# Symmetric States and Dynamics of Three Quantum Bits

Francesca Albertini Dipartimento di Tecnica e Gestione del Sistemi
Industriali, Universita’ di Padova, Vicenza, Italy<EMAIL_ADDRESS>Domenico D’Alessandro Corresponding Author, Department of Mathematics, Iowa
State University, Ames IA 50011, U.S.A<EMAIL_ADDRESS>
###### Abstract

The unitary group acting on the Hilbert space ${\cal
H}:=(\mathbb{C}^{2})^{\otimes 3}$ of three quantum bits admits a Lie subgroup,
$U^{S_{3}}(8)$, of elements which permute with the symmetric group of
permutations. Under the action of such Lie subgroup, the Hilbert space ${\cal
H}$ splits into three invariant subspaces of dimensions $4$, $2$ and $2$
respectively, each corresponding to an irreducible representation of $su(2)$.
The subspace of dimension $4$ is uniquely determined and corresponds to states
that are themselves invariant under the action of the symmetric group. This is
the so called symmetric sector.

We provide an analysis of pure states in the symmetric sector of three quantum
bits for what concerns their entanglement properties, separability criteria
and dynamics under the Lie subgroup $U^{S_{3}}(8)$. We parametrize all the
possible invariant two-dimensional subspaces and extend the previous analysis
to these subspaces as well. We propose a physical set up for the states and
dynamics we study which consists of a symmetric network of three spin
$\frac{1}{2}$ particles under a common driving electro-magnetic field.  For
such set up, we solve a control theoretic problem which consists of driving a
separable state to a state with maximal distributed entanglement.

Keywords Quantum entanglement, symmetric states, quantum symmetric evolution,
spin networks, quantum control.

##  1 Introduction

The study of quantum states is a current line of research in quantum physics
(see, e.g., [3]), in particular for what concerns their entanglement
properties. Entanglement is considered a resource in quantum information
processing and classifying states according to the amount and type of
entanglement is a problem of both fundamental and practical importance. A
related problem is to study how quantum dynamical evolution changes the
entanglement of states (see,e g., [21] for the two qubits case) and in the
quantum control setting [5] how to induce such dynamics in a specific physical
setup. The two qubits case is fairly well understood, while the case of three
qubits requires further exploration. In particular, three qubits are the
simplest type of systems which display two types of entanglement: a pairwise
entanglement quantifying the entanglement between pairs of qubits and a
distributed entanglement [4].

In this paper we are concerned with three qubits systems whose dynamics are
subject to a permutation symmetry among the three qubits. The possible unitary
evolutions on the Hilbert space ${\cal H}:=(\mathbb{C}^{2})^{\otimes 3}$ of
the system consists of unitaries which commute with the permutation group
$S_{3}$. Such a Lie subgroup of $U(8)$, which we denote by $U^{S_{3}}(8)$ has
dimension $20$ [1]. Its Lie algebra, $u^{S_{3}}(8)$, is spanned by the
matrices

$i\Pi(\sigma_{1}\otimes\sigma_{2}\otimes\sigma_{3}).$ (1)

where $\Pi$ denotes the symmetrization operator $\Pi:=\frac{1}{3!}\sum_{P\in
S_{3}}P$ and $\sigma_{1,2,3}$ are chosen to be the $2\times 2$ identity or one
of the Pauli matrices

$\sigma_{x}:=\begin{pmatrix}0&1\cr
1&0\end{pmatrix},\qquad\sigma_{y}:=\begin{pmatrix}0&i\cr-i&0\end{pmatrix},\qquad\sigma_{z}:=\begin{pmatrix}1&0\cr
0&-1\end{pmatrix}.$ (2)

The number of the matrices (1) is equal to the way of choosing the number of
occurrences of the identity and $\sigma_{x,y,z}$ out of three positions, which
is equal to $20$.111The general argument for $n$ qubit is presented in [1].

The three qubit subspace ${\cal H}$ under the action of the Lie algebra
$u^{S_{3}}(8)$, or of the Lie group $U^{S_{3}}(8)$, splits into three
invariant subspaces two of which have dimension $2$ and one of which has
dimension $4$. These correspond to irreducible representations of $su(2)$ [1]
[6]. The subspace of dimension $4$ is uniquely determined. It is the so-called
symmetric sector [16], that is, the space of the states that do not change
under permutation of two qubits. For the three qubits case, we shall use the
orthogonal basis (not normalized)

$\phi_{0}:=|000\rangle$ (3) $\phi_{1}:=|100\rangle+|010\rangle+|001\rangle$
$\phi_{2}:=|110\rangle+|110\rangle+|101\rangle$ $\phi_{3}:=|111\rangle.$

This notation follows the number of $1$’s that appear in each state, in the
sense that $\phi_{j}$ is the sum of the states in the computational basis
which have $j$ $1$’s. States in the symmetric tensor where studied in [16] and
a complete list of invariants under local unitary and symmetric
transformations was given there.

We will write a general state in the symmetric sector as

$\psi:=c_{0}\phi_{0}+c_{1}\phi_{1}+c_{2}\phi_{2}+c_{3}\phi_{3},$ (4)

for complex coefficients $c_{0},c_{1},c_{2},c_{3}$, with
$|c_{0}|^{2}+3|c_{1}|^{2}+3|c_{2}|^{2}+|c_{3}|^{2}=1$. This can also be seen
as a four level system which can be used to implement two qubits.

