thus we may

write
hOiρ = Tr (ρO)
with
K
X
ρ= pk |ψk ihψk |.
k=1
ρ is a mathematical object, the density operator, that describes all
the statistical information about the system in the case the source has
the mixed characterization {ψk , pk }.
• Properties of the density operator:

– ρ≥0.
– Tr(ρ) = 1 .
– ρ2 ≤ ρ .

Special cases:

(i) ρ is a pure state if pk = δkk0 for some given k 0 ;

1
(ii) ρ is a maximally mixed state if pk = d
for a qudit.

Note: ρ is a pure state iff ρ2 = ρ.

• The density operator of a single qubit takes the form

ρ = p0 |ψ0 ihψ0 | + p1 |ψ1 ihψ1 |,

where |ψ0 i = cos 2θ |0i + eiϕ sin 2θ |1i and |ψ1 i = −e−iϕ sin 2θ |0i + cos 2θ |1i.
In terms of Pauli operators σ = (σx , σy , σz ):

1

ρ= 1̂ + r · σ ,
2
where r = (p0 − p1 )(sin θ cos ϕ, sin θ sin ϕ, cos θ) is the Bloch vector.
Note that all states with |r| = 1 are pure. These lie on the Bloch
sphere. The interior |r| < 1 contains all non-pure states. Pure and
non-pure states form the Bloch ball. In particular, the origin |r| = 0
of the Bloch ball is the maximally mixed qubit state ρ = 12 1̂.

• Decomposition freedom: Given the mixture {ψk , pk } one can construct

uniquely the density operator ρ, but the converse is not true unless the
state is a pure state (contains a single term).
Example: For a single qubit, √ consider an equal mixing of the non-
orthogonal states |ψ± i = ± 23 |0i + 12 |1i. Geometrically this mixture
can be viewed as the intersection point between a straight lie connecting
these two points and the z axis in the Bloch ball. Thus, we may equally
well represent it as an unequal mixture of the states |0i and |1i.
Mixing theorem: [L. P. Hughston et al., Phys. Lett. A 183, 14 (1993)]
Consider the density operators ρ = K
P
PL k=1 pk |ek ihek |, hek |el i = δkl , and
σ = l=1 ql |φl ihφl |. Then, σ = ρ iff there exists a unitary L × L matrix
v such that
K
√ X √
ql |φl i = pk |ek ivkl .
k=1
√ P √
Proof: If there exists a unitary vkl such that ql |φl i = k pk |ek ivkl ,
then
X√ √ XX√ X√
σ = ql |φl ihφl | ql = pk |ek ivkl pk0 hek0 |vk∗0 l
l l k k0
X√ X †
X√
= pk pk0 |ek ihek0 | vkl vlk 0 = pk pk0 |ek ihek0 |δkk0
kk0 l kk0
X
= pk |ek ihek | = ρ.
k
√
ql
Conversely, if σ = ρ then the matrix vkl = √ hek |φl i is unitary:
pk

L L
X †
X ql 1
vkl vlk 0 = √ hek |φl ihφl |ek0 i = √ hek |ρ|ek0 i = δkk0
l=1 l=1
pk pk 0 p k pk 0

such that
K K
X √ X √ √
pk |ek ivkl = ql |ek ihek |φl i = ql |φl i
k=1 k=1

Note: If L > K then only the K × K submatrix of v is used. The

remaining L − K rows are arbitrary up to the condition that v should
be unitary.
Lecture 2

Composite systems
• A quantum system with two or more degrees of freedom that can be
separately measured = composite system.
Examples: Two distant atoms in an optical lattice, polarization + mo-
mentum of a photon, . . .

• States of composite systems are given by tensor products |ki ⊗ |li ⊗

. . ., which span a vector space HA ⊗ HB ⊗ . . ., whose dimension, by
combinatorics, is:

dim (HA ⊗ HB ⊗ . . .) = dim (HA ) · dim (HB ) · . . .

Example: For n qubits, dim (H1 ⊗ . . . ⊗ Hn ) = 2n .

These vectors can be superposed:
X
|Ψi = akl... |ki ⊗ |li ⊗ . . .
kl...

If N degrees of freedom, the system is said to be N -partite. Simpli-

fying notation:
|ki ⊗ |li ⊗ . . . = |kl . . .i.

• Bipartite case: N = 2, for which we may write:

nA X
X nB
AB
|Ψ i= akl |ki ⊗ |li,
k=1 l=1

where nA = dim HA and nB = dim HB .

Schmidt’s theorem: Pure bipartite states can always be written as:
min{nA ,nB }
X p
AB
|Ψ i= dm |Am i ⊗ |Bm i,
m=1

where hAm |An i = δmn and hBm |Bn i = δmn .

Mathematical note: Let a be an nA × nB matrix. Then there ex-
ists unitary nA × nA and nB × nB matrices u and v, respectively, and
a diagonal nA × nB matrix d ≥ 0, defining the singular value de-
composition (SVD) a = udv. The non-negative entries of d are the
singular values of a.
√
Proof: View akl as a nA × nB matrix a and set dmm = dm , where
m = 1, . . . , min{nA , nB }]. SVD of a reads
min{nA ,nB }
X p
akl = ukm dm vml ,
m=1

which implies:
nB min{n
nA X A ,nB }
X X p
AB
|Ψ i= ukm dm vml |ki ⊗ |li.
k=1 l=1 m=1

Let
nA
X
|Am i = |kiukm ,
k=1
nB
X
|Bm i = vml |li,
l=1

which yields
Xp
|ΨAB i = dm |Am i ⊗ |Bm i.
m

It remains to prove orthogonality:

X X
hAm |An i = u∗km uk0 n hk|k 0 i = u∗km ukn
kk0 k
X
= u†mk ukn = δmn ,
k
X X
∗
hBm |Bn i = vml vnl0 hl|l0 i = ∗
vml vnl
ll0 l
X †
= vnl vlm = δnm .
l


Note: Schmidt decompositions do not exist for more than two subsys-
tems (N ≥ 3) except for special states.
Example: The three-qubit W P √ |W i = |001i + |010i + |100i cannot
-state
be turned into Schmidt form k dk |Ak i ⊗ |Bk i ⊗ |Ck i; the GHZ state
|GHZi = |000i + |111i is explicitly on Schmidt form.
• Consider bipartite system A ⊗ (BC . . .) prepared in the pure state
|ΨA(BC...) i. Suppose we measure an arbitrary O on A. What is the
expectation value? We use the Schmidt form:
Xp
hOiΨA(BC...) = dk dl hAk |O|Al ih(BC . . .)k |(BC . . .)l i
kl
Xp
= dk dl hAk |O|Al iδkl
kl
X
= dk hAk |O|Ak i
k
!
X
= Tr dk |Ak ihAk |O
k
= Tr (ρA O) .

