# Entanglement of assistance in three-qubit systems

Klée Pollock  Department of Electrical and Computer Engineering, Coordinated
Science Laboratory, University of Illinois at Urbana-Champaign, Urbana,
Illinois 61801, USA  Department of Physics and Astronomy, Iowa State
University of Science and Technology, Ames, Iowa 50014-3160, USA  Ge Wang
Department of Physics, University of Illinois at Urbana-Champaign, Urbana, IL
61801, USA  Eric Chitambar  Department of Electrical and Computer Engineering,
Coordinated Science Laboratory, University of Illinois at Urbana-Champaign,
Urbana, Illinois 61801, USA

(August 27, 2024)

###### Abstract

The entanglement of assistance quantifies the amount of entanglement that can
be concentrated among a group of spatially-separated parties using the
assistance of some auxiliary system. In this paper we study the entanglement
of assistance in the simplest scenario of three qubits. We consider how much
entanglement is lost when the helper party becomes uncoupled by local
measurement and classical communication. For three-qubit pure states, it is
found that lossless decoupling is always possible when entanglement is
quantified by the maximum probability of locally generating a Bell state.
However, with respect to most other entanglement measures, such as concurrence
and entanglement entropy, a lossless decoupling of the helper is possible only
for a special family of states that we fully characterize.

##  I Introduction

A quantum system prepared in a pure state is uncoupled from any other system
in the universe. Operationally, this means that if $p(a)$ is the outcome
probability distribution when measuring a pure state of system $A$ and $p(c)$
is the outcome distribution when measuring any other system $C$, then their
joint distribution will necessarily factorize: $p(a,c)=p(a)p(c)$. When this
fact is extended to pure states of bipartite systems, a monogamy relationship
follows [1, 2, 3]. That is, if systems $A$ and $B$ are in some pure entangled
state, then $C$ must be uncoupled from them both. In particular, $A$ and $C$
cannot also be entangled.

Figure 1: A pure-state decoupling transformation.

While the “correlation shielding” provided by the monogamy of entanglement is
essential for performing cryptographic tasks like quantum key distribution [4,
5], it creates fundamental challenges to quantum information processing using
multipartite entangled states. As the simplest case, consider the three-node
quantum network depicted in Fig. 1. Suppose the parties stationed at each node
share some tripartite entangled pure state $|\psi\rangle^{ABC}$, and they wish
to transform it into a bipartite entangled state $|\varphi\rangle^{AB}$ held
by parties Alice ($A$) and Bob ($B$). Due to the monogamy of entanglement,
Charlie ($C$) must become completely uncoupled from both Alice and Bob in the
transformation $|\psi\rangle^{ABC}\to|\varphi\rangle^{AB}$. In other words,
the final state is implicitly understood to have the tripartite form
$|\varphi\rangle^{AB}|0\rangle^{C}$ (or more generally
$|\varphi\rangle\langle\varphi|^{AB}\otimes\rho^{C}$). The uncoupling of
Charlie is not modeled simply by tracing out his system, i.e.
$|\varphi\rangle\langle\varphi|^{AB}\not=\textrm{Tr}_{C}|\psi\rangle\langle\psi|^{ABC}$,
and more elaborate protocols are needed to achieve transformations of the form
$|\psi\rangle^{ABC}\to|\varphi\rangle^{AB}$.

In one scenario, Charlie could be seen as an unwanted party, either an
eavesdropper or a noisy environment. With Charlie being adversarial, the
decoupling protocol would be driven exclusively by Alice and Bob, and the
problem then reduces to the task of entanglement purification [6, 7]. In
another scenario, Charlie is seen as a helper, and he provides assistance to
Alice and Bob with the state transformation. This assistance is provided by
Charlie making local measurements on his subsystem and communicating his
outcome to Alice and Bob. Due to the stochastic nature of quantum
measurements, it becomes more appropriate to consider multi-outcome
transformations, in which Charlie is decoupled for each possible outcome. We
can denote such a process as $|\psi\rangle^{ABC}\to|\varphi_{i}\rangle^{AB}$,
where $|\varphi_{i}\rangle^{AB}$ is obtained with probability $p_{i}$.

The amount of help that Charlie can provide in generating bipartite pure-state
entanglement for Alice and Bob is known as the Entanglement of Assistance
(EoA) [8]. It can be defined as the maximum average entanglement, $\langle
E_{A|B}\rangle:=\sum_{i}p_{i}E(\varphi_{i})$, over all transformations
$|\psi\rangle^{ABC}\to|\varphi_{i}\rangle^{AB}$ in which Charlie simply
decouples himself by performing a single measurement. Here, $E(\varphi_{i})$
is the entanglement of state $|\varphi_{i}\rangle^{AB}$ for some chosen
entanglement monotone $E$. Note, the measurement of Charlie’s which maximizes
EoA for one monotone may not be the same as that which maximizes EoA for
another, and we will see an example of this below. Nevertheless, all
entanglement monotones satisfy the monotonicity constraint

$\langle E_{A|B}\rangle\leq\min\\{E_{A|BC}(\psi),E_{B|AC}(\psi)\\},$ (1)

where $E_{A|BC}(\psi)$ is the bipartite entanglement in the initial state
$|\psi\rangle^{ABC}$ when parties $B$ and $C$ are grouped together, and
similarly for $E_{B|AC}(\psi)$. Ineq. (1) holds since monotonicity prevents
both the $A|BC$ entanglement and the $B|AC$ entanglement from increasing on
average.

For a given tripartite entangled state $|\psi\rangle^{ABC}$, a natural
question is whether Charlie can perform a measurement that decouples him in a
“lossless” fashion. This means that there exists a multi-outcome
transformation $|\psi\rangle^{ABC}\to|\varphi_{i}\rangle^{AB}$ such that Ineq.
(1) is tight, i.e. $\langle
E_{A|B}\rangle=\min\\{E_{A|BC}(\psi),E_{B|AC}(\psi)\\}$. Remarkably, it has
been shown that for asymptotic pure state distillation, an equality in Ineq.
(1) indeed holds after the LHS is replaced by the optimal rate of bipartite
entanglement yield [9]. The process of lossless asymptotic decoupling has been
generalized to the multi-party setting in the form of entanglement combing
[10]. However, in the single-copy setting, Ineq. (1) is generally strict.

