Bachelor's Thesis

Detection of bound entanglement using
semideﬁnite programming
Author: S.P.H.M. Frerix
Supervisor: Prof. dr. J.D.M. Maassen
Second reader: Dr. S.J. Brain
Radboud University
Nijmegen, the Netherlands
Bachelor’s Thesis
January 2016

Abstract
This thesis concerns with detection of bound entanglement, i.e. entangled quantum
systems of which the density matrix has positive partial transposes. Detection of bound
entanglement can be done in a number of diﬀerent ways. In this thesis we detect bound
entanglement based on the existence of PPT symmetric extensions. Testing whether
such a PPT symmetric extension does exist for a given state ρ is done using semideﬁnite
programming. The systems that we consider in this text are bipartite and tripartite
quantum systems. We see that for the considered states the criterion based on the
existence of PPT symmetric extensions performs better than the test solely based on the
partial transposes of a density matrix. Furthermore, we present some explicit
entanglement witnesses for certain considered states.
i

Contents
Abstract i
Contents ii
Preface iv
1 Preliminaries 1
1.1 Linear algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 Quantum mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3 Semideﬁnite programming . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
2 Detecting entanglement 7
2.1 PPT criterion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
2.2 Detecting bound engtanglement . . . . . . . . . . . . . . . . . . . . . . . . . 8
2.3 Multipartite bound entanglement . . . . . . . . . . . . . . . . . . . . . . . . 14
2.4 Creating a hierarchy of PPT symmetric extensions . . . . . . . . . . . . . . 15
2.5 Computational resources . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
3 Bipartite states 17
3.1 A system of two qutrits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18
3.2 A 2 ⊗ 4 qubit qudit system . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
4 Tripartite quantum states 27
4.1 Tripartite qubit system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
4.2 Another tripartite qubit system . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.3 Qubit qutrit qubit system . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.4 System of three qutrits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
ii

Preface
Quantum mechanics gives rise to a lot of fascinating and counter-intuitive concepts.
These concepts have stunned the most brilliant minds. For example, Einstein (along
with Podolsky, Rosen and many more) could not wrap his head around the concept of
entanglement. They argued that quantum mechanics was an incomplete theory by
creating a thought experiment1 which became famous as the EPR paradox [3]. I will not
go into detail about the EPR-paradox in its technical formulation, but to illustrate the
fascinating concept of entanglement I will use a slightly diﬀerent version of the paradox.
In order to understand the paradox it is important to know what is meant be the spin of
a particle. The spin of a particle can be seen of as a rotational direction around an axis
of a particle. The spin can have two directions; spin up and spin down.
Let us have two observers, say Alice and Bob. Alice lives on earth, and Bob lives on
Gliese 6742. In between Alice and Bob there is a source emitting electrons with opposite
spin. Let electron eA be travelling in the direction of Alice and eB be the electron
travelling in the direction of Bob. Quantum mechanics dictates that before a
measurement is made, the spin of an electron can be up as well as down. However, when
Alice measures the spin of eA, the spin of eB is immediately determined as well (namely,
it is the opposite spin of eA). It seems that when the measurement is made, eA sent
some information to eB about which spin eB should have. But the distance between the
electrons is larger than 50 lightyears, and special relativity dictates that no information
can travel faster than the speed of light. Therefore, the information could not have
travelled instantly between the two electrons. Einstein described this transmission of
information as “spooky action at a distance”3. He argued that there must be some
hidden variables in the system of the electrons that determines the behaviour of the
spins in advance, i.e. quantum mechanics does not completely describe the physical
reality. However, Bell’s theorem tells that there could not be such a thing as a hidden
variable theory.
Entanglement is not only fascinating, it is also very useful. Quantum entanglement
gives rise to loads of applications in quantum information theory. Applications of
quantum information theory (and especially entangled states) are for example the
quantum processor and quantum cryptography. Although these applications are today
not yet very common, they are very promising because they exceed their classical
equivalent in terms of processing power (for the processor) and safety (for the
cryptography).
For applications in need of entangled systems, it would be very handy to detect
1
Or as Einstein would say in German: gedankenexperiment
2
One of the nearest exoplanets to the earth.
3
Or as he would say in German: spukhafte Fernwirkung
iii

entanglement for quantum states in a simple manner. A quantum system is entangled if
its combined state does not factorize into a convex combination of product states of its
constituents. Testing whether such a factorization exists turns out to be NP-hard [4].
There are however some conditions that prove to be necessary (but not sufficient) for
separability. If these conditions do not hold, the state is entangled. In this text we
concern ourselves with the PPT criterion [6] and with the criterion based on PPT
symmetric extensions [1], which is a criterion that can be done efficiently in the form of a
semidefinite program.
The first chapter concerns with some basic definitions and theorems regarding
quantum mechanics and semidefinite programming. In the second chapter we briefly
discuss the PPT criterion, and we define the PPT symmetric extension of a state. We
cast the the test for existence of PPT symmetric extensions in the form of a semidefinite
program. In chapter three we discuss two bipartite quantum states. Furthermore the
entanglement detection algorithm is cast as semidefinite program for bipartite as well as
tripartite systems. The text [1] concerns with bipartite systems of dimension 3 ⊗ 3 and
4 ⊗ 4. The example given in [1] is studied, as well as some examples from other texts. In
chapter four we consider some tripartite systems.
I would like to thank my supervisor Prof. Maassen for his guidance in writing this
thesis. Besides my supervisor I would like to thank my parents for their continuous
support during my time at Radboud University Nijmegen.
Bastiaan Frerix
’s-Heerenberg, January 7, 2016
iv

1Preliminaries
In this chapter we will introduce some general definitions and theorems regarding
quantum mechanics and the mathematics used to describe it. We define what is called a
semidefinite program. It turns out that the mathematical representation of mixed
quantum states has useful similarity with the form of the semidefinite program. The
semidefinite programming formalism comes with two duality theorems, which we use in
later chapters in order to identify entanglement witnesses. During this chapter it is
assumed that n and m are positive integers.
1.1 Linear algebra
In order to describe a quantum mechanical system A it necessary to define the
appropriate mathematical space. This space is called the Hilbert space.
Definition 1.1.1. The state space of the system A is a complex inner product space
that is complete with respect to the metric induced by the inner product. Such a space
is called a Hilbert space HA. This text is only concerned with finite dimensional
Hilbert spaces.
Remark 1.1.2. Throughout this text we denote the space of self-adjoint operators
acting on H by B(H).
Linear algebra is necessary to do quantum mechanics. Operators on Hilbert spaces as
well as states can be represented by matrices. These matrices have some special
properties that are useful when we consider semidefinite programs. For the sake of
clarity and readability, some tedious proofs in this section are omitted.
Definition 1.1.3. Let A ∈ Cn×m. The conjugate transpose of the matrix A is the
matrix A† ∈ Cm×n obtained from taking the tranpose of A and the complex conjugate of
each entry, i.e.
(A†
)ij = Aji.
1

Chapter 1. Preliminaries
A matrix is called self-adjoint iff the equality A = A† holds.
Definition 1.1.4. Let A ∈ Cn×n. The matrix A is positive semidefinite (PSD) iff for
every 0 = z ∈ Cn the inequality z†Az ≥ 0 holds. If A is PSD we denote this by A 0. If
for every 0 = z ∈ Cn the inequality z†Az > 0 holds, the matrix A is positive definite
(PD), and this is denoted by A 0.
Theorem 1.1.5. Let A ∈ Cn×n be a self-adjoint matrix. The matrix A is positive
semidefinite iff the eigenvalues of A are non-negative.
Proof. Since A is self-adjoint, we can rewrite A such that A = V −1ΛV , i.e. A has an
eigendecomposition in which the rows of V form an orthonormal basis of the eigenvectors
of A. The matrix Λ is a diagonal matrix whose diagonal elements are the eigenvalues of
A. Furthermore we note that for any self-adjoint matrix the eigenvalues are real
according to the spectral theorem. We see that for any y ∈ Cn there exists a z ∈ Cn
such that y = V z. Since V is a complex unitary matrix, y† = (V z)† = z†V † = z†V −1.
We have the following equivalences for any z ∈ Cn
∀z z†
Az ≥ 0 ⇐⇒ z†
V −1
ΛV z ∀z
⇐⇒ y†
Dy ≥ 0 ∀y
⇐⇒ λi ≥ 0 ∀i,
where the last equivalency follows from the fact that D is a diagonal matrix with the
diagonal entries λi.
Definition 1.1.6. Let A ∈ Cn×n. The trace of A is defined as Tr[A] = n
i=1 Aii.
Lemma 1.1.7. Let A, B ∈ Cn×n. If A and B are positive semidefinite, then
Tr[AB] ≥ 0.
Proof. Let A, B ∈ Cn×n be positive semidefinite matrices. Using the Cholesky
decomposition [8] we find X, Y such that A = XX† and B = Y Y †. Define Q = X†Y .
Tr[AB] = Tr[XX†
Y Y †
]
= Tr[Y †
XX†
Y ]
= Tr[(X†
Y )†
(X†
Y )]
= Tr[Q†
Q] =
n
i=1,j=1
q∗
ijqij =
n
i=1,j=1
|qij|2
≥ 0,
where qij are the coefficients of the matrix Q.
Definition 1.1.8. Let A ∈ Cn×n and B ∈ Cm×m. The direct sum of A and B, A ⊕ B
is defined as:
A ⊕ B =
A 0
0 B
.
2

Lemma 1.1.9. Let A ∈ Cn×n and B ∈ Cm×m. A and B are positive semidefinite iff the
direct sum A ⊕ B ∈ C(n+m)×(n+m) is positive semidefinite.
Definition 1.1.10. Let A ∈ Cn×m and B ∈ Cp×q. The Kronecker product of A and
B, A ⊗ B is the product:
A ⊗ B =





