Transformations of Entangled Mixed States Of Two Qubits
ÜMİT ALKUŞ
1 INTRODUCTION
Consider the
superposition postulate of Quantum Mechanics. It leads to the fact that composite quantum systems can be in
entangled states, i.e., states that are unfactorizable. In the history of quantum physics, the
entanglement is considered and discussed in depth as the
main reason of nonlocality and quantum correlations. Especially, it started with the criticism of Einstein, Podolsky, and Rosen on the Copenhagen interpretation of quantum mechanics[
5.2↓] in 1935. In 1964, J. S. Bell has shown that entangled states violate some inequalities, which are called as
Bell inequalities, which test the nonlocality of quantum mechanics which cannot be described classically[
5.2↓]. After Bell, quantum correlations and nonlocality associated with entanglement have been considered as the singlemost important feature of quantum mechanics that distinguishes it from the classical theories[
5.2↓
5.2↓].
On the other hand, entanglement does not appear to be only a subject of discussion or a philosophical issue, but it also appears as an ingredient as a potential resource of the quantum information processing and quantum computation [
5.2↓
5.2↓] in quantum state teleportation [
5.2↓
5.2↓], superdense coding [
5.2↓,
5.2↓], and quantum cryptography [
5.2↓,
5.2↓]. These works are mainly based on
bipartite pure state entanglement. The nature, however, also displays multipartite pure state entanglement, about which less is known. In addition to these,
mixed state entanglement (bipartite or multipartite) occurs naturally as a result of decoherence processes and, mainly for this reason, it is extensively studied. This thesis is concerned some features of the bipartite entangled mixed states. In the following sections, some of the mathematical tools that are used in this thesis are introduced.
In classical information and computation, the fundamental unit of information is
binary digit (with the abbreviation
bit), which could be either
0 or
1. In quantum information and computation the fundamental unit is called a
quantum bit (with the abbreviation
qubit)[
5.2↓]. A qubit not only takes the values of either
0 or
1, but also it can take on values which are superpositions of both
0 and
1. In other words, both values are possible for one qubit due to the
superposition principle in quantum mechanics since qubit is a twolevel quantum system [
5.2↓].
Geometrically, if a bit is a scalar quantity, qubit might be thought as a twodimensional vectorial quantity which can be decomposed into the orthogonal unit vectors
0⟩ and
1⟩. The vector space of any qubit is a twodimensional
Hilbert space denoted by
ℂ^{2}. Using Dirac
braket notation, after P.A.M. Dirac [
5.2↓], any qubit is represented by a
ket which is denoted by a symbol like
ψ⟩ with its dual correspondence
bra denoted by
⟨ψ. Then, an arbitrary qubit in a state
ψ⟩ can be written as a linear combination or superposition of the unit vectors
0⟩ and
1⟩ as follows
where the coefficients
α and
β are complex numbers. In that case, the
bra vector is
⟨ψ = α^{*}⟨0 + β^{*}⟨1 where “
*” represents the complex conjugate. Alternatively, in a onecolumn matrix (or column vector) notation, mutually orthogonal unit vectors
0⟩ and
1⟩ are taken conventionally as
which form an
orthonormal basis set for a single qubit. This set
{0⟩, 1⟩} is called
the computational basis set in quantum computation. Note that the
bra’s are then
⟨0 = (
1
0
) and
⟨1 = (
0
1
) which are onerow matrices or row vectors. Thus,
ψ⟩ would be written as
Suppose that
N qubits are identically prepared in the
pure state ψ⟩. If identical measurements on each qubit are made in the computational basis
{0⟩, 1⟩}, then the probability to obtain the outcome
0 is
α^{2} = α^{*}α and the probability to obtain the outcome
1 is
β^{2} = β^{*}β. These probabilities are called
quantum probabilities, which are obeying the restriction
α^{2} + β^{2} = 1 [
5.2↓].
1.2 Multiple qubits
When two or more qubits are considered, the tool
tensor product is used to construct the Hilbert space of the composite system. Let A and B be two qubits with their Hilbert spaces being
ℋ_{1} and
ℋ_{2}, which are two dimensional. Then the composite system AB for these qubits has the four dimensional Hilbert space
ℋ_{AB} = ℋ_{1}⊗ℋ_{2}. Now, let A and B be two local operators (or equivalently matrices) that act separately on each qubit. An operator acting on the whole space
ℋ_{1}⊗ℋ_{2} can be constructed by the tensor product as
A⊗B. In matrix notation, the tensor product of these operators is calculated by the formula
which is called as the
Kronecker product. Notice that this is a
4 × 4 matrix.
Therefore, the new fourdimensional computational basis set in
ℋ_{1}⊗ℋ_{2} are constructed by the tensor products of the unit vectors
0⟩ and
1⟩
00⟩_{AB} = ⎛⎜⎝
1
0
⎞⎟⎠_{A}⊗⎛⎜⎝
1
0
⎞⎟⎠_{B} = ⎛⎜⎝
1⎛⎜⎝
1
0
⎞⎟⎠
0⎛⎜⎝
1
0
⎞⎟⎠
⎞⎟⎠ = ⎛⎜⎝
1
0
0
0
⎞⎟⎠,
So, an orthonormal basis set in
ℋ_{1}⊗ℋ_{2} is
More generally, if
N ≥ 3 qubits are considered, the Hilbert space of the composite system is denoted by
ℋ_{1}⊗ℋ_{2}⊗⋯⊗ℋ_{N} which is a
2^{N}dimensional Hilbert space. The natural orthonormal basis for this system is constructed as
This basis is usually called as the computational basis in quantum computation studies.
1.3 Density operator
Quantum mechanics is constructed by
measurements performed on the quantum systems. The measurement outcomes are obtained with some
probabilities [
5.2↓]. Determining the probabilities corresponding to
observables are achieved only by welldefined representations [
5.2↓]. The difficulty arisen is that each state corresponding to different measurement outcomes have not only relative weights but also relative phases compared to each other. Therefore, if one has only statistical results about a quantum system, construction of the
quantum state vector in the form of a
ket may fail [
5.2↓]. The knowledge of relative phases is also needed.
The set
is called the
ensemble of states where each state
ψ_{i}⟩ has the
fractional probability
p_{i} to be
found in the ensemble. The difference between quantum probability and fractional probability is that while the former is
a posteriori result
obtained only by measurements performed on the system, the latter is
a priori result
prepared by
someone. Another interpretation is that fractional probabilities is constructed in the brain of the
observer, while quantum probabilities are ingrident of the
observable.
To illustrate this, let an experimenter order 20 electrons from an imaginary company. Let 5 of them be in a spinup state in
x direction shown by
↑_{x}⟩ and other 15 be in a spindown state in
z direction shown by
↓_{z}⟩. The total system is constructed by a
single ket such as
↑_{x}⟩^{⊗5}⊗↓_{z}⟩^{⊗15}, i.e., it is a pure state. However, the
state of each electron can not be written in terms of a pure state like
√(0.25)↑_{x}⟩ + √(0.75)↓_{z}⟩. This is not so, because there is a relationship between
↑_{x}⟩ and
↓_{z}⟩ as
meaning that
↑_{x}⟩ is not orthogonal to
↓_{z}⟩. Meanwhile, the possible state of type
√(0.25)↑_{x}⟩ + √(0.75)↓_{z}⟩ ignores the
relative phase between the states
↑_{x}⟩ and
↓_{z}⟩. Morever, the probabilities
0.25 and
0.75 give different interpretations about the system. For example, the probability
0.25 is not the quantum probability when a possible measurement is performed on a single electron. This is because each electron is actually represented by a single ket, i.e.,
↑_{x}⟩ or
↓_{z}⟩. Therefore, we need a different tool than
ket or
wavefunction to represent the states of individual quantum systems.
It is not obligatory that set
{ψ_{i}⟩} in the ensemble
ℰ = {p_{i}, ψ_{i}⟩} is an orthogonal set but all
ψ_{i}⟩ must be normalized. If an ensemble for a quantum system is given, the density operator of the quantum system, which is developed independently by Landau [
5.2↓] and von Neumann [
5.2↓,
5.2↓], can be formed as
It is easily seen that if
ψ_{i}⟩ is an
Ndimensional column vector then the density matrix is
N × N matrix because of the matrix multiplication of column vector
ψ_{i}⟩ from the right with a row vector
⟨ψ_{i}. Note that
n is the number of elements in
ℰ = {p_{i}, ψ_{i}⟩} and there is no obvious relation between the number
n and the dimension
N, so both
n ≤ N and
N ≤ n are possible [
5.2↓].
If some ρ is given, it is easy to test whether it is a density matrix or not. There are two necessary and sufficient conditions for any ρ to be a density matrix:

ρ is positive; that is, for all ψ⟩, one has ⟨ψρψ⟩ ≥ 0 since
for all ψ⟩ because of the positivity of p_{i} and of the magnitude ⟨ψψ_{i}⟩.

trρ = 1 since
where ⟨jψ_{i}⟩ = α_{ij} for some orthonormal set {j⟩} and since ψ_{i}⟩ can be written as a linear combination of j⟩ by means of the completeness relation
where I_{N} is the N × N identity matrix. Then,
for normalized ψ_{i}⟩ such that
where the orthonormality condition for {j⟩} is ⟨j∣k⟩ = δ_{jk} where δ_{jk} is the Kronecker delta.
1.4 Mixed states versus pure states
In quantum mechanics, a pure state corresponds to a single
wavefunction. If the state of a quantum system can be expressed as a
ket, then it is called
a pure state[
5.2↓]. In other words, if in the ensemble given by Eq. (
1.8↑), all possible states are identical, i.e.,
ψ_{i}⟩ = ψ⟩ for all
i, then this ensemble and the corresponding density matrix represents the pure state
ψ⟩. In that case,
ρ = ψ⟩⟨ψ. However, when it is impossible to construct the state of the system as a
ket [
5.2↓]
, the state of the system is called
mixed. In that case,
ρ = ∑^{n}_{i = 1}p_{i}ψ_{i}⟩⟨ψ_{i} where
p_{i} is the probability of
ψ_{i}⟩ in the ensemble where
ℰ = {p_{i}, ψ_{i}⟩}. Alternative definitions of mixed states can easily be made since its analysis is very easy. For example, for a mixed state there are more than one nonzero eigenvalues of the corresponding density matrix.