Here, X
ρA = dk |Ak ihAk |
k

is the reduced density operator. Since ρA determines all expectation

values for measurements of observables pertaining to subsystem A, it
is the state of subsystem A.
The reduced density operator can be found by computing the partial
trace:
X
ρA = TrBC... ρABC... = h(BC . . .)n |ρABC... |(BC . . .)n i.
n

Note: ρA is a non-pure state unless all dk except one vanish. This is an

important feature underlying the theory of entanglement.
Lecture 3

Completely positive maps

• Generalizes Schrödinger evolution ρ 7→ U (t, 0)ρU † (t, 0) valid for a closed
quantum system.

• A completely positive map (CPM) ρ 7→ E(ρ) relies on the following

physically/experimentally motivated axioms:

1. Tr [E(ρ)] is the probability that E occurs when ρ is the initial state.

Thus, 0 ≤ TrE(ρ) ≤ 1. With upper equality, the map is said to
be trace-preserving, or CPTP.
P  P
2. E k p k ρk = k pk E(ρk ) (linearity).

3. A ≥ 0 ⇒ E(A) ≥ 0 (positivity) and à ≥ 0 ⇒ (E ⊗ I) (Ã) ≥ 0 for

any extension of the system (complete positivity).

The third axiom ensures that remote systems cannot influence the phys-
ical nature of the map E.
The difference between a positive and a completely positive map can
be illustrated by the following example:
Example: Partial transposition
 TA
|Ak ihAk0 | ⊗ |Bl ihBl0 | = |Ak0 ihAk | ⊗ |Bl ihBl0 |

is a positive but not completely positive map. Such (non-physical)

operations play an important role in entanglement theory (Lecture 4).

• For practical applications, the following theorem is central:

Theorem: (Without proof)PE is a CPM iff there exists a set of linear
operators {Ek } such that k Ek† Ek ≤ 1̂ (with equality for CPTP) in
terms of which X
E(ρ) = Ek ρEk†
k

for all ρ of the system.

This way to write E is called the operator-sum or Kraus represen-
tation. The operators {Ek } are called Kraus operators.
Example: Amplitude damping of a single qubit.
 p   p 
ρ 7→ Ead (ρ) = |0ih0| + 1 − γ|1ih1| ρ |0ih0| + 1 − γ|1ih1|
+γ|0ih1|ρ|1ih0|, 0 ≤ γ ≤ 1.

This models ‘decay’ of a system: when γ increases towards 1, the

density operator tends to |0ih0| for all input ρ. For γ = 1, the out-
put ensemble √is described by the ‘ground state’ |0i. Kraus operators:
√
E0 = |0ih0| + 1 − γ|1ih1| and E1 = γ|0ih1|.
Stinespring dilation
• Physical model for a CPM: Consider a system s and its environment
e, formally defined as any extension of s such that s + e is a closed
system. Let {|n i}n≥0 be an orthonormal basis of He and assume the
input s + e state is:
%se = ρ ⊗ |0 ih0 |.
Let Use be the time evolution operator (s + e is a closed system) acting
on Hs ⊗ He . Thus,
† †
%se 7→ Use %se Use = Use ρ ⊗ |0 ih0 |Use .
The transformation of ρ is obtained by tracing over the environment:
X
†
ρ 7→ hn |Use |0 iρh0 |Use |n i.
n

This construction is called Stinespring dilation and defines the operator-

sum representation of a CPM with Kraus operators:
En ≡ hn |Use |0 i.
In fact, it is a CPTP:
X X †
En† En = hn |Use |0 i hn |Use |0 i
n n
X
†
= h0 |Use |n ihn |Use |0 i
n
†
= h0 |Use Use |0 i = h0 |0 i · 1̂ = 1̂.

• The preceding model has an interesting implication: Since trace is basis

independent the operator-sum representation cannot be unique.
To see this, consider the basis transformation:
X
|n i 7→ |fn i = |m iu†mn ,
m

umn being unitary. Since this is a basis transformation, the Kraus

operators Fn = hfn |Use |0 i define the same CPTP. Fn can be related to
En : X X
Fn = hfn |Use |0 i = unm hm |Use |0 i = unm En .
m m
Note the similarity with the mixing theorem.
Lecture 4

Measurements
• In the process of measurement, a measuring device interacts with the
system and outcomes mk are read out. This can be viewed as a CPTP
with the additional interpretations:
(i) The Kraus operators represent different measurement outcomes:
E k ↔ mk .
(ii) Born’s probability interpretation: Given system in state ρ,
the probability for the outcome mk is:
 
P (mk |ρ) = Tr Ek† Ek ρ .
P
CPTP implies that k P (mk |ρ) = 1. The positive operators
†
{Πk } = {Ek Ek } form a positive operator valued measure
(POVM).
(iii) Updating probability distribution given outcome mk :
ρ 7→ Ek ρEk† .
• Important special case: Projective measurements defined as Ek =
|kihk|. Here, Πk = Ek .
• Theorem: Non-orthogonal states cannot be reliably distinguished.
Proof: We prove the theorem by reductio ad absurdum: the assumption
that two non-orthogonal state |ψi and |φi can be disinguished leads to
a contradiction with quantum mechanics.
Suppose such a measurement is possible. Then there must exist Π1 and
Π2 such that hψ|Π1 |ψi = 1 and hφ|Π2 |φi = 1. Since probabilities sum
up to one, we have
p
hψ|Π2 |ψi =k Π2 |ψi k2 = 0,
√
which implies Π2 |ψi = 0. We may write |φi = a|ψi + b|ψ ⊥ i, where
|a|2 + |b|2 =
√ √ 1 and |b| < 1 (non-orthogonality assumption). Thus,
⊥
Π2 |φi = b Π2 |ψ i, which implies
hφ|Π2 |φi = |b|2 hψ ⊥ |Π2 |ψ ⊥ i ≤ |b|2 < 1,
which contradicts the assumption hφ|Π2 |φi = 1.

Application: Quantum key distribution

• The impossibility to distinguish non-orthogonal quantum states is the

underlying principle used to ensure security of quantum key distri-
bution.

• We describe the BB84 protocol [C. H. Bennett, G. Brassard, Proc.

