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BSc: Physics Department, Middle East Technical University (METU), Ankara, Turkey
MSc: Physics Department, METU
PhD: Qualified, Physics Department, METU

## Sunday, November 26, 2017

Transformations of Entangled Mixed States Of Two Qubits

# Transformations of Entangled Mixed States Of Two Qubits

List of Algorithms

# 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 non-locality 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 non-locality of quantum mechanics which cannot be described classically[5.2↓]. After Bell, quantum correlations and non-locality 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.

## 1.1 Qubit

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 two-level quantum system [5.2↓].
Geometrically, if a bit is a scalar quantity, qubit might be thought as a two-dimensional vectorial quantity which can be decomposed into the orthogonal unit vectors |0 and |1. The vector space of any qubit is a two-dimensional Hilbert space denoted by 2. Using Dirac bra-ket 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
(1.1) |ψ = α|0 + β|1
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 one-column matrix (or column vector) notation, mutually orthogonal unit vectors |0 and |1 are taken conventionally as
(1.2) |0 =  1 0   and |1 =  0 1
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 one-row matrices or row vectors. Thus, |ψ would be written as
(1.3) |ψ =  α β  = α 1 0  + β 0 1 .
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 1 and 2, which are two dimensional. Then the composite system AB for these qubits has the four dimensional Hilbert space AB = ℋ1⊗ℋ2. 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 1⊗ℋ2 can be constructed by the tensor product as AB. In matrix notation, the tensor product of these operators is calculated by the formula
(1.4) AB ≡  A11B A12B A21B A22B  =  A11B11 A11B12 A12B11 A12B12 A11B21 A11B22 A12B21 A12B22 A21B11 A21B12 A22B11 A22B12 A21B21 A21B22 A22B21 A22B22
which is called as the Kronecker product. Notice that this is a 4 × 4 matrix.
Therefore, the new four-dimensional computational basis set in 1⊗ℋ2 are constructed by the tensor products of the unit vectors |0 and |1
|00AB =  1 0 A 1 0 B =  1 1 0 0 1 0  =  1 0 0 0 ,
(1.5) |01AB =  0 1 0 0 ,  |10AB =  0 0 1 0 ,  |11AB =  0 0 0 1 .
So, an orthonormal basis set in 1⊗ℋ2 is
(1.6) Orthonormal basis set for 2 qubits  = {|00, |01, |10, |11}.
More generally, if N ≥ 3 qubits are considered, the Hilbert space of the composite system is denoted by 1⊗ℋ2⊗⋯⊗ℋN which is a 2N-dimensional Hilbert space. The natural orthonormal basis for this system is constructed as
(1.7) Orthonormal basis set for N qubits  = {|0⋯00, |0⋯01, ⋯, |11⋯1}.
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 well-defined 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
(1.8)  = {pi, |ψi}
is called the ensemble of states where each state |ψi has the fractional probability pi 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 spin-up state in x direction shown by |x and other 15 be in a spin-down 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
(1.9) |x = (1)/((2))(|z + |z)
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 {|ψi} in the ensemble  = {pi, |ψi} is an orthogonal set but all |ψi 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
(1.10) ρ = ni = 1pi|ψiψi|.
It is easily seen that if |ψi is an N-dimensional column vector then the density matrix is N × N matrix because of the matrix multiplication of column vector |ψi from the right with a row vector ψi|. Note that n is the number of elements in  = {pi, |ψi} 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:
1. ρ is positive; that is, for all |ψ, one has ψ|ρ|ψ ≥ 0 since
(1.11) ψ|ρ|ψ  = ni = 1piψ|ψiψi|ψ  = ni = 1piψ|ψiψ|ψi*  = ni = 1pi|ψ|ψi|2 ≥ 0
for all |ψ because of the positivity of pi and of the magnitude |ψ|ψi|.
2. trρ = 1 since
(1.12) trρ  = Nj = 1ni = 1pij|ψiψi|j  = ni = 1piNj = 1|j|ψi|2 = Nj = 1|αij|2 = 1
where j|ψi = αij for some orthonormal set {|j} and since |ψi can be written as a linear combination of |j by means of the completeness relation
(1.13) IN = Nj = 1|jj|
where IN is the N × N identity matrix. Then,
(1.14) |ψi = IN|ψi = (Nj = 1|jj|)|ψi = Nj = 1j|ψi|j = Nj = 1αij|j
for normalized |ψi such that
(1.15) |ψi  = (ψi|ψi) = (Nj, k = 1α*ijαikj|k)  = (Nj, k = 1α*ijαikδjk) = (Nj = 1α*ijαij)  = (Nj = 1|αij|2) = 1
where the orthonormality condition for {|j} is jk = δjk where δjk 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., |ψi = |ψ for all i, then this ensemble and the corresponding density matrix represents the pure state |ψ. In that case, ρ = |ψψ|. 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 = ni = 1pi|ψiψi| where pi is the probability of |ψi in the ensemble where  = {pi, |ψi}. 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 non-zero 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
(1.16) trρ2 = 1
since if the system is in pure state, say |ψ, then
(1.17) tr(ρ2)  =  tr{(|ψψ|)(|ψψ|)}  = tr(|ψψ|) = trρ = 1
and mixed if
(1.18) trρ2 < 1
since
(1.19) tr(ρ2)  =  tr(ni = 1pi|ψiψi|)2 = ni, j = 1pipjψj|ψiψi|ψj  < ni, j = 1pipj|ψi2|ψj2 = ni = 1pinj = 1pj = 1
where the Cauchy-Schwartz inequality [A]  [A] Let |v and |w be two vectors , then the Cauchy-Schwarz inequality is
|w|v|2 ≤ w|wv|v
where the equality is satisfied if only if |v and |w are parallel.
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 ρA defined by
(1.20) ρA = trBCNρ
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
(1.21) |ψAB = (1)/((2))(|01 − |10),
then the reduced density matrix ρA is calculated to be
(1.22) ρA  = trB(|ψABψAB|) (1.23)  = trB(1)/(2)[|0101| − |0110| − |1001| + |1010|]  = (1)/(2){|00| tr(|11|) − |01| tr(|10|) (1.24)  − |10| tr(|01|) + |11| tr(|00|)}  = (1)/(2)(|00| + |11|) = (I)/(2)
where I is the identity matrix. Note that in Eq. (1.22↑), the property of trace operation
(1.25) tr(|ab|) = ba
is used for any quantum states |a and |b,  i.e.,
(1.26) tr(|01|) = 1∣0 = 0 =  tr(|10|) = 0∣1
and
(1.27) tr(|00|) = 0∣0 = 1 =  tr(|11|) = 1∣1.

## 1.6 Entanglement

Let the pure state of a composite system AB be |ψAB where the corresponding Hilbert spaces being A and B. Let m and n be the dimensions of the respective spaces. The state |ψAB can be decomposed as
(1.28) |ψAB = nψi = 1(μi)|iA|iB (herenψ ≤ min(m, n))
where {|iA} forms an orthonormal set in A and {|iB} forms an orthonormal set in B. Here the real-valued coefficients (μi) are called the Schmidt coefficients, after Erhard Schmidt[5.2↓]. The numbers μi satisfy the following normalization condition
(1.29) nψiμi = 1
in which nψ is called the Schmidt rank of the state |ψAB. It is easy to show that the numbers μi are eigenvalues of both of the reduced density matrices ρA and ρB with the corresponding eigenvectors being |iA and |iB,  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 trρA = trρB = 1, they satisfy the Schmidt’s condition on positivity and reality of μi and their normalization such that nψiμi = 1. Hence, the spectral decompositions of ρA and ρB are obtained as the following:
(1.30) ρA  = trB(|ψABψAB|) = nψi, j = 1(μiμj)|iAjA|tr(|iBjB|)  = nψi, j = 1(μiμj)|iAjA|jB|iB  = nψi = 1μi|iAiA|
since |iB’s form an orthonormal basis set, i.e., jB|iB = δij. With similar calculations, ρB is obtained as
(1.31) ρB = trA(|ψABψAB|) = nψi = 1μi|iBiB|.
Accordingly, one can say that μi’s are eigenvalues of ρA and ρB with the corresponding orthonormal set of eigenvectors |iA and |iB,  respectively [5.2↓,5.2↓].
Let two spin-1/2 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 spin-1/2 particles. In terms of Eq. (1.6↑), this maximally entangled state has the form
(1.32) |ψAB = (1)/((2))(|01 − |10)
where |0 = |z and |1 = |z in this case. Particles in such entangled states are called EPR pairs by Einsten-Podolsky-Rosen [5.2↓]. An orthonormal basis formed by maximally entangled states can be found, for example the set of vectors |βxy where
(1.33) |βxy = (1)/((2))(|0, y + ( − 1)x|1, y)
where y is the negation of y; for example |0 = |1 and |1 = |0. The set of vectors {|βxy} 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 |aA|bB. 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 ρA = (1)/(2)(|00| + |11|) in the case of |ψAB = (|01 − |10) ⁄ (2), the Schmidt rank is nψ = 2, i.e., |ψAB is entangled. Also, note that eigenvectors {|0, |1} of ρA corresponds to the positive eigenvalues μ1, 2 = (1)/(2). These eigenvalues satisfy the Eq. (1.29↑) since 2iμi = 1. Now, consider the product state |ψAB = |00, then ρA = |00|. 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
(1.34) |W = (|100 + |010 + |001)/((3))
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
(1.35) |ΨW = (x0)| 0 + pk = 1(xk)|1k for k = 1, 2, ⋯, p
where |0 = |00⋯0 and |1k is the state when the kth 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, x0 and all xk are real and positive that obey the sum (or normalization)
(1.36) x0 + pk = 1xk = 1.
Note that |W defined by Eq. (1.34↑) can not be written in a product state like
(1.37) |W ≠ |a1|a2|a3
so |W is necessarily an entangled state.
Another type of tripartite entangled state can be obtained as follows
(1.38) |GHZ = (|000 + |111)/((2))
which is known as the Greenberger-Horne-Zeilinger (GHZ) state[5.2↓]. This is closely related to the “GHZ-type” entangled state which is defined as
(1.39) |ΨGHZ = (1)/(N)(|0p + z|β1|β2⊗⋯⊗|βp)
where p is the number of parties involved, N is normalization constant, z is an arbitrary complex number. It is also known as p-particle Cat (p-Cat) state because of Schrödinger’s cat [5.2↓] when |βk = |1 for all k. Also, note that each state |βk in Eq. (1.39↑) has the form
(1.40) |βk = ck|0 + sk|1 ≡  ck sk
for each k with the obvious condition that sk ≠ 0 for all k at the same time. Here, note that all |βk’s are unit vector and ck and sk can be seen as cosine and sine functions of some real angle θk,  respectively, then it is obvious that
(1.41) c2k + s2k = 1
for all k’s. Again, it is clear that |ΨGHZ in Eq. (1.39↑) can not be written as a product state such that
(1.42) |ΨGHZ ≠ |a1|a2⊗⋯⊗|ap,
if at least two sines are non-zero.