The subspaces of dimension 2, which represent isomorphic representations of
$su(2)$ are not uniquely. .

The results presented in this paper and a plan for the following sections is
as follows. In section 2 we perform the decomposition of the Hilbert space
${\cal H}:=\mathbb{C}^{\otimes 3}$ into invariant subspaces for
$u^{S_{3}}(8)$. We see that while the four dimensional invariant subspace is
uniquely determined (it is the symmetric sector) the pair of $2$-dimensional
invariant subspaces is not. We obtain a parametrization of all the possible
decompositions. In section 3 we recall the general measures of entanglement
for three qubits introduced in [4], in particular the pairwise entanglement
(concurrence), quantifying the entanglement between two qubits when the third
one is traced out and the distributed entanglement. These measures are
invariant under local unitary transformations. In section 4, we calculate the
entanglement measures for states in the symmetric sector and give conditions
of separability. The expressions we find complement the ones found in [16]
which are based on the Majorana polynomial representations of states [14]. We
also briefly recall such a representation which has an elegant geometric
representation and provide a complete set of local invariant for these states.
In section 5 we analyze the entanglement of the states in the two dimensional
invariant subspaces. We then turn our attention to the dynamics on the
invariant subspaces. The dynamics we are interested in are the ones in the Lie
subgroup of the unitary group $U(8)$ which permutes with the symmetric group,
i.e., $U^{S_{3}}(8)$ and that, as we have recalled, leaves the above subspaces
invariant. In section 6 we study the dynamics on the invariant subspaces. In
particular for the symmetric sector we prove that the group of local
(symmetric) unitary transformations is a maximal Lie subgroup of
$U^{S_{3}}(8)$ which leaves the adopted measures of entanglement unchanged. We
then study in general how the elements of the group $U^{S_{3}}(8)$ change the
entanglement in this subspace.  We give a factorization of possible unitary
evolutions on the symmetric sector in evolutions that modify the entanglement
and evolutions that do not. In section 7 we give a physical application of the
analysis described in the previous sections. We consider a symmetric network
of three spin $\frac{1}{2}$ particles coupled via identical Ising interaction
and driven by a common electro-magnetic field. The dynamics of this model
satisfies the symmetry assumptions considered in this paper. We propose an
algorithm to drive such a system from a separable state (with zero
entanglement) to a state with maximum distributed entanglement. A summary of
the results is given in section 8.

##  2 Decomposition into Invariant Subspaces

Consider the Lie algebra $su^{S_{3}}(8)$, that is, the subalgebra of $su(8)$
of matrices which commute with the permutation matrices in $S_{3}$. For
$su(8)$, we consider the standard representation on $\mathbb{C}^{8}\simeq{\cal
H}$ and therefore the matrices in $su^{S_{3}}(8)$ are also $8\times 8$. In
appropriate coordinates such matrices take a block diagonal form with blocks
of dimension $4\times 4$, $2\times 2$ and $2\times 2$, which correspond to
invariant subspaces of ${\cal H}$ of dimensions $4$, $2$ and $2$, respectively
[1]. Such subspaces correspond to irreducible representations of $su(2)$ of
dimensions $4$, $2$ and $2$ respectively (the ones of dimension $2$ being
isomorphic representations). To obtain a basis for such subspaces, in terms of
the computational basis $\\{|jkl\rangle\,\,,j,k,l=0,1\\}$ one may apply
standard methods of the quantum theory of angular momentum (see, e.g., [9],
[17]) which overlap with representation theory and the theory of Young tableau
and representations of the symmetric group (see, e.g., [8]). For instance, the
Clebsch-Gordan coefficients described in [9] (in the table on pg. 375), give
one possible change of coordinates to obtain the bases of such invariant
subspaces. Another method is given by the use of Young symmetrizers which was
reviewed in [6] in the form that uses Hermitian Young symmetrizers as
described in [2]. It must be stressed however that such a decomposition in
invariant subspaces is not unique and it is of interest to find all the
possible decompositions. In order to achieve this goal we will use in this
section a technique that was described in [5] (see Chapter 4 section 4.3.4).
One considers the commutant ${\cal C}$ of $su^{S_{3}}(8)$ in $u(8)$ which is a
reductive Lie algebra and therefore it admits Cartan subalgebras, i.e., a
maximal Abelian subalgebra. The main observation is that, if $W\oplus
V_{1}\oplus V_{2}$ (with $\dim(W)=4$ and $\dim(V_{1})=\dim(V_{2})=2$) is a
decomposition in invariant subspaces for $su^{S_{3}}(8)$, then ${\cal C}$
admits a Cartan subalgebra which, in the appropriate coordinates, has a basis
given by $A_{1}:=\texttt{diag}(i{\bf 1}_{4},{\bf 0}_{2},{\bf 0}_{2})$,
$A_{2}:=\texttt{diag}({\bf 0}_{4},i{\bf 1}_{2},{\bf 0}_{2})$,
$A_{3}:=\texttt{diag}({\bf 0}_{4},{\bf 0}_{2},i{\bf 1}_{2})$, where ${\bf
1}_{r}$ (${\bf 0}_{r}$) is the $r\times r$ identity (zero) matrix, and diag
here refers to block diagonal matrices. Possible decompositions are therefore
in correspondence with Cartan subalgebras of the commutant ${\cal C}$. Thus, a
method to obtain all the possible decompositions is the following algorithm.