Again, there is a frequently used condition for a given density matrix to be
mixed or
pure. The state is pure if
since if the system is in pure state, say
ψ⟩, then
and mixed if
since
where
the CauchySchwartz inequality is used.
1.5 Partial trace
Consider a state
ρ of a composite system formed by systems A, B, ..., N. All local measurements on A can be expressed by using a density matrix
ρ_{A} defined by
which is an operator that acts on A’s Hilbert space. Here, the
trace is taken over the whole system
excluding A.
As an example, let the particles A and B be in the state
then the reduced density matrix
ρ_{A} is calculated to be
where
I is the identity matrix. Note that in Eq. (
1.22↑), the property of trace operation
is used for any quantum states
a⟩ and
b⟩, i.e.,
and
1.6 Entanglement
Let the pure state of a composite system AB be
ψ_{AB}⟩ where the corresponding Hilbert spaces being
ℋ_{A} and
ℋ_{B}. Let
m and
n be the dimensions of the respective spaces. The state
ψ_{AB}⟩ can be decomposed as
where
{i_{A}⟩} forms an orthonormal set in
ℋ_{A} and
{i_{B}⟩} forms an orthonormal set in
ℋ_{B}. Here the realvalued coefficients
√(μ_{i}) are called the Schmidt coefficients, after Erhard Schmidt[
5.2↓]. The numbers
μ_{i} satisfy the following normalization condition
in which
n_{ψ} is called the Schmidt rank of the state
ψ_{AB}⟩. It is easy to show that the numbers
μ_{i} are eigenvalues of both of the reduced density matrices
ρ_{A} and
ρ_{B} with the corresponding eigenvectors being
i_{A}⟩ and
i_{B}⟩, respectively. This result stems from the fact that density matrices are positive operators which have spectral decompositions into their orthonormal set of eigenvectors of which diagonal elements are corresponding to their eigenvalues. Also, since they are positive and
trρ_{A} = trρ_{B} = 1, they satisfy the Schmidt’s condition on positivity and reality of
μ_{i} and their normalization such that
∑^{nψ}_{i}μ_{i} = 1. Hence, the spectral decompositions of
ρ_{A} and
ρ_{B} are obtained as the following:
since
i_{B}⟩’s form an orthonormal basis set, i.e.,
⟨j_{B}i_{B}⟩ = δ_{ij}. With similar calculations,
ρ_{B} is obtained as
Accordingly, one can say that
μ_{i}’s are eigenvalues of
ρ_{A} and
ρ_{B} with the corresponding orthonormal set of eigenvectors
i_{A}⟩ and
i_{B}⟩, respectively [
5.2↓,
5.2↓].
Let two spin1/2 particles labelled by A and B physically interact for some time in the past. Later, let A and B be get separated until this interaction terminates. Then, let Alice (A) be given one of them and Bob (B) be given the other. Despite the absence of interaction, there is a very special situation where the quantum states of individual systems are no longer represented
independently of each other. This phenomenon is known as
quantum entanglement, a name which is coined by E. Schrödinger [
5.2↓]. The simplest entangled state is the singlet state formed between two spin1/2 particles. In terms of Eq. (
1.6↑), this
maximally entangled state has the form
where
0⟩ = ↑_{z}⟩ and
1⟩ = ↓_{z}⟩ in this case. Particles in such entangled states are called
EPR pairs by EinstenPodolskyRosen [
5.2↓]. An orthonormal basis formed by maximally entangled states can be found, for example the set of vectors
β_{xy}⟩ where
where
y is the negation of
y; for example
0⟩ = 1⟩ and
1⟩ = 0⟩. The set of vectors
{β_{xy}⟩} forms an orthonormal basis of the
2 × 2 dimensional Hilbert space of two qubits[
5.2↓].
The states in Eq. (
1.33↑) are called
entangled since they can not be written in a
product form like
a⟩_{A}⊗b⟩_{B}. This state leads to some extraordinary and even unimaginable applications of entanglement in quantum information and computation. To see this, consider two parties Alice and Bob sharing particles A and B which is in an entangled state of the form Eq. (
1.33↑). Separate the particles so that there is no
interaction between the particles. Nevertheless, all the
local operations performed only by Alice on her qubit can change the state of the qubit B. These local operations directly affect the results of the local operations on the qubit B performed by Bob. This is because the state of each qubit can not be expressed separately.
In the language of the Schmidt decomposition, a bipartite pure state is entangled iff its Schmidt rank is 2 or more and unentangled if its Schmidt rank is 1. Looking at the Eq. (
1.22↑) for
ρ_{A} = (1)/(2)(0⟩⟨0 + 1⟩⟨1) in the case of
ψ_{AB}⟩ = (01⟩ − 10⟩) ⁄ √(2), the Schmidt rank is
n_{ψ} = 2, i.e.,
ψ_{AB}⟩ is entangled. Also, note that eigenvectors
{0⟩, 1⟩} of
ρ_{A} corresponds to the positive eigenvalues
μ_{1, 2} = (1)/(2). These eigenvalues satisfy the Eq. (
1.29↑) since
∑^{2}_{i}μ_{i} = 1. Now, consider the product state
ψ_{AB}⟩ = 00⟩, then
ρ_{A} = 0⟩⟨0. Its Schmidt rank is 1, therefore it is unentangled.
1.7 W and GHZ class entangled states of three qubits
In a similar manner with Sec.
1.6↑, when more than two qubits are involved, then entangled states of very different kinds can be obtained [
5.2↓
5.2↓,
5.2↓,
5.2↓
5.2↓]. However, Bennett and DaVincenzo have argued that despite a lot of work on multipartite entanglement, it still remains a mystery. Thus, they can not be treated as in the case of bipartite entanglement [
5.2↓]. In the case of three qubit system, there is a pure state
which is also known as the W state. This state is closely related to the general “W type” state of three or more particles which are described as follows: for
p qubits
where
0⟩ = 00⋯0⟩ and
1_{k}⟩ is the state when the
k^{th} qubit is in state
1 and the rest is in state 0 [
5.2↓]. In other words, Bruß states that a geniune W state is constructed by a superposition of the parties which are in cyclic permutations of each other and one party should be in excited state [
5.2↓]. Also,
x_{0} and all
x_{k} are real and positive that obey the sum (or normalization)
Note that
W⟩ defined by Eq. (
1.34↑) can not be written in a product state like
so
W⟩ is necessarily an entangled state.
Another type of tripartite entangled state can be obtained as follows
which is known as the GreenbergerHorneZeilinger (GHZ) state[
5.2↓]. This is closely related to the “GHZtype” entangled state which is defined as
where
p is the number of parties involved,
N is normalization constant,
z is an arbitrary complex number. It is also known as
pparticle Cat (pCat) state because of Schrödinger’s cat [
5.2↓] when
β_{k}⟩ = 1⟩ for all
k. Also, note that each state
β_{k}⟩ in Eq. (
1.39↑) has the form
for each
k with the obvious condition that
s_{k} ≠ 0 for all
k at the same time. Here, note that all
β_{k}⟩’s are unit vector and
c_{k} and
s_{k} can be seen as
cosine and
sine functions of some real angle
θ_{k}, respectively, then it is obvious that
for all
k’s. Again, it is clear that
Ψ⟩_{GHZ} in Eq. (
1.39↑) can not be written as a product state such that
if at least two sines are nonzero.
1.8 SLOCC equivalence of pure states
Consider two possible multipartite states
ψ⟩ and
φ⟩ of
N systems (particles)
A, B, ..., N. Stochastic local operations assisted with classical communications (SLOCC) transformations for these states are defined as follows:
ψ⟩ is
stochastically reducible to
φ⟩ (shown by
ψ⟩\undersetSLOCC⟶φ⟩) if there are local operators
A, B, …, N such that
Also,
ψ⟩ is
stochastically equivalent to
φ⟩ or
SLOCC equivalent to
φ⟩ (shown by
ψ⟩\undersetSLOCC ~ φ⟩) if
ψ⟩\undersetSLOCC⟶φ⟩ and
φ⟩\undersetSLOCC⟶ψ⟩. If we can find
invertible local operators (ILO),
A, B, ..., N, transforming the state
ψ⟩ into the state
φ⟩, then
ψ⟩ is said to be equivalent to
φ⟩ under
SLOCC transformations:
It is also true that
where
A^{ − 1},
B^{ − 1}, ...,
N^{ − 1} are the inverses of the ILO’s
A, B, ..., N, respectively [
5.2↓].
It can easily be shown that if
ψ⟩\undersetSLOCC ~ φ⟩, then ranks of the reduced density matrices of the corresponding local parties are equal [
5.2↓], i.e.,
In Chap. 4, SLOCC classes for rank 2 mixed states of two qubits are constructed.
2 CONCURRENCE
2.1 Partly entangled bipartite pure state
Suppose that the particles A and B are in the “partially” entangled bipartite pure state
ψ⟩_{AB}
where
α and
β are some arbitrary constants satisfying the normalization condition
α^{2} + β^{2} = 1.
Bennett
et. al. [
5.2↓] have shown that, for the asymptotic transformations of pure bipartite entangled states, there is a single measure of entanglement, which is defined as the
von Neumann entropy of the reduced density matrices of the state as follows
where
ρ_{A} and
ρ_{B} are the reduced density matrices of the parties
A and
B, respectively. Then, for the state
ψ⟩_{AB} given by the Eq. (
2.2↑), the reduced density matrix
ρ_{A} of
A is calculated to be
Putting Eq. (
2.4↑) into Eq. (
2.3↑) results in the entanglement
A related entanglement measure which appears to be useful in the discussion of bipartite entanglement of two qubits is the concurrence. The value of the concurrence for the pure state given in Eq. (
2.2↑) is simply equal to
The amount of entanglement
E for the pure state in Eq. (
2.2↑) can be expressed as
Here,
h(x) is the function
The concurrence can also be defined for mixed states of two qubits as well [
5.2↓], however, this time the calculation is more involved. In the following sections, we describe the calculation method and compute the value of
C for rank 2 mixed states of two qubits.