## 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
(1.43) |φ = AB⊗⋯⊗N|ψ.
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:
(1.44) |ψ\undersetSLOCC ~ |φ iff |φ = AB⊗⋯⊗N|ψfor~some~invertible~A, B, …, N.
It is also true that
(1.45) |ψ = A − 1B − 1⊗⋯⊗N − 1|φ
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.,
(2.1) r(ρψk) = r(ρφk) for k = A, B, ⋯, N.
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 |ψAB
(2.2) |ψAB = α|00 + β|11
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
(2.3) E(ψAB) =  − tr(ρAlog2ρA) =  − tr(ρBlog2ρB)
where ρA and ρB are the reduced density matrices of the parties A and B, respectively. Then, for the state |ψAB given by the Eq. (2.2↑), the reduced density matrix ρA of A is calculated to be
(2.4) ρA  = trB(|ψψ|)AB = |α|2|00| + |β|2|11|  =  |α|2 0 0 |β|2 .
Putting Eq. (2.4↑) into Eq. (2.3↑) results in the entanglement
(2.5) E(ψAB)  =  − tr |α|2log2|α|2 0 0 |β|2log2|β|2  =  − |α|2log2|α|2 − |β|2log2|β|2.
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
(2.6) C(ψAB) = 2|α||β| = 2(detρA) = 2(detρB).
The amount of entanglement E for the pure state in Eq. (2.2↑) can be expressed as
(2.7) E(C) = h(1 + (1 − C2))/(2).
Here, h(x) is the function
(2.8) h(x) =  − xlog2x − (1 − x)log2(1 − x).
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 = |ψAB of two qubits can be expressed as
(2.9) C(ψ) = |ψ|ψ̃|
where |ψ̃ is the spin flip state for |ψ calculated by the formula
(2.10) |ψ̃ = σ⊗2y|ψ*
where |ψ* is the complex conjugate of |ψ in the computational basis. Here, σ⊗2y is given by σ⊗2y = σyσy for the Pauli operator σy =  0  − i i 0 in the computational basis set {|0, |1}. It acts separately on each qubit as σy|0 = i|1  and σy|1 =  − i|0. Then,
(2.11) σ⊗2y|00 =  − |11  and σ⊗2y|11 =  − |00.
Now, applying these operations on Eq. (2.2↑), the spin flip state is found as
(2.12) |ψ̃ =  − α*|11 − β*|00.
Then, concurrence is calculated to be
(2.13) C(ψAB) = 2|α||β|.
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 = ρAB of two qubits is defined by
(2.14) R(ρ) = ((ρ)ρ̃(ρ))
where ρ̃ is the spin flip state given by
(2.15) ρ̃ = σ⊗2yρ*σ⊗2y,
where ρ* is obtained by taking the complex conjugate of ρ in the computational basis. Let the eigenvalues of R(ρ) be
(2.16) Eigenvalues of R(ρ) = {λ1, λ2, λ3, λ4}
in non-increasing order, i.e., λ1 ≥ λ2 ≥ λ3 ≥ λ4. Then the concurrence is calculated as
(2.17) C(ρ) = max{0, λ1 − λ2 − λ3 − λ4}.
For any pure state |ψ, the density matrix ρ is ρ = |ψψ| and the spin-flipped state ρ̃ is ρ̃ = |ψ̃ψ̃|. Note that ρ = |ψψ| is a four dimensional density matrix and is already diagonal or spectrally decomposed (i.e. a projector [B]  [B] Any projector has eigenvalues 1 and 0. ) 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)|ψψ| = ρ. Therefore, using the Dirac representations for (ρ) and ρ̃,  R(ρ) is found as
(2.18) R(ρ)  = ((ρ)ρ̃(ρ)) = ((|ψψ|)(|ψ̃ψ̃|)(|ψψ|))  = |ψ|ψ̃|ρ = C(ψ)ρ
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
(2.19) {λ1, λ2, λ3, λ4} = {C(ψ), 0, 0, 0}.
Thus, using Eq. (2.17↑), the concurrence C(ρAB) is
(2.20) C(ρAB)  = C(ψAB) = 2|α||β|.
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
(2.21) Eigenvalues of ρρ̃ = {γ1, γ2, γ3, γ4}
are used to find the concurrence so that
(2.22) C(ρ) = max{0, (γ1) − (γ2) − (γ3) − (γ4)}.
This can be shown as follows. It is a well-known 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
(2.23) A − 1(AB)A = BA.
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 non-invertible. Therefore, R2 = (ρ)ρ̃(ρ) and (ρ)(ρ)ρ̃ = ρρ̃ are isospectral. Note that ρρ̃ is not Hermitian. But, since R2 is a positive definite and hermitian, the eigenvalues of ρρ̃ are real and non-negative.
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 = ρAB of two qubits A and B. Let us suppose that ρAB has matrix rank 2, i.e., it has only 2 non-zero eigenvalues. Let us suppose that its eigenvalues are qi and eigenvectors are |ψiAB (i = 1, 2). Therefore, we have
(2.24) ρAB = 2i = 1qi(|ψiψi|)AB.
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 |ΨABC and ρAB = trC(|ΨΨ|)ABC. The state |ΨABC is usually termed a purification of ρAB. A straightforward purification of Eq. (2.24↑) is obtained as
(2.25) |ΨABC = (q1)|ψ1AB|0C + (q2)|ψ2AB|1C.
Therefore, |ΨABC is a pure state of 3 qubits when ρAB 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 ρAB separately. Namely, those states where purification is of W class and those ones whose purification is of GHZ class. Using well-known representations of these pure states, we will obtain the concurrence of the mixed state ρAB.