  1. 1.

Compute the commutant ${\cal C}$ of $su^{S_{3}}(8)$ in $u(8)$.

  2. 2.

Find all possible Cartan subalgebras of ${\cal C}$ which (in this case) all
have dimension $3$. The following steps refers to the a Cartan subalgebra
${\cal A}$. In order to deal with Hermitian matrices rather than skew-
Hermitian ones, we consider $i{\cal A}$.

  3. 3.

Take a basis of $i{\cal A}$ and (orthogonally) diagonalize its elements
simultaneously (this is possible since these are mutually commuting Hermitian
matrices).

  4. 4.

Place the elements on the diagonal in three row vectors so as to form a
$3\times 8$ matrix, which we shall denote by $M$.

  5. 5.

Perform a Gaussian row reduction algorithm to place the matrix in a Reduced
Row Echelon Form (see, e.g., [13]). This corresponds to taking linear
combinations of the matrices in the basis if $i{\cal A}$ so as to obtain a new
basis of elements which only have eigenvalues $1$ and $0$. Call these elements
(in the original coordinates) $(\tilde{A}_{1},\tilde{A}_{2},\tilde{A}_{3})$
with $\tilde{A}_{1}$ having eigenvalue $1$ with multiplicity 4,
$\tilde{A}_{2}$ and $\tilde{A}_{3}$ having eigenvalue $1$ with multiplicity 2

  6. 6.

$W$ is the eigenspace of $\tilde{A}_{1}$ corresponding to eigenvalue $1$.
$V_{1}$ and $V_{2}$ are the eigenspaces of $\tilde{A}_{2}$ and
$\tilde{A}_{3}$, respectively, corresponding to eigenvalue $1$. Notice that
once we know $W$ and $V_{1}$, the subspace $V_{2}$ is simply the orthogonal
complement.

Let us carry out the above program for our example. The commutant ${\cal C}$
is found by solving the linear system of equations $[{\cal C},B_{j}]=0$, where
$\\{B_{j}\\}$ is a basis of $su^{S_{3}}(8)$. In fact, since the matrices
$iH_{x}$, $iH_{y}$ and $iH_{zz}$, with (cf. (2))

$H_{x,y,z}:=\sigma_{x,y,z}\otimes{\bf 1}\otimes{\bf 1}+{\bf
1}\otimes\sigma_{x,y,z}\otimes{\bf 1}+{\bf 1}\otimes{\bf
1}\otimes\sigma_{x,y,z},$ (5) $H_{zz}:=\sigma_{z}\otimes\sigma_{z}\otimes{\bf
1}+{\bf 1}\otimes\sigma_{z}\otimes\sigma_{z}+\sigma_{z}\otimes{\bf
1}\otimes\sigma_{z},$ (6)

generate all of $su^{S_{3}}(8)$ [1], it is enough to solve

$[{\cal C},H_{x}]=0,\qquad[{\cal C},H_{y}]=0,\qquad[{\cal C},H_{zz}]=0.$

This computation, which was done in [5], leads to the basis
$\\{E_{1},E_{2},E_{3},E_{4},E_{5}\\}$ for ${\cal C}$, with

$iE_{1}:={\bf 1}\otimes{\bf 1}\otimes{\bf 1},$ $iE_{2}:=\sigma_{x}\otimes{\bf
1}\otimes\sigma_{x}+\sigma_{y}\otimes{\bf
1}\otimes\sigma_{y}+\sigma_{z}\otimes{\bf 1}\otimes\sigma_{z},$
$iE_{3}=\sigma_{x}\otimes\sigma_{x}\otimes{\bf
1}+\sigma_{y}\otimes\sigma_{y}\otimes{\bf
1}+\sigma_{z}\otimes\sigma_{z}\otimes{\bf 1},$ $iE_{4}={\bf
1}\otimes\sigma_{x}\otimes\sigma_{x}+{\bf
1}\otimes\sigma_{y}\otimes\sigma_{y}+{\bf
1}\otimes\sigma_{z}\otimes\sigma_{z},$
$iE_{5}=\sigma_{x}\otimes(\sigma_{y}\otimes\sigma_{z}-\sigma_{z}\otimes\sigma_{y})+\sigma_{y}\otimes(\sigma_{z}\otimes\sigma_{x}-\sigma_{x}\otimes\sigma_{z})+\sigma_{z}\otimes(\sigma_{x}\otimes\sigma_{y}-\sigma_{y}\otimes\sigma_{x}).$

An analysis of the Lie algebra ${\cal C}$ shows that it is the direct sum of
one two dimensional Abelian Lie algebra spanned by $E_{1}$ and
$E_{1}+E_{2}+E_{3}$ and a three dimensional Lie algebra isomorphic to $su(2)$
and spanned by $\\{E_{5},(E_{2}-E_{3}),(E_{2}-E_{4})\\}$. The Lie algebra
$su(2)$ has a one dimensional Cartan subalgebra which may be spanned by any
non zero element. Therefore a general Cartan subalgebra ${\cal A}$ of ${\cal
C}$ is such that an orthogonal basis of $i{\cal A}$ is given by
$\\{F_{1},F_{2},F_{3}\\}$, with

$F_{1}:=iE_{1},\qquad F_{2}:=i(E_{2}+E_{3}+E_{4}),\qquad
F_{3}:=aiE_{5}+bi(E_{2}-E_{3})+ci(E_{2}-E_{4}),$ (7)

for any, not simultaneously zero, real parameters $(a,b,c)$.