The goal of this paper is to investigate which three-qubit states
$|\psi\rangle^{ABC}$ can achieve an equality in Ineq. (1). The answer to this
question depends on the choice of monotone $E$. Our main result is showing
that every three-qubit pure state can achieve $\langle
E_{2}^{A|B}\rangle=\min\\{E_{2}^{A|BC}(\psi),E_{2}^{B|AC}(\psi)\\}$ for some
measurement of Charlie, where $E_{2}(\varphi^{AB})$ is twice the smallest
eigenvalue of $\varphi^{A}=\textrm{Tr}_{B}|\varphi\rangle\langle\varphi|^{AB}$
[11]. Operationally this quantity is important because it corresponds to the
maximum probability of converting $|\varphi\rangle^{AB}$ into the maximally
entangled state $|\Phi^{+}\rangle=\sqrt{1/2}(|00\rangle+|11\rangle)$ using
LOCC [12]. Hence our result implies that the optimal probability of converting
any tripartite pure state $|\psi\rangle^{ABC}$ into a bipartite maximally-
entangled state between $AB$ is the same whether $C$ acts either locally or
jointly with one of the other parties. It also implies that the EoA for
$E_{2}$ is a genuine entanglement monotone in three qubits, a fact which does
not hold for entanglement monotones in general [13]. We prove this result by
showing that a sufficient condition for lossless $E_{2}$ decoupling presented
in Ref. [14] actually applies to all three-qubit states. We then consider
other entanglement monotones and provide the complete family of states for
which $\langle E^{A|B}\rangle=\min\\{E^{A|BC}(\psi),E^{B|AC}(\psi)\\}$ with
respect to any entanglement monotone that is a strictly concave function of
the square-Schmidt coefficients [15]. The latter includes the entanglement
entropy and the concurrence [16], to name a few.

##  II Definitions and Preliminaries

###  II.1 Bipartite Entanglement Measures

The discovery that the nonlocal correlations generated by quantum entanglement
are useful for communication and information processing has motivated an
effort to precisely quantify entanglement. Various entanglement measures have
been proposed in the literature, and all of them satisfy the necessary
property that their value cannot be increased under LOCC [17]. That is for any
entanglement measure $E$, if $\mathcal{L}$ is a completely-positive trace-
preserving (CPTP) map generated by a protocol of local operations and
classical communication between $N$ parties, then $E\left(\rho\right)\geq
E\left(\mathcal{L}(\rho)\right)$ for all $N$-partite states $\rho$. It may be
the case that $\mathcal{L}$ outputs a classical register (or “flag”) storing
various amounts of classical data or measurement outcomes $x$ obtained in the
protocol. This requires that

$E\left(\rho\right)\geq E\left(\sum_{x}p_{x}\rho_{x}\otimes|x\rangle\langle
x|\right),$ (2)

where $\rho_{x}$ is the quantum state conditioned on obtaining classical data
$x$ and $p_{x}$ is the probability of this event. States on the RHS of (2) are
known as qc (quantum-classical) states. Most entanglement measures of interest
are convex-linear on qc states, meaning that
$E\left(\sum_{x}p_{x}\rho_{x}\otimes|x\rangle\langle
x|\right)=\sum_{x}p_{x}E(\rho_{x}\otimes|x\rangle\langle
x|)=\sum_{x}p_{x}E(\rho_{x})$. From this we see

$E\left(\rho\right)\geq\sum_{x}p_{x}E\left(\rho_{x}\right),$ (3)

and any function satisfying this relationship is known as an entanglement
monotone. Throughout this paper we will only consider measures that are
convex-linear on qc states, and so Eq. (3) can make no distinction between the
terms entanglement measure and monotone in terms of their behavior under LOCC.

The general structure of bipartite entanglement monotones has been analyzed by
Vidal in Ref. [15]. For LOCC entanglement transformations in which the initial
and final states are pure, it suffices to consider monotones that are defined
just on pure states. Vidal shows that every pure state entanglement monotone
is determined by a concave, symmetric function of the squared-Schmidt
coefficients of the given state $|\varphi\rangle^{AB}$ (equivalently of the
eigenvalues of the reduced density matrix $\varphi^{A}$). For example, the
Entanglement Entropy
$S_{1}(\varphi^{AB})=-\textrm{Tr}[\varphi^{A}\log\varphi^{A}]$, and more
generally the $\alpha$-Entanglement Entropy
$S_{\alpha}(\varphi^{AB})=\frac{1}{1-\alpha}\log{\textrm{Tr}{(\varphi^{A})^{\alpha}}}$
for $\alpha\in[0,1]$, all arise from a concave, symmetric function
$g:\Delta_{n}\rightarrow\mathbb{R}^{+}$ on the $n$-dimensional unit simplex
$\Delta_{n}$. Another family of monotones are the so-called Ky-Fan functions
$\\{E_{k}\\}_{k=1}^{d}$ [11], which are given by
$E_{k}(\varphi^{AB})=\sum_{n=k}^{d}\lambda^{\downarrow}_{k}$, where the
$\lambda^{\downarrow}_{k}$ are the eigenvalues of $\varphi^{A}$ arranged in
decreasing order and $d=\min\\{\dim(\varphi^{A}),\dim(\varphi^{B})\\}$. Note
that $E_{k}$ generalizes the $E_{2}$ measure mentioned above. A final example
is the generalized concurrence (G-concurrence)
$G(\varphi^{AB})=d\det[\varphi^{A}]^{\frac{1}{d}}$ [18], which for the two-
qubit case ($d=2$) reduces to the original concurrence [16], as well as the
robustness of entanglement [19].

###  II.2 The Entanglement of Collaboration and Assistance

As depicted in Fig. 1, the general problem we consider is the decoupling of
some party $C$ from parties $AB$ when they all share an entangled state
$|\psi\rangle^{ABC}$. Under locality constraints, the most general procedure
they can perform is three-way LOCC with multiple rounds of classical
communication exchanged between all the parties. The maximum average bipartite
entanglement they generate under such a process is called the entanglement of
collaboration (EoC) [13, 20], and for a bipartite entanglement measure $E$ it
is given by

$\displaystyle
E^{(c)}(\psi^{ABC})=\sup_{\mathcal{L}}\sum_{x}p_{x}E(\varphi_{x}^{AB}),$ (4)

with the supremum taken over all three-way LOCC maps $\mathcal{L}$
transforming $|\psi\rangle^{ABC}$ to $|\varphi_{x}\rangle^{AB}$ with
probability $p_{x}$. Note that EoC is an entanglement monotone for tripartite
pure states since it is defined in terms of an optimization over LOCC
protocols [15].