2.1.1 Pure state concurrence C(ψ)
Concurrence of a pure state
ψ⟩ = ψ⟩_{AB} of two qubits can be expressed as
where
ψ̃⟩ is the
spin flip state for
ψ⟩ calculated by the formula
where
ψ^{*}⟩ is the complex conjugate of
ψ⟩ in the computational basis. Here,
σ^{⊗2}_{y} is given by
σ^{⊗2}_{y} = σ_{y}⊗σ_{y} for the Pauli operator
σ_{y} = ⎛⎜⎝
0
− i
i
0
⎞⎟⎠ in the computational basis set
{0⟩, 1⟩}. It acts separately on each qubit as
σ_{y}0⟩ = i1⟩ and σ_{y}1⟩ = − i0⟩. Then,
Now, applying these operations on Eq. (
2.2↑), the spin flip state is found as
Then, concurrence is calculated to be
Notice that the concurrence of a maximally entangled state given by Eq. (
1.32↑) is found as
C = 1 since
α = − β = 1 ⁄ √(2). However, the concurrence of any product state like
00⟩ is found
0 since
β = 0. Therefore, concurrence is some kind of a measure of entanglement, i.e., more concurrence means a more entangled state. Note that
E = E(C) in Eq. (
2.7↑) is a monotonically increasing function of
C, which also takes values in the interval
(0, 1). Hence, by the same token,
E is a similar kind of measure.
2.1.2 Concurrence of a mixed state
For a mixed state, the concurrence is defined in an elaborate way. The
R matrix of a mixed state
ρ = ρ_{AB} of two qubits is defined by
where
ρ̃ is the spin flip state given by
where
ρ^{*} is obtained by taking the complex conjugate of
ρ in the computational basis. Let the eigenvalues of
R(ρ) be
in
nonincreasing order, i.e.,
λ_{1} ≥ λ_{2} ≥ λ_{3} ≥ λ_{4}. Then the concurrence is calculated as
For any pure state
ψ⟩, the density matrix
ρ is
ρ = ψ⟩⟨ψ and the spinflipped state
ρ̃ is
ρ̃ = ψ̃⟩⟨ψ̃. Note that
ρ = ψ⟩⟨ψ is a four dimensional density matrix and is already
diagonal or
spectrally decomposed (i.e. a
projector) with eigenvalue 1 for the eigenstate
ψ⟩ and 0 for the other three
mutually orthonormal eigenstates, which all form an orthonormal basis. Then, the square root function
√(⋯) can operate directly on
ρ as
√(ρ) = √(1)ψ⟩⟨ψ = ρ. Therefore, using the Dirac representations for
√(ρ) and
ρ̃, R(ρ) is found as
which is already in its spectral form like
ρ whose one eigenvalue is 1 and other three are 0. Then the eigenvalues, in of
R(ρ) are
Thus, using Eq. (
2.17↑), the concurrence
C(ρ_{AB}) is
As a result, the Eq. (
2.17↑) of the concurrence for mixed states can also be used to find the concurrence of a pure state
which is a rank 1 mixed state.
2.1.3 Eigenvalues of ρρ̃
In this section, instead of finding the eigenvalues
λ of
R(ρ), the square roots of the eigenvalues
γ of the matrix multiplication
ρρ̃
are used to find the concurrence so that
This can be shown as follows. It is a wellknown result in linear algebra that similar matrices have the same set of eigenvalues. For two matrices
A and
B, the matrix
AB is similar to
Hence, the set of eigenvalues of
AB and
BA are identical. By continuity of the dependence of eigenvalues on matrices, the same conclusion holds even when
A and
B are noninvertible. Therefore,
R^{2} = √(ρ)ρ̃√(ρ) and
√(ρ)√(ρ)ρ̃ = ρρ̃ are isospectral. Note that
ρρ̃ is not Hermitian. But, since
R^{2} is a positive definite and hermitian, the eigenvalues of
ρρ̃ are real and nonnegative.
In conclusion, concurrence of a bipartite pure state can be found by the Eq. (
2.9↑) for pure states, by
R(ρ) in Eq. (
2.14↑), and also by direct multiplication
ρρ̃ developed for rank 2 mixed states.
2.2 Bipartite mixed states with matrix rank 2
Consider a mixed state
ρ = ρ_{AB} of two qubits A and B. Let us suppose that
ρ_{AB} has matrix rank 2, i.e., it has only 2 nonzero eigenvalues. Let us suppose that its eigenvalues are
q_{i} and eigenvectors are
ψ_{i}⟩_{AB} (i = 1, 2). Therefore, we have
For some problems, it is useful to think of AB as being entangled to a hypothetical system C such that ABC are in a pure state
Ψ⟩_{ABC} and
ρ_{AB} = tr_{C}(Ψ⟩⟨Ψ)_{ABC}. The state
Ψ⟩_{ABC} is usually termed a purification of
ρ_{AB}. A straightforward purification of Eq. (
2.24↑) is obtained as
Therefore,
Ψ⟩_{ABC} is a pure state of 3 qubits when
ρ_{AB} has matrix rank 2. The classification by Dür
et. al. [
5.2↓] of pure states of 3 qubits enables us to classify the rank 2 mixed states of 2 qubits.
In the following, we will consider the two types of mixed states ρ_{AB} separately. Namely, those states where purification is of W class and those ones whose purification is of GHZ class. Using wellknown representations of these pure states, we will obtain the concurrence of the mixed state ρ_{AB}.
2.2.1 The case where the purification is of W class
Consider three qubits A, B and C. Let ABC be in a 3partite Wtype entangled state
Ψ⟩_{ABC} defined by Eq. (
1.35↑) for
p = 3 such that
where the coefficients
x_{0} and each
x_{k} for
k = 1, 2, 3 are some
positive real numbers which satisfy
In the following sections, from the pure state
Ψ⟩_{ABC}, first the density matrix of the composite system ABC, then, the reduced density matrix of the subsystem
AB are calculated. Finally, it is shown that there is an entanglement between the qubits belonging to
A
and B
by using Eq. (
2.17↑) for concurrence derived for mixed states of 2 qubits. For the reader to easily follow the calculations, it is more convenient to separate
Ψ⟩_{ABC} as ABC like in the purification of
ρ_{AB} given by Eq. (
2.17↑) as follows
with
which are unnormalized vectors. Here, also
q_{1} = 1 and
q_{2} = 1 which are nothing to do with the eigenvalues given by Eq. (
2.24↑). The density operator
ρ^{W}_{ABC} of
ABC system is obtained then
ρ^{W}_{ABC}
= (Ψ⟩⟨Ψ)_{ABC}
= (ψ_{1}⟩⟨ψ_{1})_{AB}⊗(0⟩⟨0)_{C} + (ψ_{1}⟩⟨ψ_{2})_{AB}⊗(0⟩⟨1)_{C}
+ (ψ_{2}⟩⟨ψ_{1})_{AB}⊗(1⟩⟨0)_{C} + (ψ_{2}⟩⟨ψ_{2})_{AB}⊗(1⟩⟨1)_{C}.
Therefore the reduced density matrix
ρ^{W}_{AB} is
Here, in the intermediate steps, Eqs. (
1.26↑) and (
1.27↑) are used for the trace operations. On the other hand,
ρ^{W}_{AB} can be represented in a matrix form in the computational basis set for two qubits substituting
ψ_{1}⟩_{AB} and
ψ_{2}⟩_{AB} defined by Eq. (
2.29↑) into Eq. (
2.30↑) as
Since
x_{k}’s in Eq. (
2.26↑) are real for all
k, then
ρ^{W}_{AB} is a real matrix. Also, the tensor product
σ^{⊗2}_{y} is obtained in a matrix representation as
Thus, the spin flip state is easily calculated by matrix multiplication as
in the computational basis set. The matrix multiplication of
ρ^{W}_{AB} with
ρ̃^{W}_{AB} is then
Then, the eigenvalues
γ_{i} of the matrix
ρ^{W}_{AB}ρ̃^{W}_{AB} can easily be calculated since they are the roots of the
characteristic equation
whose solution, i.e.
c(γ) = 0, gives the set of eigenvalues
γ_{i} as
in nonincreasing order. Square root of the eigenvalues
γ_{i} of
ρ^{W}_{AB}ρ̃^{W}_{AB} gives the set of eigenvalues
λ_{i} of
R(ρ^{W}_{AB}) as
in nonincreasing order. Substitute the eigenvalues
λ_{i} given by Eq. (
2.37↑) into Eq. (
2.17↑) for the concurrence to get
since both
x_{1} and
x_{2} are positive numbers.
2.2.2 The case where the purification is of GHZ class
Start with the general form given by Eq. (
1.39↑) for
ppartite GHZtype state. Let
p = 3 so that three qubits A, B, and C are in the following tripartite GHZtype entangled state
where
N represents the normalization constant, and
z is an arbitrary complex number. Here,
β_{i}⟩’s are some arbitrary unit state vectors defined by Eq. (
1.40↑). In order for
Ψ⟩_{ABC} to be a unit vector, the normalization constant
N should be equal to the norm of
000⟩ + zβ_{1}β_{2}β_{3}⟩ such that
To simplify the calculations note that
⟨0β_{i}⟩ = c_{i} and ⟨1β_{i}⟩ = s_{i} and substitute them into Eq. (
2.40↑) to obtain
where
ℜ(z) = (z + z^{*}) ⁄ 2 is the real part of
z.
While it was convenient to deal with the matrix representations for the example of Wtype in the Sec.
2.2.1↑, it is not in here. Due to the fact that
ρ^{GHZ}_{ABC} contains too many parameters, another approach is called for. The following quantum mechanical tool is developed for faster calculations instead of the slower matrix method.
Firstly, write
Ψ⟩_{ABC} given by Eq. (
2.39↑) as the sum of the tensor products of the subsystem AB and C as follows
Then, we note that
Secondly,
a_{1}⟩_{AB} and
a_{2}⟩_{AB} can be given a matrix representation in the computational basis set. The tensor product
β_{1}β_{2}⟩ is obtained as
where the column vector representations of
β_{1}⟩_{A} and
β_{2}⟩_{B} given by Eq. (
1.40↑) are used in Eq. (
2.45↑).