A11B A12B . . . A1nB
A21B A22B . . . A2nB
...
...
...
...
Am1B Am2B . . . AmnB





.
Lemma 1.1.11. Let A ∈ Cn×n and B ∈ Cm×m such that A, B 0. Then the Kronecker
product A ⊗ B is positive semidefinite.
Proof. Let {λ1, . . . , λn} be the eigenvalues of A, and {ν1, . . . , νm} be the eigenvalues of
B. Since A, B are positive semidefinite, the eigenvalues λi, νi ≥ 0. The eigenvalues of
A ⊗ B are λiνj where 1 ≤ i ≤ n, 1 ≤ j ≤ m, which are clearly non-negative as well.
Counting the multiplicity of all eigenvalues it is fairly straightforward to see that we
have taken all eigenvalues into account. It follows that A ⊗ B has no negative
eigenvalues and therefore A ⊗ B is positive semidefinite.
Definition 1.1.12. Let HA, HB be Hilbert spaces. The partial trace
TrB : L(HA ⊗ HB) → L(HA) is the linear map defined by TrB[R ⊗ S] = R Tr[S] for
every R ∈ L(HA) and for every S ∈ L(HB).
Remark 1.1.13. Consider the Hilbert space HA ⊗ HB ⊗ HA. We denote the partial
trace over the third component HA by TrC.
Definition 1.1.14. Let A ∈ Cn×n. The transpose map is defined as T : Cn×n → Cn×n,
such that
(TA)ij = Aji
Definition 1.1.15. Let ρ be a density matrix corresponding to a state in HA ⊗ HB.
The partial transpose with respect to A is defined as ρTA := (T ⊗ 1B)ρ.
1.2 Quantum mechanics
By doing quantum mechanics we are studying quantum systems, for example the system
A. The system A is described by a quantum state. The state of system A can either be
pure or mixed.
Definition 1.2.1. A pure quantum state ψ corresponding to a system A is
represented by |ψ ∈ HA, i.e. a vector in the Hilbert space. This vector is also known as
the ket of ψ. To every ket there is a dual called the bra, denoted by ψ| ∈ HA
∗
, which is
a linear functional which maps a ket to a complex number.
Definition 1.2.2. A quantum state in H is described by the density matrix ρ ∈ B(H),
which has the following properties:
3

1. The density matrix is self-adjoint.
2. The density matrix is positive semidefinite.
3. The trace of a density matrix Tr[ρ] is equal to one.
Remark 1.2.3. Pure quantum states can always be expressed in the form of a density
matrix, but the converse is in general not true. The states that cannot be written as
pure state are called mixed quantum states.
Definition 1.2.4. Suppose |ψ ∈ H. The norm of |ψ is ψ = ψ | ψ .
Definition 1.2.5. Let ρ be the state of a quantum system in HA ⊗ HB. The state ρ is
separable iff it is a convex combination of product states, i.e.
ρ =
n
pn |ϕn ϕn| ⊗ |ψn ψn| (1.1)
with ϕn ∈ HA, ψn ∈ HB, pn ≥ 0 and n pn = 1.
Remark 1.2.6. For the sake of clarity we define separability for tripartite quantum
systems as well. Let ρ be the state of a system in HA ⊗ HB ⊗ HC. The state ρ is
separable iff it is a convex combination of product states, i.e.
ρ =
n
pn |ϕn ϕn| ⊗ |ψn ψn| ⊗ |θn θn| (1.2)
with ϕn ∈ HA, ψn ∈ HB, θn ∈ HC, pn ≥ 0 and n pn = 1.
Definition 1.2.7. A quantum state on a combined system is entangled iff it is not
separable.
Definition 1.2.8. Let H1 ⊗ . . . ⊗ Hn−1 ⊗ Hn be a composite Hilbert space of which the
first and last component are the same, i.e. H1 = Hn. Define the swap operator P1n as
the operator that interchanges the components of H1 with the components of Hn. That
is
P1n |i1, i2, . . . , in−1, in := |in, i2, . . . , in−1, i1
1.3 Semidefinite programming
The test for entanglement that will be discussed in the next chapter is based on
semidefinite programming. There are different (although equivalent) definitions of
semidefinite programs. In this section we give the definition that we have used for the
semidefinite program and we define some of notions belonging to semidefinite
programming. To solve semidefinite programs we use Matlab with the SeDuMi package
[7].
Definition 1.3.1. (SDP) Let c ∈ Rm and let F0, . . . , Fm ∈ Cn×n, where the matrices Fi
are self-adjoint. A semidefinite program is an optimalization over the vector x under the
following constraints:
minimize cT x
subject to F(x) 0
, (1.3)
where F(x) := F0 + m
i=1 xiFi. An SDP in this form is called a primal problem.
4

With the primal SDP (1.3) arises an interesting dual problem. These dual problems
appear to be very useful in the search for entanglement witnesses. Entanglement
witnesses are very useful maps for identifying entanglement because they are easy to use.
Definition 1.3.2. (SDD) The dual problem corresponding to the SDP (1.3) is an
optimization of the self-adjoint matrices Z ∈ Cn×n under the following constraints:
maximize −Tr[F0Z]
subject to Tr[FiZ] = ci, i = 1, . . . , m
Z 0
, (1.4)
where the c and Fi are as defined in (1.3).
The semidefinite contraints in the SDP (1.3) can be used in combination with the
semidefinite contraint that holds for density matrices. This specific property of
semidefinite programs will be used extensively for the remainder of this text. However,
as we will see, the minimized function imposed by c will be of no concern to us. The
separability test introduced in the next chapter will be a feasibility problem.
Definition 1.3.3. The SDP (1.3) is called primal feasible if there exists an x such that
F(x) 0, and strict primal feasible if there exists an x such that F(x) 0. Similarly,
the dual (1.4) is dual feasible if there exists an Z such that Tr[FiZ] = ci, Z 0, and it
is called strict dual feasible if there exist such Z that Tr[FiZ] = ci, Z 0.
Definition 1.3.4. Let Sp denote the set of all feasible solutions to the primal SDP (1.3),
i.e.
Sp := {x | F(x) 0}.
Furthermore, let Sd denote the set of all feasible solutions to the dual (1.4), that is
Sd := {Z | Z 0, Tr[FiZ] = ci ∀i}.
We define optimal primal to be x∗ such that cT x∗ = minx∈Sp cT x , and the optimal
dual Z∗ such that −Tr[F0Z∗] = maxZ∈Sd
−Tr[F0Z].
For identification of entanglement witnesses we need to introduce the duality theorems
regarding semidefinite programming. The duality theorems concern with the optimal
values of the primal and the dual problem.
Theorem 1.3.5. (Weak duality theorem) Let x ∈ Cm be a feasible solution to the SDP
(1.3) and let Z ∈ Cn×n be a feasible solution to the SDD (1.4). Then we have
cT x ≥ −Tr[ZF0].
Proof. The equalities cT x + Tr[ZF0] = m
i=1 Tr[ZxiFi] + Tr[ZF0] = Tr[ZF(x)] hold.
Invoking lemma 1.1.7 gives Tr[ZF(x)] ≥ 0, proving the theorem.
The weak duality theorem tells that the value of the primal objective function on a
feasible vector x is always greater or equal to the dual objective function evaluated on
any feasible matrix Z. However, the weak duality theorem provides more information
about feasibility of the primal problem when feasibility about the dual is assumed.
5

Remark 1.3.6. Let c = 0 for the SDP (1.3) and the SDD (1.4). Suppose that there
exists a Z such that Z 0, for all i the Tr[ZFi] = 0 (so Z is a feasible solution for the
dual (1.4)) and that Tr[ZF0] < 0. Invoking the weak duality theorem 1.3.5 with c = 0,
we ﬁnd that for any feasible x the inequality Tr[ZF0] ≥ 0 holds. Therefore, if there is
exists a Z such that Z 0, for all i the Tr[ZFi] = 0 and that Tr[ZF0] < 0, then the
primal problem (1.3) is infeasible.
Theorem 1.3.7. (Strong duality theorem) If there exists an x for the SDP (1.3) such
that F(x) 0 or there exists a Z for the SDD (1.4) such that Z 0, then the optimal
value of the primal and the optimal value of the dual objective functions are equal, i.e.
cT
x∗
= −Tr[F0Z∗
].
6

2Detecting entanglement
It has been proven that determining the existence of a convex combination as in
definition 1.2.5 is NP-hard [4]. However, there are necessary conditions for separability
that can be computed in an efficient manner. An example of such a criterion is the
positive partial transpose (PPT) criterion [6]. In the bipartite 2 ⊗ 2 and 2 ⊗ 3 case the
PPT criterion can be proven to be sufficient for separability. When we consider systems
with higher dimensions or more parties, the PPT criterion is insufficient. This indicates
that there could be states with positive partial transposes that are entangled. These
states are called bound entangled. For these bound entangled states there are stronger
criteria based on semidefinite programming, for example in [1]. We shall use the criterion
based on the existence of PPT symmetric extensions, defined in [1], and we will test
some new examples with this criterion. The text [1] concerns with bipartite quantum
states. We concern ourselves with bipartite as well as tripartite quantum states. The
generalization to three systems is given in the last section of this chapter.
2.1 PPT criterion
A first criterion to test states for entanglement can be the PPT criterion [6]. This
criterion is necessary for separability. The main advantage of the criterion is that it is
very easy to calculate, but there are a lot of states for which the PPT criterion is
inconclusive. In the following chapters we will see some examples that are entangled but
do have positive partial transposes, such that they are not detected by the PPT criterion.
Theorem 2.1.1. Let ρ be the state of a system in HA ⊗ HB. If ρ is separable, then
ρTA 0 and ρTB 0.
Proof. Let ρ be a separable state of a system in HA ⊗ HB. It is then given by a convex
combination of product states ρ = n pn |φn φn| ⊗ |ψn ψn|, where |φn ∈ HA and
|ψn ∈ HB. The partial transpose with respect to party A is the transpose with respect
7

Chapter 2. Detecting entanglement
to the terms |φn φn|. Suppose |φn = (c1, . . . , cn) with respect to the basis of HA and
c1, . . . , cn ∈ C. We see that
|φn φn|
T
=



c1c∗
1 . . . c1c∗
n
...
...
...
cnc∗
1 . . . cnc∗
n



T
=



c∗
1c1 . . . c∗
1cn
...
...
...
c∗
nc1 . . . c∗
ncn


 = φ†
n φ†
n .
Thus,
ρTA
=
n
pn φ†
n φ†
n ⊗ |ψn ψn| (2.1)
=
n
pn φ†
n ⊗ ψn φ†
n ⊗ ψn .
Since this is a convex combination of product states, the density matrix ρTA must be
positive semidefinite. The proof for the partial transpose with respect to HB is
analogue.
The PPT criterion is easy to compute and provides a powerful yet simple way to test
for entanglement. However, since it is not sufficient for separability, it proves to be
inconclusive for some states.
Definition 2.1.2. Let ρ be a state in HA ⊗ HB. We call ρ bound entangled iff ρ is
not separable and ρTA , ρTB 0.
2.2 Detecting bound engtanglement
For these bound entangled states we need to think about another criterion to test
entanglement. The text [1] provides such a criterion. This separability criterion can be
computed with semidefinite progamming. Throughout the rest of this section we assume
that ρ is extended to the party HA. For an extension to another party, the construction
of the semidefinite program is completely analogue. The separability test relies on a
criterion based on the existence of PPT symmetric extensions of a state ρ.
Definition 2.2.1. Let ˜ρ be the state of a system in HA ⊗ HB ⊗ HA. It is called a PPT
symmmetric extension of ρ if ˜ρ satisfies the following conditions:
1. Tracing out the third party returns the original state ρ, i.e. TrC[˜ρ] = ρ
2. There is symmetry between the first and the third component of the extension,
that is ˜ρ = P13 ˜ρP13.
3. The extension must be positive semidefinite, so ˜ρ 0, and the extension must
satisfy the PPT criterion, i.e. ˜ρTA , ˜ρTB 0.
Remark 2.2.2. We can interpret the creation of a PPT symmetric extension of ρ as
“adding a copy of the party HA” to the quantum system. Suppose ρ is the state of a
system in HA ⊗ HB, and consider the extension ˜ρ ∈ B(HA ⊗ HB ⊗ HA). The state ˜ρ
8