### 2.2.1 The case where the purification is of W class

Consider three qubits A, B and C. Let ABC be in a 3-partite W-type entangled state |ΨABC defined by Eq. (1.35↑) for p = 3 such that
(2.26) |ΨABC = (x0)|000 + (x1)|100 + (x2)|010 + (x3)|001
where the coefficients x0 and each xk for k = 1, 2, 3 are some positive real numbers which satisfy
(2.27) x0 + 3k = 1xk = 1.
In the following sections, from the pure state |ΨABC, 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 |ΨABC as AB-C like in the purification of ρAB given by Eq. (2.17↑) as follows
(2.28) |ΨABC = ((x0)|00 + (x1)|10 + (x2)|01)AB|0C + (x3)|00|1C
with
(2.29) |ψ1AB = (x0)|00 + (x1)|10 + (x2)|01  and |ψ2AB = (x3)|00
which are unnormalized vectors. Here, also q1 = 1 and q2 = 1 which are nothing to do with the eigenvalues given by Eq. (2.24↑). The density operator ρWABC of ABC system is obtained then
ρWABC  = (|ΨΨ|)ABC  = (|ψ1ψ1|)AB(|00|)C + (|ψ1ψ2|)AB(|01|)C  + (|ψ2ψ1|)AB(|10|)C + (|ψ2ψ2|)AB(|11|)C.
Therefore the reduced density matrix ρWAB is
(2.30) ρWAB = trCρWABC = (|ψ1ψ1|)AB + (|ψ2ψ2|)AB.
Here, in the intermediate steps, Eqs. (1.26↑) and (1.27↑) are used for the trace operations. On the other hand, ρWAB can be represented in a matrix form in the computational basis set for two qubits substituting |ψ1AB and |ψ2AB defined by Eq. (2.29↑) into Eq. (2.30↑) as
(2.31) ρWAB =  x0 + x3 (x0x2) (x0x1) 0 (x0x2) x2 (x1x2) 0 (x0x1) (x1x2) x1 0 0 0 0 0 .
Since xk’s in Eq. (2.26↑) are real for all k, then ρWAB is a real matrix. Also, the tensor product σ⊗2y is obtained in a matrix representation as
(2.32) σ⊗2y =  0  − iσy iσy 0  =  0 0 0  − 1 0 0 1 0 0 1 0 0  − 1 0 0 0 .
Thus, the spin flip state is easily calculated by matrix multiplication as
(2.33) ρ̃WAB  = σ⊗2yρW*ABσ⊗2y =  0 0 0 0 0 x1 (x1x2)  − (x0x1) 0 (x1x2) x2  − (x0x2) 0  − (x0x1)  − (x0x2) (x0 + x3)
in the computational basis set. The matrix multiplication of ρWAB with ρ̃WAB is then
(2.34) ρWABρ̃WAB =  0 2x1(x0x2) 2x2(x0x1)  − 2x0(x1x2) 0 2x1x2 2x2(x1x2)  − 2x2(x0x1) 0 2x1(x1x2) 2x1x2  − 2x1(x0x2) 0 0 0 0 .
Then, the eigenvalues γi of the matrix ρWABρ̃WAB can easily be calculated since they are the roots of the characteristic equation
(2.35) c(γ) = γ3(γ − 4x1x2)
whose solution, i.e. c(γ) = 0, gives the set of eigenvalues γi as
(2.36) {γ1, γ2, γ3, γ4} = {4x1x2, 0, 0, 0}
in non-increasing order. Square root of the eigenvalues γi of ρWABρ̃WAB gives the set of eigenvalues λi of R(ρWAB) as
(2.37) {λ1, λ2, λ3, λ4} = {2(x1x2), 0, 0, 0}
in non-increasing order. Substitute the eigenvalues λi given by Eq. (2.37↑) into Eq. (2.17↑) for the concurrence to get
(2.38) C(ρWAB) = 2(x1x2)
since both x1 and x2 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 p-partite GHZ-type state. Let p = 3 so that three qubits A, B, and C are in the following tripartite GHZ-type entangled state
(2.39) |ΨABC = (1)/(N)(|000 + z|β1β2β3)
where N represents the normalization constant, and z is an arbitrary complex number. Here, |βi’s are some arbitrary unit state vectors defined by Eq. (1.40↑). In order for |ΨABC to be a unit vector, the normalization constant N should be equal to the norm of |000 + z|β1β2β3 such that
(2.40) N = ((000| + z*β1β2β3|)(|000 + z|β1β2β3)).
To simplify the calculations note that 0|βi = ci and 1|βi = si and substitute them into Eq. (2.40↑) to obtain
(2.41) N = (1 + 2c1c2c3ℜ(z) + |z|2)
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 W-type in the Sec. 2.2.1↑, it is not in here. Due to the fact that ρGHZABC 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 |ΨABC given by Eq. (2.39↑) as the sum of the tensor products of the subsystem AB and C as follows
(2.42) |ΨABC  = |a1AB|0C + |a2AB|1C (2.43)  = (1)/(N)(|00 + zc3|β1β2)AB|0C + (zs3)/(N)|β1β2AB|1C.
Then, we note that
(2.44) |a1AB = (1)/(N)(|00 + zc3|β1β2)  and |a2AB = (zs3)/(N)|β1β2.
Secondly, |a1AB and |a2AB can be given a matrix representation in the computational basis set. The tensor product |β1β2 is obtained as
(2.45) |β1β2 =  c1c2 c1s2 s1c2 s1s2
where the column vector representations of |β1A and |β2B given by Eq. (1.40↑) are used in Eq. (2.45↑).
Finally, substituting Eq. (2.45↑) for |β1β2 and Eq. (1.5↑) for |00 into Eq. (2.44↑), |a1AB and |a2AB have the column vector representations
(2.46) |a1AB = (1)/(N) 1 + zc1c2c3 zc1s2c3 zs1c2c3 zs1s2c3 ,  |a2AB = (zs3)/(N) c1c2 c1s2 s1c2 s1s2 .
Besides, by normalization of Eq. (2.42↑) and with the abbreviation |ai = |aiAB note that
(2.47) |ΨABC2 = 1 = a1|a1
of which use is made repeatedly throughout the thesis.
Thus, the reduced density matrix ρGHZAB of AB is found as
(2.48) ρGHZAB = trC(|ΨΨ|)ABC = |a1a1| + |a2a2| = 2i = 1|aiai|
which is a rank 2 mixed state. Alternatively, if ρGHZAB is written in a matrix representation putting Eq. (2.46↑) in Eq. (2.48↑), the columns of the 4 × 4 matrix ρGHZAB will be calculated as
(2.49) {1st Col.}  = (1)/(N2) 1 + (|z|c1c2)2 + 2c1c2c3ℜ(z) (|z|c1)2c2s2 + 2c1s2c3ℜ(z) (|z|c2)2c1s1 + 2s1c2c3ℜ(z) |z|2c1s1c2s2 + 2s1s2c3ℜ(z) ,  (2.50) {2nd Col.}  = (1)/(N2) (|z|c1)2c2s2 + 2c1s2c3ℜ(z) (|z|c1s2)2 |z|2c1s1c2s2 (|z|s2)2s1c2 ,  (2.51) {3rd Col.}  = (1)/(N2) (|z|c2)2c1s1 + 2s1c2c3ℜ(z) |z|2c1s1c2s2 (|z|s1c2)2 (|z|s1)2c2s2 ,  (2.52) {4th Col.}  = (1)/(N2) |z|2c1s1c2c2 + 2s1s2c3ℜ(z) (|z|s2)2s1c2 (|z|s1)2c2s2 (|z|s1s2)2 .
In order to obtain an outer product representation for ρ̃GHZAB, the following procedure is used. Firstly, find the spin flip of the orthonormal basis set given by Eq. (1.6↑) as
(2.53) Spin flip orthonormal basis set for 2 qubits :{|ij̃} for i, j = 0, 1
where
(2.54) |00̃  = σ⊗2y|00* = (i|1)(i|1) =  − |11,  (2.55) |01̃  = σ⊗2y|01* = (i|1)( − i|0) = |10,  (2.56) |10̃  = σ⊗2y|10* = ( − i|0)(i|1) = |01,  (2.57) |11̃  = σ⊗2y|11* = ( − i|0)( − i|0) =  − |00.
Secondly, use the spin flip states |β̃j = i  − sj cj with 0|β̃j =  − isj, 1|β̃j = icj,  and βj|β̃j = 0 and the tensor product |β̃1β̃2 of the spin flip states |β̃j as
(2.58) |β̃1β̃2 =   − s1s2 s1c2 c1s2  − c1c2
with
(2.59) 00|β̃1β̃2  =  − s1s2,  (2.60) 01|β̃1β̃2  = s1c2,  (2.61) 10|β̃1β̃2  = c1s2,  (2.62) 11|β̃1β̃2  =  − c1c2
as well as β1β2|β̃1β̃2 = β1|β̃1β2|β̃2 = 0.
Finally, express spin flip states |1 and |2 using Eq. (2.44↑) as
(2.63) |1 = (1)/(N)( − |11 + z*c3|β̃1β̃2)
and
(2.64) |2 = (z*s3)/(N)|β̃1β̃2.
All in all, start from Eq. (2.48↑) for ρGHZAB, then represent ρ̃GHZAB given by the Eq. (2.15↑) in the following form
(2.65) ρ̃GHZAB = |11| + |22| = 2i = 1|ii|.
A matrix representation of an operator A:V → W, where V and W are any two vector spaces, is defined to be
(2.66) A|vj = iAij|wi.
Here {|vj} and {|wi} 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, Aij are the matrix elements of the matrix representation of A. Aij is the entry in the ith row and the jth column.
If |a1 and |a2 are the two elements of the basis set in VAB (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 |a3 and |a4 can be chosen to be orthogonal to both |1 and |2, too, for convenience. Then, let ρGHZABρ̃GHZAB act on each element of the basis set {|aj} for j = 1, 2, 3, 4 to get
(2.67) ρGHZABρ̃GHZAB|aj = 2i = 1(AA)ij|ai
where Aik = ai|k. Therefore, ρGHZABρ̃GHZAB as an operator taking the vectors {|aj} from the vector space VAB to the same vectors {|ai} in the four dimensional vector space VAB represented by the 2 × 2 matrix AA.
Now, use the calculated ai|k’s given below
(2.68) a1|1  =  − 2(z*)/(N2)s1s2c3,  a1|2  =  − (z*s1s2s3)/(N2) = a2|1,  a2|2  = 0
to get
(2.69) A =  a1|1 a1|2 a2|1 a2|2 .
Then AA is obtained as
(2.70) AA =  |a1|1|2 + |a1|2|2 a1|1a1|2* a1|1*a1|2 |a1|2|2
in terms of only ai|k’s for convenience.
Eigenvalues of ρGHZABρ̃GHZAB are the solution of the characteristic equation
(3.1) c(γ) = det(AA − Iγ) = 0
with the solution
(3.2) γ1 = (|z|2s21s22)/(N4)(1 + c3)2  and γ2 = (|z|2s21s22)/(N4)(1 − c3)2.
Square roots of the eigenvalues γi found in Eq. (3.2↑) of ρGHZABρ̃GHZAB gives the set of eigenvalues λi of R(ρGHZAB) as
(3.3) {λ1, λ2, λ3, λ4} = (|z|s1s2)/(N2)(1 + c3), (|z|s1s2)/(N2)(1 − c3), 0, 0
in non-increasing order. Using Eq. (2.17↑), the concurrence C(ρGHZAB) is finally calculated to be
(3.4) C(ρGHZAB) = (2|z|s1s2c3)/(N2) = |a1|1|.