IEEE Int. Conf. (1984)]:

- Two agents, usually named Alice and Bob, wish to share a secret
key in the form of a ordered sequence of bit values by using an
unpolarized single-photon source.
- Alice performs polarization measurements on the photons by send-
ing them one-by-one through a polarizer randomly oriented either
so as to select |Hi, |V i or √12 (|Hi ± |V i).
- Bob performs random measurements also selecting |Hi, |V i or
√1 (|Hi ± |V i) on the photons he receives from Alice.
2
- Without revealing the outcomes, Alice and Bob announce in pub-
lic the chosen sequence of measurements. They keep only those
outcomes corresponding to the same measurement.
- These outcomes define a common sequence of bit values, the secret
key.
- The states |Hi, |V i and √12 (|Hi ± |V i) are mutually non-orthogonal:
   
hH| √ (|Hi ± |V i) = hV | √ (|Hi ± |V i) = √1 .
1 1
   
 2   2  2
Thus, since a potential eavesdropper Eve cannot distinguish the
states from the two groups she cannot eavesdrop the photons sent
by Alice without introducing randomness when the photons reach
Bob. This randomness will make the secret key useless.

• The security of quantum key distribution is ensured by the physical

laws: in order for Eve to be a successful eavesdropper in the above
scheme, she must be able to distinguish non-orthogonal states, which
is forbidden by the laws of quantum mechanics.
In contrast, the security of classical schemes is based on the complexity
of the coding procedure, such as in the classically hard problem to find
prime factors of large integer numbers, used in RSA key distribution.
Lindblad equation
• Conceptual framework: At each (small) time step, the system under-
goes a CPTP. This means that the environment does not memorize
its interaction with the system. This is the essence of the Markov
approximation.

• Derivation: Let us consider evolution ρt , t ∈ [0, τ ], and divide τ in N

small time steps such that tk+1 − tk = τ /N ≡ δt. Under the Markovian
assumption, we may write the evolution as a sequence of CPTPs:

ρ0 7→ Ekδt,(k−1)δt ◦ . . . ◦ Eδt,0 (ρ0 ).

For notational simplicity we assume time-translational symmetry:

Et+δt,t ≡ Eδt .

We can turn the evolution into a differential equation for ρt . Consider

the mapping:
X
ρt 7→ ρt+δt = Eδt (ρt ) = En (δt)ρt En† (δt).
n≥0

Since Eδt→0 = I, we make the ansatz

E0 (δt) = 1̂ − i(H − iG)δt = 1̂ − iHδt − Gδt,

√
En≥1 (δt) = δtLn ,

H and G being Hermitian and Ln arbitrary. G can be found from the

normalization associated with the CPTP property:
X X
1̂ = E0† (δt)E0 (δt)+ En† (δt)En (δt) = 1̂−2Gδt+δt L†n Ln +O δt2 ,


n≥1 n≥1

which implies
1X †
G= L Ln .
2 n≥1 n
Insert this and the ansatz into the mapping ρt 7→ ρt+δt , one finds
1  X 1 †

†
ρ̇t = lim ρt+δt − ρt = −i [H, ρt ] + Ln ρt Ln − {Ln Ln , ρt } .
δt→0 δt 2
n≥1

This is the Lindblad equation. Ln are Lindblad operators. [G.

Lindblad, Comm. Math. Phys. 48, 119 (1976)]
 √
• Example: Single qubit phase damping. Let L = γσz and H = 0
(‘wide open’ system). We find:
 1   
ρ̇t = γ σz ρt σz − {σz σz , ρt } = γ σz ρt σz − ρt .
2

1
Let ρ = 2
1̂ + rt · σ , yielding:

ṙt = −2γ (xt , yt , 0)

with solution:
xt = x0 e−2γt ; yt = y0 e−2γt ; zt = z0 .
Lecture 5

Definition of quantum entanglement

• Entanglement describes the separability of quantum systems. It refers

to the preparation procedure: If a bipartite state ρAB can be prepared
by using local operations and classical communication (LOCC) on a
product state ρA ⊗ ρB , then ρAB is separable. If this is not possible,
then ρAB is entangled.
Mathematically: LOCC is a distrubtion of local CPMs:

{Ek ⊗ Fk , pk },

which when acting on the product input state ρA ⊗ ρB yields:

X X
ρA ⊗ ρB 7→ pk Ek (ρA ) ⊗ Fk (ρB ) ≡ pk ρA B
k ⊗ ρk .
k k

Thus, we arrive at the conclusion that a state ρAB is separable if there

exists a decomposition such that
X
ρAB = pk ρA B
k ⊗ ρk .
k

Otherwise ρAB is said to be entangled. Entanglement theory focuses on

the converse issue, to detect and characterize the entanglement content
of ρAB .

• Special case: The pure bipartite state |ΨAB i is entangled if it is impos-

sible to write it as a product state.
One may check this by using the Schmidt form: If two or more of the
Schmidt coefficients dk are non-zero, then |ΨAB i is entangled. Equiv-
alently: if the marginal state ρA = TrB |ΨAB ihΨAB | is non-pure, then
|ΨAB i is entangled.
The distribution of the Schmidt coefficients dk provides information
about the entanglement in the state. If dk = δkk0 , then |ΨAB i is not
entangled (product state); if dk = d−1 , d = min{nA , nB }, then |ΨAB i
is said to be maximally entangled.
Detecting quantum entanglement: PPT

• Given a bipartite state ρAB how do we determine whether it is entangled

or not? If it is a pure state it is simple: Compute one of the reduced
density operators and check whether it is pure or not. However, if ρAB
is non-pure, the task looks formidable since we have to search through
an uncountably number of decompositions.

• Entanglement detection is based on the fact that some operations are

positive but not completely positive. This can be used to formulate a
precise theorem that underlies entanglement detection:
Theorem: [M. Horodecki et al. PLA 223, 1 (1996)] A mixed bipar-
tite state ρAB is separable if for every positive map A on one of the
subsystems, the output A ⊗ IB (ρAB ) ≥ 0.

• For 2 ⊗ 2 and 2 ⊗ 3 systems, the above theorem reduces to a useful test

(positive partial transpose, PPT):

TA
ρAB separable ⇔ ρAB ≥ 0.