represents the state ρ with an additional copy of the party HA. If ρ is a 2 ⊗ 2 state (i.e.
a state of two qubits), the state ˜ρ would represent a state of a system in which there are
two identical qubits belonging to HA, and one qubit belonging to HB.
Theorem 2.2.3. If the density matrix ρ of a combined system has no PPT symmetric
extension, then the state ρ of the combined system is entangled.
Proof. We prove this theorem from contradiction. Let ρ be a separable (i.e. not
entangled) state of a system in HA ⊗ HB. From 1.1 it is known that ρ can be expressed
as convex combination of product states, that is:
ρ =
n
pn |ϕn ϕn| ⊗ |ψn ψn| .
Consider the extension to the first party HA:
˜ρ =
n
pn |ϕn ϕn| ⊗ |ψn ψn| ⊗ |ϕn ϕn| . (2.2)
1. The first requirement TrC[˜ρ] = n pn |ϕn ϕn| ⊗ |ψn ψn| · Tr[|ϕn ϕn|] = ρ,
where equality holds because Tr[|ϕn ϕn|] = 1.
2. The equality ˜ρ = P13 ˜ρP13 is trivial because the extension has the same components
as the first party, i.e. this holds per construction of ˜ρ.
3. First consider the positivity of ˜ρ. It is clear that for each n the projections
|ϕn ϕn| and |ψn ψn| are positive. Invoking lemma 1.1.11, which states that the
Kronecker product of positive semidefinite matrices is again positive, the state
(2.2) can be seen to be positive semidefinite as well. The matrix inequalities for the
partial transposes ˜ρTA , ˜ρTB 0 hold as well because separable states satisfy the
PPT criterion. Let ρTA be the partial transpose of ρ as in equation (2.1). The
projection |ϕn ϕn| is positive, and the Kronecker product of positive semidefinite
matrices is again positive semidefinite. Therefore ˜ρTA is positive semidefinite. The
proof for ˜ρTB 0 is analogue.
We see that for every separable state ρ there exists a PPT symmetric extension of the
state ρ. Therefore, if a state ρ has no PPT symmetric extension, the state is
entangled.
Theorem 2.2.3 provides a necessary condition for separability. Testing whether the
state ρ has a PPT symmetric extension can be cast in the form of a semidefinite
program. Semidefinite programming is a form of convex optimalization, but we use the
semidefinite program only to test whether there exists such a PPT symmetric extension.
This means that we concern ourselves with feasibility of the linear matrix inequalities,
and we are not interested in optimizing an objective function1. For the rest of this
section we take ρ to be the state of a system in HA ⊗ HB.
In order to cast the separability test based on theorem 2.2.3 in the form of a
semidefinite program it is necessary to define some basics for the program. First we
1
This is not entirely true, but we will explain this later on.
9

define the basis to which we want the semidefinite program to compute a PPT
symmetric extension. Since ρ is an element of B(HA ⊗ HB), we need a basis for the space
B(HA ⊗ HB). Let {σA
i }
d2
A
i=1 and {σB
i }
d2
B
i=1 be bases for B(HA) and B(HB), with the
following properties:
Tr[σX
i σX
j ] = αXδij and Tr[σX
i ] = δi1, (2.3)
where αX is a constant that comes with the chosen basis. It is fairly straightforward
to see that the set {σA
i ⊗ σB
j } gives rise to a basis for B(HA ⊗ HB).
Theorem 2.2.4. Let {σA
i }
d2
A
i=1 and {σB
i }
d2
B
i=1 be bases for B(HA) and B(HB) which satisfy
(2.3). The density matrix ρ corresponding to the state of a system in HA ⊗ HB can be
expressed in the chosen basis according to
d2
A
i=1
d2
B
j=1
ρij{σA
i ⊗ σB
j } where ρij = α−1
A α−1
B Tr[ρ{σA
i ⊗ σB
j }]. (2.4)
The proof of theorem 2.2.4 is rather trivial and therefore omitted. We make the
extension of ρ to its first party, that is HA. Since the extension ˜ρ is an element of
B(HA ⊗ HB ⊗ HA), it can be expressed as linear combination of the elements forming a
basis for B(HA ⊗ HB ⊗ HA). This basis is given by {σA
i ⊗ σB
j ⊗ σA
k }. The most general
expression for ˜ρ is given by:
˜ρ =
d2
A
i=1
d2
B
j=1
d2
A
k=1
˜ρijk{σA
i ⊗ σB
j ⊗ σA
k }. (2.5)
However, definition 2.2.1 imposes conditions on equation (2.5) and its coefficients ˜ρijk.
The symmetry requirement (2) between the first and third party imposes a symmetry
condition on the coefficients, that is ˜ρijk = ˜ρkji. It is straightforward to see that this
symmetry condition reduces the general form of (2.5) to
˜ρ =
d2
A
k=1
d2
B
j=1
k−1
i=1
˜ρijk{σA
i ⊗ σB
j ⊗ σA
k + σA
k ⊗ σB
j ⊗ σA
i } +
+
d2
A
k=1
d2
B
j=1
˜ρkjk{σA
k ⊗ σB
j ⊗ σA
k }. (2.6)
Besides the symmetry condition in 2.2.1 we have a restriction on the partial trace of the
extended state ˜ρ. The PPT symmetric extension makes a copy of the first party and
adds it to the system, creating a tripartite system. Tracing out the last party is
essentially the same as getting rid of the copy created for the extension. Therefore we
would like to retrieve the original system when tracing out the third party. This imposes
another condition on the coefficients of the extended system.
Theorem 2.2.5. If ˜ρ is a PPT symmetric extension of ρ, then ˜ρij1 = ρij.
10

Proof.
d2
A
i=1
d2
B
j=1
ρij{σA
i ⊗ σB
j } = ρ = TrC[˜ρ]
= TrC[
d2
A
i=1
d2
B
j=1
d2
A
k=1
˜ρijk{σA
i ⊗ σB
j ⊗ σA
k }]
=
d2
A
i=1
d2
B
j=1
˜ρij1{σA
i ⊗ σB
j }.
The last equality follows because of the second condition in (2.3).
Remark 2.2.6. Up until now we derived conditions on the coefficients of the extension
of ρ and we combined some of the matrices due to the symmetry condition. We note
that the semidefinite program (2.8) requires a vector c and self-adjoint matrices Fi. The
expression (2.6) and theorem 2.2.5 impose conditions on the matrices Fi, since these
matrices impose constraints on the feasibility of the primal semidefinite program. From
now on we take c = (0, 0, . . . , 0, 1) ∈ Rm+1.
For the definition of the semidefinite program we need the appropriate matrices
F0, F1, ..., Fm. We do this by taking the different possible combinations (i, j, k) into
account.
For k = 1 we note that the coefficients are determined by 2.2.5, which will set the
matrix G0 in the definition 1.3.1:
G0 =
d2
A
i=2
d2
B
j=1
ρij{σA
i ⊗ σB
j ⊗ σA
1 + σA
1 ⊗ σB
j ⊗ σA
i } +
+
d2
B
j=1
ρ1j{σA
1 ⊗ σB
j ⊗ σA
1 }.
Let i ≥ 2. We see that the matrices Giji in definition 1.3.1 have to be defined as
Giji = σA
i ⊗ σB
j ⊗ σA
i .
For k > i ≥ 2 the matrices Gijk have to be defined as
Gijk = {σA
i ⊗ σB
j ⊗ σA
k + σA
k ⊗ σB
j ⊗ σA
i }.
According to definition 1.3.1 the semidefinite program is given by
˜ρ = G(x) = G0 +
J
xJ GJ 0,
with the appropriate subindices J.
11

However, we have one more requirement in definition 2.2.1 that we need to take into
account. This is the third requirement, which states that the partial transposes of the
extension of ρ have to be positive semidefinite. Taking the partial transpose of the
extension can be done by taking the partial transpose of the appropriate matrix σX
i in
the basis. The condition ˜ρTA 0 is then expressed as semidefinite program by
˜ρTA
= G(x)TA
= GTA
0 +
J
xJ GTA
J 0.
In the same manner, ˜ρTB 0 yields the semidefinite program
˜ρTB
= G(x)TB
= GTB
0 +
J
xJ GTB
J 0.
These constraints have to be satisfied simultaneously, which can be expressed through
theorem 1.1.9 in the equivalency
˜ρ 0, ˜ρTA
0, ˜ρTB
0 ⇐⇒ ˜ρ ⊕ ˜ρTA
⊕ ˜ρTB
0.
Therefore, when taking the last requirement into account, we define the matrices Fi to
be F0 = G0 ⊕ GTA
0 ⊕ GTB
0 and FJ = GJ ⊕ GTA
J ⊕ GTB
J . Taking all the requirements for the
PPT symmetric extension in account we arrive at the following expression for the
constraint matrices in the semidefinite program:
F(x) = F0 +
d2
A
i=1
d2
B
j=1
(xijiFiji +
i−1
k=1
xijkFijk) 0. (2.7)
Remark 2.2.7. Note that m =
d4
Ad2
B−d2
Ad2
B
2 . This will be justified in section 2.4.
Definition 2.2.8. Let F(x) be as defined in (2.7) and let c ∈ R(m+1) such that
c = (0, 0, . . . , 0, 1). The test for the existence of a PPT symmetric extension is the
semidefinite program
minimize xm+1
subject to xm+1 1ABA + F(x) 0.
(2.8)
Remark 2.2.9. The semidefinite program (2.8) has the property of always being
feasible due to the addition of the last component in c. The modification of adding the
last xm+1 is rewarding because of three reasons:
1. Entanglement can be very easy detected. If the program 2.8 returns an xm+1 > 0,
entanglement is directly observed because the value of xm+1 is minimized over all
possible solutions for the linear matrix inequalities.
2. The proofs for the entanglement witnesses that are found become very elegant.
Remark 2.2.10. Note that the dual to (2.8) is always strict feasible (the same goes for
the primal). Indeed, the matrix Z = 1
3(2dA+dB)1ABA is a solution that is strict feasible,
due to the fact that for every i = m + 1 we have Tr[Fi] = 0. This allows us to use the
strong duality theorem.
12