# 3 OPTIMAL ENSEMBLE REPRESENTING MIXED STATES

In this chapter, we can invoke what Kirkpatrick [5.2↓] call as Schrödinger-HJW theorem, a very useful result which has been discovered and re-discovered 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
(3.5) ni = 1|αiαi| = mj = 1|βjβj|
where |α1, ⋯, |αn, |β1, ⋯, |βm 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
(3.6) |αi  = mj = 1Uij|βj (fori = 1, 2, ⋯, n, n + 1, ⋯, m),  (3.7) |βj  = mi = 1U*ij|αi (forj = 1, 2, ⋯, m)
where we define |αn + 1 = |αn + 2 = ⋯ = |αm = 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 = ρAB with matrix rank 2 so that it has only 2 non-zero eigenvalues. Therefore, we have
(3.8) ρAB = 2i = 1βi|vivi|
where βi are the eigenvalues corresponding to the mutually orthogonal eigenvectors |vi (i = 1, 2).
In our case, consider any arbitrary ensemble ℰ = {ri, |wi} with n = 2 states that realizes ρAB, then
(3.9) ρAB = 2i = 1βi|vivi| = 2i = 1ri|wiwi|
where ri is the weight of the state |wi in the ensemble ℰ. Meanwhile, Eq. (3.9↑) can be expressed as
(3.10) ρAB = 2i = 1(|wiwi|)sub = 2i = 1(|vivi|)sub.
Here, the subscript sub stands for the subnormalization of |vi which is defined by
(3.11) (vi|vi)sub = βi
and also
(3.12) (wi|wi)sub = ri.
Therefore, by Schrödinger-HJW theorem we can find 2 × 2 unitary matrix U such that
(3.13) |wisub = 2j = 1U*ij|vjsub (for i = 1, 2).
In this chapter, the main aim is to find the optimal ensemble [C]  [C] The superscript opt is the abbreviation of the word optimal.
(3.14) opt = {pi, |ψioptAB}
that represents ρAB for the matrix rank 2 states which are studied in chapter 2↑. This ensemble satisfies
(3.15) ρAB = ipi(|ψiψi|)optAB.
Moreover, for this ensemble the average value of entanglement reaches its minimum value, i.e.,
(3.16) E(C(ρAB)) = miniqiE(ψi, AB) = ipiE(ψopti, AB)
where {qi, |ψiAB} is any ensemble having density matrix ρAB and the minimization is carried out over such ensembles. Here, E(ψi, 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(ρWAB) = 2(x1x2) > 0 and C(ρGHZAB) = 2|z|s1s2c3 ⁄ N2 > 0. For this type of density matrices, Wootters [5.2↓] proposes three successive decompositions of ρAB using unitary and orthogonal transformations. These decompositions are represented by the ensembles of n = 2 pure states which are listed as the following
(3.17) (ρAB) =  1 = {ri, |wi}    2 = {qi, |yi}    3 = {hi, |zi} .
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 {|wi} in the first ensemble 1 are obtained by a unitary matrix U from the eigenvectors {|vi} using the Schrödinger-HJW 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 tilde-inner products defined by τij = (vi|vj̃)sub. By restricting |wi as
(3.18) (wi|j)sub = λiδij
where λis are the eigenvalues of R(ρWAB) so that eigenvalues of ττ* are equal to the absolute squares of λi.
Later, the set of states |yi in the second ensemble 2 are obtained from the set of states {|wi} in the previously determined ensemble 1. First, Wootters [5.2↓] defines the preconcurrence for any pure state |ψ as follows
(3.19) c(ψ) = ((ψ|ψ̃)sub)/((ψ|ψ)sub).
Next, the average preconcurrence of the ensemble 2 = {qi, |yi} is chosen to be
(3.20) c(2) = iqic(yi) = λ1 − λ2 − λ3 − λ4 = C(ρAB).
Here, C(ρAB) is the concurrence of the mixed state ρAB. Thus, Eq. (3.20↑) restricts the elements |yi to the relations
(3.21) |y1sub = |w1sub
and
(3.22) |yjsub = i|wjsub (for j = 2, 3, 4)
because of the Eq. (3.18↑) for the elements |wi. So, |yi is obtained from |wi easily.
The third ensemble 3 = {hi, |zi}, which will be our optimal ensemble, is obtained from the elements |yi of the second ensemble 2 using real positive determinant orthogonal matrix V. In this case, preconcurrences c(zi) of each state |zi in the ensemble are obtained by equating them to the average preconcurrence c(ℰ3) of the ensemble 3, i.e., C(ρWAB) . This means that average entanglement ihiE(zi) is equal to the entanglement E(C(ρWAB)) of ρWAB 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 ρGHZAB, the known decomposition ρGHZAB = 2i = 1|aiai| given by Eq. (2.48↑) is chosen to be the starting point. It is reasonable since finding the eigenvalues and eigenvectors of ρGHZAB is somewhat cumbersome. Therefore, the tilde inner product now is calculated by τij = ai|aj̃. Now, construct the set of states {|wi} in the first ensemble 1 by the unitary matrix U from the set {|ai} using the Schrödinger-HJW theorem by Eq. (3.13↑), i.e., (ri)|wi = 2j = 1U*ij|aj for (i = 1, 2). Then, the rest, i.e. determination of |yi and |zi, is the same as the case of ρWAB.

## 3.1 Mixed states with W class purifications

The eigenvalues βi of ρWAB are the roots of the characteristic equation
(3.23) c(γ)  = det(ρWAB − βI) = β2{β2 − β + x3(x1 + x2)}
as
(3.24) {β1, β2, β3, β4}  = (1 + ())/(2), (1 − ())/(2), 0, 0
where △ = 1 − 4x3(x1 + x2).
It is now sufficient to find the eigenvectors corresponding to the two nonzero eigenvalues β1 and β2 given by the Eq. (3.24↑). The solution to the eigenvector-eigenvalue relations ρWAB|vi = βi|vi in terms of the entries of |vi is found to be
(3.25) |vi =  ai bi ci di  = ci (1)/((x0x1))(βi − (x1 + x2)) ((x2)/(x1)) 1 0 .
It can be normalized for appropriate choice of ci, but use the form given by the Eq. (3.25↑) for simplicity.
Subnormalization of |vi defined by the Eq. (3.11↑) is given by
(3.26) (vi|vi)sub = ((βi − x1 − x2)2 + x0x2)/(x0x1) + 1c2i = βi
whose solution for ci is
(3.27) ci = ((x0x1x3)/((2βi − 1)(βi − 1 + x3))).
Now, |vi is subnormalized if the value ci found in Eq. (3.27↑) is put into Eq. (3.25↑) leading to |visub. It is convenient to leave these results with ci 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 Schrödinger-HJW theorem
(3.28) |wisub = 2j = 1U*ij|vjsub (fori = 1, 2)
for a 2 × 2 unitary matrix U. Therefore, the first ensemble 1 = {ri, |wi} is obtained from the spectral decomposition of ρWAB, i.e. {βi, |vi}. The subnormalized states |visub are calculated by Eq. (3.25↑) where ci 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
(3.29) τij = (vi|vj̃)sub.
Here |vj̃sub is the spin flip state |vjsub defined by Eq. (2.10↑) such that
(3.30) |vj̃sub = σ⊗2y|v*jsub.
It is easy to show that the matrix τ defined by Eq. (3.29↑) is symmetric, i.e. τij = τji. Using Eq. (3.29↑) for |vj̃sub we have
(3.31) τij  = (vi|σ⊗2y|v*j)sub = (v*j|(σ⊗2y)|vi)*sub  = (v*j|σ⊗2y|vi)*sub = (v*j|vĩ*)*sub  = (vj|vĩ*)*sub = (vj|vĩ)sub = τji
because σ⊗2y is Hermitian and real by Eq. (2.32↑). As a result, τ is a symmetric matrix. However, it is not necessarily Hermitian since
(3.32) (τ)ij = τ*ji = (vj|vĩ)*sub = (vĩ|vj)sub ≠ τij.
For the set {λi} of the non-negative eigenvalues Eq. (2.37↑) of the R matrix, R(ρWAB), let |wisub be given by Eq. (3.28↑). It satisfies the condition (wi|j)sub = λiδij as previously assumed by Eq. (3.18↑) where the spin flip state |wĩsub is
(3.33) |wĩsub = σ⊗2y|w*isub = 2j = 1Uij|vj̃sub.
Then, Eq. (3.18↑) can also be written in terms of U and τ as the following
(3.34) wi|wj̃sub  = 2k, l = 1UikUjlvk|vl̃sub = 2k, l = 1UikτklUTlj  = (UτUT)ij = λiδij.
Eq. (3.34↑) implies that UτUT is diagonal with the diagonal elements λi and there is a unitary U that diagonalizes τ. However, the diagonalization of ττ* gives us a wider aspect. First, multiply UτUT with (UτUT)* = U*τ*U to obtain
(3.35) {UτUTU*τ*U}ij  = {Uτ(UU)*τ*U}ij = (Uττ*U)ij  = k(UτUT)ik(U*τ*U)kj  = |λi|2δij
so
(3.36) (Uττ*U)ij = |λi|2 = λ2i
since λ1 = 2(x1x2) and λ2 = 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(ρWAB). This means that ττ* has a spectral spectral decomposition
(3.37) ττ* = 2i = 1λ2i|titi|
if |tis are the orthonormal eigenvectors of ττ* corresponding to the eigenvalues λ2i.
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 λ2i as the following
(3.38) U = ( |t1 |t2 ).
So U satisfies the completeness equation UU = UU = I. Indeed, U diagonalizes ττ* to its spectral form as the following
(3.39) Uττ*U =  t1| t2| (ττ*)( |t1 |t2 ) =  λ21 0 0 λ22
where we used Eq. (3.37↑) to have ττ*|ti = λ2i|ti.
Now, we find the elements τij by Eq. (3.29↑) for |visub’s in terms of the coefficients ci by Eq. (3.27↑) and we get
(3.40) τij = vi|vj̃sub = 2(((x2)/(x1)))cicj.
Next, construct the matrix elements (ττ*)ij of ττ* as
(3.41) (ττ*)ij = 2k = 1τikτ*kj = (4x2)/(x1)(cicj)2k = 1c2k.
After detailed calculations, we find the following expression 2k = 1c2k = x1 and get (ττ*)ij = 4x2cicj such that
(3.42) ττ* = 4x2 c21 c1c2 c1c2 c22
which is a Hermitian matrix as proved in Eq. (3.37↑).
Using the fact that ττ* has only one-nonzero eigenvalue λ21 = 4x1x2 with the others being λ2, 3, 4 = 0, the eigenvector of ττ* corresponding to the eigenvalue λ21 can be found as
|t1 = (1)/((x1)) c1 c2 .
The second eigenvector corresponding to the eigenvalue λ22 = 0 will be orthogonal to |t1. Therefore, we obtain
|t2 = (1)/((x1)) c2  − c1 ,
an therefore the unitary matrix U is
(3.43) U =  t1| t2|  = (1)/((x1)) c1 c2 c2  − c1 .
As a result, using Eq. (3.43↑) for U, then |w1sub of the first decomposition 1 is calculated as
(3.44) |w1sub  = 2j = 1U*1j|vjsub  = (c1)/((x1))|v1sub + (c2)/((x1))|v2sub.
After detailed calculations, we get
(3.45) |w1sub = (x0)|00 + (x2)|01 + (x1)|10.
Therefore, putting Eq. (3.45↑) into the subnormalization formula for |w1 by Eq. (3.12↑), we find the probability r1 to obtain |w1 in the ensemble 1 = {ri, |wi} as
(3.46) w1|w1sub = r1 = x0 + x1 + x2 = 1 − x3
and |w1 is found to be
(3.47) |w1  = (1)/((r1))|w1sub  = (1)/((1 − x3))((x0)|00 + (x2)|01 + (x1)|10).
Next, to find the second member |w2 of the ensemble 1, use the preceding procedure to get
|w2sub  = 2j = 1U*2j|vjsub (3.48)  = (c2)/((x1))|v1sub − (c1)/((x1))|v2sub (3.49)  = (x3)|00
and by Eq. (3.12↑), we find the probability r2 for |w2 in the ensemble 1 as
(3.50) w2|w2sub = r2 = x3.
Finally |w2 is found to be
(3.51) |w2 ≡ (1)/((x3))|w2sub = |00.
For consistency, note that the sum of probabilities {r1, r2} found in Eqs. (3.46↑) and (3.50↑) is 1, i.e., r1 + r2 = 1 − x3 + x3 = 1.