We now proceed to step $3$ of the above algorithm. The matrix $F_{1}$ is just
the identity matrix which is diagonal in every basis. The matrix $F_{2}$ has
eigenvalue $\lambda_{1}=3$ with eigenspace $Q_{3}$ spanned by
$\\{\vec{e}_{1},\vec{e}_{2}+\vec{e}_{3}+\vec{e}_{5},\vec{e}_{4}+\vec{e}_{6}+\vec{e}_{7},\vec{e}_{8}\\}$
and eigenvalue $\lambda_{2}=-3$ with eigenspace $Q_{-3}$ spanned by
$\\{\vec{f}_{1},\vec{f}_{2},\vec{f}_{3},\vec{f}_{4}\\}$ , with
$\vec{f}_{1}:=\vec{e}_{3}-\vec{e}_{2},$
$\vec{f}_{2}:=\vec{e}_{2}+\vec{e}_{3}-2\vec{e}_{5}$,
$\vec{f}_{3}:=\vec{e}_{6}-\vec{e}_{4},$
$\vec{f}_{4}:=-2\vec{e}_{7}+\vec{e}_{6}+\vec{e}_{4}$. Consider now $F_{3}$
acting on $Q_{3}$ and $Q_{-3}$. Direct verification shows that $F_{3}$ is zero
on $Q_{3}$. On $Q_{-3}$ we have

$F_{3}\vec{f}_{1}=3c\vec{f}_{1}+(2ia+2b+c)\vec{f}_{2},$
$F_{3}\vec{f}_{2}=(3c+6b+6ia)\vec{f}_{1}-3c\vec{f}_{2},$
$F_{3}\vec{f}_{3}=3b\vec{f}_{3}+(b+2c+2ia)\vec{f}_{4},$
$F_{3}\vec{f}_{4}=(3b+6c-6ia)\vec{f}_{3}-3b\vec{f}_{4}.$

which shows that the subspace $Q_{-3}$ splits into two invariant subspaces for
$F_{3}$ spanned by $\\{\vec{f}_{1},\vec{f}_{2}\\}$ and
$\\{\vec{f}_{3},\vec{f}_{4}\\}$ respectively. Calculating the spectrum of
$F_{3}$ on such subspaces we find that $F_{3}$ has eigenvalues $\pm\lambda$ on
both subspaces where

$\lambda:=2\sqrt{3c^{2}+3cb+3b^{2}+3a^{2}}.$ (8)

Notice that $\lambda$ is never zero otherwise we would have $a=b=c=0$ which we
have excluded. Listing the eigenvalues of $F_{1}$, $F_{2}$ and $F_{3}$ and
constructing the $M$-matrix of the above algorithm, we have

$M=\begin{pmatrix}1&1&1&1&1&1&1&1\cr 3&3&3&3&-3&-3&-3&-3\cr
0&0&0&0&\lambda&\lambda&-\lambda&-\lambda\end{pmatrix}.$

Row reduction to transform this matrix in its Reduced Row Echelon Form which
is

$RREF(M)=\begin{pmatrix}1&1&1&1&0&0&0&0\cr 0&0&0&0&1&1&0&0\cr
0&0&0&0&0&0&1&1\end{pmatrix},$

corresponds to multiplication of $M$ on the left by the matrix

$R:=\frac{1}{2}\begin{pmatrix}1&\frac{1}{3}&0\cr\frac{1}{2}&-\frac{1}{6}&\frac{1}{\lambda}\cr\frac{1}{2}&-\frac{1}{6}&-\frac{1}{\lambda}\end{pmatrix}.$

The rows of these matrices give the coefficients for the linear combinations
of $\\{F_{1},F_{2},F_{3}\\}$ whose eigenspaces are the sought vector spaces.
In particular consider

$\Pi_{1}:=\frac{1}{2}\left(F_{1}+\frac{1}{3}F_{2}\right),$ (9)
$\Pi_{2}:=\frac{1}{2}\left(\frac{1}{2}F_{1}-\frac{1}{6}F_{2}+\frac{1}{\lambda}F_{3}\right),$
(10)
$\Pi_{3}:=\frac{1}{2}\left(\frac{1}{2}F_{1}-\frac{1}{6}F_{2}-\frac{1}{\lambda}F_{3}\right).$
(11)

All these projections form a complete set of, symmetric, mutually orthogonal
idempotents (which are also called generalized Young symmetrizers [6]). In
fact, one can verify directly that

$\Pi_{1}+\Pi_{2}+\Pi_{3}={\bf 1},$ (12) $\Pi_{j}\Pi_{k}=\delta_{j,k}\Pi_{j}.$
(13)

The eigenspaces corresponding to the eigenvalue $1$ of these matrices are the
spaces $W$ (for $\Pi_{1}$), $V_{1}$ (for $\Pi_{2}$), $V_{2}$ (for $\Pi_{3}$).
They coincide with the images of these matrices. The result is explicitly
given in the following theorem.