In this paper we are interested in more restricted decoupling protocols in
which only one-way communication from the helper is allowed. In particular we
consider a protocol where Charlie (the helper) first makes a decoupling
measurement on his part of $|\psi\rangle^{ABC}$, and then broadcasts his
measurement outcome to both Alice and Bob. In this process, Charlie’s
measurement “collapses” the tripartite state into some bipartite entangled
state shared by Alice and Bob, the form of which they are made aware when they
receive Charlie’s classical information. Alice and Bob can then proceed with
bipartite LOCC processing of their post-measurement state.

The Entanglement of Assistance is defined with respect to these types of
protocols. If $\vec{\mathcal{L}}$ denotes the set of one-way LOCC maps from
Charlie to Alice and Bob, the EoA is given by

$\displaystyle
E^{(a)}(\psi^{ABC})=\max_{\vec{\mathcal{L}}}\sum_{x}p_{x}E(\varphi_{x}^{AB}).$
(5)

From Ineq. (1), one can immediately see that any bipartite entanglement
monotone $E$ satisfies the hierarchy

$E^{(a)}(\psi)\leq E^{(c)}(\psi)\leq\min\\{E^{A|BC}(\psi),E^{B|AC}(\psi)\\}.$
(6)

In this paper we are viewing the entanglement of assistance as a function of
tripartite pure states [13]. However, this is not the only perspective one
could take, and originally [8] the EoA was introduced as a function of
bipartite mixed states given by

$\widetilde{E}^{(a)}(\rho^{AB})=\max_{\\{p_{k},|\varphi_{k}\rangle\\}_{k}}\sum_{k}p_{k}E(\varphi_{k}),$
(7)

where the maximization is taken over all pure state ensembles
$\\{p_{k},|\varphi_{k}\rangle\\}_{k}$ such that
$\rho^{AB}=\sum_{k}p_{k}|\varphi_{k}\rangle\langle\varphi_{k}|$. Since every
possible pure state ensemble for $\rho^{AB}$ can be realized as the post-
measurement ensemble generated by some measurement of Charlie on any
purification of $|\psi\rangle^{ABC}$ or $\rho^{AB}$ (i.e.
$\rho^{AB}=\textrm{Tr}_{C}|\psi\rangle\langle\psi|^{ABC}$) [21], we clearly
have

$E^{(a)}(\psi^{ABC})=\widetilde{E}^{(a)}(\rho^{AB}).$ (8)

The reason we prefer to interpret EoA as a tripartite pure-state measure is
because its operational meaning refers to a process conducted on three
systems. Stated differently, if an experimenter only has access to the
bipartite state $\rho^{AB}$ and not a purifying third system, then the
physical meaning of $\widetilde{E}^{(a)}(\rho^{AB})$ is not clear since it no
longer describes how much entanglement can be concentrated between systems $A$
and $B$.

Unlike the EoC, the EoA is not a tripartite entanglement monotone in general
[13]. This means that when starting with some state $|\psi\rangle^{ABC}$,
there exists LOCC protocols in which communication from Alice and Bob to
Charlie before he makes his decoupling measurement can increase their average
post-measurement entanglement. A known exception to this is for $2\otimes
2\otimes n$ systems when concurrence is used as the two-qubit entanglement
measure [14], as well as for higher-dimensional tripartite systems when the
$G$-concurrence is the bipartite entanglement quantifier [20]. A main result
of this paper establishes that $E_{2}^{(a)}$ is also an entanglement monotone
for three-qubit pure states, which is the EoA for $E_{2}$ entanglement.

##  III Main Results

The central question we ask is when
$E^{(a)}(\psi)=\min\\{E^{A|BC}(\psi),E^{B|AC}(\psi)\\}$ in Ineq. (6) for a
three-qubit state $|\psi\rangle$. We begin our analysis by stating a
structural result for three qubit pure states, which was proven in Ref. [14]
as a method for computing $E^{(a)}_{2}$. In the following, we let
$|\tilde{\Phi}^{+}\rangle^{AB}$ denote the unnormalized maximally entangled
state $|00\rangle+|11\rangle$ for some a priori fixed bases of systems $A$ and
$B$, and complex conjugation is understood to be taken with respect to these
bases.

######  Proposition III.1 ([14]).

Let $|\psi\rangle^{ABC}$ be an arbitrary $2\otimes 2\otimes n$ state. Then
there exists two orthonormal bases $\\{|k\rangle^{C}\\}_{k=1}^{n}$ and
$\\{|k^{\prime}\rangle^{C}\\}_{k=1}^{n}$ for system $C$ such that

$\displaystyle|\psi\rangle$
$\displaystyle=\sum_{k}A_{k}\otimes\mathbb{I}^{BC}|\tilde{\Phi}^{+}\rangle^{AB}|k\rangle^{C}$
(9) $\displaystyle|\psi\rangle$ $\displaystyle=\sum_{k}\mathbb{I}^{AC}\otimes
B_{k}|\tilde{\Phi}^{+}\rangle^{AB}|k^{\prime}\rangle^{C}$ (10)

with both $\\{A_{k}A_{k}^{\dagger}\\}_{k}$ and
$\\{B_{k}B_{k}^{\dagger}\\}_{k}$ being sets of pairwise commuting operators.

Using this proposition, we prove our first result.

######  Theorem III.2.

For any three-qubit state $|\psi\rangle^{ABC}$

$E_{2}^{(a)}(\psi)=\min\\{E_{2}^{A|BC}(\psi),E_{2}^{B|AC}(\psi)\\}.$ (11)

###### Proof.

Assume that $\textrm{Tr}_{C}{|\psi\rangle\langle\psi|}^{ABC}$ is not rank-one,
or else Charlie is already decoupled and the statement trivially holds.
Consider the operators $\\{A_{k}\\}_{k=1}^{2}$ and $\\{B_{k}\\}_{k=1}^{2}$
given by Eqns. (9) and (10) in Proposition III.1, where we have taken $n=2$.
For $|\psi_{k}\rangle:=(A_{k}\otimes\mathbb{I})|\tilde{\Phi}^{+}\rangle$ and
$p_{k}=\textrm{Tr}[A_{k}A_{k}^{\dagger}]$, we can write