### 3.1.2 The second ensemble

The set of states |yi in the second ensemble 2 = {qi, |yi} (i = 1, 2) are obtained from the first ensemble 1 = {ri, |wi} found in Sec. 3.1.1↑ so that
(3.52) 2i = 1(|yiyi|)sub = 2i = 1(|wiwi|)sub = ρWAB
where |yisub = (qi)|yi.
Using the preconcurrence formula given by Eq. (3.19↑) for |yi, we get
(3.53) c(yi) = ((yi|yĩ)sub)/((yi|yi)sub) = ((yi|yĩ)sub)/(qi).
Next, the average preconcurrence of the ensemble 2 = {qi, |yi} is chosen to be equal to
(3.54) c(2) = 2i = qic(yi) = λ1 − λ2 = 2(x1x2) = C(ρAB).
Here, C(ρAB) is the concurrence of the mixed state ρWAB. Thus, Eq. (3.54↑) restricts the elements |yi to such relations
(3.55) |y1sub = |w1sub
and
(3.56) |y2sub = i|w2sub
since (wi|wj̃)sub = λiδij by Eq. (3.18↑). Therefore, 2 satisfies c(2) = C(ρAB) given by Eq. (3.54↑) such that
(3.57) c(2)  = (y1|1)sub + (y2|2)sub  = (w1|1)sub − (w2|2)sub  = λ1 − λ2 = 2(x1x2) = C(ρAB).
Now, we derive the elements {|yi}i = 1, 2 of the second decomposition 2 = {qi, |yi} realizing ρWAB from the first ensemble 1 = {ri, |wi} obtained is the Sec.(3.1.1↑) as
(3.58) |y1sub  = (q1)|y1 = |w1sub  = (x0)|00 + (x2)|01 + (x1)|10,
then
(3.59) |y1 = (1)/((1 − x3))((x0)|00 + (x2)|01 + (x1)|10)
and
(3.60) |y2sub = (q2)|y2 = i|w2sub = i(x3)|00
then
(3.61) |y2 = (1)/((q2))|y2sub = i|00.
Note that the probabilities qi of |yi in the ensemble 2 are equal to the probabilities of ri of |wi in the ensemble 1 given by , i.e. q1 = r1 = 1 − x3  and q2 = r2 = x3.

### 3.1.3 The optimal ensemble

#### 3.1.3.1 Prescription

Compared to the other first two ensembles 1 and 2 discussed in the previous sections, now let the last ensemble, 3 = {hi, |zi}, be the optimal ensemble as Wootters proposes [5.2↓]. This decomposition is arranged in such a way that preconcurrences c(zi) of each states |zi in the ensemble are equal to the average preconcurrence c(ℰ3) of the ensemble 3 which is equal to C(ρWAB). Accordingly, entanglement E(zi) of each states |zi in the ensemble will be equal to the average entanglement E(ℰ3) = 2i = 1hiE(zi) of the ensemble which is equal to E(C(ρAB)).
Now, we make use of the formula for the first ensemble 1 given by Eq. (3.13↑) of ρWAB so that
(3.62) |zisub = 2j = 1V*ij|yjsub
for a 2 × 2 unitary matrix V.
For further discussions, Wootters [5.2↓] defines a 2 × 2 matrix Y whose elements are given by
(3.63) Yij = yi|yj̃sub
which are found to be
(3.64) Y11 = y1|y1̃sub = 2(x1x2)
with the other elements being zero. Consequently, written in a matrix form, Y becomes
(3.65) Y =  2(x1x2) 0 0 0
which is a real diagonal matrix.
Thus, the average preconcurrence is
(3.66) c(ℰ3)  = 2i = 1hic(zi) = 2i = 1zi|zĩsub  = 2i = 12k, l = 1VikYklVTli = 2i = 1(VYVT)ii (3.67)  = tr(VYVT).
In this decomposition, let the unitary matrix V be also a real matrix, that is, it is an orthogonal matrix with the property V = VT ⇒ VTV = I ⇒ VT = V − 1. In such a case, the average preconcurrence c(ℰ3) 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
(3.68) c(ℰ3)  =  tr(VYVT) = tr(VYV − 1)  =  tr(V − 1VY) = trY = 2(x1x2) = C(ρWAB).
Put it in another way, the last decomposition 3 = {hi, |zi} derived by the orthogonal transformations of {|yi} is the optimal ensemble opt. However, as described in the paragraph above, we are interested only in the transformation that makes the individual preconcurrences c(zi) equal to the average preconcurrence c(ℰ3) of the ensemble.

#### 3.1.3.2 Orthogonal matrix V

As discussed in the previous section, the preconcurrence of |z1 must be equal to C(ρWAB) such that
(3.69) c(z1)  = (z1|z1̃sub)/(z1|z1sub) = ((VYVT)11)/(h1) = 2(x1x2)
and of |z2:
(3.70) c(z2) = (z2|z2̃)/(z2|z2) = ((VYVT)22)/(h2) = 2(x1x2).
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.,
(3.71) det(V) = V11V22 − V12V21 > 0,
the following matrix can be obtained
(3.72) V =  (h1)  − (h2) (h2) (h1) .

#### 3.1.3.3 The third ensemble

Ultimately, one can construct the set of states {|zi} in the third ensemble 3 = {hi, |zi} from the set of states {|yi} in the second ensemble 2 = {qi, |yi} by the orthogonal matrix V given by Eq. (3.72↑). We use Eq. (3.62↑) to get
(3.73) |z1sub  = V*11|y1sub + V*12|y2sub  = ((h1x0) − i(h2x3))|00  + (h1)((x2)|01 + (x1)|10)
and
(3.74) |z2sub = ((h2x0) + i(h1x3))|00 + (h2)((x2)|01 + (x1)|10).
Finally optimal states which are equal to {|zi} become
(3.75) |ψopt1AB = ((x0) − i((p2x3)/(p1)))|00 + (x2)|01 + (x1)|10
with probability p1 and
(3.76) |ψopt2AB = ((x0) + i((p1x3)/(p2)))|00 + (x2)|01 + (x1)|10
with probability p2 for being in the optimal ensemble opt = {pi, |ψoptiAB}. However, by normalization of |ψoptiAB and the equation 3m = 0xm = 1, the values of pi’s are obtained easily as
(3.77) p1 = p2 = (1)/(2).
Therefore, the following results are obtained
(3.78) V =  1 ⁄ (2)  − 1 ⁄ (2) 1 ⁄ (2) 1 ⁄ (2) ,
(3.79) |ψopt1AB = ((x0) − i(x3))|00 + (x2)|01 + (x1)|10,
and
(3.80) |ψopt2AB = ((x0) + i(x3))|00 + (x2)|01 + (x1)|10.