Finally, substituting Eq. (
2.45↑) for
β_{1}β_{2}⟩ and Eq. (
1.5↑) for
00⟩ into Eq. (
2.44↑),
a_{1}⟩_{AB} and
a_{2}⟩_{AB} have the column vector representations
Besides, by normalization of Eq. (
2.42↑) and with the abbreviation
a_{i}⟩ = a_{i}⟩_{AB} note that
of which use is made repeatedly throughout the thesis.
Thus, the reduced density matrix
ρ^{GHZ}_{AB} of AB is found as
which is a
rank 2
mixed state. Alternatively, if
ρ^{GHZ}_{AB} is written in a matrix representation putting Eq. (
2.46↑) in Eq. (
2.48↑), the columns of the
4 × 4 matrix
ρ^{GHZ}_{AB} will be calculated as
In order to obtain an outer product representation for
ρ̃^{GHZ}_{AB}, the following procedure is used. Firstly, find the spin flip of the orthonormal basis set given by Eq. (
1.6↑) as
where
Secondly, use the spin flip states
β̃_{j}⟩ = i⎛⎜⎝
− s_{j}
c_{j}
⎞⎟⎠ with
⟨0β̃_{j}⟩ = − is_{j}, ⟨1β̃_{j}⟩ = ic_{j}, and ⟨β_{j}β̃_{j}⟩ = 0 and the tensor product
β̃_{1}β̃_{2}⟩ of the spin flip states
β̃_{j}⟩ as
with
as well as
⟨β_{1}β_{2}β̃_{1}β̃_{2}⟩ = ⟨β_{1}β̃_{1}⟩⟨β_{2}β̃_{2}⟩ = 0.
Finally, express spin flip states
ã_{1}⟩ and
ã_{2}⟩ using Eq. (
2.44↑) as
and
All in all, start from Eq. (
2.48↑) for
ρ^{GHZ}_{AB}, then represent
ρ̃^{GHZ}_{AB} given by the Eq. (
2.15↑) in the following form
A matrix representation of an operator
A:V → W, where
V and
W are any two vector spaces, is defined to be
Here
{v_{j}⟩} and
{w_{i}⟩} are bases (not necessarily orthonormal) in spaces
V and
W respectively. The number of vectors in the basis set should be the same as the dimension of the corresponding vector space [
5.2↓]. Also,
A_{ij} are the matrix elements of the matrix representation of
A.
A_{ij} is the entry in the
i^{th} row and the
j^{th} column.
If
a_{1}⟩ and
a_{2}⟩ are the two elements of the basis set in
V_{AB} (which is a four dimensional vector space since
AB is a two qubit system) the rest of the basis set, say some mutually orthogonal states
a_{3}⟩ and
a_{4}⟩ can be chosen to be orthogonal to both
ã_{1}⟩ and
ã_{2}⟩, too, for convenience. Then, let
ρ^{GHZ}_{AB}ρ̃^{GHZ}_{AB} act on each element of the basis set
{a_{j}⟩} for
j = 1, 2, 3, 4 to get
where
A_{ik} = ⟨a_{i}ã_{k}⟩. Therefore,
ρ^{GHZ}_{AB}ρ̃^{GHZ}_{AB} as an operator taking the vectors
{a_{j}⟩} from the vector space
V_{AB} to the same vectors
{a_{i}⟩} in the four dimensional vector space
V_{AB} represented by the
2 × 2 matrix
AA^{†}.
Now, use the calculated
⟨a_{i}ã_{k}⟩’s given below
to get
Then
AA^{†} is obtained as
in terms of only
⟨a_{i}ã_{k}⟩’s for convenience.
Eigenvalues of
ρ^{GHZ}_{AB}ρ̃^{GHZ}_{AB} are the solution of the characteristic equation
with the solution
Square roots of the eigenvalues
γ_{i} found in Eq. (
3.2↑) of
ρ^{GHZ}_{AB}ρ̃^{GHZ}_{AB} gives the set of eigenvalues
λ_{i} of
R(ρ^{GHZ}_{AB}) as
in nonincreasing order. Using Eq. (
2.17↑), the concurrence
C(ρ^{GHZ}_{AB}) is finally calculated to be
3 OPTIMAL ENSEMBLE REPRESENTING MIXED STATES
In this chapter, we can invoke what Kirkpatrick [
5.2↓] call as SchrödingerHJW theorem, a very useful result which has been discovered and rediscovered many times. It is first shown by Schrödinger in 1936 [
5.2↓], later by Jaynes in 1957 [
5.2↓], and by Hughston, Jozsa and Wootters in 1993 [
5.2↓]. The theorem can be stated as follows: Suppose that we have an equality
where
α_{1}⟩, ⋯, α_{n}⟩, β_{1}⟩, ⋯, β_{m}⟩ are some, possibly unnormalized vectors. The numbers n and m may be equal, but they may also be different. Suppose that
n ≤ m, without loss of generality. Then, there is an
m × m unitary matrix
U such that
where we define
α_{n + 1}⟩ = α_{n + 2}⟩ = ⋯ = α_{m}⟩ = 0.
It is straightforward to check that the opposite is also true, i.e., Eq. (
3.6↑) [or Eq. (
3.6↑)] implies Eq. (
3.5↑). The proof of the actual theorem, i.e., Eq. (
3.5↑) implies Eqs. (
3.6↑) and (
3.6↑), can be found in Nielsen [
5.2↓] as well.
As in the discussion in Sec.
2.2↑, consider a mixed state
ρ = ρ_{AB} with matrix rank 2 so that it has only 2 nonzero eigenvalues. Therefore, we have
where
β_{i} are the eigenvalues corresponding to the mutually orthogonal eigenvectors
v_{i}⟩ (i = 1, 2).
In our case, consider any arbitrary ensemble
ℰ = {r_{i}, w_{i}⟩} with
n = 2 states that realizes
ρ_{AB}, then
where
r_{i} is the weight of the state
w_{i}⟩ in the ensemble
ℰ. Meanwhile, Eq. (
3.9↑) can be expressed as
Here, the subscript
sub stands for the
subnormalization of
v_{i}⟩ which is defined by
and also
Therefore, by SchrödingerHJW theorem we can find
2 × 2 unitary matrix
U such that
In this chapter, the main aim is to find the optimal ensemble
that represents
ρ_{AB} for the matrix rank 2 states which are studied in chapter
2↑. This ensemble satisfies
Moreover, for this ensemble the average value of entanglement reaches its minimum value, i.e.,
where
{q_{i}, ψ_{i}⟩_{AB}} is any ensemble having density matrix
ρ_{AB} and the minimization is carried out over such ensembles. Here,
E(ψ_{i, AB}) is defined earlier as the von Neumann entropy by Eq. (
2.3↑).
To find the optimal ensembles, only the types of density matrices for which
C(ρ) > 0 are used since
C(ρ^{W}_{AB}) = 2√(x_{1}x_{2}) > 0 and
C(ρ^{GHZ}_{AB}) = 2zs_{1}s_{2}c_{3} ⁄ N^{2} > 0. For this type of density matrices, Wootters [
5.2↓] proposes three successive decompositions of
ρ_{AB} using
unitary and
orthogonal transformations. These decompositions are represented by the ensembles of
n = 2 pure states which are listed as the following
The way to find these ensembles are described in the subsections of the following sections, in detail.
Briefly, in Sec.
3.1↓, the tool developed by Wootters [
5.2↓] is utilized to obtain the optimal ensemble for mixed states with W class purifications. The set of states
{w_{i}⟩} in the first ensemble
ℰ_{1} are obtained by a unitary matrix
U from the eigenvectors
{v_{i}⟩} using the SchrödingerHJW theorem by Eq. (
3.13↑). Then, the unitary matrix
U is obtained easily so that it diagonalizes the matrix
τ whose matrix elements are obtained by tildeinner products defined by
τ_{ij} = (⟨v_{i}v_{j}̃⟩)_{sub}. By restricting
w_{i}⟩ as
where
λ_{i}’s are the eigenvalues of
R(ρ^{W}_{AB}) so that eigenvalues of
ττ^{*} are equal to the absolute squares of
λ_{i}.
Later, the set of states
y_{i}⟩ in the second ensemble
ℰ_{2} are obtained from the set of states
{w_{i}⟩} in the previously determined ensemble
ℰ_{1}. First, Wootters [
5.2↓] defines the
preconcurrence for any pure state
ψ⟩ as follows
Next, the average preconcurrence of the ensemble
ℰ_{2} = {q_{i}, y_{i}⟩} is chosen to be
Here,
C(ρ_{AB}) is the concurrence of the mixed state
ρ_{AB}. Thus, Eq. (
3.20↑) restricts the elements
y_{i}⟩ to the relations
and
because of the Eq. (
3.18↑) for the elements
w_{i}⟩. So,
y_{i}⟩ is obtained from
w_{i}⟩ easily.
The third ensemble
ℰ_{3} = {h_{i}, z_{i}⟩}, which will be our optimal ensemble, is obtained from the elements
y_{i}⟩ of the second ensemble
ℰ_{2} using
real positive determinant orthogonal matrix V. In this case, preconcurrences
c(z_{i}) of each state
z_{i}⟩ in the ensemble are obtained by equating them to the average preconcurrence
⟨c(ℰ_{3})⟩ of the ensemble
ℰ_{3}, i.e.,
C(ρ^{W}_{AB}) . This means that average entanglement
∑_{i}h_{i}E(z_{i}) is equal to the entanglement
E(C(ρ^{W}_{AB})) of
ρ^{W}_{AB} as given by Eq. (
3.16↑).
However, in Sec.
3.2↓, a new approach is developed to do same for mixed states with GHZ class purifications. In this case, instead of finding the spectral decomposition of the
ρ^{GHZ}_{AB}, the known decomposition
ρ^{GHZ}_{AB} = ∑^{2}_{i = 1}a_{i}⟩⟨a_{i} given by Eq. (
2.48↑) is chosen to be the starting point. It is reasonable since finding the eigenvalues and eigenvectors of
ρ^{GHZ}_{AB} is somewhat cumbersome. Therefore, the tilde inner product now is calculated by
τ_{ij} = ⟨a_{i}a_{j}̃⟩. Now, construct the set of states
{w_{i}⟩} in the first ensemble
ℰ_{1} by the unitary matrix
U from the set
{a_{i}⟩} using the SchrödingerHJW theorem by Eq. (
3.13↑), i.e.,
√(r_{i})w_{i}⟩ = ∑^{2}_{j = 1}U^{*}_{ij}a_{j}⟩ for
(i = 1, 2). Then, the rest, i.e. determination of
y_{i}⟩ and
z_{i}⟩, is the same as the case of
ρ^{W}_{AB}.