######  Theorem 1.

Every decomposition of ${\cal H}:={\cal H}_{2}^{\otimes 3}=W\oplus V_{1}\oplus
V_{2}$ in invariant subspaces for $su^{S_{3}}(8)$ ($SU^{S_{3}}(8)$) or
$u^{S_{3}}(8)$ ($U^{S_{3}}(8)$) corresponds to a triple $(a,b,c)\neq(0,0,0)$.
An orthogonal basis of $W$ is given by

${\cal
W}=\\{\vec{e}_{1},\vec{e}_{2}+\vec{e}_{3}+\vec{e}_{5},\vec{e}_{4}+\vec{e}_{6}+\vec{e}_{7},\vec{e}_{8}\\},$
(14)

which coincides with the basis $\\{\phi_{0},\phi_{1},\phi_{2},\phi_{3}\\}$ in
(3) and uniquely determines $W$ which coincides with the symmetric sector. An
orthogonal basis of $V_{1}$ is given by $\\{|v_{1}\rangle,|w_{1}\rangle\\}$
with

$\begin{array}[]{ccl}|v_{1}\rangle&=&x_{2}|001\rangle+x_{3}|010\rangle+x_{5}|100\rangle\\\
|w_{1}\rangle&=&x_{4}|011\rangle+x_{6}|101\rangle+x_{7}|110\rangle\end{array}$
(15)

with $x_{2}:=\frac{5}{3}\lambda-6b-2ia-2c$,
$x_{3}:=-\frac{\lambda}{3}+6ia-4c+2b$,
$x_{5}=-(x_{2}+x_{3})=-\frac{4}{3}\lambda-4ia+4b+6c$,
$x_{4}=\frac{5}{3}\lambda-6c+2ia-2b$, $x_{6}=-\frac{\lambda}{3}-6ia-4b+2c$,
$x_{7}=-(x_{4}+x_{6})=-\frac{4}{3}\lambda+4ia+4c+6b$. An orthogonal basis of
$V_{2}$ is given by

$\begin{array}[]{ccl}|v_{2}\rangle&=&y_{2}|001\rangle+y_{3}|010\rangle+y_{5}|100\rangle\\\
|w_{2}\rangle&=&y_{4}|011\rangle+y_{6}|101\rangle+y_{7}|110\rangle\end{array}$
(16)

with $y_{2}=\frac{5\lambda}{3}+6b+2ia+2c$,
$y_{3}=-\frac{\lambda}{3}-6ia+4c-2b$,
$y_{5}=-(y_{2}+y_{3})=-\frac{4}{3}\lambda-4b-6c+4ia$,
$y_{4}=\frac{5}{3}\lambda+6c-2ia+2b$, $y_{6}=-\frac{\lambda}{3}+6ia+4b-2c$,
$y_{7}=-(y_{4}+y_{6})=-\frac{4}{3}\lambda-4c-4ia-6b$.

A direct computation using (8) shows that the decomposition is orthogonal.

###### Proof.

The theorem follows by explicitly writing the matrices $\Pi_{1}$, $\Pi_{2}$
and $\Pi_{3}$ in (9), (10) (11).

The matrix $\Pi_{1}$ is the following:

$\Pi_{1}:=\begin{pmatrix}1&0&0&0&0&0&0&0\cr
0&\frac{1}{3}&\frac{1}{3}&0&\frac{1}{3}&0&0&0\cr
0&\frac{1}{3}&\frac{1}{3}&0&\frac{1}{3}&0&0&0\cr
0&0&0&\frac{1}{3}&0&\frac{1}{3}&\frac{1}{3}&0\cr
0&\frac{1}{3}&\frac{1}{3}&0&\frac{1}{3}&0&0&0\cr
0&0&0&\frac{1}{3}&0&\frac{1}{3}&\frac{1}{3}&0\cr
0&0&0&\frac{1}{3}&0&\frac{1}{3}&\frac{1}{3}&0\cr 0&0&0&0&0&0&0&1\end{pmatrix}$

Thus the orthogonal basis for the subspace $W$, is the one given by equation
(14).