## 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 ρGHZAB. However, as shown in the Appendix A, the eigenvalues and eigenvectors of ρGHZAB are not manipulated easily like in the case of ρWAB (See Sec. 3.1↑). Then, we claim that since we have an available ensemble representing ρGHZAB, i.e., ρGHZAB = 2i = 1|aiai| 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 ρGHZAB 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
(3.81) a = {ai, |ai}
realizing ρGHZAB given by Eq. (2.48↑). Namely,
(3.82) ρGHZAB = 2i = 1|aiai| = 2i = 1pi|aiai|
for the already subnormalized states |ai = (ai)|ai with the probability ai being the square of norm of |ai.
We argue that the set of states {|wi} in the first ensemble 1 = {ri, |wi} defined by Eq. (3.13↑) can directly be obtained by the set of states {|ai} in the ensemble a = {ai, |ai} given by Eq. (3.81↑). This is to say that there is a unitary matrix U between the states {|wi} and {|ai} with the following relation
(3.83) |wisub = 2j = 1U*ij|aj for i = 1, 2
where |wisub = (ri)|wi. Following the procedure used in the previous section, there is a constraint defined by Eq. (3.18↑) on the set of states {|wi} such that
(3.84) (wi|j)sub = λiδij.
Here, {λi} is the set of eigenvalues of the matrix R(ρGHZAB) given by Eq. (3.3↑). Hence, the elements of the matrix τ is obtained by the tilde inner product by
(3.85) τij = ai|j
so that
(3.86) Uττ*U = λ2iδij.
We will then find the unitary matrix with U =  t1| t2| which diagonalizes the matrix ττ*. Remember that |ti is the eigenvector of ττ* corresponding to the eigenvalue λ2i.
Now, we use the values of ai|j’s which are given by Eq. (2.68↑) which are τij’s defined by Eq. (3.85↑). Therefore, we have
(3.87) τ =  a1|1 a1|2 a2|1 0  = (a1|1)/(2c3) 2c3 s3 s3 0
since
(3.88) (a1|2)/(a1|1) = ( − z*s1s2s3 ⁄ N2)/( − 2z*s1s2c3 ⁄ N2) = (s3)/(2c3).
Notice that τ found in Eq. (3.87↑) is a real Hermitian matrix
(3.89) H =  2c3 s3 s3 0
multiplied by a complex number a1|1 ⁄ 2c3. In other words, τ* is also the same Hermitian matrix multiplied by a1|1* ⁄ 2c3. 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 α1, 2 = c3±1. Thus, the eigenvectors of H are obtained as
(3.90) |t1 = (1)/((2(1 − c3))) 1 − c3 s3 ,  |t2 = (1)/((2(1 − c3))) s3 c3 − 1
corresponding to the eigenvalues α1, 2 respectively.
Consequently, the unitary matrix U can be expressed as
(3.91) U = (1)/((2(1 − c3))) 1 − c3 s3 s3 c3 − 1
which is also a real Hermitian matrix. Hence, by the Eq. (3.83↑), the subnormalized forms of the states |wi of the first ensemble 1 are obtained from the states {|ai} in the ensemble a as
(3.92) |w1sub = (1)/((2(1 − c3))){(1 − c3)|a1 + s3|a2}
and
(3.93) |w2sub = (1)/((2(1 − c3))){s3|a1 + (c3 − 1)|a2}.
It can be shown that the probability ri of |wi in the ensemble 1 = {ri, |wi} is equal to the probability ai of |ai in a = {ai, |ai} so that
(3.94) r1 = a1 = |a12 = (N2 − s23|z|2)/(N2),  r2 = |a22 = (s23|z|2)/(N2)
where {|ai} are defined by Eq. (2.44↑). Finally, by definition |wisub = (ri)|wi and substituting the open forms of {|ai} given by Eq. (2.44↑) into the result, we have
(3.95) |w1 = κ1|00 + zc3 + (s23)/(1 − c3)|β1β2
and
(3.96) |w2 = κ2{|00 + z(2c3 − 1)|β1β2}
where
(3.97) κ1 = ((1 − c3)/(2(N2 − s23|z|2))),  κ2 = ((1)/(2|z|(1 − c3))).

### 3.2.2 The second ensemble

Similar discussions with the case of W class applies here to obtain the second ensemble 2 = {qi, |yi} (i = 1, 2) from the first ensemble 1 = {ri, |wi} found in Sec. 3.2.1↑. Therefore,
(3.98) |y1 = |w1 = κ1|00 + zc3 + (s23)/(1 − c3)|β1β2
and
(3.99) |y2 = i|w2 = iκ2{|00 + z(2c3 − 1)|β1β2}
where κ1 and κ2 are given by Eq. (3.97↑). Also, since |yisub = (qi)|yi and
(3.100) |yisub = |wisub,
then the probabilities qi of |yi in 2 are the same as the probabilities ri of |wi in 1, namely
(3.101) q1 = r1 = (N2 − s23|z|2)/(N2),  q2 = r2 = (s23|z|2)/(N2).

### 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
(3.102) Y =  λ1 0 0  − λ2 .
Here, λi’s are the eigenvalues of the matrix R(ρGHZAB) in non-increasing order given by the Eq. (3.3↑). Note that
(3.103) trY = λ1 − λ2 = C(ρGHZAB)
as also proved by Eq (3.68↑).
Now, in order for each preconcurrence of the states |zi in the third ensemble 3 = {hi, |zi} to be equal to C(ρGHZAB), we solve the following
(4.1) c(zi) = ((VYVT)ii)/(hi) = λ1 − λ2.
Therefore,
(4.2) λ1V211 − λ2V212 = h1(λ1 − λ2),  λ1V221 − λ2V222 = h2(λ1 − λ2).
By the orthonormality of the rows of V, we have
(4.3) V211 + V212 = 1,  V221 + V222 = 1,  V11V21 =  − V22V12.
Solving Eqs. (4.2↑) and (4.3↑) regarding the positivity of V, i.e., detV > 0,  we get
(4.4) V = (1)/((2)) (c3(2h1 − 1) + 1)  − (c3(1 − 2h1) + 1) (c3(2h2 − 1) + 1) (c3(1 − 2h2) + 1) .
Thus, the states |zi in 3 are obtained from the states |yi in 2 given by the Eqs.  (3.98↑) and (3.99↑). Since the third ensemble is the optimal ensemble, we get
(4.5) |ψoptj = μ1j|00 + μ2j|β1β2 for j = 1, 2
with probability pj = hj in the ensemble opt = {pj, |ψoptj}. Here, in terms of the entries Vij of V, the complex coefficients μ1j and μ2j are calculated as
(4.6) μ1j = ((1 + s3 − c3)(Vj1 − iVj2))/(N(2pj(1 − c3))),
(4.7) μ2j = (z)/(N(2pj(1 − c3))){Vj1[c3(1 − c3) + s23] + iVj2s3[2c3 − 1]}.

# 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: ρAB is stochastically reducible to ρAB (shown by ρ\undersetSLOCCρ) if there are operators A and B such that
(4.8) ρAB’ = (AB)ρ(AB)
and ρAB is stochastically equivalent to ρAB or SLOCC equivalent to ρAB (shown by ρ\undersetSLOCC ~ ρ) if ρ\undersetSLOCCρ and ρ\undersetSLOCCρ. It can be shown that ρAB is SLOCC equivalent to ρAB iff Eq. (4.8↑) holds for some invertible A and B.
Let ρAB be a matrix-rank 2 state of two qubits and ρ\undersetSLOCCρ’,  then matrix rank of ρAB is less than 2. If matrix rank of ρAB is 1 (i.e., if ρAB is a pure state) then one of A and B is not invertible and therefore ρAB 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 = c|αα| for some |α ∈ ℋA and constant c ∈ ℂ and therefore
(4.9) ρAB’ = (|αα|)AσB
for some 2 × 2 density matrix σ. It is apparent that ρ is unentangled and uncorrelated.
Therefore we can state the following: if ρ\undersetSLOCCρ with Eq. (4.8↑), and if ρAB has matrix rank 2, then
(4.10) (ρAB has matrix rank 2) ⇔ (Aand Bare invertible).
From now on, consider all transformations ρ\undersetSLOCCρ where both ρAB and ρAB have rank 2. In that case, ρ\undersetSLOCCρ as well, because only for invertible A and B one can satisfy Eq. (4.8↑).

## 4.1 Support of ρAB as a subspace

Let ρAB be a rank 2 mixed state of a two qubit system AB. ρAB has a spectral decomposition as
(4.11) ρAB = 2i = 1pi|ψiψi|
where {|ψi} is the orthonormal set of eigenvectors of ρAB corresponding to the eigenvalues {pi} of ρAB. Then, there is a 2-dimensional subspace of HAB = HAHB called as the support of ρAB or supp(ρAB), which is defined as the linear span of its eigenvectors
(4.12) supp(ρAB) = span{|ψ1, |ψ2}.
Any two-dimensional subspace of HAB 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:
1. Class 1. All the product states supp(ρAB) contains are parallel. Hence if
(4.13) |a = |α1A|α2B
is the product state in supp(ρAB), then supp(ρAB) 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(ρAB).
2. Class 2. In this case, supp(ρAB) contains two linearly independent product states. Call these
(4.14) |a = |α1A|α2B and |b = |β1A|β2B.
Therefore, any vector in supp(ρAB) can be written as a superposition of these.
In short, in both cases, one can find a basis {|a, |φ} of supp(ρAB) such that |a is a product state.
Now, we consider alternative representations of ρAB in the form
ρAB = 2i = 1|uiui = 2i = 1|uiui
where Eq. (4.11↑) is a special example. In all of these examples, {|u1, |u2} (similarly, {|u1, |u2}) spans the support of ρAB. In fact they form alternative bases for the support. By Schrödinger-HJW theorem, these alternative representations are related by |ui = jVij|uj. 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 |ψsubi = (pi)|ψi where |ψi are the eigenvectors and pi are the eigenvalues of ρAB, then
(4.15) 2i = 1|ψsubiψsubi| = 2i = 1pi|ψiψi| = ρAB.
Now, suppose that an unnormalized product state |P in supp(ρAB) is parallel to the product state |a defined by Eq. (4.13↑). |P can be written as a linear combination of {|ψsub1, |ψsub2} such that
(4.16) |P = κ|a = d1|ψsub1 + d2|ψsub2
for some complex numbers κ, d1 and d2. Here, we require that {d1, d2} satisfies |d1|2 + |d2|2 = 1. Note that
(4.17) |P2 = p1|d1|2 + p2|d2|2 ≤ 1
since |ψsub1 = (pi). Now, define a unitary
(4.18) V =  d1 d2  − d*2 d*1 .
It can be seen that this is a unitary matrix because its columns (also rows) form an orthonormal set. For another set of states {|ui} which realizes ρAB, it is true that |ui = nj = 1Vij|ψsubj then |u1 = |P = κ|a. Therefore, one can always find a representation {|ui} of ρAB where one of the states |u1 is parallel to a given product state |a. Also, |u2 will be some other state in supp(ρAB) which is
(4.19) |u2 =  − d*2|ψsub1 + d*1|ψsub2
so that
(4.20) 2i = 1|uiui| = 2i = 1|ψsubiψsubi| = ρAB.