Note that for other dimensionalities there can be entangled states with

T
ρAB A ≥ 0.
Quantifying quantum entanglement

• Idea: Given two bipartite states ρAB and ρ̃AB , which one is ‘most’
entangled? That is,

we wish to find physically proper entanglement
AB
measures µ ρ ≥ 0 that quantify the amount of entanglement in

a given state. By ‘physically proper’ essentially means that µ ρAB
vanishes for separable states and is non-increasing under LOCC. We
describe two important proper entanglement measures:

– PPT suggests negativity [G. Vidal and R. F. Werner, PRA 65,

032314 (2002)]:

TA
AB k ρAB k −1
Ne (ρ )= ,
2
√
with the trace norm kAk= Tr A† A.
– For qubit pairs, an entanglement measure can be based on the
universal flip operation

Θ (α|0i + β|1i) = −β ∗ |0i + α∗ |1i,

which takes any input qubit state into an orthogonal state. Note
that
Θ = −iσy K,
where K is complex conjugation with respect to the computational
basis {|0i, |1i}.
To see how Θ can be used to quantify entanglement, we first con-
sider a pure two-qubit state |ΨAB i. Acting with the universal flip
on each qubit, yields:
p p
|ΨAB i = d0 |A0 i ⊗ |B0 i + d1 |A1 i ⊗ |B1 i
p p
7→ |Ψ̃AB i = d0 |A1 i ⊗ |B1 i + d1 |A0 i ⊗ |B0 i.

The distinguishability of the input and output states, as measured

by the overlap, defines the concurrence C(ΨAB ) of the state:
  p
AB  AB AB 
C(Ψ ) = hΨ̃ |Ψ i = 2 d0 d1 .
p
Since d0 + d1 = 1, we find C(ΨAB ) = 2 d0 (1 − d0 ), from which
we deduce that 0 ≤ C(ΨAB ) ≤ 1. The lower bound is attained for
product states d0 (1 − d0 ) = 0; the upper bound is attained for the
maximally entangled states d0 = 12 .
Concurrence can be generalized to mixed two-qubit states ρAB
according to the following procedure [W. K. Wootters, Phys. Rev.
Lett. 80, 2245 (1998)]:


- Define the qubit-flipped state ρ̃AB = σy ⊗ σy K ρAB σy ⊗ σy .
- Compute the eigenvalues λk of ρAB ρ̃AB . These eigenvalues
turn out to be real and positive and can be ordered λ1 ≥
. . . ≥ λ4 .
- Concurrence of ρAB takes the form:
( 4 p
)
p X
AB
C(ρ ) = max 0, λ1 − λk .
k=2

One can show that C(ρAB ) = 0 iff there exists a separable de-
composition of the density operator. We shall see later (lecture 8)
that C(ρAB ) has an operational meaning.
Lecture 6

Application: Quantum teleportation

• The key point with entanglement is that it can be used as a resource

for communicating quantum information between distant parties.
• Important communication scheme: quantum teleportation.
[Theoretical proposal: C. H. Bennett et al., Phys. Rev. Lett. 70, 1895
(1993). Experiment: D. Bouwmeester et al. Nature 390, 575 (1997).]
Alice wishes to communicate an unknown qubit state |φi to Bob. They
share a maximally entangled state, say |Φ+ i = √12 (|00i + |11i), where
the first (second) entry belongs to Alice (Bob). Thus, the full initial
state reads |Γi = |φi ⊗ |Φ+ i.
Given |Γi, Alice and Bob can teleport |φi by using LOCC:

- LO (Alice): Alice performs a Bell measurement meaning a pro-

jective measurement onto the four mutually orthogonal maximally
entangled Bell states:
1  
|Φ± i = √ |00i ± |11i ,
2
1  
|Ψ± i = √ |01i ± |10i .
2
- CC (Alice and Bob): For each measurement outcome, Alice sends
two classical bits xy of information to Bob:
if Φ+ is obtained then she sends xy = 00
if Φ− is obtained then she sends xy = 01
if Ψ+ is obtained then she sends xy = 10
if Ψ− is obtained then she sends xy = 11
- LO (Bob): For each communicated bit pair, Bob performs a uni-
tary map Uxy on his part of the original Bell pair:
if xy = 00 is received then he performs U00 = 1̂
if xy = 01 is received then he performs U01 = σz
if xy = 10 is received then he performs U10 = σx
if xy = 11 is received then he performs U11 = σy
In all four cases, Bob’s qubit ends up in the state |φi thereby
completing the teleportation.
Multipartite entanglement

• How can one characterize and quantify multipartite entanglement?

What is the meaning of genuine multipartite entanglement?

• We limit to pure states. We may write a general multipartite state as

X
|ΨABC... i = anA nB nC ... |nA nB nC . . .i,
nA nB nC ...

where |nA nB nC . . .i span the Hilbert space of the full system. One can
characterize entanglement in terms of:

- bipartite partitioning of the full state |ΨA(BC...) i, |ΨB(AC...) i, |Ψ(AB)(C...) i

etc;
- bipartite mixed state entanglement ρAB = TrC... |ΨABC... ihΨABC... |;
- SL-invariant polynomials: Let Hn = (C)⊗n and G = SL(2, C)⊗n .
I : Hn 7→ C is SL-invariant if I(gΨAB... ) = I(ΨAB... ) for all g ∈
G. SL(2, C) is the group of complex 2 × 2 matrices with unit
determinant.

• The simplest multipartite case: three qubits prepared in |ΨABC i.

Consider for instance the three-qubit state |W ABC i = √13 |001i+|010i+
|100i and the state |GHZABC i = √12 |000i + |111i . One finds:

r r √
A(BC) 2 1 A(BC)

2 2
|W i = |0i ⊗ |Ψ+ i + |1i ⊗ |00i → C W = ,
3 3 3
r r
1 1
|GHZA(BC) i = |0i ⊗ |00i + |1i ⊗ |11i → CGHZ;A(BC) = 1,
2 2
2 1
2
ρAB
W = |Ψ+ ihΨ+ | + |00ih00| → C ρAB W = ,
3 3 3
1 1
ρAB |00ih00| + |11ih11| → C ρAB


GHZ = GHZ = 0,
2 2

These numbers give apparently conflicting views on the entanglement

in these two states.

• Residual entanglement τABC is a way to combine the above concur-

rences so as to provide a measure of genuine multipartite entanglement
 in a pure tripartite state ΨABC . It is defined as [V. Coffman, J. Kundu,
W.K. Wootters Phys. Rev. A 61, 052306 (2000)]:

τ ΨABC = CA(BC)2 2 2


− CAB − CAC .