$\displaystyle\rho_{k}^{A}=\textrm{Tr}_{B}|\psi_{k}\rangle\langle\psi_{k}|=\frac{1}{p_{k}}A_{k}A_{k}^{\dagger}=\frac{1}{2}(\mathbb{I}+\mathbf{r}_{k}\cdot\vec{\sigma}),$
(12)

where $\mathbf{r}_{k}$ are the Bloch vectors of the $\rho_{k}^{A}$. Since
$\rho_{k}^{A}$ and
$\rho_{k}^{B}=\textrm{Tr}_{A}|\psi_{k}\rangle\langle\psi_{k}|$ have the same
eigenvalue spectrum, they are unitarily related. This means we have
$\rho_{k}^{B}=\frac{1}{2}(\mathbb{I}+\mathbf{r}^{\prime}_{k}\cdot\vec{\sigma})$
for Bloch vectors $\mathbf{r}^{\prime}_{k}$ for which
$|\mathbf{r}^{\prime}_{k}|=|\mathbf{r}_{k}|$. Similarly, we can write

$\displaystyle\sigma_{k}^{B}=\frac{1}{q_{k}}B_{k}B_{k}^{\dagger}=\frac{1}{2}(\mathbb{I}+\mathbf{s}_{k}\cdot\vec{\sigma}),$
(13)

where $q_{k}=\textrm{Tr}[B_{k}B_{k}^{\dagger}]$ and
$\sigma_{k}^{A}=\frac{1}{2}(\mathbb{I}+\mathbf{s}_{k}^{\prime}\cdot\vec{\sigma})$
with $|\mathbf{s}^{\prime}_{k}|=|\mathbf{s}_{k}|$. Expressing the commutators
in terms of the Bloch vectors,

$\displaystyle[\rho_{1}^{A},\rho_{2}^{A}]$
$\displaystyle=\frac{i}{2}(\mathbf{r}_{2}\times\mathbf{r}_{1})\cdot\vec{\sigma}$
(14) $\displaystyle[\sigma_{1}^{B},\sigma_{2}^{B}]$
$\displaystyle=\frac{i}{2}(\mathbf{s}_{2}\times\mathbf{s}_{1})\cdot\vec{\sigma},$
(15)

the vanishing commutators imply that $\\{\mathbf{r}_{1},\mathbf{r}_{2}\\}$ and
$\\{\mathbf{s}_{1},\mathbf{s}_{2}\\}$ are pairs of vectors that are either
parallel (denoted $\uparrow\uparrow)$ or anti-parallel (denoted
$\uparrow\downarrow$). We identify the pair
$\\{\mathbf{r}_{1},\mathbf{r}_{2}\\}$ ($\\{\mathbf{s}_{1},\mathbf{s}_{2}\\}$)
as being parallel when one of them is zero.

The reduced density operators of the given state $|\psi\rangle^{ABC}$ can also
be expressed in terms of their Bloch vectors,

$\displaystyle\rho^{A}$
$\displaystyle=\textrm{Tr}_{BC}|\psi\rangle\langle\psi|^{ABC}=\frac{1}{2}(\mathbb{I}+\mathbf{R}\cdot\vec{\sigma})$
(16) $\displaystyle\quad\sigma^{B}$
$\displaystyle=\textrm{Tr}_{AC}|\psi\rangle\langle\psi|^{ABC}=\frac{1}{2}(\mathbb{I}+\mathbf{S}\cdot\vec{\sigma}).$
(17)

It is straightforward to compute that

$\displaystyle E^{A|BC}(\psi)$ $\displaystyle=1-|\mathbf{R}|,$ $\displaystyle
E^{B|AC}(\psi)$ $\displaystyle=1-|\mathbf{S}|.$ (18)

Now suppose that $\mathbf{r}_{1}\uparrow\uparrow\mathbf{r}_{2}$. Then since
$\mathbf{R}=\sum_{k}p_{k}\mathbf{r}_{k}$, when Charlie measures $|\psi\rangle$
in the basis $|k\rangle^{C}$, the average post-measurement $E_{2}$
entanglement for Alice and Bob will be

$\displaystyle\langle
E_{2}^{A|B}\rangle=1-\sum_{k}p_{k}|\mathbf{r}_{k}|=1-|\mathbf{R}|=E_{2}^{A|BC}.$
(19)

Similarly, when $\mathbf{s}_{1}\uparrow\uparrow\mathbf{s}_{2}$, Charlie can
measure in the basis $|k^{\prime}\rangle^{C}$ to assist in generating
entanglement $E_{2}^{B|AC}$ for Alice and Bob. Therefore, we have proven
Theorem III.2 whenever $\mathbf{r}_{1}\uparrow\uparrow\mathbf{r}_{2}$ or
$\mathbf{s}_{1}\uparrow\uparrow\mathbf{s}_{2}$. It remains to consider the
case when both $\mathbf{r}_{1}\uparrow\downarrow\mathbf{r}_{2}$ and
$\mathbf{s}_{1}\uparrow\downarrow\mathbf{s}_{2}$ with none of these Bloch
vectors being zero.

Suppose that $\mathbf{r}_{1}\uparrow\downarrow\mathbf{r}_{2}$. Since

$\sigma^{B}=\textrm{Tr}_{AC}|\psi\rangle\langle\psi|=\sum_{k}p_{k}\rho_{k}^{B}=\sum_{k}\frac{p_{k}}{2}(\mathbb{I}+\mathbf{r}_{k}^{\prime}\cdot\vec{\sigma}),$

we have

$\displaystyle|\mathbf{S}|=|p_{1}\mathbf{r}_{1}^{\prime}+p_{2}\mathbf{r}_{2}^{\prime}|$
$\displaystyle\geq|p_{1}|\mathbf{r}_{1}^{\prime}|-p_{2}|\mathbf{r}^{\prime}_{2}||$
$\displaystyle=|p_{1}|\mathbf{r}_{1}|-p_{2}|\mathbf{r}_{2}||$
$\displaystyle=|p_{1}\mathbf{r}_{1}+p_{2}\mathbf{r}_{2}|=|\mathbf{R}|,$ (20)

where the first inequality is the triangle inequality and the last line
follows from the vectors being anti-parallel. Similarly, since