3.1 Mixed states with W class purifications
The eigenvalues
β_{i} of
ρ^{W}_{AB} are the roots of the characteristic equation
as
where
△ = 1 − 4x_{3}(x_{1} + x_{2}).
It is now sufficient to find the eigenvectors corresponding to the two nonzero eigenvalues
β_{1} and
β_{2} given by the Eq. (
3.24↑). The solution to the eigenvectoreigenvalue relations
ρ^{W}_{AB}v_{i}⟩ = β_{i}v_{i}⟩ in terms of the entries of
v_{i}⟩ is found to be
It can be normalized for appropriate choice of
c_{i}, but use the form given by the Eq. (
3.25↑) for simplicity.
Subnormalization of
v_{i}⟩ defined by the Eq. (
3.11↑) is given by
whose solution for
c_{i} is
Now,
v_{i}⟩ is subnormalized if the value
c_{i} found in Eq. (
3.27↑) is put into Eq. (
3.25↑) leading to
v_{i}⟩_{sub}. It is convenient to leave these results with
c_{i} given by Eq. (
3.27↑) due to the reading convenience.
3.1.1 The first ensemble
Begin with the results of Eq. (
3.13↑) obtained by SchrödingerHJW theorem
for a
2 × 2 unitary matrix
U. Therefore, the first ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} is obtained from the spectral decomposition of
ρ^{W}_{AB}, i.e.
{β_{i}, v_{i}⟩}. The subnormalized states
v_{i}⟩_{sub} are calculated by Eq. (
3.25↑) where
c_{i} is given by Eq. (
3.27↑).
Now, define a
2 × 2 symmetric, but
not necessarily Hermitian, matrix
τ whose elements are constructed by
tilde inner products
Here
v_{j}̃⟩_{sub} is the spin flip state
v_{j}⟩_{sub} defined by Eq. (
2.10↑) such that
It is easy to show that the matrix
τ defined by Eq. (
3.29↑) is symmetric, i.e.
τ_{ij} = τ_{ji}. Using Eq. (
3.29↑) for
v_{j}̃⟩_{sub} we have
because
σ^{⊗2}_{y} is Hermitian and real by Eq. (
2.32↑). As a result,
τ is a symmetric matrix. However, it is not necessarily Hermitian since
For the set
{λ_{i}} of the nonnegative eigenvalues Eq. (
2.37↑) of the
R matrix, R(ρ^{W}_{AB}), let
w_{i}⟩_{sub} be given by Eq. (
3.28↑). It satisfies the condition
(⟨w_{i}w̃_{j}⟩)_{sub} = λ_{i}δ_{ij} as previously assumed by Eq. (
3.18↑) where the spin flip state
w_{i}̃⟩_{sub} is
Then, Eq. (
3.18↑) can also be written in terms of
U and
τ as the following
Eq. (
3.34↑) implies that
UτU^{T} is diagonal with the diagonal elements
λ_{i} and there is a unitary
U that diagonalizes
τ. However, the diagonalization of
ττ^{*} gives us a wider aspect. First, multiply
UτU^{T} with
(UτU^{T})^{*} = U^{*}τ^{*}U^{†} to obtain
so
since
λ_{1} = 2√(x_{1}x_{2}) and
λ_{2} = 0 are real.
Thus, in general, we deduce that
ττ^{*} is Hermitian because it is diagonalized by the unitary matrix
U and its eigenvalues are the absolute squares of the eigenvalues of the matrix
R(ρ^{W}_{AB}). This means that
ττ^{*} has a spectral spectral decomposition
if
t_{i}⟩’s are the orthonormal eigenvectors of
ττ^{*} corresponding to the eigenvalues
λ^{2}_{i}.
Using the results above, we claim that the rows (or the columns) of a unitary
U (or
U^{†}) that diagonalizes
ττ^{*} can be chosen as the eigenbras (or the eigenkets) of
ττ^{*} in the proper order of the eigenvectors corresponding to the eigenvalues
λ^{2}_{i} as the following
So
U satisfies the completeness equation
U^{†}U = UU^{†} = I. Indeed,
U diagonalizes
ττ^{*} to its spectral form as the following
where we used Eq. (
3.37↑) to have
ττ^{*}t_{i}⟩ = λ^{2}_{i}t_{i}⟩.
Now, we find the elements
τ_{ij} by Eq. (
3.29↑) for
v_{i}⟩_{sub}’s in terms of the coefficients
c_{i} by Eq. (
3.27↑) and we get
Next, construct the matrix elements
(ττ^{*})_{ij} of
ττ^{*} as
After detailed calculations, we find the following expression
∑^{2}_{k = 1}c^{2}_{k} = x_{1} and get
(ττ^{*})_{ij} = 4x_{2}c_{i}c_{j} such that
which is a Hermitian matrix as proved in Eq. (
3.37↑).
Using the fact that
ττ^{*} has only onenonzero eigenvalue
λ^{2}_{1} = 4x_{1}x_{2} with the others being
λ_{2, 3, 4} = 0, the eigenvector of
ττ^{*} corresponding to the eigenvalue
λ^{2}_{1} can be found as
t_{1}⟩ = (1)/(√(x_{1}))⎛⎜⎝
c_{1}
c_{2}
⎞⎟⎠.
The second eigenvector corresponding to the eigenvalue
λ^{2}_{2} = 0 will be orthogonal to
t_{1}⟩. Therefore, we obtain
t_{2}⟩ = (1)/(√(x_{1}))⎛⎜⎝
c_{2}
− c_{1}
⎞⎟⎠,
an therefore the unitary matrix
U is
As a result, using Eq. (
3.43↑) for
U, then
w_{1}⟩_{sub} of the first decomposition
ℰ_{1} is calculated as
After detailed calculations, we get
Therefore, putting Eq. (
3.45↑) into the subnormalization formula for
w_{1}⟩ by Eq. (
3.12↑), we find the probability
r_{1} to obtain
w_{1}⟩ in the ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} as
and
w_{1}⟩ is found to be
Next, to find the second member
w_{2}⟩ of the ensemble
ℰ_{1}, use the preceding procedure to get
and by Eq. (
3.12↑), we find the probability
r_{2} for
w_{2}⟩ in the ensemble
ℰ_{1} as
Finally
w_{2}⟩ is found to be
For consistency, note that the sum of probabilities
{r_{1}, r_{2}} found in Eqs. (
3.46↑) and (
3.50↑) is 1, i.e.,
r_{1} + r_{2} = 1 − x_{3} + x_{3} = 1.
3.1.2 The second ensemble
The set of states
y_{i}⟩ in the second ensemble
ℰ_{2} = {q_{i}, y_{i}⟩} (i = 1, 2) are obtained from the first ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} found in Sec.
3.1.1↑ so that
where
y_{i}⟩_{sub} = √(q_{i})y_{i}⟩.
Using the preconcurrence formula given by Eq. (
3.19↑) for
y_{i}⟩, we get
Next, the average preconcurrence of the ensemble
ℰ_{2} = {q_{i}, y_{i}⟩} is chosen to be equal to
Here,
C(ρ_{AB}) is the concurrence of the mixed state
ρ^{W}_{AB}. Thus, Eq. (
3.54↑) restricts the elements
y_{i}⟩ to such relations
and
since
(⟨w_{i}w_{j}̃⟩)_{sub} = λ_{i}δ_{ij} by Eq. (
3.18↑). Therefore,
ℰ_{2} satisfies
⟨c(ℰ_{2})⟩ = C(ρ_{AB}) given by Eq. (
3.54↑) such that
Now, we derive the elements
{y_{i}⟩}_{i = 1, 2} of the second decomposition
ℰ_{2} = {q_{i}, y_{i}⟩} realizing
ρ^{W}_{AB} from the first ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} obtained is the Sec.(
3.1.1↑) as
then
and
then
Note that the probabilities
q_{i} of
y_{i}⟩ in the ensemble
ℰ_{2} are equal to the probabilities of
r_{i} of
w_{i}⟩ in the ensemble
ℰ_{1} given by , i.e.
q_{1} = r_{1} = 1 − x_{3} and q_{2} = r_{2} = x_{3}.
3.1.3 The optimal ensemble
Compared to the other first two ensembles
ℰ_{1} and
ℰ_{2} discussed in the previous sections, now let the last ensemble,
ℰ_{3} = {h_{i}, z_{i}⟩}, be the optimal ensemble as Wootters proposes [
5.2↓]. This decomposition is arranged in such a way that preconcurrences
c(z_{i}) of each states
z_{i}⟩ in the ensemble are equal to the average preconcurrence
⟨c(ℰ_{3})⟩ of the ensemble
ℰ_{3} which is equal to
C(ρ^{W}_{AB}). Accordingly, entanglement
E(z_{i}) of each states
z_{i}⟩ in the ensemble will be equal to the average entanglement
⟨E(ℰ_{3})⟩ = ∑^{2}_{i = 1}h_{i}E(z_{i}) of the ensemble which is equal to
E(C(ρ_{AB})).
Now, we make use of the formula for the first ensemble
ℰ_{1} given by Eq. (
3.13↑) of
ρ^{W}_{AB} so that
for a
2 × 2 unitary matrix
V.
For further discussions, Wootters [
5.2↓] defines a
2 × 2 matrix
Y whose elements are given by
which are found to be
with the other elements being zero. Consequently, written in a matrix form,
Y becomes
which is a
real diagonal matrix.