It has the following properties:

- τABC ≥ 0.
- τABC is invariant under permutation of A, B, and C.
- τABC is invariant under SL(2, C)⊗3 ; transformations that belong
to the group of complex 2 × 2 matrices A with det A = 1.
- τABC defines a SL-invariant polynomial: For |ΨABC i = |0i|ψ0 i +
|1i|ψ1 i, the function
 
ABC hψ̃0 |ψ0 i hψ̃0 |ψ1 i
I(Ψ ) = det
hψ̃1 |ψ0 i hψ̃1 |ψ1 i

is an SL-invariant degree four polynomial in the expansion coeffi-

cients hklm|ΨABC i for which τABC = |I(ΨABC )|.
ABC



• One finds: τ W ABC = 0 and
τ GHZ = 1. W ABC has bi-
AB
partite entanglement (C ρW 6= 0) but no tripartite entanglement
= 0); GHZABC has no bipartite entanglement
ABC



(τ W
(C ρAB
W =
ABC
0) but has tripartite entanglement (τ W 6= 0).

• There are two types of pure state entanglement for three qubits: W
and GHZ: single copies of states of these classes cannot be converted
into each other by means of local invertible transformations (SLOCC).

• There is no straightforward extension of these considerations to more

than three qubits. Each number of qubits has its special characters
and must therefore be treated separately.
Lecture 6

Bell’s inequalities

• The Bell inequalities [J. S. Bell, Physics 1, 195 (1964)] put restric-
tions on the strength of classical correlations. We focus on the Clauser-
Horne-Shimony-Holt (CHSH) inequality [J. F. Clauser et al., Phys.
Rev. Lett. 23, 880 (1969)], which provides a bound on the correla-
tions between two pairs of two-valued classical variables a, a0 = ±1 and
b, b0 = ±1. It reads:

|habi + hab0 i + ha0 bi − ha0 b0 i| ≤ 2.

• Derivation: The assumption a, a0 , b, b0 = ±1 implies the identity:

a(b + b0 ) + a0 (b − b0 ) = ±2.

Averaging gives:

−2 ≤ ha(b + b0 ) + a0 (b − b0 )i ≤ 2 ⇒ |ha(b + b0 ) + a0 (b − b0 )i| ≤ 2.

• We now show that quantum mechanics violate CHSH. Suppose we have

two qubits prepared in a state ρAB and distributed on Alice (A) and Bob
(B). Alice performs local projective measurements of the observables
a · σ, a0 · σ and Bob similarly of b · σ, b0 · σ; a, . . . , b0 being unit vectors
so that each observable has two possible outcomes ±1. Define the Bell
operator:

BCHSH = (a · σ) ⊗ (b · σ) + (a · σ) ⊗ (b0 · σ)
+(a0 · σ) ⊗ (b · σ) − (a0 · σ) ⊗ (b0 · σ),

which clearly has the same structure as in the CHSH inequality. In

other words, the validity of the CHSH inequality in quantum mechanics
can be tested by evaluating the expectation value (average) of BCHSH
for ρAB .
For ρAB = |Ψ− ihΨ− |, one finds:
 
hBCHSH iΨ−  = |Tr (|Ψ− ihΨ− |BCHSH )|
= |hΨ− |BCHSH |Ψ− i|
= |a · b + a · b0 + a0 · b − a0 · b0 | .
 By putting a, . . . , b0 in a plane, making angles ϕa , . . . , ϕb0 relative to
some fixed direction in this plane, we find:
  


hBCHSH iΨ−  = cos ϕa − ϕb + cos ϕa − ϕb0



+ cos ϕa0 − ϕb − cos ϕa0 − ϕb0  .

By choosing ϕa = 0, ϕb = π4 , ϕa0 = π2 , and ϕb0 = − π4 , we find:

√
 
   π π π 3π
hBCHSH iΨ−  = cos + cos + cos − cos  = 2 2 > 2.

 4 4 4 4

Thus, quantum √mechanics violates the CHSH inequality. In fact, one

can show that 2 2 is the largest possible violation of CHSH (Cirelson
bound).

• The reason why we cannot expect the CHSH inequality to hold in

quantum mechanics is that although the outomes of each of the pairwise
qubit measurements are ±1, we cannot expect the eigenvalues of BCHSH
to be ±2. This would only hold if the four terms in BCHSH had the same
eigenvectors, i.e., that they were mutually commuting. Clearly this is
not the case for the above chosen ϕa , ϕa0 , ϕb , ϕb0 . Thus, the assumption
a(b + b0 ) + a0 (b − b0 ) = ±2 is not a priori valid, basically due to the
complementarity feature of observables in quantum mechanics. The
violation of the CHSH inequality demonstrates that the underlying
assumption behind the CHSH inequality underestimates the potential
strength of correlations in quantum mechanics.

• Note: ρAB violates CHSH ⇒ ρAB is entangled. The converse does

not hold: there are entangled states that do not violate CHSH. Thus,
violation of CHSH is a stronger correlation criterion than entanglement.
Quantum cloning

• Cloning can be used for superluminal signalling.

Suppose Alice and Bob share the Bell state |Φ+ i. Alice encodes a bit
of information into a filtering measurement of either σz (bit value 0)
or σx (bit value 1). Bob task is to tell which bit value (σz or σx ) Alice
had chosen. Is this possible?
We show that it would indeed be possible if cloning was possible.
Suppose Alice choose σz and she obtains |0i. This changes Bob’s
state to |0i, which is cloned into |ψi = |0i⊗n . Suppose Alice instead
chooses σx and she obtains √12 (|0i + |1i). This changes Bob’s state to
√1 (|0i + |1i), which is cloned into |φi = n/2
2 2
1
(|0i + |1i)⊗n . Bob can
asymptotically distinguish these two cloned states:
1
|hψ|φi|2 = ,
2n
as this overlap tends to zero when the number of copies n tends to
infinity.
Now, the problem with this is that the above cloning scheme would
allow for superluminal signalling since Alice and Bob could be very far
apart. This is to say that quantum states cannot be cloned:
• Theorem: (No-cloning) Non-orthogonal quantum states cannot be cloned.
Proof: We prove the theorem by reductio ad absurdum: the assumption
that a quantum cloning device of non-orthogonal states exists leads to
a contradiction with quantum mechanics.
Suppose we have a quantum device with three slots, the data and the
target slots, as well as an ancilla system whose role is to make the
data+target+ancilla a closed system. The data slot starts out in the
unknown pure state |ψi. The target slot is prepared in some standard
pure state |0i and the ancilla in |ai.
Assume now the device can perform perfect cloning. This means that
there should exist a unitary U (since data+target+ancilla is a closed
system) such that
U (|ψi ⊗ |0i ⊗ |ai) = |ψi ⊗ |ψi ⊗ |aψ i.
Since U is by assumption a perfect cloner, it should also clone any other
state |φi in the same way, i.e.:
U (|φi ⊗ |0i ⊗ |ai) = |φi ⊗ |φi ⊗ |aφ i.
 We assume that |φi 6= |ψi and non-orthogonal, i.e., 0 < |hφ|ψi| < 1.
By taking the scalar products of the left- and right-hand sides of the
cloning transformations, we find:

(hφ| ⊗ h0| ⊗ ha|) U † U (|ψi ⊗ |0i ⊗ |ai) = hφ|ψih0|0iha|ai = hφ|ψi

= (hφ| ⊗ hφ| ⊗ haφ |) (|ψi ⊗ |ψi ⊗ |aψ i) = (hφ|ψi)2 haφ |aψ i.