$\rho^{A}=\textrm{Tr}_{BC}|\psi\rangle\langle\psi|=\sum_{k}q_{k}\sigma_{k}^{A}=\sum_{k}\frac{q_{k}}{2}(\mathbb{I}+\mathbf{s}_{k}^{\prime}\cdot\vec{\sigma}),$

the assumption that $\mathbf{s}_{1}\uparrow\downarrow\mathbf{s}_{2}$ yields
$|\mathbf{R}|\geq|\mathbf{S}|$. Hence, we have that
$|\mathbf{R}|=|\mathbf{S}|$ or equivalently
$E_{2}^{A|BC}(\psi)=E_{2}^{B|AC}(\psi)$. Furthermore, the equality condition
in Ineq. (III) shows that
$\mathbf{r}^{\prime}_{1}\uparrow\downarrow\mathbf{r}^{\prime}_{2}$, and so we
obtain the commutator relation $[\rho_{1}^{B},\rho_{2}^{B}]=0$. Therefore,
when comparing with Eq. (14), we see that the reduced density matrices for
$|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ are pairwise commuting for both
parties. Since $\mathbf{r}^{\prime}_{1}$ and $\mathbf{r}_{2}^{\prime}$ are
nonzero, the reduced density matrices have no degeneracy in their eigenvalue
spectrum, and so the states $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$ have
unique Schmidt bases for both parties. The pairwise commuting relations then
imply that either
$|\psi\rangle\overset{\mathrm{LU}}{\simeq}(a|00\rangle+b|11\rangle)|0\rangle+(c|01\rangle+d|10\rangle)|1\rangle$
or
$|\psi\rangle\overset{\mathrm{LU}}{\simeq}(a|00\rangle+b|11\rangle)|0\rangle+(c|00\rangle+d|11\rangle)|1\rangle$.
In the first case, the condition $E_{2}^{A|BC}=E_{2}^{B|AC}$ requires that
either $a=b$ or $c=d$, which is a contradiction since we assume that the Bloch
vectors are nonzero. In the second case, we can rewrite the state as

$|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}\sqrt{p}|00\rangle|\eta_{0}\rangle+\sqrt{1-p}|11\rangle|\eta_{1}\rangle,$
(21)

where the $|\eta_{0}\rangle$ and $|\eta_{1}\rangle$ on Charlie’s system are
normalized but not necessarily orthogonal. However, we can always find a basis
$\\{|e_{0}\rangle,|e_{1}\rangle\\}$ such that

$\displaystyle|\eta_{0}\rangle$
$\displaystyle=\cos\theta|e_{0}\rangle+\sin\theta|e_{1}\rangle$
$\displaystyle|\eta_{1}\rangle$
$\displaystyle=\cos\theta|e_{0}\rangle-\sin\theta|e_{1}\rangle$ (22)

for some $\theta$ determined by $\langle\eta_{0}|\eta_{1}\rangle$. Then

$\displaystyle|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}$
$\displaystyle\cos\theta(\sqrt{p}|00\rangle+\sqrt{1-p}|11\rangle)|e_{0}\rangle$
(23) $\displaystyle+$
$\displaystyle\sin\theta(\sqrt{p}|00\rangle-\sqrt{1-p}|11\rangle)|e_{1}\rangle.$

When Charlie measures in the basis $|e_{k}\rangle$, we get the optimal post-
measurement entanglement. ∎

We now ask for conditions under which we achieve lossless decoupling when EoA
is measured with respect to other entanglement monotones. As noted above, any
two-qubit pure state entanglement monotone $E$ is given by
$E(\varphi^{AB})=f(\lambda_{\min}(\varphi^{A}))$ with
$f:[0,\frac{1}{2}]\rightarrow\mathbb{R}^{+}$ being concave and
$\lambda_{\min}(\varphi^{A})$ the smallest eigenvalue of $\varphi^{A}$. Via
Eq. (8), there exists an optimal ensemble
$\\{p_{x},|\varphi_{x}\rangle\\}_{x}$ (with all $p_{x}>0$) for
$\textrm{Tr}_{C}{|\psi\rangle\langle\psi|}^{ABC}$ such that

$\displaystyle E^{(a)}(\psi^{ABC})$
$\displaystyle=\sum_{x}p_{x}E(\varphi_{x}^{AB})$
$\displaystyle=\sum_{x}p_{x}f(\lambda_{\min}(\varphi_{x}^{A}))$
$\displaystyle\overset{\text{(a)}}{\leq}f\left(\sum_{x}p_{x}\lambda_{\min}(\varphi_{x}^{A})\right)$
$\displaystyle\overset{\text{(b)}}{\leq}f(\lambda_{\min}(\psi^{A}))=E^{A|BC}(\psi^{ABC}).$
(24)

Inequality (a) is the concavity of $f$, and (b) follows from $f$ monotonically
increasing on the interval $[0,\frac{1}{2}]$ along with the fact that

$\sum_{x}p_{x}\lambda_{\min}(\varphi_{x}^{A})\leq\lambda_{\min}\left(\sum_{x}p_{x}\varphi_{x}^{A}\right)=\lambda_{\min}(\psi^{A}).$
(25)

A similar chain of inequalities allows us to replace the RHS with
$E^{B|AC}(\psi^{ABC})$.

Motivated by the fact that some common entanglement monotones such as
entanglement entropy and concurrence are determined by an $f$ strictly
concave, we restrict $f$ to be as such. In this case, $f$ is then continuous
and strictly increasing on the interval $(0,1/2]$. If we find for example that
$E^{(a)}(\psi^{ABC})=E^{A|BC}(\psi^{ABC})$, then the strict concavity of $f$
implies that the optimal ensemble $\\{p_{x},|\varphi_{x}\rangle\\}_{x}$ in Eq.
(24) either consists of just a single nonzero element, or
$\lambda_{\min}(\varphi_{x}^{A})$ is the same for every $\varphi_{x}$. If
there is just a single element in the pure-state decomposition of
$\varphi^{AB}$, then system $C$ is already decoupled. Otherwise, if the
ensemble has multiple elements with $\lambda_{\min}(\varphi_{x}^{A})$ being
the same for every $\varphi_{x}$, then because these are qubit operators,
there exist unitary matrices $U_{x}$ such that

$\varphi_{x}^{A}=U_{x}\varphi_{0}^{A}U_{x}^{\dagger}$ (26)

for some fixed $\varphi_{0}^{A}$. Tightness in Ineq. (25) then requires

$\lambda_{\min}(\varphi_{0}^{A})=\lambda_{\min}\left(\sum_{x}p_{x}U_{x}\varphi_{0}^{A}U_{x}^{\dagger}\right)=\lambda_{\min}(\psi^{A}),$
(27)

and the following proposition characterizes when such a condition holds.