Definition 2.2.11. Let ρ be an entangled state. The operator ˜Z is called an
entanglement witness for ρ iff for ρ the inequality Tr[ρ ˜Z] < 0 holds and for any
separable state ρsep the inequality Tr[ρsep
˜Z] ≥ 0 holds.
We claim that our program returns entanglement witnesses if a given state is
entangled. First we note that the dual problem to the problem (2.8) provides a matrix
solution,
Z = Z0 ⊕ ZTA
1 ⊕ ZTB
2 ,
that is the solution to the dual problem of (2.8). To see that Z provides an entanglement
witness for a state ρ, we define the maps:
Λ(ρ) = ρ ⊗ 1A/dA + P13(ρ ⊗ 1A)P13/dA − 1A ⊗ TrA[ρ] ⊗ 1A/d2
A, (2.9)
Λ∗
(Z) = TrC[Z]/dA + TrC[P13ZP13]/dA − 1A ⊗ TrAC[Z]/d2
A. (2.10)
Remark 2.2.12. These maps satisfy Tr[Λ(ρ)Z] = Tr[ρΛ∗(Z)].
Remark 2.2.13. For ρ a system in HA ⊗ HB described by the density matrix ρ, the
equality Λ(ρ) = G0 holds [1]. Furthermore, we note that Tr[GTX
0 ZTX
i ] = Tr[G0Zi] for
X = A, B and i = 1, 2.
We now formulate a very important theorem regarding entanglement witnesses. This
theorem is used extensively in the detection of entanglement and finding entanglement
witnesses.
Theorem 2.2.14. Let ρ be an entangled state. If Z = Z0 ⊕ ZTA
1 ⊕ ZTB
2 is a solution to
the dual of the semidefinite program (2.8), then the operator ˜Z = Λ∗(Z0 + Z1 + Z2) is an
entanglement witness for the state ρ.
Proof. We prove that ˜Z satisfies both conditions in definition 2.2.11.
1. Suppose ρsep is a separable state. From theorem 2.2.1 we know that there is a PPT
symmetric extension of ρsep. Therefore, the semidefinite program (2.7) returns an
xm+1 ≤ 0 in (2.8). The weak duality theorem 1.3.5 gives that
0 ≥ xm+1 ≥ −Tr[F0Z] for any dual feasible, resulting in
0 ≥ −Tr[F0Z] = −Tr[G0(Z0 + Z1 + Z2)]
= −Tr[Λ(ρsep)(Z0 + Z1 + Z2)]
= −Tr[ρsepΛ∗
(Z0 + Z1 + Z2)]
= −Tr[ρsep
˜Z].
Therefore, Tr[ρsep
˜Z] ≥ 0 for any separable state ρsep and for any feasible solution
Z to the dual of (2.8), satisfying the first inequality in definition 2.2.11.
2. For the entangled state ρ we exploit the power of the strong duality theorem 1.3.7.
Since the state ρ is entangled, the coefficient xm+1 > 0. The strong duality
theorem tells us that 0 > xm+1 = −Tr[F0Z] for the dual solution Z, implying that
0 > Tr[F0Z] = Tr[ρ ˜Z],
where the last equality follows the same way as for the separable case. Therefore,
the inequality Tr[ρ ˜Z] < 0 holds, satisfying the second condition of definition 2.2.11.
13

Since both conditions in definition 2.2.11 are satisfied, the operator ˜Z is an entanglement
witness for the state ρ.
Remark 2.2.15. Note that the solution Z∗ that is found in (2) of the above proof is
also a feasible solution to the dual of the semidefinite program for any separable state
ρsep. Therefore, we can use the weak duality theorem as we have done in (1).
The semidefinite program provides us with a set of entanglement witnesses obtained
for different states ρ. We would like to find analytic expressions for these entanglement
witnesses corresponding to the states that are tested. However, since SeDuMi solves the
semidefinite programs numerically, we obtain for each different state different numerical
witness. When considering a state with one (or more) parameter a, the entanglement
witnesses ˜Za returned by the semidefinite program are in general not the same.
Throughout this text the set of entanglement witnesses returned by the semidefinite
program for a state ρ will be denoted by EWρ := { ˜Za}, where a is the parameter
corresponding at which the witness was found. Furthermore, we denote entanglement
witnesses for which we found an analytic expression by ˜ZEW .
2.3 Multipartite bound entanglement
The semidefinite program as described in [1] and the previous section concerns with
bipartite systems. It is not hard to generalize the bipartite criterion based on the
existence of a PPT symmetric extension to the case for which the systems are
multipartite. This section concerns with some elemantary changes in the criterion for
multipartite systems. The proofs of the theorems in this section will be omitted because
they are completely analogue to the proofs for the bipartite criterion. Throughout this
section ρ is the state of a system in HA ⊗ HB ⊗ HC.
Theorem 2.3.1. Let ρ be the state of a system in HA ⊗ HB ⊗ HC. If ρ is separable, then
ρTA 0, ρTB 0 and ρTC 0.
Theorem 2.3.2. Let {σA
i ⊗ σB
j ⊗ σC
k } be a basis for B(HA ⊗ HB ⊗ HC). The density
matrix ρ corresponding to a system in HA ⊗ HB ⊗ HC can be expressed in the chosen
basis according to
d2
A
i=1
d2
B
j=1
d2
C
k=1
ρijk{σA
i ⊗ σB
j ⊗ σC
k } where ρij = α−1
A α−1
B α−1
C Tr[ρ{σA
i ⊗ σB
j ⊗ σC
k }]. (2.11)
Theorem 2.3.3. If ˜ρ is a PPT symmetric extension of ρ, the equality ˜ρijk1 = ρijk holds
for all i, j, k.
The last modification is made regarding the PPT criterion for multipartite systems.
The partial transpose with respect to the third subsystem has to be positive semidefinite
as well.
˜ρ 0, ˜ρTA
0, ˜ρTB
0, ˜ρTC
0 ⇐⇒ ˜ρ ⊕ ˜ρTA
⊕ ˜ρTB
⊕ ˜ρTC
0.
The definition of the semidefinite program for tripartite systems is straightforward.
14

Remark 2.3.4. The derivation for entanglement witnesses in multipartite states is the
same as in bipartite states. This can be seen by consideration of the maps (2.9). Note
that the properties of the second party HB are nowhere used. If we write
HB = HC1 ⊗ HC2 the maps (2.9) do not change.
Remark 2.3.5. Although the density matrix of a 2 ⊗ 4 or 4 ⊗ 2 bipartite state has the
same dimensions as a 2 ⊗ 2 ⊗ 2 system for example, the test based on the multipartite
system is stronger. This means that if a system is entangled as 2 ⊗ 4 state it is entangled
as 2 ⊗ 2 ⊗ 2 state as well, but the converse is in general not true. We have included an
example of such a state in the next chapter.
2.4 Creating a hierarchy of PPT symmetric extensions
This text concerns only with PPT symmetric extensions that make one copy of HA, but
the text [1] proposes an hierarchical structure of PPT symmetric extensions of ρ. Denote
the i-th test in the hierarchy (i.e. the extension of ρ to i copies of HA) by
˜ρ =
n
pn |ϕn ϕn| ⊗ |ψn ψn| ⊗ |ϕn ϕn|⊗i
. (2.12)
Remark 2.4.1. The PPT criterion can be seen of as the 0th test in the hierarchy.
Theorem 2.4.2. Let ρ be an entangled state. Then there exists a natural number I such
that for every i ≥ I the state ρ has no PPT symmetric extension to i copies of HA.
The proof for this theorem can be found in [1]. This theorem implies that every
entangled state is guaranteed to fail the test for existence of a PPT symmetric extension
for some finite I in the hierarchy.
2.5 Computational resources
One can think of the computational resources needed for the test based one the existence
of a PPT symmetric extension. The needed resources grow fast when considering
multipartite systems and extensions to high dimensional hilbert spaces HA. In this
section we briefly concern ourselves with computational resources needed for the
semidefinite program (2.8).
Remark 2.5.1. For the bipartite SDP (2.8) there are
d4
Ad2
B−d2
Ad2
B
2 free variables. This can
be seen by counting the matrices F1, . . . , Fm that are used in the semidefinite program
(2.8). Note that this are the matrices for which k = 1. This results in d2
Ad2
B(d2
A − 1)
different matrices. However, the symmetry constraint in the form of k ≥ i reduces the
number of matrices by a factor 2, so we see that there are m =
d4
Ad2
B−d2
Ad2
B
2 free variables.
Besides the number of free variables we are interested in the size of the matrices that
are used. For the bipartite system in the dA ⊗ dB state, we see that the matrices are
3d4
Ad2
B × 3d4
Ad2
B. When considering a 3 ⊗ 3 system the storage of the matrices needs
around 300 megabytes of ram memory, which does not give any problems. However,
when considering tripartite systems the resources needed grow quickly.
15

Remark 2.5.2. For the semidefinite program corresponding to tripartite systems there
are
d2
Ad2
Bd2
C(d2
A−1)
2 free variables. This can again be seen by counting the number of
matrices F1, . . . Fm in which the extension is expressed.
However, the number of free variables does not affect the needed ram as much as the
size of the matrices F0, . . . Fm. These matrices are 4d4
Ad2
Bd2
C × 4d4
Ad2
Bd2
C matrices, and
they use a lot ram memory. When we consider a tripartite 3 ⊗ 3 ⊗ 3 state, the storage of
the matrices need 7 gigabytes of ram memory. Therefore, we let the 3 ⊗ 3 ⊗ 3 be our
final goal.
16

3Bipartite states
Throughout the rest of the text we will concern ourselves with examples of quantum
states that are bound entangled. Some of these states are found in the literature and
some states are states that we thought of ourselves. However, all the considered
quantum states depend on a parameter. Entanglement is studied for diﬀerent values of
the parameter. We mostly study values of the parameter for which the state is bound
entangled.
In this chapter we study bipartite quantum states, i.e. combined systems with two
constituents. For example, the combined system of two qubits is a bipartite state with
dimension 2 ⊗ 2. The states that are considered are the state of a 3 ⊗ 3 system and the
state of a 2 ⊗ 4 system. The criterion based on the existence of PPT symmetric
extensions detects entanglement for almost all the considered states. The semideﬁnite
program also provides us with entanglement witnesses. Unfortunately we have not been
able to extract an explicit entanglement witness for the bipartite states that are
concerned, and we are left with numerical results. However, it turns out that numerical
entanglement witnesses do vary in terms of performance. This behaviour is also studied
in this chapter.
Remark 3.0.1. Throughout the rest of this text we write · where the matrix entries are
actually 0.
17