Since |hφ|ψi| > 0 ⇒ hφ|ψi =

6 0, which implies:

hφ|ψihaφ |aψ i = 1 ⇒ |hφ|ψihaφ |aψ i| = 1. (1)

Since |hφ|ψi| < 1, we obtain:

|hφ|ψihaφ |aψ i| < |haφ |aψ i| ≤ 1,

which contradicts Eq. (1).

• Note: Quantum teleportation is not cloning, since Alice copy of the

unknown state |φi is lost after her Bell measurement.
Application: Quantum copying
• Even if perfect cloning is impossible, one may ask how well an unknown
state can be copied. More precisely, we wish to construct a scheme that
copies all pure input states of a single qubit with the same and largest
possible quality (fidelity). This is the idea of optimal cloning or
quantum copying.
Suppose we wish to copy the unknown state |φi = a|0i + b|1i of a data
qubit as well as possible on a target qubit. The copying quality should
be independent of |φi. This can be done with the help of an extra
ancilla qubit as follows:

- Implement a unitary transformation U acting on all three qubits

such that:
r
2 1  
U : |000i 7→ |000i − √ |011i + |101i ,
3 6
r
2 1  
U : |100i 7→ − |111i + √ |010i + |100i
3 6
with entries from left to right corresponding to the data, target,
and ancilla qubit, respectively.
- By linearity, we find:
r 
2  a  
U : |φ00i 7→ |Ψi = a|000i − b|111i − √ |011i + |101i ,
3 6
b  
+ √ |010i + |100i .
6
- The output state is unchanged when permuting the data and tar-
get qubits; thus, the reduced states ρdata and ρtarget are identical.
In other words, we have obtained two copies of the same state.
We find:
5 1 
ρdata = ρtarget = |φihφ| + 1̂ − |φihφ| ≡ ρ.
6 6
- The quality of the cloning is measured by the (squared) fidelity:
5
F 2 (|φi, ρ) = hφ|ρ|φi =
6
Thus, the quality of the cloning procedure is about 83 % inde-
pendent of the input state |φi. It can be shown that this is the
optimal fidelity for a single qubit.
Lecture 8

Shannon entropy

• Entropy measures the information-carrying capacity of a source. It

quantifies how much information is gained on average when we learn
the value of a random variable.

• The information gain for an outcome mk occuring with probability pk ,

is defined as log(1/pk ) = − log pk motivated by:

1. A very common event (high probability pk ) does not carry much

information;
2. ‘log’ ensures extensivity: log(a1 a2 ) = log a1 + log a2

Averaging the information gain leads to the Shannon entropy:

X
H({pk }) = − pk log pk
k

given the probability distribution {pk }. The extensive property can be

checked by considering an uncorrelated jointPprobability distribution
P
pk ql , with marginals {pk } and {ql } such that k pk = 1 and l ql = 1,
which yields
X
H({pk ql }) = − pk ql (log pk + log ql )
kl
X X X X
= − ql pk log pk − pk ql log ql
l k k l
= H({pk }) + H({ql }).

By convention, we take log ≡ log2 throughout, since the maximum

capacity for a single bit
 
max H(p0 , p1 ) = max − p0 log p0 − p1 log p1 = 1,
p0 ,p1 p0 ,p1

which occurs for p0 = p1 = 21 (e.g., a ‘fair’ coin). Thus, with this

convention, in a fair binary system each symbol carries one bit of in-
formation.
Note: pk log pk → 0 when pk → 0. Thus, we define 0 log 0 ≡ 0.
von Neumann and relative entropy

• The information-carrying capacity of a quantum source is fully con-

tained in the density operator. This ρ describes two sources of ran-
domness:

(i) The randomness in the preparation procedure, as captured by the

‘classical’ probabilities {pk }.
(ii) The ‘intrinsic’ randomness in the quantum states {|ψk i}, as cap-
tured by Born’s probability rule: P (mn |ψk ) = hψk |Πn |ψk i.

To take care of both these types of randomness in a quantum me-

chanical source described by a density operator ρ, one uses the von
Neumann entropy:
 
S(ρ) = −Tr ρ log ρ .

This quantity is the Shannon entropy of the eigenvalues of ρ. In par-

ticular, it follows that S(ρ) vanishes for pure states, since such states
have eigenvalues pk = δkl . But S(ρ) is more subtle than this due to the
non-commuting nature of quantum probability distributions (density
operators).
• To examine the consequences of the non-commuting feature of density
operators, the relative entropy is a useful quantity. For two arbitrary
density operators ρ1 and ρ2 of a quantum system, the relative entropy
is defined as
S(ρ1 k ρ2 ) = Tr (ρ1 log ρ1 ) − Tr (ρ1 log ρ2 ) .
Its usefulness derives from Klein’s inequality:
S(ρ1 k ρ2 ) ≥ 0
with equality iff ρ1 = ρ2 .
Roughly, this inequality can be understood from the operator version
of the inequality log(p/q) ≤ p/q − 1 ⇒ q(log p − log q) ≤ p − q, which
reads
A(log B − log A) ≤ B − A,
by putting A = ρ1 , B = ρ2 , and taking the trace.
One can derive several important results from Klein’s inequality, illus-
trating the non-commuting feature of density operators:
 P
– Consider
P an unselective projective measurement ρ →
7 k Πk ρΠk =
k |ψ iρ
k kk hψ k | ≡ ρ D with Π k = |ψk ihψ k | and ρ kk = hψk |ρ|ψk i.
Klein’s inequality implies:
X
0 ≤ S(ρ k ρD ) = Tr(ρ log ρ) − Tr(ρ|ψk ihψk | log ρkk )
k
X
= −S(ρ) − hψl |ρ|ψk ihψk |ψl i log ρkk
kl
X
= −S(ρ) − ρkk log ρkk = −S(ρ) + S(ρD )
k
⇒ S(ρ) ≤ S(ρD ).