######  Proposition III.3.

A qubit Hermitian operator $H$ satisfies
$\lambda_{\min}(H)=\lambda_{\min}\left(\sum_{x=0}^{t}p_{x}U_{x}HU_{x}^{\dagger}\right)$
for unitary operators $U_{x}$ such that $U_{0}=\mathbb{I}$ and non-zero
probabilities $p_{x}$ if and only if $[H,U_{x}]=0$ for all $x$.

###### Proof.

(Here we present a simple proof based on an external result. A slightly longer
but from first principles one is presented in the appendix). Sufficiency is
obvious. To prove necessity, note that the operators $\sqrt{p_{x}}U_{x}$
described above may be thought of as Kraus operators for a unital qubit
channel. In [22], it is shown that the fixed point set of a unital channel is
exactly the set of all operators commuting with all Kraus operators in any
operator-sum representation of the channel. With this in mind, suppose that
$\lambda_{\min}(H)=\lambda_{\min}\left(\sum_{x}^{t}p_{x}U_{x}HU_{x}^{\dagger}\right)$.
Then we have for some unitary $V$,
$V^{\dagger}HV=\sum_{x}^{t}p_{x}U_{x}HU_{x}^{\dagger}$, so that $H$ is a fixed
point of a new unital channel whose Kraus operators are $\sqrt{p_{x}}VU_{x}$.
From here it follows that $[H,VU_{x}]=0$ for all $x$. The requirement that
$U_{0}=\mathbb{I}$ implies $[H,V]=0$, so that $[H,U_{x}]=0$ for all $x$. ∎

We combine this proposition with Eq. (27) to conclude that

######  Theorem III.4.

Let $E$ be an entanglement monotone for two qubits determined by a strictly
concave function of the Schmidt coefficients. Then a three qubit state
$|\psi\rangle^{ABC}$ will satisfy $E^{(a)}(\psi)=E^{A|BC}(\psi)$ if and only
if

$\displaystyle|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}\sum_{x}\sqrt{p_{x}}(U_{x}^{A}\otimes
V_{x}^{B})|\lambda_{\min}\rangle^{AB}|x\rangle^{C}$ (28)

where the $p_{x}$ are understood to be non-zero,
$|\lambda_{\min}\rangle=\sqrt{\lambda_{\min}}|00\rangle+\sqrt{1-\lambda_{\min}}|11\rangle$
for $\lambda_{\min}:=\lambda_{\min}(\psi^{A})$, and either each $U_{x}^{A}$ is
diagonal in the $\\{|0\rangle,|1\rangle\\}$ basis or $\lambda_{\min}=1/2$. An
analogous statement holds under an interchange of systems $A$ and $B$.

The statement of Theorem III.4 also holds for $2\otimes 2\otimes n$ states, as
there is nothing restricting the dimension of Charlie’s system in the
preceding analysis. Note that in any case, it suffices to consider an index
$x$ in Eq. (28) that has a range of no more than four distinct values [23].

When Charlie’s system is a qubit, it is instructive to compare Eq. (28) with
other three-qubit decompositions known in the literature. Wootters has shown
that every two-qubit mixed state $\rho^{AB}$ admits a pure-state decomposition
with each state having the same concurrence [16]. Hence every three-qubit
state $|\psi\rangle$ can be written as

$\displaystyle|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}\sum_{x}\sqrt{p_{x}}(U_{x}^{A}\otimes
V_{x}^{B})|\lambda\rangle^{AB}|x\rangle^{C}$ (29)

where $|\lambda\rangle=\sqrt{\lambda}|00\rangle+\sqrt{1-\lambda}|11\rangle$
and the local unitaries are unrestricted. De Vicente et al. later showed [24]
that for three-qubit pure states, a decomposition always exists of the form

$\displaystyle|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}\frac{1}{\sqrt{2}}\left(|\lambda\rangle^{AB}|1\rangle^{C}+U^{A}\otimes
V^{B}|\lambda\rangle^{AB}|2\rangle^{C}\right),$ (30)

where again the local unitaries are unrestricted. The crucial difference
between our Eq. (28) and these two decompositions is that
$\lambda=\lambda_{\min}(\psi^{A})$ and the unitaries $U_{x}^{A}$ are required
to commute with $\textrm{Tr}_{B}|\lambda\rangle\langle\lambda|^{AB}$. Only a
special subset of states $|\psi\rangle$ will admit such a decomposition, and
so most states satisfy
$E^{(a)}(\psi)<\min\\{E^{A|BC}(\psi),E^{B|AC}(\psi)\\}$. In contrast, Theorem
III.2 shows that
$E_{2}^{(a)}(\psi)=\min\\{E_{2}^{A|BC}(\psi),E_{2}^{B|AC}(\psi)\\}$ for all
three-qubit states. The reason that we have universality for $E_{2}$ is that
it is an entanglement monotone not based on a strictly concave eigenvalue
function, and in fact, inequality (a) in (24) will always be tight since $f$
is linear in $\lambda_{\min}$ for $E_{2}$.

As a final remark, one might wonder whether or not Theorem III.4 also applies
to the entanglement of collaboration. That is, whether
$E^{(c)}(\psi)=E^{A|BC}(\psi)$ if and only if $|\psi\rangle$ has the form of
Eq. (28). In fact it does. To see this, suppose that
$E^{(c)}(\psi)=E^{A|BC}(\psi)$ and consider a general multi-outcome LOCC
transformation $|\psi\rangle^{ABC}\to|\varphi_{i}\rangle$ that attains an
average post-measurement entanglement of $E^{A|BC}(\psi)$ (a similar argument
holds for average post-measurement entanglement of $E^{B|AC}(\psi)$).
Following the reasoning of Theorem III.4, suppose that Bob performs a
measurement that induces the transformation
$|\psi\rangle^{ABC}\to|\psi_{x}\rangle^{ABC}$, with $|\psi_{x}\rangle^{ABC}$
having probability $p_{x}$. Note that the $A|BC$ entanglement cannot decrease
on average. Then by strict concavity of $f$, we must have that
$\lambda_{\min}(\psi_{x}^{A})$ is the same for every $\psi_{x}^{ABC}$. Hence
Eq. (27) holds and the conclusion of Proposition III.3 implies that
$\psi^{A}=\psi_{x}^{A}$ for all $x$. In other words, if Bob’s measurement is
described by Kraus operators $\\{M_{x}\\}_{x}$, then we have that