Chapter 3. Bipartite states
3.1 A system of two qutrits
We start by considering a bipartite qutrit system. Let ρa be a 3 ⊗ 3 system in the
Hilbert space HA ⊗ HB, and let its state be deﬁned as
ρa =
1
8a + 1















a · · · a · · · a
· a · · · · · · ·
· · a · · · · · ·
· · · a · · · · ·
a · · · a · · · a
· · · · · a · · ·
· · · · · · 1+a
2 ·
√
1−a2
2
· · · · · · · a ·
a · · · a ·
√
1−a2
2 · 1+a
2















. (3.1)
This state can be found in [5]. Note that ρa (3.1) is not a quantum state for a ∈ [0, 1],
because for a ∈ [−1, 0) the state is not positive, and for a ∈ [−∞, −1) ∪ (1, ∞] the state
is clearly not self-adjoint.
We therefore restrict ourselves to values of a ∈ [0, 1]. We note that the trace of the
density matrix ρa is equal to one, and that (3.1) is a self-adjoint matrix for values of
a ∈ [0, 1]. We try to determine entanglement for the state (3.1) for every a ∈ [0, 1].
Claim 3.1.1. The state ρa (3.1) is separable for a = 0 or a = 1.
Proof. We make the distinction between a = 0 and a = 1.
• For a = 0: Let p1 = 1, |ϕ1 = |2 ∈ HA, |ψ1 = 1√
2
(|0 + |2 ) ∈ HB. The claim
follows directly from 1.2.5, since ρ0 = p1 |ϕ1 ϕ1| ⊗ |ψ1 ψ1|.
• For a = 1, the matrix (3.1) is
ρa =
1
9














1 · · · 1 · · · 1
· 1 · · · · · · ·
· · 1 · · · · · ·
· · · 1 · · · · ·
1 · · · 1 · · · 1
· · · · · 1 · · ·
· · · · · · 1 · ·
· · · · · · · 1 ·
1 · · · 1 · · · 1














.
Let pn = 1
8 for every n ∈ {1, . . . , 8}. Furthermore, let |ϕn ∈ HA and |ψn ∈ HB be
deﬁned as follows:
|ϕn =
1
√
3



1
e
3
4
iπn
e
1
4
iπn


 , |ψn =
1
√
3



1
e−3
4
iπn
e−1
4
iπn


 . (3.2)
18

The density matrices belonging to ϕn and ψn are
|ϕn ϕn| =



1 e−3
4
iπn
e−1
4
iπn
e
3
4
iπn
1 e
2
4
iπn
e
1
4
iπn
e−2
4
iπn
1


 ,
|ψn ψn| =



1 e
3
4
iπn
e
1
4
iπn
e−3
4
iπn
1 e−2
4
iπn
e−1
4
iπn
e
2
4
iπn
1


 .
The Kronecker product |ϕn ϕn| ⊗ |ψn ψn| is the matrix

















1 e
3
4
iπn
e
1
4
iπn
e−3
4
iπn
1 (−1)
n
2 e−1
4
iπn
(−1)
n
2 1
e−3
4
iπn
1 (−1)
n
2 (−1)
n
2 e−3
4
iπn
e−5
4
iπn
(−1)n e−1
4
iπn
e−3
4
iπn
e−1
4
iπn
(−1)
n
2 1 (−1)n e−1
4
iπn
e
3
4
iπn
(−1)
n
2 e
1
4
iπn
e−1
4
iπn
e
3
4
iπn
(−1)
n
2 (−1)n 1 e
3
4
iπn
e
1
4
iπn
(−1)
n
2 e
5
4
iπn
e
3
4
iπn
1 e
3
4
iπn
e
1
4
iπn
e−3
4
iπn
1 (−1)
n
2 e−1
4
iπn
(−1)
n
2 1
(−1)
n
2 e
5
4
iπn
e
3
4
iπn
e−1
4
iπn
(−1)
n
2 1 e
1
4
iπn
(−1)n (−1)
n
2
e
1
4
iπn
(−1)n (−1)
n
2 (−1)
n
2 e
1
4
iπn
e− 1
4
iπn
1 e
3
4
iπn
e
1
4
iπn
(−1)
n
2 e
1
4
iπn
e−1
4
iπn
e−5
4
iπn
(−1)
n
2 (−1)n e−3
4
iπn
1 (−1)
n
2
1 e
3
4
iπn
e
1
4
iπn
e−3
4
iπn
1 (−1)
n
2 e−1
4
iπn
(−1)
n
2 1

















where we remember the normalization factor 1
9 and take it into account in the last
step. We immediately make the observation that the entries of the Kronecker
product that are independent of n are equal to 1, and this happens at the same
coordinates at where the entries of ρ1 (3.1) are equal to 1. Therefore, if the entries
that do depend on n will sum up to 0, we have proven separability for a = 1. The
entries of n pn |ϕn ϕn| ⊗ |ψn ψn| that are depended on n enlisted in the
following list, where we impose the requirement that these entries sum up to zero:
1. N
n=1(−1)n = 0. This equality can easily be seen to hold for any N that is an
even number.
2. N
n=1(−1)
n
2 = 0. For this equality to hold we take N
2 to be an even number.
Note that N is an even number for sure, leaving the ﬁrst equality to hold.
3. N
n=1 e
1
4
iπn
= 0. For this equality to hold take N
4 to be an even number.
Since the last term of this sum is of the form e2πiP where P is an integer, the
sum equals zero. If the sum is taken with the complex conjugates of the
terms, equality still holds due to the properties of the complex powers of e.
Furthermore, note that the property required in 2. is also satisﬁed.
4. The equalities N
n=1 e
3
4
iπn
= 0, N
n=1 e
5
4
iπn
= 0 hold for N such that N
4 is an
even number.
Therefore, we choose the smallest number such that N
4 is even, that is N = 8. We
see that for pn = 1
8 the convex combination n pn |ϕn ϕn| ⊗ |ψn ψn| equals the
state (3.1) for a = 1 and thus the state is separable for a = 1.
19

Therefore, the state ρa (3.1) is separable for a = 0 and for a = 1.
This does not yield information about states ρa for which the parameters a ∈ (0, 1).
As a first criterion of entanglement we try the PPT criterion.
Claim 3.1.2. The state ρa (3.1) has a positive partial transpose for every a ∈ [0, 1].
Proof. First we have to calculate the partial transpose of ρa.
ρTA
a =
1
8a + 1















a · · · · · · · ·
· a · a · · · · ·
· · a · · · a · ·
· a · a · · · · ·
· · · · a · · · ·
· · · · · a · a ·
· · a · · · 1+a
2 ·
√
1−a2
2
· · · · · a · a ·
· · · · · ·
√
1−a2
2 · 1+a
2















(3.3)
Note that ρTA
a = ρTB
a . By inspection of (3.3) we find the eigenvalues and corresponding
eigenvectors of ρTA :
1. The first eigenvalue λ1 = 0. The corresponding eigenvectors are:
v1 = (0, −1, 0, 1, 0, 0, 0, 0, 0),
v2 = (0, 0, 0, 0, 0, −1, 0, 1, 0),
v3 = (0, 0, −
√
1 − a2
a − 1
, 0, 0, 0,
√
1 − a2
a − 1
, 0, 1).
2. The second eigenvalue λ2 = a
8a+1. The eigenvectors corresponding to this
eigenvalue are:
v4 = (1, 0, 0, 0, 0, 0, 0, 0, 0),
v5 = (0, 0, 0, 0, 1, 0, 0, 0, 0).
3. The third eigenvalue λ3 = 2a
8a+1. The corresponding eigenvectors are:
v6 = (0, 1, 0, 1, 0, 0, 0, 0, 0),
v7 = (0, 0, 0, 0, 0, 1, 0, 1, 0).
4. The fourth eigenvalue λ4 = 2a+
√
2a2−2a+1+1
2(8a+1) . The corresponding eigenvector is:
v8 = (0, 0,
2a − 1 +
√
2a2 − 2a + 1
√
1 − a2
, 0, 0, 0,
a +
√
2a2 − 2a + 1
√
1 − a2
, 0, 1).
20

5. The fifth and last eigenvalue λ5 = 2a−
√
2a2−2a+1+1
2(8a+1) , with the corresponding
eigenvector:
v9 = (0, 0,
2a − 1 −
√
2a2 − 2a + 1
√
1 − a2
, 0, 0, 0,
a −
√
2a2 − 2a + 1
√
1 − a2
, 0, 1).
It not hard to see that for every a ∈ [0, 1] the eigenvalues λ1,...,4 are greater or equal to
zero. Moreover, we see that for every a ∈ [0, 1] the inequality 2a + 1 ≥
√
2a2 − 2a + 1
holds, implying that λ5 ≥ 0. Thus, the state ρa has positive partial transposes.
Since the PPT criterion does not detect entanglement for these states ρa of the 3 ⊗ 3
system, we need to consider a different criterion. The state ρa (3.1) is a 3 ⊗ 3 state. We
can consider determining the existence of PPT symmetric extensions of the state ρa (3.1)
to the first party HA. The semidefinite program that we used is given in (2.8). Our
computer solves the SDP in about 1.38 seconds with the SeDuMi [7] package. For
inspection we solve the semidefinite program belonging to the state (3.1) for N = 1000
different values of a ∈ [0, 1]. Before solving the SDP we expect the state to be separable
for a ∈ {0, 1}, and therefore to have a PPT symmetric extension for these values of a.
0 0.2 0.4 0.6 0.8 1
−2
−1
0
·10−3
value of a
Tr[ρa
˜Za]
Entanglement of ρa
Figure 3.1: Trace for N = 1000 different values of a ∈ [0, 1].
Figure 3.1 provides the information necessary to detect entanglement of ρa (3.1). The
program returns Tr[ρ0
˜Z0] = 7.9 · 10−14 ≥ 0 and Tr[ρ1
˜Z1] = 4.0 · 10−11 ≥ 0 as expected
since these states are separable and therefore have a PPT symmetric extension. For
every a ∈ (0, 1) the trace Tr[ρa
˜Za] < 0, indicating entanglement for these values of a.
We would like to extract an explicit uniform witness from the results returned by the
semidefinite program. It is important to note that such a witness does not exist in
21

general. We start by counting the number of entangled states detected by each of the
witnesses { ˜Za} given by the semidefinite program. These results can be seen in figure 3.2.
We see that the witnesses detecting the most states are found for a around 0.75. Out
of the 1000 states ρa the best performing witness detects 913 entangled states, which
results in a performance slightly over 91%. Unfortunately, inspection of the results does
not lead to an analytic witness. The entanglement witnesses { ˜Za} do not seem to
converge to one matrix at a specific parameter.
0 0.2 0.4 0.6 0.8 1
0
200
400
600
800
1,000
value of a
Numberofdetectionsby˜Za
Test optimality of witnesses
Figure 3.2: Performance of witnesses found by the SDP.
3.2 A 2 ⊗ 4 qubit qudit system
As a second example we consider a 2 ⊗ 4 system in the state ρb, which can be found in [2].
The state ρb is a state of a system in HA ⊗ HB. Let the density matrix of the system be
ρb =
1
7b + 1













b · · · · b · ·
· b · · · · b ·
· · b · · · · b
· · · b · · · ·
· · · · 1+b
2 · ·
√
1−b2
2
b · · · · b · ·
· b · · · · b ·
· · b ·
√
1−b2
2 · · 1+b
2