Thus, the initial randomness in the preparation process is in-

creased by the intrinsic randomness induced when the measuring
device gets entangled with the system.
– For a bipartite state ρAB and its marginals ρA = TrB ρAB and
ρB = TrA ρAB , we have S(ρAB ) ≤ S(ρA ) + S(ρB ) (sub-additivity):

0 ≤ S(ρAB k ρA ⊗ ρB )
= Tr(ρAB log ρAB ) − Tr (ρA ⊗ ρB log [ρA ⊗ ρB ])
= Tr(ρAB log ρAB ) − Tr(ρA log ρA )TrρB − TrρA Tr(ρB log ρB )
= −S(ρAB ) + S(ρA ) + S(ρB ).

Note: The right-hand side S(ρA ) + S(ρB ) − S(ρAB ) is called the

quantum mutual information that measures correlations in bipar-
tite quantum states.
Application: Quantifying entanglement by means of entropy

• Quantum entropy can be used to quantify the amount of entanglement

in composite quantum systems. We discuss this in the bipartite case
for the resource-based measure entanglement of formation.
• Entanglement can be viewed as a resource that can be consumed (used)
to perform certain tasks (e.g., teleporting unknown quantum states).
It is natural to ask how to quantify this resource.
Entanglement of formation EF quantifies entanglement in terms of how
many copies of a given bipartite state can be formed from copies of a
maximally entangled state by using LOCC. In the two-qubit case, it
turns out that EF has an interesting relation to concurrence.
Suppose we are given m copies of a maximally entangled state:
d
1 X
|MESi = √ |Ak i ⊗ |Bk i
d k=1

with d = min{nA , nB }. We are asked to produce as many high-

fidelity copies of a given pure bipartite state |ψ AB i by using LOCC
on |MESi⊗m . If n is the number of copies of |ψ AB i that can be pro-
duced in this way, then we need roughly nEF (ψ AB ) copies of MES,
i.e., m ∼ nEF (ψ AB ), which yields the definition of entanglement of
formation:
m
EF (ψ AB ) = lim .
n→∞ n

One can prove that the optimal LOCC protocol yields:

EF (ψ AB ) = S(ρA )

with ρA = TrB |ψ AB ihψ AB | the reduced density operator.

Examples:

– We expect EF to vanish for product states, i.e., states of the form

|ψ AB i = |ψ A i⊗|ψ B i. Indeed, the reduced states of |ψ AB i are pure,
i.e., ρA = |ψ A ihψ A | and ρB = |ψ B ihψ B |, which implies EF (ψ AB ) =
0 since the von Neumann entropy of a pure state vanishes.
– For a maximally entangled state, we find
d
1X
ρA = TrB (|MESihMES|) = |Am ihAm |.
d m=1
 This implies
d
X 1 1
EF (MES) = − log = log d,
m=1
d d
which is the maximal value for qudits. For qubit-pairs (d = 2) we
find: EF (MES) = 1. One says that a maximally entangled qubit
state represents one unit of entanglement, i.e., one e-bit.
– For a qubit-pair state with Schmidt coefficients d0 and d1 :
EF (ψ AB ) = −d0 log d0 − d1 log d1 = H(d0 , d1 = 1 − d0 ) ≡ h(d0 ),
h(d0 ) = −d0 log d0 − (1 − d0 ) log(1 − d0 ) being the one-bit Shannon
entropy. From normalization d0 + d1 = 1, one finds:
q
1 + 1 − [C(ψ AB )]2
d0 = ,
2
√
where C(ψ AB ) = 2 d0 d1 is the concurrence of ψ AB and we have
assumed that d0 ≥ 12 ≥ d1 (without loss of generality). We thus
obtain:  q 
1 + 1 − [C(ψ AB )]2
EF (ψ AB ) = h  ,
2
which demonstrates a simple relation between concurrence and
entanglement of formation.
• Entanglement of formation, general case. Consider the bipartite state
ρAB . We define:
X
EF (ρAB ) = min pk EF (ψkAB ),
{pk ,ψkAB }
k

where minimum is taken over all decompositions of ρAB into pure states.
– This kind of extension of a quantity defined on pure states to
mixed states is called a convex roof construction.
– For qubit pairs one can show [W. K. Wootters, Phys. Rev. Lett.
80, 2245 (1998)]:
 q 
1 + 1 − [C(ρ AB )]2
EF (ρAB ) = h  .
2

This provides an operational meaning to concurrence for arbitrary

two-qubit states.
Lecture 9

Distance measures between probability distributions

• What does it mean to say that two items of information are similar?
And how is the information preserved in some process? These two
questions can be tackled by introducing distance measures between
quantum probability distributions as represented by density operators.

• Let us first consider the ‘classical’ case. Let {pk } = (p1 , p2 , . . . , pK )

and {qk } = (q1 , q2 , . . . , qK ) be two probability distributions over the
same index k. How similar are they? Two measures that answer this
question:

– The l1 distance:
K
1 X  
D({pk }, {qk }) = pk − qk  ,
2 k=1

which vanishes iff pk = qk .

– The fidelity: X√
F ({pk }, {qk }) = pk qk ,
k

which takes its maximal value Fmax = 1 iff pk = qk .

Trace distance

• The quantum analog of the l1 distance is the trace distance:

1  
D(ρ1 , ρ2 ) = Trρ1 − ρ2 .
2

• D(ρ1 , ρ2 ) satisfies all criteria for being a proper distance:

 
(i) Non-negative: D(ρ  1 , ρ2 ) ≥ 0.
P ρ1 − ρ2 has eigenvalues λk ≥ 0,
 
which implies Trρ1 − ρ2  = k λk ≥ 0.
(ii) Non-degenerate: D(ρ1 , ρ2 ) = 0 implies ρ2 = ρ1 . Since D(ρ1 , ρ2 ) =
1
P
2
λ
k k with λ k ≥ 0, D(ρ1 , ρ2 ) = 0 can only happen if all λk = 0,
which implies |ρ1 − ρ2 | = 0.
(iii) Symmetric: D(ρ1 , ρ2 ) = D(ρ2 , ρ1 ). Follows directly from defini-
tion.
(iv) Triangle inequality:

D(ρ1 , ρ2 ) ≤ D(ρ1 , ρ3 ) + D(ρ3 , ρ2 )

for any density operators ρ1 , ρ2 , ρ3 of the system.