$\textrm{Tr}_{B}[\mathbb{I}^{A}\otimes(\mathbb{I}-p_{x}^{-1}M_{x}^{\dagger}M_{x})^{B}\psi^{AB}]=0$

for all outcomes $x$ with nonzero probability. This is equivalent to the
condition

$\sqrt{\psi^{A}}(\mathbb{I}-p_{x}^{-1}M_{x}^{*}M_{x}^{T})\sqrt{\psi^{A}}=0.$
(31)

Since the rank of $\psi^{A}$ is two (or else $|\psi\rangle^{ABC}$ would be a
product state), $\psi^{A}$ is invertible and we thus obtain
$\mathbb{I}=p_{x}^{-1}M_{x}^{\dagger}M_{x}$, which means that Bob’s
measurement is the application of a random unitary. The same reasoning applies
for a measurement by Alice when interpreting $|\psi\rangle^{A(BC)}$ as a two-
qubit state; i.e. $\psi^{BC}$ can be expressed as a $2\times 2$ positive
matrix in the Schmidt basis of $|\psi\rangle^{A(BC)}$. In summary, a general
LOCC protocol that converts $|\psi\rangle^{ABC}\to|\varphi_{i}\rangle$ can be
replaced by one in which Alice and Bob only apply random local unitaries.
Since these local unitaries can always be delayed until the end of the
protocol, we conclude that

$\displaystyle E^{(c)}(\psi)$
$\displaystyle=\min\\{E^{A|BC}(\psi),E^{B|AC}(\psi)\\}$
$\displaystyle\Rightarrow E^{(a)}(\psi)$ $\displaystyle=E^{(c)}(\psi).$ (32)

The argument given here is a variation of the one given in Ref. [25] for
entropy-preserving LOCC transformations.

Finally, we close by a general observation that follows from the analysis
carried out in Theorem III.2. We suspect this might have independent interest
in the study of three-qubit networks. If one labels the vertices of a graph by
the bipartite pure state concurrences of $|\psi\rangle^{ABC}$, and the edges
by the mixed state concurrences as shown in Fig. 2, a symmetry relationship
follows which is made precise in the following corollary.

Figure 2: Symmetry of entanglement across bipartite “cuts” implies symmetry of
pairwise entanglement. For almost every state, it also implies symmetry under
exchange of the respective parties.

######  Corollary III.5.

For any three-qubit pure state $|\psi\rangle^{ABC}$ such that
$\rho_{k}^{A}\neq\frac{1}{2}\mathbb{I}$ for $k=1,2$, the following are
equivalent.

(i) $\displaystyle\quad E_{2}^{A|BC}=E_{2}^{B|AC}$ (33) (ii)
$\displaystyle\quad|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}|\psi\rangle^{BAC}$
(iii) $\displaystyle\quad\mathcal{C}(\rho^{AC})=\mathcal{C}(\rho^{BC})$

where $\mathcal{C}(\rho)$ is the two-qubit mixed state concurrence [26].

###### Proof.

If Charlie is decoupled from Alice and Bob, then Alice and Bob can always swap
their state via LU’s, which follows from Schmidt decomposition. If Charlie is
not decoupled, we can see from the analysis in Theorem III.2 that
$E_{2}^{A|BC}=E_{2}^{B|AC}$ if and only if both $[\rho_{1}^{A},\rho_{2}^{A}]$
and $[\rho_{1}^{B},\rho_{2}^{B}]$ vanish. The additional restriction that none
of these reduced density matrices is $\frac{1}{2}\mathbb{I}$ implies that
$|\psi\rangle^{ABC}$ has the form of Eq. (21), which is symmetric under
exchange of Alice and Bob. Therefore (i) implies (ii), and the converse is
trivial. Equivalence of (i) and (iii) follows from the CKW monogamy relations
[2]

$\displaystyle\mathcal{C}^{2}_{A|BC}$
$\displaystyle=\tau_{ABC}+\mathcal{C}^{2}(\rho^{AB})+\mathcal{C}^{2}(\rho^{AC})$
(34) $\displaystyle\mathcal{C}^{2}_{B|AC}$
$\displaystyle=\tau_{BAC}+\mathcal{C}^{2}(\rho^{BA})+\mathcal{C}^{2}(\rho^{BC}).$
(35)

In particular, the three-tangle $\tau$ and the mixed state concurrence are
symmetric under exchange of parties, so taking the difference of the above
equations gives

$\displaystyle\mathcal{C}^{2}(\rho^{AC})-\mathcal{C}^{2}(\rho^{BC})=\mathcal{C}^{2}_{A|BC}-\mathcal{C}^{2}_{B|AC}.$
(36)

The pure state concurrence is however in one-to-one correspondence with
$\lambda_{\text{min}}$, so equivalence of (i) and (iii) follows. ∎

We note here that the restriction $\rho_{k}^{A}\neq\frac{1}{2}\mathbb{I}$ is
not actually very restrictive because when choosing three-qubit states at
random, $\rho_{k}^{A}=\frac{1}{2}\mathbb{I}$ occurs with probability zero.
Nonetheless, there are certain states with $\rho_{k}^{A}=\frac{\mathbb{I}}{2}$
for some $k$ that can be included in the statement of Corollary III.5. For
example, any state of the form

$|\psi\rangle^{ABC}\overset{\mathrm{LU}}{\simeq}\sqrt{p}|\Phi^{+}\rangle|0\rangle+\sqrt{1-p}|\varphi\rangle|1\rangle,$
(37)

with $|\varphi\rangle^{BA}=(U\otimes U^{*})|\varphi\rangle^{AB}$ for some
unitary $U$ will satisfy the conclusion of Corollary III.5.

##  IV Conclusion and Discussion

In Theorem III.2 we have established for three-qubit pure states that imposing
locality constraints on a helper system does not decrease the probability of
obtaining a maximally entangled state $|\Phi^{+}\rangle$ between any two fixed
parties. One may wonder whether this holds for tripartite systems in other
dimensions. It can already be seen that it does not when Alice and Bob have
qutrit ($d=3$) systems. Consider density matrices $\rho^{AB}$ in which
$\rho^{A}$ and $\rho^{B}$ are both maximally mixed; i.e.
$\frac{\mathbb{I}}{3}$. Then $\rho^{AB}$ is the Choi matrix for a qutrit
unital channel; however, it is known that not every qutrit unital channel can
be expressed as a probabilistic mixture of unitary operations [27]. This means
that it is not possible to find a pure-state decomposition of $\rho^{AB}$
consisting of only maximally entangled states, which would be required if
Ineq. (1) were tight. For general $2\otimes 2\otimes n$ states with $n>2$, we
leave it as an open problem to decide whether
$E^{(a)}_{2}=\min\\{E_{2}^{A|BC},E_{2}^{B|AC}\\}$, but we conjecture it to be
true.