. (3.4)
Note that ρb (3.4) is not a quantum state for b ∈ [0, 1]. The state ρb is positive
semidefinite for b ∈ [0, 1] and the trace Tr[ρb] = 1. Thus ρb is a valid quantum state for
22

b ∈ [0, 1]. As with state (3.1) we try to extract as much information about entanglement
of the state as possible with the semidefinite program, but we first try to identify
separability.
Claim 3.2.1. The state ρb is separable for b = 1 and for b = 0.
Proof. We again prove this claim by separating the cases b = 0 and b = 1.
1. For b = 0, define p1 = 1, |ϕ1 = |1 ∈ HA, |ψ1 = 1√
2
(|0 + |3 ) ∈ HB. From 1.1 we
see immediately that the state ρ0 is separable, since ρ0 = p1 |ϕ1 ϕ1| ⊗ |ψ1 ψ1|.
2. Let b = 1 we define pn = 1
12 for every n ∈ {1, . . . , 12}, and we define the state
vectors |ϕn ∈ HA and |ψn ∈ HB as
|ϕn =
1
√
2
1
e
1
6
iπn , |ψn =
1
√
4





1
e−1
6
iπn
e−1
3
iπn
e−1
2
iπn





=
1
e−1
3
iπn ⊗
1
e−1
6
iπn .
The proof for separability is analogue to the proof given for the state 3.1. Note
that this state is separable as 2 ⊗ 2 ⊗ 2 system and thus it is also separable as 2 ⊗ 4
and 4 ⊗ 2 system
Thus for b = 0 or b = 1 the state ρb (3.4) is separable.
Claim 3.2.2. For every b ∈ [0, 1], the state ρb has positive partial transposes.
Proof. To prove this claim we first note that ρTA
b = ρTB
b .
ρTA
b =
1
7b + 1













b · · · · · · ·
· b · · b · · ·
· · b · · b · ·
· · · b · · b ·
· b · · 1+b
2 · ·
√
1−b2
2
· · b · · b · ·
· · · b · · b ·
· · · ·
√
1−b2
2 · · 1+b
2













. (3.5)
By inspection of ρTA
b we find the eigenvalues of the density matrix.
1. The first eigenvalue λ1 = 0, with the corresponding eigenvectors:
v1 = (0, 0, −1, 0, 0, 1, 0, 0),
v2 = (0, 0, 0, −1, 0, 0, 1, 0),
v3 = (0, −
√
1 − b2
b − 1
, 0, 0,
√
1 − b2
b − 1
, 0, 0, 1).
23

2. The second eigenvalue λ2 = b
7b+1 has the eigenvector:
v4 = (1, 0, 0, 0, 0, 0, 0, 0).
3. The third eigenvalue λ3 = 2b
7b+1 with the corresponding eigenvectors:
v5 = (0, 0, 1, 0, 0, 1, 0, 0),
v6 = (0, 0, 0, 1, 0, 0, 1, 0).
4. The fourth eigenvalue λ4 = 2b+
√
2b2−2b+1+1
2(7b+1) corresponds to the eigenvector:
v7 = (0,
2b − 1 +
√
2b2 − 2b + 1
√
1 − b2
, 0, 0,
b +
√
2b2 − 2b + 1
√
1 − b2
, 0, 0, 1).
5. The fifth and last eigenvalue λ5 = 2b−
√
2b2−2b+1+1
2(7b+1) corresponds to the eigenvector:
v8 = (0,
2b − 1 −
√
2b2 − 2b + 1
√
1 − b2
, 0, 0,
b −
√
2b2 − 2b + 1
√
1 − b2
, 0, 0, 1).
It is not hard to see that for every b ∈ [0, 1] the inequalities for the eigenvalues λ1,...,4 ≥ 0
hold. For the eigenvalue λ5 we have the inequality 2b + 1 ≥
√
2b2 − 2b + 1, implying that
λ5 ≥ 0. Since the eigenvalues λ1,...,5 ≥ 0, the state ρb (3.4) has positive semidefinite
partial transposes.
Since the partial transposes of the density matrix ρb are positive, the PPT criterion is
inconclusive. We therefore try the criterion based on semidefinite programming to detect
entanglement. Entanglement is detected for all b ∈ (0, 1) as can be seen in figure 3.3.
Matlab returns values of Tr[ρ0
˜Z0] = 9.1 · 10−16 ≥ 0 and Tr[ρ1
˜Z1] = 5.2 · 10−14 ≥ 0. This
means that the semidefinite program did find a PPT symmetric extension for b ∈ {0, 1}.
Since we have proven separability for these values of b, this is what we expected.
However, it is easy to see from figure 3.4 that for values of b ∈ (0, 1) the state ρb is
entangled. We have seen that for state ρa (3.1) the program did not come up with a
witness that detects all entangled states for a ∈ (0, 1). We are interested in the
behaviour of the witnesses found for the state ρb (3.4). These witnesses seem to behave
better than the ones found for thet state ρa (3.1), as they detect all witnesses for b close
to 1. Therefore, we inspect the witnesses that are found when the program is provided a
density matrix with b close to one. However, when considering the witnesses that are
found for b close to 1, the witnesses seem to converge to a zero map. The problem is of
course that for ˜Z to be a valid entanglement witness for the state ρb, the strict inequality
Tr[ρb
˜Z] < 0 has to hold at least for one value of b. This requirement is ofcourse not
satisfied for a zero map.
24

0 0.2 0.4 0.6 0.8 1
−2
−1.5
−1
−0.5
0
0.5
·10−3
value of b
Tr[ρb
˜Zb]
Entanglement of ρb
Figure 3.3: Trace of the witness N = 1000 diﬀerent values of b.
0 0.2 0.4 0.6 0.8 1
0
200
400
600
800
1,000
value of b
Numberofdetectionsby˜Zb
Performance of witnesses ˜Zb
0 0.2 0.4 0.6 0.8 1
−2
0
2
·10−3
value of b
Tr[ρbZb1]
Trace of witness
b1 =0
b1 =0.25
b1 =0.5
b1 =0.75
b1 =1
Figure 3.4: Identiﬁcation of analytic entanglement witnesses for ρb.
Peculiar behaviour
While the state ρb (3.4) is separable for b ∈ {0, 1} when taken as 2 ⊗ 4 state, ρb (3.4) is
clearly not separable for b = 0 when taken as 4 ⊗ 2 state. To see this, let b = 0 in (3.4)
and divide the matrix (3.4) in sixteen 2 × 2 matrix blocks, as done in (3.6). Now
consider the partial transpose with respect to the two dimensional party HB and note
25

that the lower right matrix block does not have a positive partial transpose for every b
(consider b = 0 for example). Indeed, when the state (3.4) is provided to the semideﬁnite
program as 4 ⊗ 2 state, it is detected as an entangled state for b = 0.
ρb =
1
7b + 1













b · · · · b · ·
· b · · · · b ·
· · b · · · · b
· · · b · · · ·
· · · · 1+b
2 · ·
√
1−b2
2
b · · · · b · ·
· b · · · · b ·
· · b ·
√
1−b2
2 · · 1+b
2













(3.6)
ρTB
0 =












· · · · · · · ·
· · · · · · · ·
· · · · · · · ·
· · · · · · · ·
· · · · 1
2 · · ·
· · · · · · 1
2 ·
· · · · · 1
2 · ·
· · · · · · · 1
2












26

4Tripartite quantum states
In this chapter we will discuss some results obtained by the criterion based on the
existence of a PPT symmetric extension for tripartite systems, i.e. systems in
HA ⊗ HB ⊗ HC. Tripartite systems in an entangled state are of interest in quantum
information theory because they provide exciting opportunities for new applications. For
example, there are tripartite states which are still entangled when a bipartite subsystem
is taken (that is, one party is traced out of the density matrix).
When we work with more parties it can be rewarding to consider multidimensional
parameters in the state ρ. In this chapter we consider one, two and three dimensional
parameters. For some of the considered states we extract an explicit entanglement
witness. The numerical resources that are used when considering tripartite systems grow
significantly faster than when we consider bipartite states. For example, the semidefinite
program takes about one second for the 2 ⊗ 2 ⊗ 2 system, but takes more than a minute
for states of a 3 ⊗ 3 ⊗ 3 system. We take the 3 ⊗ 3 ⊗ 3 system as our final goal because
when we consider higher dimensional systems we need more computational resources
than available at this time. Note that all the states that are considered in the chapter
are self thought states.
4.1 Tripartite qubit system
To start with the the study of tripartite quantum systems, we consider a system
consisting of three qubits, i.e. a 2 ⊗ 2 ⊗ 2 system, of which the computation is handled
by the computer pretty good. Its state is described by the following density matrix
ρa =
1
16
|GHZ GHZ| +
a
16
σ+ +
5 − a
16
σ−. (4.1)
27

Chapter 4. Tripartite quantum states
The state |GHZ 1 is defined as
|GHZ =
|000 + |111
√
2
.
The matrices σ are given by σ+ = |100 100| + |010 010| + |001 001|, and
σ− = |110 110| + |011 011| + |101 101|. Furthermore, for ρa to be a valid quantum
state we have a ∈ [0, 5]. That the density matrix ρa (4.1) is self-adjoint and positive
semidefinite for all b ∈ [0, 5] is immediately clear.
Claim 4.1.1. The state ρa (4.1) is separable for a ∈ [0.5, 4.5].
Proof. Let a ∈ [0.5, 4.5]. It is possible to rewrite (4.1) as
ρa =
1
16
|ψ+ ψ+| +
1
2
(σ+ + σ−) + (a − 0.5)σ+ + (4.5 − a)σ− . (4.2)
It is easy to see that the last two terms of equation (4.2) are separable, because they are
a combination of product states. Thus we need to consider the first two terms,
|ψ+ ψ+| + 1
2(σ+ + σ−). This part of the density matrix is again separable, as can be
seen by considering equation (1.2) with pn = 1
12 for n ∈ {1, ..., 12},
|ψn =
1
e
iπn
3
, |ϕn =
1
e
iπn
6
, |φn =
1
e
−iπn
2
. (4.3)
Combination of these results gives the desired separability of (4.1) for a ∈ [0.5, 4.5].
Claim 4.1.2. The state ρa (4.1) has positive partial transposes for
0.05 ≈
5
2
−
√
6 ≤ a ≤
5
2
+
√
6 ≈ 4.95.
Proof. It is not hard to see that the partial transposes of ρa (4.1) have the same
eigenvalues. The eigenvalues and corresponding eigenvectors of ρTA
a are:
1. The first eigenvalue λ1 = 1
32, with the corresponding eigenvectors
v1 = (1, 0, 0, 0, 0, 0, 0, 0)
v2 = (0, 0, 0, 0, 0, 0, 0, 1)
2. The second eigenvalue λ2 = 5−a
16 corresponds to the eigenvectors
v3 = (0, 0, 0, 0, 0, 1, 0, 0)
v4 = (0, 0, 0, 0, 0, 0, 1, 0)
3. The third eigenvalue λ3 = a
16 corresponds to the eigenvectors
v5 = (0, 1, 0, 0, 0, 0, 0, 0)
v6 = (0, 0, 1, 0, 0, 0, 0, 0)
1
Greenberger–Horne–Zeilinger
28