The work for the matrices $\Pi_{2}$ and $\Pi_{3}$ which depend on the
parameters $a$, $b$, and $c$, requires some extra considerations. Let us
consider the discussion for $\Pi_{2}$ and $V_{1}$. The matrix $\Pi_{2}$ in
(10) is $\Pi_{2}:=\frac{1}{12\lambda}(\Pi_{2,1},\Pi_{2,2})$ with

$\Pi_{2,1}:=\begin{pmatrix}0&0&0&0\cr 0&4\lambda-12b&-2\lambda-12ia-12c&0\cr
0&-2\lambda+12ia-12c&4\lambda+12b+12c&0\cr 0&0&0&4\lambda-12c\cr
0&-2\lambda-12ia+12b+12c&-2\lambda+12ia-12b&0\cr 0&0&0&-2\lambda-12ia-12b\cr
0&0&0&-2\lambda+12ia+12b+12c\cr 0&0&0&0\end{pmatrix}$
$\Pi_{2,2}:=\begin{pmatrix}0&0&0&0\cr-2\lambda+12ia+12b+12c&0&0&0\cr-2\lambda-12ia-12b&0&0&0\cr
0&-2\lambda+12ia-12b&-2\lambda-12ia+12b+12c&0\cr 4\lambda-12c&0&0&0\cr
0&4\lambda+12b+12c&-2\lambda+12ia-12c&0\cr
0&-2\lambda-12ia-12c&4\lambda-12b&0\cr 0&0&0&0\end{pmatrix}$

Considering the columns $2,3,$ and $5$ of $\Pi_{2}$, one sees that the sum of
second, third and fifth row is zero. Therefore at the most two of these
columns are linearly independent. In fact using the definition of $\lambda$ in
(8) it follows that only one column is linearly independent.

Taking the first one multiplied by ${6\lambda}$ plus the second one multiplied
by $2\lambda$ one obtains the first element in the (orthogonal) basis of
${\cal V}_{1}$. This choice is made because this vector is never zero if
$(a,b,c)\neq 0$. 222Set all the components equal to zero. This gives $a=0$,
and multiplying the component along $\vec{e}_{3}$ by $5$ and summing the
$\vec{e}_{2}$-component and the $\vec{e}_{3}$-component multiplied by $5$ one
obtains $b=\frac{11}{2}c$. Plugging this in the $\vec{e}_{2}$-component (set
to zero), one gets $\sqrt{199c^{2}}=21c$ which implies $c=0$. However this
would imply $b=0$ as well, which along with $a=0$, gives a contradiction.
Analogously one obtains the second vector of the basis. In fact the form of
this vector can be found by noticing that the coefficients of $\vec{e}_{4}$,
$\vec{e}_{6}$ and $\vec{e}_{7}$ can be found by the ones of $\vec{e}_{2}$,
$\vec{e}_{3}$ and $\vec{e}_{5}$, by exchanging $b\leftrightarrow c$
$a\leftrightarrow-a$.

The discussion of $\Pi_{3}$ and ${\cal V}_{2}$ is analogous. In fact, an
explicit calculation shows that $\Pi_{3}$ can be obtained from $\Pi_{2}$ with
the exchanges $a\leftrightarrow-a$, $b\leftrightarrow-b$,
$c\leftrightarrow-c$. ∎

This decomposition includes, as special cases, decompositions found in the
standard quantum physics literature. For examples, the two dimensional space
obtained with the Young symmetrizers in [6], which can also be obtained with
the recursive use of the Clebsch-Gordan coefficients [9], is spanned by

$\hat{\psi}_{1}:=\frac{1}{\sqrt{2}}|010\rangle-\frac{1}{\sqrt{2}}|100\rangle:=\frac{1}{\sqrt{2}}\vec{e}_{3}-\frac{1}{\sqrt{2}}\vec{e}_{5}$
$\hat{\psi}_{2}:=-\frac{1}{\sqrt{2}}|011\rangle+\frac{1}{\sqrt{2}}|101\rangle=-\frac{1}{\sqrt{2}}\vec{e}_{4}+\frac{1}{\sqrt{2}}\vec{e}_{6},$

and it is obtained as a special case of $V_{1}$ by choosing $a=0$.
$b=\frac{1}{3\sqrt{2}}$, $c=-\frac{1}{6\sqrt{2}}$ (which give
$\lambda=\frac{1}{\sqrt{2}}$).333This choice gives $\hat{\psi}_{1}$ for the
first element of the basis ${\cal V}_{1}$ and a multiple of $\hat{\psi}_{2}$
for the second element. Notice that the elements of the bases we describe are
not necessarily normalized.

######  Remark 2.1.

Consider the basis $\\{|v_{1}\rangle,|w_{1}\rangle\\}$ of $V_{1}$ (similar
consideration can be done for $V_{2}$) and assume $x_{6}=0$. Using the
definition of $\lambda$ (8), this gives, beside $a=0$,

$\sqrt{3b^{2}+3c^{2}+3bc}=3c-6b.$

Squaring both terms and solving the corresponding quadratic equation we obtain
that if $x_{6}=0$ $c=b$ and $b<0$ or $c=\frac{11}{2}b$ and $b>0$. The second
case gives $x_{7}=0$ which also gives $x_{4}=0$ which is not possible since it
would make one of the basis vectors equal to zero. Therefore not all the
values of $(a,b,c)$ are possible. In the case $c=b<0$ we have besides
$x_{6}=0$ also $x_{3}=0$. Assume now $x_{7}=0$. A similar reasoning as above
leads to conclude that the only possibilities are $a=0$ and $c=\frac{11}{2}b$
and $c=-\frac{b}{2}$, in both cases, $b>0$. The case $c=\frac{11}{2}b$ is
however not possible because, as we have seen, this would imply $x_{6}=0$. On
the other hand, the case $c=-\frac{b}{2}$ with $b>0$ also gives $x_{2}=0$.
Analogous reasoning shows that $x_{4}=0$ implies $x_{5}=0$. In fact one has
$x_{4}=0\leftrightarrow x_{5}=0$, $x_{6}=0\leftrightarrow x_{3}=0$,
$x_{7}=0\leftrightarrow x_{2}=0$. We shall use these considerations in section
5.