Thus, the average preconcurrence is
In this decomposition, let the unitary matrix
V be also a real matrix, that is, it is an orthogonal matrix with the property
V^{†} = V^{T} ⇒ V^{T}V = I ⇒ V^{T} = V^{ − 1}. In such a case, the average preconcurrence
⟨c(ℰ_{3})⟩ remains invariant under transformations by the
2 × 2 real orthogonal matrices. Since the trace of
Y is preserved due to the cyclic property of the trace operation,
tr(ABC) = tr(BCA) = tr(CAB), then
Put it in another way, the last decomposition ℰ_{3} = {h_{i}, z_{i}⟩} derived by the orthogonal transformations of {y_{i}⟩} is the optimal ensemble ℰ^{opt}. However, as described in the paragraph above, we are interested only in the transformation that makes the individual preconcurrences c(z_{i}) equal to the average preconcurrence ⟨c(ℰ_{3})⟩ of the ensemble.
3.1.3.2 Orthogonal matrix V
As discussed in the previous section, the preconcurrence of
z_{1}⟩ must be equal to
C(ρ^{W}_{AB}) such that
and of
z_{2}⟩:
Next, consider a real orthogonal matrix
V with a positive determinant. The columns and rows of
V form orthonormal sets. Then, in order to obtain a positive determinant for
V, i.e.,
the following matrix can be obtained
3.1.3.3 The third ensemble
Ultimately, one can construct the set of states
{z_{i}⟩} in the third ensemble
ℰ_{3} = {h_{i}, z_{i}⟩} from the set of states
{y_{i}⟩} in the second ensemble
ℰ_{2} = {q_{i}, y_{i}⟩} by the orthogonal matrix
V given by
Eq. (
3.72↑). We use Eq. (
3.62↑) to get
and
Finally optimal states which are equal to
{z_{i}⟩} become
with probability
p_{1} and
with probability
p_{2} for being in the optimal ensemble
ℰ^{opt} = {p_{i}, ψ^{opt}_{i}⟩_{AB}}. However, by normalization of
ψ^{opt}_{i}⟩_{AB} and the equation
∑^{3}_{m = 0}x_{m} = 1, the values of
p_{i}’s are obtained easily as
Therefore, the following results are obtained
and
3.2 Mixed states with GHZ class purifications
In this section, we use the same procedure prescribed by Wootters [(
5.2↓)] to find the optimal ensemble representing
ρ^{GHZ}_{AB}. However, as shown in the Appendix A, the eigenvalues and eigenvectors of
ρ^{GHZ}_{AB} are not manipulated easily like in the case of
ρ^{W}_{AB} (See Sec.
3.1↑). Then, we claim that since we have an available ensemble representing
ρ^{GHZ}_{AB}, i.e.,
ρ^{GHZ}_{AB} = ∑^{2}_{i = 1}a_{i}⟩⟨a_{i} which is given by Eq. (
2.48↑), then it is better to start with this ensemble. In this case, there is no loss in generality. In other words, it is unnecessary to start with the spectral decomposition of
ρ^{GHZ}_{AB} in order to obtain the optimal ensemble. The following sections describe this new method.
3.2.1 The first ensemble
Looking from the ensemble picture defined by Eq. (
1.8↑), there is an ensemble
realizing
ρ^{GHZ}_{AB} given by Eq. (
2.48↑). Namely,
for the already
subnormalized states a_{i}⟩ = √(a_{i})a_{i}’⟩ with the probability
a_{i} being the square of norm of
a_{i}⟩.
We argue that the set of states
{w_{i}⟩} in the first ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} defined by Eq. (
3.13↑) can directly be obtained by the set of states
{a_{i}⟩} in the ensemble
ℰ_{a} = {a_{i}, a_{i}’⟩} given by Eq. (
3.81↑). This is to say that there is a unitary matrix
U between the states
{w_{i}⟩} and
{a_{i}⟩} with the following relation
where
w_{i}⟩_{sub} = √(r_{i})w_{i}⟩. Following the procedure used in the previous section, there is a constraint defined by Eq. (
3.18↑) on the set of states
{w_{i}⟩} such that
Here,
{λ_{i}} is the set of eigenvalues of the matrix
R(ρ^{GHZ}_{AB}) given by Eq. (
3.3↑). Hence, the elements of the matrix
τ is obtained by the tilde inner product by
so that
We will then find the unitary matrix with
U = ⎛⎜⎝
⟨t_{1}
⟨t_{2}
⎞⎟⎠ which diagonalizes the matrix
ττ^{*}. Remember that
t_{i}⟩ is the eigenvector of
ττ^{*} corresponding to the eigenvalue
λ^{2}_{i}.
Now, we use the values of
⟨a_{i}ã_{j}⟩’s which are given by Eq. (
2.68↑) which are
τ_{ij}’s defined by Eq. (
3.85↑). Therefore, we have
since
Notice that
τ found in Eq. (
3.87↑) is a real Hermitian matrix
multiplied by a complex number
⟨a_{1}ã_{1}⟩ ⁄ 2c_{3}. In other words,
τ^{*} is also the same Hermitian matrix multiplied by
⟨a_{1}ã_{1}⟩^{*} ⁄ 2c_{3}. Then, it is straightforward to show that
ττ^{*} has the same eigenvectors as those of
H. If calculated, the eigenvalues of
H are found to be
α_{1, 2} = c_{3}±1. Thus, the eigenvectors of
H are obtained as
corresponding to the eigenvalues
α_{1, 2} respectively.
Consequently, the unitary matrix
U can be expressed as
which is also a real Hermitian matrix. Hence, by the Eq. (
3.83↑), the subnormalized forms of the states
w_{i}⟩ of the first ensemble
ℰ_{1} are obtained from the states
{a_{i}⟩} in the ensemble
ℰ_{a} as
and
It can be shown that the probability
r_{i} of
w_{i}⟩ in the ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} is equal to the probability
a_{i} of
a_{i}’⟩ in
ℰ_{a} = {a_{i}, a_{i}’⟩} so that
where
{a_{i}⟩} are defined by Eq. (
2.44↑). Finally, by definition
w_{i}⟩_{sub} = √(r_{i})w_{i}⟩ and substituting the open forms of
{a_{i}⟩} given by Eq. (
2.44↑) into the result, we have
and
where
3.2.2 The second ensemble
Similar discussions with the case of W class applies here to obtain the second ensemble
ℰ_{2} = {q_{i}, y_{i}⟩} (i = 1, 2) from the first ensemble
ℰ_{1} = {r_{i}, w_{i}⟩} found in Sec.
3.2.1↑. Therefore,
and
where
κ_{1} and
κ_{2} are given by Eq. (
3.97↑). Also, since
y_{i}⟩_{sub} = √(q_{i})y_{i}⟩ and
then the probabilities
q_{i} of
y_{i}⟩ in
ℰ_{2} are the same as the probabilities
r_{i} of
w_{i}⟩ in
ℰ_{1}, namely
3.2.3 The optimal ensemble
As summarized in Sec.
3.1.3↑ for the W case, we now find a real diagonal
2 × 2 matrix
Y whose diagonal elements are obtained by the Eq. (
3.63↑) so that
Here,
λ_{i}’s are the eigenvalues of the matrix
R(ρ^{GHZ}_{AB}) in nonincreasing order given by the Eq. (
3.3↑). Note that
as also proved by Eq (
3.68↑).
Now, in order for each preconcurrence of the states
z_{i}⟩ in the third ensemble
ℰ_{3} = {h_{i}, z_{i}⟩} to be equal to
C(ρ^{GHZ}_{AB}), we solve the following
Therefore,
By the orthonormality of the rows of
V, we have
Solving Eqs. (
4.2↑) and (
4.3↑) regarding the positivity of
V, i.e.,
detV > 0, we get
Thus, the states
z_{i}⟩ in
ℰ_{3} are obtained from the states
y_{i}⟩ in
ℰ_{2} given by the Eqs. (
3.98↑) and (
3.99↑). Since the third ensemble is the optimal ensemble, we get
with probability
p_{j} = h_{j} in the ensemble
ℰ^{opt} = {p_{j}, ψ^{opt}_{j}⟩}. Here, in terms of the entries
V_{ij} of
V, the complex coefficients
μ_{1j} and
μ_{2j} are calculated as
4 TRANSFORMATIONS OF MIXED STATES
As introduced in Sec.
1.8↑ for pure states, SLOCC transformations for mixed states can also be defined as follows:
ρ_{AB} is
stochastically reducible to
ρ_{AB}’ (shown by
ρ\undersetSLOCC⟶ρ’) if there are operators
A and
B such that
and
ρ_{AB} is
stochastically equivalent to
ρ_{AB}’ or
SLOCC equivalent to
ρ_{AB}’ (shown by
ρ\undersetSLOCC ~ ρ’) if
ρ\undersetSLOCC⟶ρ’ and
ρ’\undersetSLOCC⟶ρ. It can be shown that
ρ_{AB} is SLOCC equivalent
to
ρ_{AB}’ iff Eq. (
4.8↑) holds for some invertible
A and
B.
Let
ρ_{AB} be a matrixrank
2 state of two qubits and
ρ\undersetSLOCC⟶ρ’, then matrix rank of
ρ_{AB}’ is less than 2. If matrix rank of
ρ_{AB}’ is
1 (i.e., if
ρ_{AB}’ is a pure state) then one of
A and
B is not invertible and therefore
ρ_{AB}’ must unentangled. This can be shown as follows: Suppose that
A is not invertible. As
A is a “
2 × 2 matrix”, this implies that
A = cα⟩⟨α for some
α⟩ ∈ ℋ_{A} and constant
c ∈ ℂ and therefore
for some
2 × 2 density matrix
σ. It is apparent that
ρ’ is unentangled and uncorrelated.
Therefore we can state the following: if
ρ\undersetSLOCC⟶ρ’ with Eq. (
4.8↑), and if
ρ_{AB}’ has matrix rank 2, then
From now on, consider all transformations
ρ\undersetSLOCC⟶ρ’ where both
ρ_{AB} and
ρ_{AB}’ have rank
2. In that case,
ρ’\undersetSLOCC⟶ρ as well, because only for invertible
A and
B one can satisfy Eq. (
4.8↑).
4.1 Support of ρ_{AB} as a subspace
Let
ρ_{AB} be a rank 2 mixed state of a two qubit system
AB.