4. The fourth eigenvalue λ4 = 5+
√
4a2−20a+26
32 corresponds to the eigenvector
v7 = (0, 0, 0, 5 − 2a + 4a2 − 20a + 26, 1, 0, 0, 0)
5. The fifth eigenvalue λ5 = 5−
√
4a2−20a+26
32 corresponds to the eigenvector
v8 = (0, 0, 0, 5 − 2a − 4a2 − 20a + 26, 1, 0, 0, 0).
The eigenvalues λ1,...4 are easily seen to be positive. Furthermore, we have the
equivalency
λ5 ≥ 0 ⇐⇒ 0.05 ≈
5
2
−
√
6 ≤ a ≤
5
2
+
√
6 ≈ 4.95.
For the partial transpose with respect to B or C the eigenvalues are the same, so the
same bounds on a hold. We see that for a ∈ (5
2 −
√
6, 5
2 +
√
6), the partial transposes ρTA
a
and ρTB
a are positive semidefinite. Therefore, the PPT criterion is inconclusive.
Since the PPT criterion proves to be inconclusive for some values of a we consider the
criterion based on the existence of a PPT symmetric extension of the state ρa. Indeed,
the criterion based on semidefinite programming detects entanglement for 0 ≤ a < 0.5
and 4.5 < a ≤ 5. Inspection of the results provides an entanglement witness ˜ZEW for the
state ρa for a ∈ [0, 0.5):
˜ZEW = −2(|000 111| + |111 000|) + |001 001| +
+ |010 010| + |100 100| + |111 111| . (4.4)
Claim 4.1.3. The operator ˜ZEW (4.4) is an entanglement witness for the state ρa (4.1)
for a ∈ [0, 5).
Proof. To prove the claim we first prove that ˜ZEW (4.4) is nonnegative on separable
states. We also prove that the inequality Tr[ρa
˜ZEW ] < 0 holds.
1. Let |xyz be any separable state in HA ⊗ HB ⊗ HC. The product
xyz| ˜ZEW |xyz x | x = |x1x1y1z2|2
+ |x2y2z2x1|2
− 2x2y2z2x1x∗
1y∗
1z∗
1x∗
1
+ |x1y2z1x1|2
+ |x2y1z1x1|2
− 2x1y1z1x1x∗
2y∗
2z∗
2x∗
1
+ |x1y1z2x2|2
+ |x2y2z2x2|2
− 2x2y2z2x2x∗
1x∗
2y∗
1z∗
1
+ |x1y2z1x2|2
+ |x2y2z1x2|2
− 2x1y1z1x2x∗
2y∗
2z∗
2x∗
2
= |x1z2x∗
1y∗
1 − x1z1x∗
2y∗
2|2
+ |y2z2x∗
1x∗
2 − y1z1x∗
2x∗
2|2
+ |x1y2x∗
1z∗
1 − x1y1x∗
2z∗
2|2
+ |x∗
1x∗
2 + y∗
1z∗
1 − x2y2z2x2|2
≥ 0.
Thus, for every separable state the entanglement witness ˜ZEW (4.4) has
nonnegative trace Tr[ρa
˜ZEW ].
2. We have the inequality Tr[ ˜ZEW ] = 3
32(2a − 1) < 0 for every a ∈ [0, 0.5).
29

The operator ˜ZEW (4.4) satisfies the requirements in definition 2.2.11, and thus it is a
valid entanglement witness for a ∈ [0, 5).
Remark 4.1.4. While the operator ˜ZEW (4.4) provides an entanglement witness for
a ∈ [0, 5), it is not a valid witness for a ∈ (4.5, 5]. A witness for the latter case can be
found in an analogue way to the witness (4.4). This means that we inspect a witness
close to a > 4.5.
4.2 Another tripartite qubit system
We have come up with another state of three qubits, which is a superposition of state
that are well known. This state is represented by a mixture of the GHZ, the W and
some product states. It turns out that by mixing these states for different parameters
there are a lot of combinations for a, b such that the mixture is bound entangled. We
define the system ρa,b to be a system in HA ⊗ HB ⊗ HC, where its density matrix is given
by
ρa,b =
1
8
(|W W| + a |GHZ GHZ| + (3 − b) |011 011| +
+ b |100 100| + |101 101| + (3 − a) |110 110| . (4.5)
The W state is defined as
|W =
1
√
3
(|100 + |010 + |001 ).
The state ρa,b (4.5) is clearly self-adjoint and the trace can easily be seen to be equal
to one. For ρa,b (4.5) to be a valid quantum state it has to be positive semidefinite, so
the eigenvalues have to be nonnegative.
Claim 4.2.1. The state ρa,b (4.5) has nonnegative eigenvalues.
Proof. We calculate the eigenvalues and the corresponding eigenvectors.
1. The first eigenvalue λ1 = 0, with the corresponding eigenvectors:
v1 = (0, −1, 1, 0, 0, 0, 0, 0),
v2 = (−1, 0, 0, 0, 0, 0, 0, 1).
2. The second eigenvalue is λ2 = 1
8 corresponds to the eigenvector:
v3 = (0, 0, 0, 0, 0, 0, 1, 0).
3. The third eigenvalue is λ3 = 1
8(3 − a), which corresponds to the eigenvector:
v4 = (0, 0, 0, 0, 0, 1, 0, 0).
4. The fourth eigenvalue is λ4 = 1
8(3 − b), with the corresponding eigenvector:
v5 = (0, 0, 0, 1, 0, 0, 0, 0).
30

5. The fifth eigenvalue is λ5 = a
8 , corresponding to the eigenvector:
v6 = (1, 0, 0, 0, 0, 0, 0, 1).
6. The sixth eigenvalue is λ6 = 1
16(b + 1 −
√
9b2−6b+9
3 ), corresponding to the
eigenvector:
v7 =
1
4
(0, 1 − 3b − 9b2 − 6b + 9, 1 − 3b − 9b2 − 6b + 9, 0, 4, 0, 0, 0).
7. The seventh and last eigenvalue is λ7 = 1
16(b + 1 +
√
9b2−6b+9
3 ) with the
corresponding eigenvector:
v8 =
1
4
(0, 1 − 3b + 9b2 − 6b + 9, 1 − 3b + 9b2 − 6b + 9, 0, 4, 0, 0, 0).
These eigenvalues are clearly nonnegative for 0 ≤ a, b ≤ 3.
The partial transposes of the state (4.5) are calculated numerically because the
density matrix is not very pretty for the most values of the parameters a and b.
We consider parameters 0 ≤ a ≤ 1.5 and 0 ≤ b ≤ 3, since these parameters give a nice
illustration of different kinds of entanglement. In figure 4.1 we see a scatter plot in which
the red dots represent states for which at least one of the partial transposes is not
positive semidefinite. The blue dots represent states that are bound entangled, i.e. for
which there does not exist a PPT symmetric extension semidefinite programming. At
last, the green dots represent states for which the semidefinite program could find a PPT
symmetric extension. Therefore, for the green dots we can not draw any conclusions
about separability other than the fact that the state is PPT symmetric extendible.
As can be seen from figure 4.1, the state ρ1,2 (4.5) has a PPT symmetric extension.
Therefore we try to find an analytic expression or ρ1,2 as in equation (1.2) in order to
prove separability. Unfortunately, we did not succeed in finding such an expression. This
motivates for a test about the existence of a PPT symmetric extension that is higher in
the hierarchy.
31

0 0.2 0.4 0.6 0.8 1 1.2 1.4
0
0.5
1
1.5
2
2.5
3
value of a
valueofb
Entanglement of ρa,b
Figure 4.1: Entanglement of ρa,b (4.5) for various parameters (a, b)
32

4.3 Qubit qutrit qubit system
In this section we again consider a state with two different parameters, of which the
results can again be interpreted using scatter plots. Consider the following system in
HA ⊗ HB ⊗ HC, that is a 2 ⊗ 3 ⊗ 2 system of which its state is described by the density
matrix
ρa,b =
1
18





















1 · · · · · · · · · · 1
· 1 · · · · · · · · · ·
· · 1 · · · · · · · · ·
· · · 1 · · · · · · · ·
· · · · b · · · · · · ·
· · · · · 5 − a · · · · · ·
· · · · · · a · · · · ·
· · · · · · · 5 − b · · · ·
· · · · · · · · 1 · · ·
· · · · · · · · · 1 · ·
· · · · · · · · · · 1 ·
1 · · · · · · · · · · 1





















. (4.6)
Where a, b ∈ [0, 5] because ρa,b has to be positive semidefinite.
Claim 4.3.1. The state ρa,b (4.6) is separable for 1 ≤ a, b ≤ 4.
Proof. Let a, b ∈ [1, 4]. Rewrite ρa,b as
ρa,b = 112 + |121 000| + |000 121| + (4 − a) |021 021| + (4 − b) |101 101| +
+ (a − 1) |100 100| + (b − 1) |020 020| (4.7)
We consider the first three terms of ρa,b as written in (4.7), because the remainder of ρa,b
are simply product states which are separable per definition. The separability is proven
in the same way as for the other states. We define the following product vectors:
|ψn =
1
e
iπn
6
, |ϕn =



1
e
5iπn
6
e
iπn
12


 , |φn =
1
e
−iπn
4
. (4.8)
Furthermore, take pn = 1
24 for n ∈ {1, . . . , 24}. Now equation (1.2) gives the separability
of 112 + |121 000| + |000 121|. Note that we do not concern ourselves with
normalization constants because we can change normalization in the vectors that we
have used in (4.8).
Claim 4.3.2. The state ρa,b (4.6) has positive partial transposes if the inequalities
5
2
−
√
21
2
≤ a, b ≤
5
2
+
√
21
2
hold for a and b.
33