## 4.2 States with class ℙ1 supports

Consider all representations of ρAB
(4.21) ρAB = |u1u1| + |u2u2|
where |u1 = κ|a for some product state |a and |u2 is necessarily entangled if supp(ρAB) is of class 1. Moreover, |u2 + z|u1 are always entangled for all z ∈ ℂ. Let
(4.22) ρAB = |u1u1| + |u2u2|
be another representation of ρAB such that |u1 is a product state. In this case, since supp(ρAB) contains only one non-parallel product state, we necessarily have |u1 = κ|a for some number κ. Because of the Schrödinger-HJW theorem we also know that then there is a unitary V such that |u1 = V11|u1 + V12|u2. But, if V12 were nonzero, then |u1 would have been entangled. However, it is chosen as a product state, so V12 = 0 necessarily and therefore V is diagonal. Then
(4.23) |ui = eiθi|ui (i = 1, 2)
The states are identical up to an overall phase factor.
In conclusion, if ρAB is such that supp(ρAB) contains at most one linearly independent product state, then, the decomposition of ρAB given by Eq. (4.21↑) such that |u1 is a product state is unique up to overall phases of |u1 and |u2.
Let ρAB be of class 1. One can then find a basis {|α0, |α1} of A = ℂ2 and a basis {|β0, |β1} of B = ℂ2 such that ρAB = |u1u1| + |u2u2| where
(4.24) |u1AB  = |α0A|β0B,  |u2AB  = |α0A|β1B + |α1A|β0B.
Unfortunately, this representation is not unique. Let ρAB = |u1u1| + |u2u2| where
(4.25) |u1  = eiθ1|u1 = |α0|β0,  |u2  = eiθ2|u2 = |α0|β1 + |α1|β0
then
(4.26) |α0  = z|α0,  |β0  = (eiθ1)/(z)|β0,  |β1  = (eiθ2|β1 − λ|β0)/(z),  |α1  = ze − iθ1(eiθ2|α1 + λ|α0)
where z ≠ 0, λ ∈ ℂ, θ1, θ2 ∈ ℝ are arbitrary.
Let ρAB and ρAB be of class 1. Can we stochastically reduce ρAB to ρAB, i.e., ρ\undersetSLOCC\overset?⟶ρ? We will show below that the answer is affirmative. Let ρAB = 2i = 1|uiui|, ρAB’ = 2i = 1|uiui| and let
(4.27) |u1  = |α0|β0,  |u2  = |α0|β1 + |α1|β0,  |u1  = |α0|β0,  |u2  = |α0|β1 + |α1|β0.
Can one find local operators A and B such that (AB)|ui = |ui(i, j = 2)? The problem can be solved by finding A and B such that
(4.28) A|αi = |αi,  B|βi = |βi.
The local operators A and B that satisfy these are not unique. In general, one can find A and B such that
(4.29) A|α0  = z|α0,  A|α1  = ze − iθ1(eiθ2|α1 + λ|α0),  B|β0  = (eiθ1)/(z)|β0,  B|β1  = (eiθ2|β1 − λ|β0)/(z)
for any given z ≠ 0, λ ∈ ℂ,  θ1, θ2 ∈ ℝ. For all of such A and B the relation ρAB’ = (AB)ρ(AB) is satisfied. This proves the claim, i.e., any two states ρAB and ρAB having supports of class 2 are SLOCC equivalent. Therefore, all density matrices with supports of class 2 form a SLOCC class.

## 4.3 States with class ℙ2 supports

Let |a and |b be two linearly independent product states given by Eq. (4.14↑). Then, either {|α1A, |β1A} is linearly independent or {|α2B, |β2B} is linearly independent or both. There are three cases that one can distinguish.
1. Class 2B (only B is mixed): While {|α1A, |β1A} is linearly dependent, {|α2B, |β2B} is linearly independent. For this case, |α2 is parallel to |α1, and all states in supp(ρAB) is of the form |α1A|ψB where |ψB is arbitrary. Therefore, ρAB is unentangled and uncorrelated, with A being in a pure state |α1A and B is in a mixed state (say σ)
(4.30) ρAB = (|α1α1|)AσB.
2. Class 2A (only A is mixed): While {|α2B, |β2B} is linearly dependent, {|α1A, |β1A} is linearly independent. For this case, |β2 is parallel to |α2 and all states in supp(ρAB) is of the form |ψA|α2B where |ψA is arbitrary. ρAB is unentangled and uncorrelated, with B being in a pure state |α2B and A is in a mixed state (say σ)
(4.31) ρAB = σA(|α2α2|)B.
3. Class 2AB (both A and B are mixed): Both {|α1A, |β1A} and {|α2B, |β2B} are linearly independent. This implies that the generic state
(4.32) |ψ  = c1|a + c2|b = c1|α1A|β1B + c2|α2A|β2B
is always entangled for c1 ≠ 0,  c2 ≠ 0.
It can be shown that all states having supports of class 2A and those that have supports of class 2B do form separate SLOCC classes. The case of 2AB is highly non-trivial and is studied in detail below.
If |u ∈ supp(ρAB) and |u is a product state then either |u = N|a or |u = N|b for some complex number N. So if ρAB is of type given by Eq. (4.21↑) and |u1 is a product state, then either |u1 is parallel to |a or to |b. Consider two possible representations of ρAB,
(4.33) ρAB = |u1u1| + |u2u2| = |u1u1| + |u2 ´u2|
where |u1 and |u1 are product states. If |u1 is parallel to |u1 then we have
(4.34) |ui = eiθi|ui,
i.e., the states in the two ensembles are identical up to an overall phase factor.
Now, consider the case where |u1 is parallel to |a and |u1 is parallel to |b. Let
(4.35) |u1 = x|a
(where x is real and x > 0) and
(4.36) |u2 = y|a + z|b.
We can choose y to be real and y ≥ 0. Obviously z ≠ 0. Thus,
(4.37) ρAB  = 2i = 1|uiui|  = (x2 + y2)|aa| + yz|ba|  + yz*|ab| + |z|2|bb|.
Similarly, let
(4.38) |u1 = x|b
and
(4.39) |u2 = y|b + z|a
where x and y are real, x’ > 0 and y’ ≥ 0, and again, z’ ≠ 0. Thus,
(4.40) ρAB  = 2i = 1|uiui|  = |z|2|aa| + yz|ab|  + yz*|ba| + (x2 + y2)|bb|
then
(4.41) |z|2 = x2 + y2,  yz’ = yz*,  x2 + y2 = |z|2.
If the expansion of ρAB is known, one can select the overall phase of |b (relative to that of |a) such that z is real and positive.
Let
(4.42) ρAB = R11|aa| + R12(|ab| + |ba|) + R22|ab|
where the overall phases of |b and |a have been redefined such that R12 is real and positive. Then,
(4.43) x2 + y2 = z2 = R11,  yz = yz’ = R12,  z2 = x2 + y2 = R22
⇒ z ≡ (R12) , z’ ≡ (R11),
(4.44) y ≡ (R12)/((R22)), y’ ≡ (R12)/((R11)),
x ≡ (R11 − (R212)/(R22)) = ((detR)/(R22)), x’ ≡ ((detR)/(R11)).
Note that even though the phases of |a and |b are adjusted such that R12 is real and positive, the inner product a|b might still have a phase.