ρ_{AB} has a spectral decomposition as
where
{ψ_{i}⟩} is the orthonormal set of eigenvectors of
ρ_{AB} corresponding to the eigenvalues
{p_{i}} of
ρ_{AB}. Then, there is a 2dimensional subspace of
H_{AB} = H_{A}⊗H_{B} called as the
support of
ρ_{AB} or
supp(ρ_{AB}), which is defined as the linear span of its eigenvectors
Any twodimensional subspace of
H_{AB} always contains a product state, of which detailed proof is given by Sampera, Tarrach, and Vidal [
5.2↓].
There are two nontrivial cases we can think of:

Class ℙ_{1}. All the product states supp(ρ_{AB}) contains are parallel. Hence if
is the product state in supp(ρ_{AB}), then supp(ρ_{AB}) does not contain any other product state that is linearly independent of a⟩. In such a case, we can find an entangled state b⟩ such that {a⟩, b⟩} is a basis of supp(ρ_{AB}).

Class ℙ_{2}. In this case, supp(ρ_{AB}) contains two linearly independent product states. Call these
Therefore, any vector in supp(ρ_{AB}) can be written as a superposition of these.
In short, in both cases, one can find a basis {a⟩, φ⟩} of supp(ρ_{AB}) such that a⟩ is a product state.
Now, we consider alternative representations of
ρ_{AB} in the form
ρ_{AB} = ^{2}⎲⎳_{i = 1}u_{i}⟩⟨u_{i}⟩ = ^{2}⎲⎳_{i = 1}u’_{i}⟩⟨u’_{i}⟩
where Eq. (
4.11↑) is a special example. In all of these examples,
{u_{1}⟩, u_{2}⟩} (similarly,
{u_{1}’⟩, u_{2}’⟩}) spans the support of
ρ_{AB}. In fact they form alternative bases for the support. By SchrödingerHJW theorem, these alternative representations are related by
u’_{i}⟩ = ∑_{j}V_{ij}u’_{j}⟩. Recall that the dimension of the unitary
V depends on the number of elements in the corresponding sets and therefore, it is taken
2 × 2 in this case. This type of ensembles are called
minimal ensembles which contains only
n states to represent rank
n mixed states [
5.2↓].
A special case is the subnormalized states
ψ^{sub}_{i}⟩ = √(p_{i})ψ_{i}⟩ where
ψ_{i}⟩ are the eigenvectors and
p_{i} are the eigenvalues of
ρ_{AB}, then
Now, suppose that an unnormalized product state
P⟩ in
supp(ρ_{AB}) is parallel to the product state
a⟩ defined by Eq. (
4.13↑).
P⟩ can be written as a linear combination of
{ψ^{sub}_{1}⟩, ψ^{sub}_{2}⟩} such that
for some complex numbers
κ,
d_{1} and
d_{2}. Here, we require that
{d_{1}, d_{2}} satisfies
d_{1}^{2} + d_{2}^{2} = 1. Note that
since
∥ψ^{sub}_{1}⟩∥ = √(p_{i}). Now, define a unitary
It can be seen that this is a unitary matrix because its columns (also rows) form an orthonormal set. For another set of states
{u_{i}⟩} which realizes
ρ_{AB}, it is true that
u_{i}⟩ = ∑^{n}_{j = 1}V_{ij}ψ^{sub}_{j}⟩ then
u_{1}⟩ = P⟩ = κa⟩. Therefore, one can always find a representation
{u_{i}⟩} of
ρ_{AB} where one of the states
u_{1}⟩ is parallel to a given product state
a⟩. Also,
u_{2}⟩ will be some other state in
supp(ρ_{AB}) which is
so that
4.2 States with class ℙ_{1} supports
Consider all representations of
ρ_{AB}
where
u_{1}⟩ = κa⟩ for some product state
a⟩ and
u_{2}⟩ is necessarily entangled if
supp(ρ_{AB}) is of class
ℙ_{1}. Moreover,
u_{2}⟩ + zu_{1}⟩ are always entangled for all
z ∈ ℂ. Let
be another representation of
ρ_{AB} such that
u_{1}’⟩ is a product state. In this case, since
supp(ρ_{AB}) contains only one nonparallel product state, we necessarily have
u_{1}’⟩ = κ’a⟩ for some number
κ’. Because of the SchrödingerHJW theorem we also know that then there is a unitary
V such that
u_{1}’⟩ = V_{11}u_{1}⟩ + V_{12}u_{2}⟩. But, if
V_{12} were nonzero, then
u_{1}’⟩ would have been entangled. However, it is chosen as a product state, so
V_{12} = 0 necessarily and therefore
V is diagonal. Then
The states are identical up to an overall phase factor.
In conclusion, if
ρ_{AB} is such that
supp(ρ_{AB}) contains at most one linearly independent product state, then, the decomposition of
ρ_{AB} given by Eq. (
4.21↑) such that
u_{1}⟩ is a product state is unique up to overall phases of
u_{1}⟩ and
u_{2}⟩.
Let
ρ_{AB} be of class
ℙ_{1}. One can then find a basis
{α_{0}⟩, α_{1}⟩} of
ℋ_{A} = ℂ^{2} and a basis
{β_{0}⟩, β_{1}⟩} of
ℋ_{B} = ℂ^{2} such that
ρ_{AB} = u_{1}⟩⟨u_{1} + u_{2}⟩⟨u_{2} where
Unfortunately, this representation is not unique. Let
ρ_{AB} = u_{1}’⟩⟨u_{1}’ + u_{2}’⟩⟨u_{2}’ where
then
where
z ≠ 0,
λ ∈ ℂ,
θ_{1}, θ_{2} ∈ ℝ are arbitrary.
Let
ρ_{AB} and
ρ_{AB}’ be of class
ℙ_{1}. Can we stochastically reduce
ρ_{AB} to
ρ_{AB}’, i.e.,
ρ\undersetSLOCC\overset?⟶ρ’? We will show below that the answer is affirmative. Let
ρ_{AB} = ∑^{2}_{i = 1}u_{i}⟩⟨u_{i},
ρ_{AB}’ = ∑^{2}_{i = 1}u_{i}’⟩⟨u_{i}’ and let
Can one find local operators
A and
B such that
(A⊗B)u_{i}⟩ = u_{i}’⟩(i, j = 2)? The problem can be solved by finding
A and
B such that
The local operators
A and
B that satisfy these are not unique. In general, one can find
A and
B such that
for any given
z ≠ 0, λ ∈ ℂ, θ_{1}, θ_{2} ∈ ℝ. For all of such
A and
B the relation
ρ_{AB}’ = (A⊗B)ρ(A^{†}⊗B^{†}) is satisfied. This proves the claim, i.e., any two states
ρ_{AB} and
ρ_{AB}’ having supports of class
ℙ_{2} are SLOCC equivalent. Therefore, all density matrices with supports of class
ℙ_{2} form a SLOCC class.
4.3 States with class ℙ_{2} supports
Let
a⟩ and
b⟩ be two linearly independent product states given by Eq. (
4.14↑). Then, either
{α_{1}⟩_{A}, β_{1}⟩_{A}} is linearly independent or
{α_{2}⟩_{B}, β_{2}⟩_{B}} is linearly independent or both. There are three cases that one can distinguish.

Class ℙ_{2B} (only B is mixed): While {α_{1}⟩_{A}, β_{1}⟩_{A}} is linearly dependent, {α_{2}⟩_{B}, β_{2}⟩_{B}} is linearly independent. For this case, α_{2}⟩ is parallel to α_{1}⟩, and all states in supp(ρ_{AB}) is of the form α_{1}⟩_{A}⊗ψ⟩_{B} where ψ⟩_{B} is arbitrary. Therefore, ρ_{AB} is unentangled and uncorrelated, with A being in a pure state α_{1}⟩_{A} and B is in a mixed state (say σ)

Class ℙ_{2A} (only A is mixed): While {α_{2}⟩_{B}, β_{2}⟩_{B}} is linearly dependent, {α_{1}⟩_{A}, β_{1}⟩_{A}} is linearly independent. For this case, β_{2}⟩ is parallel to α_{2}⟩ and all states in supp(ρ_{AB}) is of the form ψ⟩_{A}⊗α_{2}⟩_{B} where ψ⟩_{A} is arbitrary. ρ_{AB} is unentangled and uncorrelated, with B being in a pure state α_{2}⟩_{B} and A is in a mixed state (say σ)

Class ℙ_{2AB} (both A and B are mixed): Both {α_{1}⟩_{A}, β_{1}⟩_{A}} and {α_{2}⟩_{B}, β_{2}⟩_{B}} are linearly independent. This implies that the generic state
is always entangled for c_{1} ≠ 0, c_{2} ≠ 0.
It can be shown that all states having supports of class ℙ_{2A} and those that have supports of class ℙ_{2B} do form separate SLOCC classes. The case of ℙ_{2AB} is highly nontrivial and is studied in detail below.
If
u⟩ ∈ supp(ρ_{AB}) and
u⟩ is a product state then either
u⟩ = Na⟩ or
u⟩ = Nb⟩ for some complex number
N. So if
ρ_{AB} is of type given by Eq. (
4.21↑) and
u_{1}⟩ is a product state, then either
u_{1}⟩ is parallel to
a⟩ or to
b⟩. Consider two possible representations of
ρ_{AB},
where
u_{1}⟩ and
u_{1}’⟩ are product states. If
u_{1}⟩ is parallel to
u_{1}’⟩ then we have
i.e., the states in the two ensembles are identical up to an overall phase factor.
Now, consider the case where
u_{1}⟩ is parallel to
a⟩ and
u_{1}’⟩ is parallel to
b⟩. Let
(where
x is real and
x > 0) and
We can choose
y to be real and
y ≥ 0. Obviously
z ≠ 0. Thus,
Similarly, let
and
where
x’ and
y’ are real,
x’ > 0 and
y’ ≥ 0, and again,
z’ ≠ 0. Thus,
then
If the expansion of
ρ_{AB} is known, one can select the overall phase of
b⟩ (relative to that of
a⟩) such that
z is real and positive.