We omit the calculation of the eigenvalues since it is analogue to the calculation of the
eigenvalues for the state (4.1).
The state ρa,b (4.6) satisfies the PPT criterion for almost any a, b ∈ [0, 5], as can be
seen by consideration of the eigenvalues. For these values of a and b we would like to
determine entanglement of the state (4.6). It turns out that bound entanglement is
detected for quite a lot combinations of a and b. Our simulation picks N = 5000 random
points in the plane S := {(a, b) | 0 ≤ a, b ≤ 5}. Of all the states in S, 4156 (≈ 83%)
turned out to satisfy the PPT criterion. Out of the 4156 states that have positive partial
transposes, there turned out to be 846 states that are not PPT symmetric extendible,
and thus they are (bound) entangled. The test based on the PPT symmetric extension
detects more than twice as many entangled states than the test based solely on the PPT
criterion.
Figure 4.2 is a scatter plot for different values of (a, b) ∈ S. The red markers are
values of (a, b) for which the state has negative partial transpose, and thus does not
satisfy the PPT criterion. The green markers represent states of (a, b) where the
semidefinite program did find a PPT symmetric extension, and thus is inconclusive. The
blue markers represent states for (a, b) such that the semidefinite program was not
successful in computing a PPT symmetric extension, implying (bound) entanglement.
In figure 4.3 we have drawn the separating line between PPT symmetric extendible
states of (4.6) and states of (4.6) that are not extendible. This line is the line
corresponding where the trace of the state multiplied with the entanglement witness
multiplied that we have found equals zero. The witness detects the entangled states for
the parameters E := {(a, b) | b < 2 − a}.
Claim 4.3.3. The operator
˜ZEW = |001 001| + |020 020| + |100 100| +
+ |121 121| − 2 |000 121| − 2 |121 000| (4.9)
is an entanglement witness for every state ρa,b with (a, b) ∈ E.
Proof. We have to prove that Tr[ρsep
˜ZEW ] is nonnegative for all separable states,
whereas Tr[ρ ˜ZEW ] < 0 for all state ρa,b with (a, b) ∈ E.
1. Let ρ be a separable 2 ⊗ 3 ⊗ 2 state. We have that ρ can be expressed as product
state, that is ρ = |xyz . We have the inequalities
xyz| ˜ZEW |xyz x | x = |x1x1y1z2|2
+ |x2y3z2x1|2
− 2x2y3z2x1x∗
1y∗
1z∗
1x∗
1
+ |x1y3z1x1|2
+ |x2y1z1x1|2
− 2x1y1z1x1x∗
2y∗
3z∗
2x∗
1
+ |x1y1z2x2|2
+ |x2y3z2x2|2
− 2x2y3z2x2x∗
1x∗
2y∗
1z∗
1
+ |x1y3z1x2|2
+ |x2y3z1x2|2
− 2x1y1z1x2x∗
2y∗
3z∗
2x∗
2
= |x1z2x∗
1y∗
1 − x1z1x∗
2y∗
3|2
+ |y3z2x∗
1x∗
2 − y1z1x∗
2x∗
2|2
+ |x1y3x∗
1z∗
1 − x1y1x∗
2z∗
2|2
+ |x∗
1x∗
2 + y∗
1z∗
1 − x2y3z2x2|2
≥ 0.
Therefore, Tr[ρsep
˜ZEW ] ≥ 0 for any separable 2 ⊗ 3 ⊗ 2 state ρsep.
34

2. The negativity constraint is fairly straightforward to prove. We know that
Tr[ρa,b
˜ZEW ] = a + b − 2, which is clearly negative for all ρa,b with (a,b) in E.
Thus, ˜ZEW is an entanglement witness for every ρa,b with (a, b) ∈ E.
For every entangled state it is possible to find such a witness. However, we note the
nonlinear boundary of the PPT symmetric extendable states in figure 4.2. This makes it
hard to find an explicit uniform witness that detects every entangled state.
0 1 2 3 4 5
0
1
2
3
4
5
value of a
valueofb
Entanglement of ρa,b
Figure 4.2: Results of the semidefinite program for a two dimensional parameter
35

0 1 2 3 4 5
0
1
2
3
4
5
value of a
valueofb
Separation by witness ˜ZEW
Figure 4.3: Entangelment of states below the magenta line are detected by ˜ZEW
4.4 System of three qutrits
At last we consider a system of three qutrits. We see this state as our final goal mostly
because of computational resources: where our computer takes about 1.3 seconds to
solve the semidefinite program for a 3 ⊗ 3 system, the computer takes often more than a
minute to solve the program for a 3 ⊗ 3 ⊗ 3 system. Especially when considering monte
carlo simulations, we have to consider computing time as well as generating sufficient
data in order to draw significant conclusions.
Let ρ be the state of a system in HA ⊗ HB ⊗ HC. Since the system is a 3 ⊗ 3 ⊗ 3
system, its density matrix has 27 rows and columns. Therefore we do not write the
matrix down, but we define the state by coefficient. The coefficients that are not defined
are taken to be zero. We define the state ρ by the following coefficients:
ρ(1, 27) = 1, ρ(27, 1) = 1, ρ(2, 2) = 5 − a, ρ(3, 3) = a, ρ(4, 4) = 5 − b,
ρ(6, 6) = 5 − b, ρ(7, 7) = 5 − b, ρ(8, 8) = b, ρ(9, 9) = c, ρ(11, 11) = b,
ρ(13, 13) = a, ρ(15, 15) = 5 − a, ρ(16, 16) = c, ρ(17, 17) = c, ρ(18, 18) = 5 − c,
ρ(19, 19) = 5 − c, ρ(20, 20) = 5 − c, ρ(21, 21) = b, ρ(23, 23) = c, ρ(24, 24) = 5 − c,
ρ(25, 25) = 5 − a, ρ(26, 26) = a, ρ(1, 1) = 1, ρ(27, 27) = 1, ρ(5, 5) = 1,
ρ(12, 12) = 1, ρ(14, 14) = 1, ρ(22, 22) = 1. (4.10)
The matrix (4.10) does not satisfy the requirement that its trace equals one. Therefore,
we define the state ρa,b,c to be
ρa,b,c =
1
18
ρ (4.11)
36

Claim 4.4.1. The density matrix corresponding to the state ρa,b,c (4.11) is positive
semidefinite for 0 ≤ a, b, c ≤ 5, it is self-adjoint and Tr[ρa,b,c] = 1. Therefore, it is a
valid quantum state.
Proof. It is not hard to see that this matrix is positive semidefinite for 0 ≤ a, b, c ≤ 5.
The eigenvalues of (4.11) are the diagonal elements λ2 = ρ(2, 2), . . . , λ26 = ρ(26, 26), and
λ1 = 0 = λ27. Furthermore, the state is clearly self-adjoint since the coefficients of ρa,b,c
are real and the density matrix is symmetric. Per construction the equality Tr[ρ] = 1
holds. Therefore, the state is a valid quantum state.
We do not explicitly write down the partial transposes of the state (4.11) because they
are fairly easy to calculate due to the simple form of ρa,b,c. Furthermore, the state is
constructed such that all the positive partial transposes have eigenvalues depending on a
single parameter. The eigenvalues of the partial transposes are all positive for every
a, b, c ≥ 0, with exception of one eigenvalue. This eigenvalue can be seen to be
λ =
1
2
(5 − 29 − 20x + 4x2),
where x can be a, b or c. This is clearly positive for
5 −
√
21
2
≤ x ≤
5 +
√
21
2
.
For the state ρa,b,c (4.11) with the three dimensional parameter the computer takes
quite some time to solve the SDP: often more than a minute. Since we perform monte
carlo analysis we would like to generate sufficient data to draw significant conclusions.
We therefore omit the states where the state does not have a positive partial transpose
since we know that such a state is necessarily entangled. In figure 4.4 the blue dots
represent bound entangled states. For the green dots the SDP for the state ρa,b,c (4.11)
did find a PPT symmetric extension and we cannot draw conclusions about
entanglement at this point. It is clear (not necessary from the printed figure, but Matlab
allows rotation of the three dimensional plot) that the green states form a convex volume
in the cube. Furthermore, the blue states are located as we could have expected from the
lower dimensional results as in figure 4.2 and the witness (4.4).
Claim 4.4.2. The operator
˜ZEW = |000 000| − |111 000| − |222 000| + |022 022| +
+ |100 100| − |000 111| + |111 111| − |222 111| +
+ |211 211| − |000 222| − |111 222| + |222 222| (4.12)
is an entanglement witness for the state ρa,b,c as defined in (4.11).
In figure 4.5 the magenta plane separates states that have a PPT symmetric extension
from states that do not have such an extension, i.e. entangled states. This surface is
described by the plane where the trace Tr[ ˜ZEW ρa,b,c] = 2−c = 0. This means that points
below the surface have a trace less than 0, and therefore are entangled. Such a surface
can be drawn for c ≥ 4 as well. For c ≥ 4 we need another entanglement witness, but
this witness can be extracted in the same way as the witness ˜ZEW (4.12) was extracted.
37

0 1 2 3 4 5
0
2
4
0
2
4
value of a
value of b
valueofc
Entanglement of ρa,b,c
Figure 4.4: Detection of entangled states of 4.11
0 2 4 6
024
0
1
2
3
4
5
value of avalue of b
valueofc
Entanglement witness for ρa,b,c
Figure 4.5: The separating surface given by Tr[ρa,b,c
˜ZEW ] = 0.
38

Bibliography
[1] F.M. Spedalieri A.C. Doherty, P.A. Parrilo. A complete family of separability
criteria. Phys. Rev. A, 69(022308), 2004.
[2] G. Alber, Th. Beth, M. Horodecki, P. Horodecki, R. Horodecki, M. R¨otteler,
H. Weinfurter, R. Werner, and A. Zeilinger. Quantum Information. Springer, 2001.
[3] A. Einstein, B. Podolsky, and N. Rosen. Can quantum-mechanical description of
physical reality be considered complete? Phys. Rev., 47:777–780, May 1935.
[4] Leonid Gurvits. Classical deterministic complexity of edmonds’ problem and
quantum entanglement. In Proceedings of the Thirty-ﬁfth Annual ACM Symposium
on Theory of Computing, STOC ’03, pages 10–19, New York, NY, USA, 2003. ACM.
[5] Ryszard Horodecki, Pawel Horodecki, Michal Horodecki, and Karol Horodecki.
Quantum entanglement. Rev. Mod. Phys., 81:887, Jun 2009.
[6] Asher Peres. Separability criterion for density matrices. Phys. Rev. Lett.,
77:1413–1415, Aug 1996.
[7] J. Sturm. SeDuMi version 1.05.
[8] David S. Watkins. Gaussian Elimination and Its Variants. John Wiley & Sons, Inc.,
2005.
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