##  3 Measures of Entanglement for General Three Qubits States

For a general multi-partite quantum system, a measure of entanglement is a
nonnegative real function on the space of density matrices which satisfies
certain axioms. In particular it does not increase under local operations and
classical communication (LOCC), it is zero on separable states (that is,
statistical mixtures of product states), it is unchanged by local unitary
operations, and it is usually normalized to one (cf., e.g., [3] and [11] for a
detailed introduction to entanglement measures). For the case of two qubits
$A$ and $B$, a very common measure is the concurrence [19] whose square is
called the 2-tangle, $\tau_{AB}$. This can be defined from the density matrix
$\rho_{AB}$, by calculating the spectrum of
$\rho_{AB}\sigma_{y}\otimes\sigma_{y}\rho^{*}_{AB}\sigma_{y}\otimes\sigma_{y}$444Given
a matrix $A$ or a complex constant $c$, we always denote by $A^{*}$ and by
$c^{*}$ the complex conjugate. which can be shown to be made of real and
nonnegative values
$\lambda_{1}^{2}\geq\lambda_{2}^{2}\geq\lambda_{3}^{2}\geq\lambda_{4}^{2}$,
and $\tau_{AB}$ is defined as

$\tau_{AB}=\left[\max\\{\lambda_{1}-\lambda_{2}-\lambda_{3}-\lambda_{4},0\\}\right]^{2}.$
(17)

Consider now a pure state where three qubits ($A$, $B$, and $C$) are present,
which is the case of this paper. One can consider after tracing out $C$, the
entanglement between $A$ and $B$, $\tau_{AB}$, and analogously $\tau_{AC}$ and
$\tau_{BC}$. In this case a monogamy relation holds [18] [20]: If $A$ is fully
entangled with $B$, that is $\tau_{AB}=1$ then we must have $\tau_{AC}=0$,
that is, the state $\rho_{AC}$ is separable (where $\rho_{AC}$ is the partial
trace with respect to subsystem $B$). In fact, a more refined inequality holds
[4]. Consider a pure state $\rho$ and consider the system as a bipartite
system $A-(BC)$. Even though $BC$ is four dimensional, it follows from the
Schmidt decomposition (see, e.g., [15] pg. 109) that only two (orthogonal)
directions are necessary to express the full state. Therefore, we can treat
effectively $(BC)$ as a two level system and define the entanglement
$\tau_{A(BC)}$ between $A$ and $(BC)$. Then one has the following inequality
which was one of the main results of [4]

$\tau_{AB}+\tau_{AC}\leq\tau_{A(BC)}.$ (18)

The difference between $\tau_{A(BC)}$ and $\tau_{AB}+\tau_{AC}$ is by
definition, the distributed entanglement or $3-$tangle, which we denote simply
by $\tau$, that is, the amount of entanglement not due to pairwise
entanglement between the quantum bits. Explicit formulas were given in [4] for
$\tau_{A(BC)}$ and $\tau$. We report them below because we shall use them in
our analysis.  Let $\rho_{A}$ be the partial trace with respect to subsystems
$B$ and $C$.

$\tau_{A(BC)}=4\det(\rho_{A});$ (19)
$\tau=\tau_{A(BC)}-\tau_{AB}-\tau_{AC}=4\left|t_{000}^{2}t_{111}^{2}+t_{001}^{2}t_{110}^{2}+t_{010}^{2}t_{101}^{2}+t_{100}^{2}t_{011}^{2}-2d_{1}+4d_{2}\right|,$
(20)

with

$d_{1}:=t_{000}t_{111}t_{011}t_{100}+t_{000}t_{111}t_{101}t_{010}+t_{000}t_{111}t_{110}t_{001}+$
$t_{011}t_{100}t_{101}t_{010}+t_{011}t_{100}t_{110}t_{001}+t_{101}t_{010}t_{110}t_{001},$
$d_{2}:=t_{000}t_{110}t_{101}t_{011}+t_{111}t_{001}t_{010}t_{100},$

for a state

$|\psi\rangle=\sum_{ijk}t_{ijk}|ijk\rangle,$

with $ijk=0,1$.

##  4 States in the Symmetric Sector and their Entanglement

For any state in the symmetric sector, because of symmetry, we have
$\rho_{A}=\rho_{B}=\rho_{C}$, so for these states,
$\tau_{AB}=\tau_{AC}=\tau_{BC}$ and therefore we have, using (19) (20),

$\tau_{AB}=\tau_{AC}=\tau_{BC}=\frac{4\det(\rho_{A})-\tau}{2}.$ (21)

In order to express the entanglement measures $\tau$ and
$\tau_{AB}=\tau_{AC}=\tau_{BC}$ in a compact fashion, we introduce an extra
piece of notation. Define

$X_{2}:=c_{0}c_{2}-c_{1}^{2},\qquad X_{3}:=c_{0}c_{3}-c_{1}c_{2},\qquad
X_{4}:=c_{1}c_{3}-c_{2}^{2}.$ (22)

The quantities $X_{2}$, $X_{3}$ and $X_{4}$ give a quick test of separability
for states in the symmetric sector as described in the following proposition.