Let
where the overall phases of
b⟩ and
a⟩ have been redefined such that
R_{12} is real and positive. Then,
⇒ z ≡ √(R_{12}) , z’ ≡ √(R_{11}),
x ≡ √(R_{11} − (R^{2}_{12})/(R_{22})) = √((detR)/(R_{22})), x’ ≡ √((detR)/(R_{11})).
Note that even though the phases of
a⟩ and
b⟩ are adjusted such that
R_{12} is real and positive, the inner product
⟨ab⟩ might still have a phase.
4.4 SLOCC classes
Let
ρ\undersetSLOCC ~ ρ’, then
Let us show this implication. Let
ρ_{AB} = ∑^{2}_{i = 1}u_{i}⟩⟨u_{i} and
ρ_{AB}’ = ∑^{2}_{i = 1}u_{i}’⟩⟨u_{i}’ be representatives where
u_{1}⟩ and
u_{1}’⟩ are product states and let
ρ_{AB}’ = (A⊗B)ρ(A^{†}⊗B^{†}). Let us first show the forward implication (
( ⇒ :) Suppose
ρ_{AB} is of class
ℙ_{2}, so
where
a⟩ and
b⟩ are product states. Define
that is,
ã⟩ and
b̃⟩ are product states. Then,
so
ρ_{AB} is also in class
ℙ_{2}. The reverse implication
(⇐:) is obvious as this is equivalent to
ρ’\undersetSLOCC⟶ρ.
If
ρ\undersetSLOCC ~ ρ’, then
This has been shown somewhere above. In general, if ρ\undersetSLOCC ~ ρ’ then

ρ_{AB} ∈ ℙ_{1} ⇔ ρ_{AB}’ ∈ ℙ_{1},

ρ_{AB} ∈ ℙ_{2} ⇔ ρ_{AB}’ ∈ ℙ_{2},

ρ_{AB} ∈ ℙ_{2A} ⇔ ρ_{AB}’ ∈ ℙ_{2A},

ρ_{AB} ∈ ℙ_{2B} ⇔ ρ_{AB}’ ∈ ℙ_{2B},

ρ_{AB} ∈ ℙ_{2AB} ⇔ ρ_{AB}’ ∈ ℙ_{2AB}.
Let
ρ_{AB} be of class
ℙ_{1},
ρ_{AB} = ∑^{2}_{i = 1}u_{i}⟩⟨u_{i} and
Now, redefine
α_{1}⟩ or
α_{2}⟩ such that
c = 1, then
Let
{α_{1}⟩_{A}, β_{1}⟩_{A}} be a basis of
H_{A} = ℂ^{2} and
{α_{2}⟩_{B}, β_{2}⟩_{B}} be a basis of
H_{B} = ℂ^{2}. Let
Here
This implies that
g = 0 (otherwise
ρ_{AB} is of class
ℙ_{2}). Then
Redefine
β_{1}⟩ and
β_{2}⟩ as follows:
for given
λ, select
μ = d − λ which leads to the following corollary:
Let
ρ_{AB} be of class
ℙ_{2AB}. There is a basis
{α_{0}⟩, α_{1}⟩} of
H_{A} = ℂ^{2} and a basis
{β_{0}⟩, β_{1}⟩} of
H_{B} = ℂ^{2} such that
ρ_{AB} = u_{1}⟩⟨u_{1} + u_{2}⟩⟨u_{2} where
In general, one can absorb
c_{1}, c_{2}, and
c_{3} into the definitions of
α_{i}⟩, β_{i}⟩. But,
c_{2} ⁄ c_{1} can not be changed by such redefinitions.
where
k is real with
k ≥ 0.
Note that
a purification of
ρ_{AB} is
where
requirement is
k ≠ ∞ (
c_{3} ≠ 1). Hence the parameter
k is related to the cosine
c_{3} of
3^{rd} party
C. Let
ρ_{AB}, ρ_{AB}’ ∈ ℙ_{2AB}.
ρ\undersetSLOCC⟶ρ’ iff
k ≠ k’. The parameter
k can not change in stochastic transformations.
5 OPTIMAL ENSEMBLE REPRESENTING $\rho_{AB}^{GHZ}$ STARTING WITH SPECTRAL DECOMPOSITION
To find the eigenvalues and the eigenstates of the mixed state ρ^{GHZ}_{AB} of the composite system AB, consult the tool of Schmidt decomposition of Ψ⟩_{ABC} as AB − C which is used in the following sections.
5.1 Schmidt decomposition of Ψ⟩_{ABC} as AB − C
Consider the tripartite system
ABC as the composition of the
AB − C where
AB and
C are defined on the Hilbert spaces
ℋ^{⊗4}_{AB} and
ℋ^{⊗2}_{C}, respectively. Therefore, it is to say that eigenvalues of the subsystems
AB and
C are the same and then that
which means that ranks of
ρ^{GHZ}_{AB} and
ρ^{GHZ}_{C} are at most
2. For this reason, it is more convenient to deal with the eigenvalues and eigenvectors of
2 × 2 matrix
ρ^{GHZ}_{C} rather than the
4 × 4 matrix
ρ_{AB} [
5.2↓,
5.2↓,
5.2↓].
ρ^{GHZ}_{C} is calculated by partial tracing the party
AB from the total state
Ψ⟩_{ABC}, namely
which can also be given a matrix representation
in the computational basis set
{0⟩, 1⟩}. The set of eigenvalues
μ_{i} are the solution of the characteristic equation
so that
where
△ is the discriminant defined by
Schmidt decomposition given by Eq. (
1.28↑) for
Ψ⟩_{ABC} of the system
ABC into the subsystems
AB − C gives
where
i_{AB}⟩ and
i_{C}⟩ are orthonormal set of the eigenstates of
ρ^{GHZ}_{AB} and
ρ^{GHZ}_{C}, respectively, corresponding to the eigenvalues
μ_{i}’s in the decreasing order. One can represent the states
0⟩ and
1⟩ in the orthonormal basis set
1_{C}⟩ and
2_{C}⟩ such that
and
noting that
Putting these values into Eq.
2.42↑ and then expanding also Eq. (
5.9↑), we get
equating Eqn.s (
5.13↑) and (
5.13↑) gives the eigenstates
i_{AB}⟩ of
ρ^{GHZ}_{AB} in terms of the coefficients of the eigenstates
i_{C}⟩ of
ρ^{GHZ}_{C} and the vectors
a_{1}⟩ and
a_{2}⟩ as the following
or
Eigenstates
i_{C}⟩ of
ρ^{GHZ}_{C} are determined by the eigenvectoreigenvalue relation
as
where
N_{i} is found as the following:
adding
c(μ_{i}) = μ^{2}_{i} − μ_{i} + ∥a_{1}⟩∥^{2}∥a_{2}⟩∥^{2} − ⟨a_{1}a_{2}⟩^{2} = 0 which is characteristic equation given by Eq. (
5.6↑) to the right side gives
Finally, the eigenvectors
i_{AB}⟩ of
ρ^{GHZ}_{AB} become
Subnormalization of
i_{AB}⟩ is defined by
⟨i_{AB}i_{AB}⟩ = μ_{i} which results in the subnormalized eigenstates
or simply
where
5.2 The first decomposition of ρ^{GHZ}_{AB}
Start with the spectral decomposition of
ρ^{GHZ}_{AB}
or in terms of the subnormalized eigenstates
i_{AB}⟩_{sub} ≡ √(μ_{i})i_{AB}⟩
There is a unitary matrix
U that transforms the states
i_{AB}⟩_{sub} to another set of states
w_{i}⟩_{sub} with the formula
so that
Thus, it means that there is an another ensemble
{r_{i}, w_{i}⟩}, where
r_{i} is the weight of
w_{i}⟩ in the ensemble, which represent the mixed state
ρ^{GHZ}_{AB} such that
Meanwhile, since
C(ρ^{GHZ}_{AB}) ≥ 0, then the density matrix
ρ^{GHZ}_{AB} is of the first class which leads to consider the following procedure to find the optimal ensemble.
Begin with the general decomposition defined by Eq. (
5.27↑) for the subnormalized states
{w_{i}⟩} of
ρ^{GHZ}_{AB} where the unitary matrix is chosen to diagonalize the Hermitian matrix
ττ^{*} with the eigenvalues square of the absolute values of the eigenvalues of the
R matrix. In the same sense, it is sufficient to determine the eigenvectors
t_{i}⟩ of the
ττ^{*} which construct the columns of
U^{†}.
The matrix elements
τ_{ij} of
τ are formed by
“tilde inner products”:
where
\widetildej_{AB}⟩ is the spin flipped state of the eigenstate
j_{AB}⟩ of
ρ^{GHZ}_{AB} defined by
\widetildej_{AB}⟩ = σ^{⊗2}_{y}j^{*}_{AB}⟩ by means of the PauliY operator
σ^{⊗2}_{y} = σ_{y}⊗σ_{y} acting separately on each qubit such that
or simply
Since, the eigenvalues
λ^{2}_{i} = γ_{i} of
ττ^{*} are known, it is possible to find the eigenvectors
t_{i}⟩ of
ττ^{*} corresponding to them by the eigenvectoreigenvalue relation:
or equivalently in matrix notation
where
(ττ^{*})_{ij}’s are the matrix element of
ττ^{*}; e_{i} and
f_{i} are the vector elements of
t_{i}⟩ which can be chosen as
e_{i} = (ττ^{*})_{12} and
f_{i} = γ_{i} − (ττ^{*})_{11} then
t_{i}⟩ can be written
with the normalization constant
t_{i} defined by
t_{i} = g^{ − 1}_{i} = √(e_{i}^{2} + f_{i}^{2}). Therefore, the unitary matrix
U can be written as
By means of the complex conjugate of the unitary matrix
U^{*}
the subnormalized states
w_{i}⟩_{sub} can be formed as
and or in terms of
{a_{i}⟩}
or simply
where
and
Then construct
put them into
to obtain
{w_{i}⟩_{sub}}.
If one can find anything about
{w_{i}⟩_{sub}} after the above calculations, next s/he has to find the second set of states
{y_{i}⟩} and finally, if it is possible, it is time to find the real orthogonal matrix
V transforming the second set
{y_{i}⟩} to the optimal set
{z_{i}⟩}. Therefore, it is better to apply the new method to those type examples (see Sec.
3.2↑).
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