Note that Theorem III.2 can be phrased entirely in terms of the two-qubit
density matrix as follows: every rank-two, two-qubit density matrix
$\rho^{AB}$ satisfies

$\widetilde{E}^{(a)}_{2}(\rho^{AB})=2\min\\{\lambda_{\min}(\rho^{A}),\lambda_{\min}(\rho^{B})\\}.$
(38)

Interestingly this relationship also applies for any function that varies
linearly in $\lambda_{\min}$ over the interval $(0,\frac{1}{2}]$. As an
extreme case, consider the monotone

$S_{0}(\lambda)=\begin{cases}0&\text{if}\quad\lambda=0\\\ 1&\text{if}\quad
0<\lambda\leq\frac{1}{2},\end{cases}$ (39)

which is the $\alpha$-entropy of entanglement for $\alpha=0$. This function
also satisfies Eq. (38). To see this, observe that it trivially holds for any
$\rho^{AB}$ in which either reduced state $\rho^{A}$ or $\rho^{B}$ is pure. On
the other hand, if both $\rho^{A}$ and $\rho^{B}$ are genuinely mixed, then it
is always possible to find a pure-state decomposition for $\rho^{AB}$
consisting of entangled states. This follows from the fact that in such a case
the set of product states in the support of $\rho^{AB}$ is measure zero. We
give a rigorous argument of this in the appendix.

## Acknowledgment

This work was supported by NSF Award No. 1914440.

##  V Appendix

###  V.1 Alternative Proof of Prop. III.3

Sufficiency is obvious. To prove necessity, we note that for Hermitian
operators $A$ and $B$,
$\lambda_{\text{max}}(A+B)=\lambda_{\text{max}}(A)+\lambda_{\text{max}}(B)$ if
and only if $A$, $B$, and $A+B$ share a principle eigenvector which has the
maximal eigenvalue for $A$, $B$, and $A+B$ [28]. We can extend this by
induction to a finite set of operators, taking them to be the
$p_{x}U_{x}HU_{x}^{\dagger}$. Because these are qubit operators and
$\textrm{Tr}{(\sum_{x}^{t}p_{x}U_{x}HU_{x}^{\dagger})}=\textrm{Tr}{(H)}$, we
can replace max with min in the above fact. Therefore suppose that
$\lambda_{\min}\left(\sum_{x}^{t}p_{x}U_{x}HU_{x}^{\dagger}\right)=\lambda_{\min}(H)$.
Then there is a non-zero vector $|h\rangle$ such that $\forall\ x$,

$\displaystyle H|h\rangle$ $\displaystyle=\lambda_{\max}(H)|h\rangle$ (40)
$\displaystyle U_{x}HU_{x}^{\dagger}|h\rangle$
$\displaystyle=\lambda_{\max}(H)|h\rangle,$

where the first line follows from the assumption that $U_{0}=\mathbb{I}$.
These jointly imply that $[H,U_{x}^{\dagger}]|h\rangle=0.$ Let
$|h^{\perp}\rangle$ be a vector orthogonal to $|h\rangle$, which is uniquely
determined up to an overall phase. Then, $|h^{\perp}\rangle$ is also an
eigenvector of $H$ and the $U_{x}HU_{x}^{\dagger}$ (with eigenvalue possibly
zero), and for the same reason as above,
$[H,U_{x}^{\dagger}]|h^{\perp}\rangle=0.$ Hence

$[H,U_{x}^{\dagger}](\alpha|h\rangle+\beta|h^{\perp}\rangle)=0$ (41)

for arbitrary $\alpha,\beta$, and so we have $[H,U_{x}]=0$ for all $x$.

###  V.2 Decomposition of Certain Density Matrices into Entangled States

Consider a two-qubit separable state
$\rho^{AB}=\sum_{k=1}^{4}|\alpha_{k}\rangle\langle\alpha_{k}|^{A}\otimes|\beta_{k}\rangle\langle\beta_{k}|^{B}$.
Suppose that both $\rho^{A}$ are $\rho^{B}$ are genuinely mixed. Then we can
always find a pair $|\alpha_{k_{1}}\rangle|\beta_{k_{1}}\rangle$ and
$|\alpha_{k_{2}}\rangle|\beta_{k_{2}}\rangle$ such that
$\\{|\alpha_{k_{1}}\rangle,|\alpha_{k_{2}}\rangle\\}$ and
$\\{|\beta_{k_{1}}\rangle,|\beta_{k_{2}}\rangle\\}$ span two-dimensional
spaces. This means that the two-dimensional subspace of
$\mathbb{C}^{2}\otimes\mathbb{C}^{2}$ subspace spanned by
$\\{|\alpha_{k_{1}}\rangle|\beta_{k_{1}}\rangle,|\alpha_{k_{2}}\rangle|\beta_{k_{2}}\rangle\\}$
contains at most two product states [29]. Hence, we can always find a $2\times
2$ unitary matrix with components $u_{ij}$ such that
$u_{11}|\alpha_{k_{1}}\rangle|\beta_{k_{1}}\rangle+u_{12}|\alpha_{k_{2}}\rangle|\beta_{k_{2}}\rangle$
and
$u_{21}|\alpha_{k_{1}}\rangle|\beta_{k_{1}}\rangle+u_{12}|\alpha_{k_{2}}\rangle|\beta_{k_{2}}\rangle$
are both entangled states. We can thus replace product states $k_{1}$ and
$k_{2}$ in the decomposition of $\rho^{AB}$ with these entangled states. We
then choose a remaining product state
$|\alpha_{k_{3}}\rangle|\beta_{k_{3}}\rangle$ in the ensemble and mix it with
one of the entangled states by another $2\times 2$ unitary matrix so to
generate two new entangled states. This allows us to replace
$|\alpha_{k_{3}}\rangle|\beta_{k_{3}}\rangle$ with an entangled state in the
decomposition of $\rho^{AB}$. Proceeding in this way allows us to eliminate
all product states in the decomposition. Note, this argument also applies for
separable states in arbitrary dimensions since a generic two dimensional
subspace in $\mathbb{C}^{d_{A}}\otimes\mathbb{C}^{d_{B}}$ will always contain
a finite number of product states.

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