## 4.4 SLOCC classes

Let ρ\undersetSLOCC ~ ρ, then
(4.45) supp(ρAB) contains 2  linearly independent product states (i.e., ρABis of class 2)  ⇔  supp(ρAB) contains 2  linearly independent product states (i.e., ρABis of class 2) .
Let us show this implication. Let ρAB = 2i = 1|uiui| and ρAB’ = 2i = 1|uiui| be representatives where |u1 and |u1 are product states and let ρAB’ = (AB)ρ(AB). Let us first show the forward implication (( ⇒ :) Suppose ρAB is of class 2,  so
(4.46) |u1  = x|a,  |u2  = y|a + z|b
where |a and |b are product states. Define
(4.47) |  = AB|a,  |  = AB|b,
that is, | and | are product states. Then,
(4.48) |1  = x|,  |2  = y| + z|
(4.49)  ⇒ ρAB’ = 2i = 1|\widetildeui\widetildeui|
so ρAB is also in class 2. The reverse implication (⇐:) is obvious as this is equivalent to ρ\undersetSLOCCρ.
If ρ\undersetSLOCC ~ ρ, then
(4.50) (ρABis of class 1) ⇔ (ρAB is of class 1).
This has been shown somewhere above. In general, if ρ\undersetSLOCC ~ ρ then
1. ρAB ∈ 1 ⇔ ρAB’ ∈ 1,
2. ρAB ∈ 2 ⇔ ρAB’ ∈ 2,
3. ρAB ∈ 2A ⇔ ρAB’ ∈ 2A,
4. ρAB ∈ 2B ⇔ ρAB’ ∈ 2B,
5. ρAB ∈ 2AB ⇔ ρAB’ ∈ 2AB.
Let ρAB be of class 1, ρAB = 2i = 1|uiui| and
(4.51) |u1 = c|aAB = c|α1A|α2B.
Now, redefine |α1 or |α2 such that c = 1,  then
(4.52) |u1 = |aAB = |α1A|α2B.
Let {|α1A, |β1A} be a basis of HA = ℂ2 and {|α2B, |β2B} be a basis of HB = ℂ2. Let
(4.53) |u2  = d|α1α2 + e|β1α2  + f|α1β2 + g|β1β2.
Here
(4.54) det d + z e f g  ≠ 0  for all z.
This implies that g = 0 (otherwise ρAB is of class 2). Then
(4.55) |u2 = d|α1α2 + e|β1α2 + f|α1β2.
Redefine |β1 and |β2 as follows:
(4.56) |β̃1  = e|β1 + λ|α1,  |β̃2  = f|β2 + μ|α2
(4.57)  ⇒ |u2 = (d − λ − μ)|α1α2 + |β̃1α2 + |α1β̃2
for given λ, select μ = d − λ which leads to the following corollary:
Let ρAB be of class 2AB. There is a basis {|α0, |α1} of HA = ℂ2 and a basis {|β0, |β1} of HB = ℂ2 such that ρAB = |u1u1| + |u2u2| where
(4.58) |u1  = c1|α0|β0,  |u2  = c2|α0|β0 + c3|α1|β1.
In general, one can absorb c1,  c2,  and c3 into the definitions of |αi,  |βi. But, |c2 ⁄ c1| can not be changed by such redefinitions.
(4.59) |u1  = |α0|β0 |u2  = k|α0|β0 + |α1|β1
where k is real with k ≥ 0.
Note that a purification of ρAB is
(5.1) |ψABC  = |u1AB|1C + |u2AB|0C  = |α1β1⊗0 + |α1β1(k|0 + |1)
where
(5.2) (k)/((k2 + 1)) = c3  or k = (c3)/((1 − c23)) = (c3)/(s3)
requirement is k ≠ ∞ (c3 ≠ 1). Hence the parameter k is related to the cosine c3 of 3rd party C. Let ρAB,  ρAB’ ∈ ℙ2AB. ρ\undersetSLOCCρ 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 ρGHZAB of the composite system AB, consult the tool of Schmidt decomposition of |ΨABC 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 ⊗4AB and ⊗2C, respectively. Therefore, it is to say that eigenvalues of the subsystems AB and C are the same and then that
(5.3) nψ ≤ min(4, 2) = 2
which means that ranks of ρGHZAB and ρGHZC are at most 2. For this reason, it is more convenient to deal with the eigenvalues and eigenvectors of 2 × 2 matrix ρGHZC rather than the 4 × 4 matrix ρAB [5.2↓,5.2↓,5.2↓].
ρGHZC is calculated by partial tracing the party AB from the total state |ΨABC, namely
(5.4) ρGHZC  = trC(|ΨΨ|)ABC  = |a12|00| + a2|a1|01|  + a1|a2|10| + |a22|11|
which can also be given a matrix representation
(5.5) ρGHZC =  |a12 a1|a2* a1|a2 |a22
in the computational basis set {|0, |1}. The set of eigenvalues μi are the solution of the characteristic equation
(5.6) c(μ)  = det(ρGHZC − μI)  = μ2 − μ + |a12|a22 − |a1|a2|2 = 0
so that
(5.7) μ1, 2 = (())/(2).
where is the discriminant defined by
(5.8) △ = 1 − 4(|z|2s23)/(N4){1 − c21c22}.
Schmidt decomposition given by Eq. (1.28↑) for |ΨABC of the system ABC into the subsystems AB − C gives
(5.9) |ΨABC = 2i = 1(μi)|iAB|iC
where |iAB and |iC are orthonormal set of the eigenstates of ρGHZAB and ρGHZC, respectively, corresponding to the eigenvalues μis in the decreasing order. One can represent the states |0 and |1 in the orthonormal basis set |1C and |2C such that
(5.10) |0  = \undersetx11C|0|1C + \undersetx22C|0|2C  = x1|1C + x2|2C
and
(5.11) |1  = \undersety11C|1|1C + \undersety22C|1|2C  = y1|1C + y2|2C
noting that
(5.12) |iC =  mi ni  =  0|iC 1|iC  =  x*i y*i .
Putting these values into Eq. 2.42↑ and then expanding also Eq. (5.9↑), we get
|ΨABC  = |a1|0 + |a2|1  = (x1|a1 + y1|a2)|1C (5.13)  + (x2|a1 + y2|a2)|2C (5.14)  = (μ1)|1AB|1C + (μ2)|2AB|2C
equating Eqn.s (5.13↑) and (5.13↑) gives the eigenstates |iAB of ρGHZAB in terms of the coefficients of the eigenstates |iC of ρGHZC and the vectors |a1 and |a2 as the following
(5.15) |iAB = (1)/((μi)){xi|a1 + yi|a2}
or
(5.16) |iAB = (1)/((μi)){m*i|a1 + n*i|a2}.
Eigenstates |iC of ρGHZC are determined by the eigenvector-eigenvalue relation
(5.17) ρGHZC|iC = μi|iC
as
(5.18) |iC = (1)/(Ni) mi n  = (1)/((±()(μi − |a22))) μi − |a22 a1|a2
where Ni is found as the following:
(5.19) N2i  = |mi|2 + |n|2  = μ2i + |a24 − 2μi|a22 + |a1|a2|2
adding c(μi) = μ2i − μi + |a12|a22 − |a1|a2|2 = 0 which is characteristic equation given by Eq. (5.6↑) to the right side gives
(5.20) N2i  = (μi − |a22)\underset±()(2μi − 1)  = ±()(μi − |a22).
Finally, the eigenvectors |iAB of ρGHZAB become
(5.21) |iAB = ((μi − |a22)|a1 + a1|a2*|a2)/((±μi()(μi − |a22)))
Subnormalization of |iAB is defined by iAB|iAB = μi which results in the subnormalized eigenstates
(5.22) |iABsub = ((μi − |a22)|a1 + a1|a2*|a2)/((±()(μi − |a22)))
or simply
(5.23) |iABsub = di(mi|a1 + n*|a2)
where
(5.24) di = N − 1i.

## 5.2 The first decomposition of ρGHZAB

(5.25) ρGHZAB = 2i = 1μi|iABiAB|
or in terms of the subnormalized eigenstates |iABsub ≡ (μi)|iAB
(5.26) ρGHZAB = 2i = 1|iABsubiAB|sub.
There is a unitary matrix U that transforms the states |iABsub to another set of states |wisub with the formula
(5.27) |wisub = 2i = 1U*ij|jABsub
so that
(5.28) 2i = 1|wisubwi|sub = 2j = 1|jABsubjAB|sub = ρGHZAB.
Thus, it means that there is an another ensemble {ri, |wi}, where ri is the weight of |wi in the ensemble, which represent the mixed state ρGHZAB such that
(5.29) ρGHZAB = 2i = 1ri|wiwi|.
Meanwhile, since C(ρGHZAB) ≥ 0,  then the density matrix ρGHZAB 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 {|wi} of ρGHZAB 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 |ti of the ττ* which construct the columns of U.
The matrix elements τij of τ are formed by “tilde inner products”:
(5.30) τij = iAB|\widetildejAB
where |\widetildejAB is the spin flipped state of the eigenstate |jAB of ρGHZAB defined by |\widetildejAB = σ⊗2y|j*AB by means of the Pauli-Y operator σ⊗2y = σyσy acting separately on each qubit such that
(5.31) |\widetildejAB = ((μj − |a22)|1 + a1|a2|2)/((±()(μj − |a22)))
or simply
(5.32) |\widetildejAB = di(mi|1 + n|2).
Since, the eigenvalues λ2i = γi of ττ* are known, it is possible to find the eigenvectors |ti of ττ* corresponding to them by the eigenvector-eigenvalue relation:
(5.33) (ττ* − γiI)|ti = 0
or equivalently in matrix notation
(5.34) (ττ*)11 − γi (ττ*)12 (ττ*)21 (ττ*)22 − γi ei fi  = 0
where (ττ*)ij’s are the matrix element of ττ*; ei and fi are the vector elements of |ti which can be chosen as ei = (ττ*)12 and fi = γi − (ττ*)11 then |ti can be written
(5.35) |ti = (1)/(ti) (ττ*)12 γi − (ττ*)11  = gi (ττ*)12 γi − (ττ*)11
with the normalization constant ti defined by ti = g − 1i = (|ei|2 + |fi|2). Therefore, the unitary matrix U can be written as
(5.36) U =  t1| t2|  =  g1(τ*τ)12 g1[γ1 − (τ*τ)11] g2(τ*τ)12 g2[γ2 − (τ*τ)11] .
By means of the complex conjugate of the unitary matrix U*
(5.37) U* =  t*1| t*2|  =  g1(ττ*)12 g1[γ1 − (ττ*)11] g2(ττ*)12 g2[γ2 − (ττ*)11] ,
the subnormalized states |wisub can be formed as
(5.38) |wisub = 2j = 1U*ij|jAB = gi{(ττ*)12|1AB + [γi − (ττ*)11]|2AB}
and or in terms of {|ai}
(5.39) |wisub  = gi{(d1m1(ττ*)12 + d2m2γi − d2m2(ττ*)11)|a1  + n*[d2γi + (ττ*)12 − (ττ*)11]|a2}
or simply
(5.40) |wisub = wi1|a1 + wi2|a2
where
(5.41) wi1 = gi[m2γi + m1(ττ*)12 − m2(ττ*)11]
and
(5.42) wi2 = (n*)/(gi)[γi + (ττ*)12 − (ττ*)11].
Then construct
(5.43) (ττ*)11 = τ11τ*11 + τ21τ*21,
(5.44) (ττ*)12 = τ11τ*12 + τ12τ*22,
(5.45) (ττ*)12 − (ττ*)11 = τ11(τ*12 − τ*11) + τ12(τ*22 − τ*21)
put them into
(5.46) m1(ττ*)12 − m2(ττ*)11 = τ11(m1τ*12 − m2τ*11) + τ12(m1τ*22 − m2τ*21)
to obtain {|wisub}.
If one can find anything about {|wisub} after the above calculations, next s/he has to find the second set of states {|yi} and finally, if it is possible, it is time to find the real orthogonal matrix V transforming the second set {|yi} to the optimal set {|zi}. Therefore, it is better to apply the new method to those type examples (see Sec. 3.2↑).

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