Proof: ρ1 − ρ2 is an Hermitian operator with spectrum on [−1, 1].
We may thus divide it into a positive and a negative part:

ρ1 − ρ2 = V12+ − V12− with V12± ≥ 0

⇒ ρ1 − ρ2  = V12+ + V12−
 

and introduce the projection Π12 onto V12+ , i.e., V12+ = Π12 (ρ1 − ρ2 )
and V12− = (Π12 − 1̂)(ρ1 − ρ2 ). By taking the trace we find:

Tr(ρ1 − ρ2 ) = 0 = Tr(V12− − V12+ ) ⇒ TrV12− = TrV12+ .

We obtain:
1   1
Tr ρ1 − ρ2  = Tr V12+ + V12− = TrV12+


D(ρ1 , ρ2 ) =
2 2
= Tr [Π12 (ρ1 − ρ2 )] .

Since trace is linear, we may write:

D(ρ1 , ρ2 ) = Tr [Π12 (ρ1 − ρ3 )] + Tr [Π12 (ρ3 − ρ2 )] .

 Now,

Tr [Π12 (ρ1 − ρ3 )] = Tr Π12 (V13+ − V13− )

 

= Tr Π12 V13+ − Tr Π12 V13−

   

≤ Tr Π12 V13+ ≤ TrV13+ = D(ρ1 , ρ3 ),

 

where we have used Tr(ΠA) ≤ Tr(A) for any A ≥ 0. We conclude:

D(ρ1 , ρ2 ) ≤ D(ρ1 , ρ3 ) + D(ρ3 , ρ2 ).

• Physical interpretation of trace distance: We have found that

[Π(ρ1 − ρ2 )] ≤ TrD(ρ1 , ρ2 )

with equality if Π = Π12 . The left-hand side may be interpreted as the

probability difference for obtaining the POVM element Π ≥ 0, given
the state ρ1 or the state ρ2 . Trace distance is thus the maximal value
of this probability difference.

• Theorem: Trace distance is contractive under CPTP, i.e.,

D(E(ρ1 ), E(ρ2 )) ≤ D(ρ1 , ρ2 )

for any CPTP ρ 7→ E(ρ).

Proof: Same technique as when proving the triangle inequality above
and by using that TrE(ρ) = Trρ for CPTP.

• Application: The contractive nature of trace distance under CPTP can

be used to quantify the non-Markovianity of a given evolution [H.-
P. Breuer et al., Phys. Rev. Lett. 103, 042103 (2010)]. The idea is
based on the observation that any Markovian evolution can be viewed
as a composite map of near-unity CPTPs, from which follows, accord-
ing to the theorem above, that trace distance cannot increase when
a system undergoes Markovian evolution. Thus, if the trace distance
would increase for some time interval, there must be a non-Markovian
component involved in the equations of motion of the system.
Fidelity

• The fidelity for two quantum states ρ1 and ρ2 is defined as:

q 
√ √
F (ρ1 , ρ2 ) = Tr ρ1 ρ2 ρ1 ,

√ √ 2
where A is defined as A = A. One can prove that 0 ≤ F (ρ1 , ρ2 ) ≤
1 with equality in the first inequality iff ρ1 and ρ2 have orthogonal sup-
port and equality in the second inequality iff ρ2 = ρ1 .

• Fidelity is symmetric: F (ρ2 , ρ1 ) = F (ρ1 , ρ2 ). To see this, we need to

introduce the concept of purification.
Consider a density operator ρ acting on Hilbert space H. A pure state
|ψi ∈ H ⊗ Ha is a purification of ρ if

ρ = Tra |ψihψ|.

Theorem: (Uhlmann) [A. Uhlmann, Rep. Math. Phys. 9, 273 (1976)]

F (ρ1 , ρ2 ) = max |hψ1 |ψ2 i| ,

|ψ1 i,|ψ2 i

where the maximization is over all purifications |ψ1 i and |ψ2 i of ρ1 and
ρ2 , respectively.
Proof: The proof uses some results in linear algebra: for A arbitrary
and V unitary, we have:
 p p 
|Tr(AV )| = |Tr(|A|U V )| = Tr( |A| |A|U V )
 
p
≤ Tr|A|Tr(V † U † |A|V U ) = Tr|A| (2)

by using the right polar decomposition A = |A|U and Cauchy-Schwarz

for
p the Hilbert-Schmidt inner productP (A, B) = Tr(A† B), i.e., |(A, B)| ≤
|(A, A)| · |(B, B)|; for |µi = k |kki ∈ H ⊗ Ha , we have:
X X X
hµ|(A ⊗ B)|µi = hll|A ⊗ B |kki = hl|A|kihl|B|ki
l k kl
X X
T
= hk|A |lihl|B|ki = hk|AT B|ki
kl k
T
= Tr(A B) (3)
 We may write arbitrary purifications of ρ1 and ρ2 as:
√
|ψ1 i = (U ⊗ ρ1 Ua )|µi,
√
|ψ2 i = (V ⊗ ρ2 Va )|µi.
Thus, we find:
√ √
|hψ1 |ψ2 i| = hµ|U † V ⊗ Ua ρ1 ρ2 Va |µi
 
√ √ √ √

= Tr V T U ∗ Ua ρ1 ρ2 Va  = Tr Va V T U ∗ Ua ρ1 ρ2 

 

by using Eq. (3) and cyclicity of trace. Now, by using Eq. (2), we find:
√ √ √ √
q
|hψ1 |ψ2 i| ≤ Tr | ρ1 ρ2 | = Tr ρ1 ρ2 ρ1 .

Equality is obtained by setting U = Ua = V = 1̂ and Va = W † , where

√ √ √ √
ρ1 ρ2 =  ρ1 ρ2  W . This last point further shows that
F (ρ1 , ρ2 ) = max |hψ1 |ψ2 i| .
|ψ2 i

• Uhlmann’s theorem provides a physical interpretation of fidelity as the

worst case measure of distinguishability (orthogonality) of purifications
of the states ρ1 and ρ2 .
• Fidelity is not a proper distance since it does not satisfy the triangle
inequality. However, fidelity can be used to introduce a proper distance
measure in terms of the angle between states:
A(ρ1 , ρ2 ) = arccos F (ρ1 , ρ2 ).

• Monotonicity of fidelity:
F (E(ρ1 ), E(ρ2 )) ≥ F (ρ1 , ρ2 )
for any CPTP map E.
• Application: Quality Q of ideal quantum gates |ψi 7→ U |ψi for quan-
tum computation, under the factual evolution |ψihψ| 7→ E(|ψihψ|), is
naturally measured by the ‘worst-case’ fidelity:
Q ≡ min F U |ψihψ|U † , E(|ψihψ|)


|ψi
p
= min hψ|U † E(|ψihψ|)U |ψi.
|ψi