######  Proposition 4.1.

A state $\psi$ (4) in the symmetric sector is separable if and only if

$X_{2}=X_{3}=X_{4}=0.$ (23)

In this case $\psi$ is a symmetric product state of the form

$\psi=\phi\otimes\phi\otimes\phi,$ (24)

with $\phi$ a one qubit state.

###### Proof.

Assume that $\psi$ in (4) is a product state, i.e.,

$\psi=(\alpha_{0}|0\rangle+\alpha_{1}|1\rangle)\otimes(\beta_{0}|0\rangle+\beta_{1}|1\rangle)\otimes(\gamma_{0}|0\rangle+\gamma_{1}|1\rangle).$

Expanding and comparing with (4), we have

$c_{0}=\alpha_{0}\beta_{0}\gamma_{0},$
$c_{1}=\alpha_{1}\beta_{0}\gamma_{0}=\alpha_{0}\beta_{1}\gamma_{0}=\alpha_{0}\beta_{0}\gamma_{1},$
$c_{2}=\alpha_{1}\beta_{1}\gamma_{0}=\alpha_{1}\beta_{0}\gamma_{1}=\alpha_{0}\beta_{1}\gamma_{1},$
$c_{3}=\alpha_{1}\beta_{1}\gamma_{1}.$

Using these in (22) one verifies (23). For example, for $X_{2}$ we have

$c_{0}c_{2}=\alpha_{0}\beta_{0}\gamma_{0}\alpha_{1}\beta_{1}\gamma_{0}=(\alpha_{1}\beta_{0}\gamma_{0})(\alpha_{0}\beta_{1}\gamma_{0})=c^{2}_{1}.$

Viceversa, assume (23) is verified and consider the state (4). If $c_{0}=0$,
then, from (23), (22), it follows that $c_{1}=0$ and $c_{2}=0$. Therefore the
state coincides with $|111\rangle$ which is separable and of the symmetric
form (24). If $c_{0}\neq 0$ we can assume $c_{0}=1$, without loss of
generality keeping the state not normalized. Conditions (23) (22) give
$c_{2}=c_{1}^{2}$, $c_{3}=c_{1}^{3}$. Therefore, the state $\psi$ in (4) is of
the form (24) with $\phi=|0\rangle+c_{1}|1\rangle$. ∎

With the notation (22), the entanglement measures $\tau$ and
$\tau_{AB}=\tau_{AC}=\tau_{BC}$ take a compact form as described in the
following two propositions.

######  Proposition 4.2.

The distributed entanglement $\tau$ on the symmetric sector is given by

$\tau=4|X_{3}^{2}-4X_{2}X_{4}|.$ (25)

###### Proof.

Applying formula (20) we obtain

$\tau=4|c_{0}^{2}c_{3}^{2}-3c_{1}^{2}c_{2}^{2}-6c_{0}c_{1}c_{2}c_{3}+4c_{0}c_{2}^{3}+4c_{3}c_{1}^{3}|.$
(26)

Direct verification using formulas (22) in (25) shows that $\tau$ in (25)
coincides with (26). ∎

######  Proposition 4.3.

The pairwise entanglement $\tau_{AB}=\tau_{AC}=\tau_{BC}$ is given by

$\tau_{AB}=\tau_{AC}=\tau_{BC}=2\left(\det(\rho_{A})-|X_{3}^{2}-4X_{2}X_{4}|\right),$
(27)

where

$\det(\rho_{A})=|X_{3}|^{2}+2|X_{2}|^{2}+2|X_{4}|^{2}.$ (28)

Therefore the expression for the pairwise entanglement is

$\tau_{AB}=\tau_{AC}=\tau_{BC}=2\left(|X_{3}|^{2}+2|X_{2}|^{2}+2|X_{4}|^{2}-|X_{3}^{2}-4X_{2}X_{4}|\right).$
(29)

###### Proof.

We explicitly write the state $\psi$ in (4) as
$\psi=(c_{0},c_{1},c_{1},c_{2},c_{1},c_{2},c_{2},c_{3})^{T}$ and the
associated density matrix $\rho=\psi\psi^{\dagger}$. By taking the partial
trace with respect to $B$ and $C$, we obtain,

$\rho_{A}:=\begin{pmatrix}|c_{0}|^{2}+2|c_{1}|^{2}+|c_{2}|^{2}&c_{0}c_{1}^{*}+2c_{1}c_{2}^{*}+c_{2}c_{3}^{*}\cr
c_{0}^{*}c_{1}+2c_{1}^{*}c_{2}+c_{2}^{*}c_{3}&|c_{1}|^{2}+2|c_{2}|^{2}+|c_{3}|^{2}\end{pmatrix},$
(30)

and, after simplifications,

$\det(\rho_{A})=2|c_{0}|^{2}|c_{2}|^{2}+|c_{0}|^{2}|c_{3}|^{2}+2|c_{1}|^{4}+2|c_{1}|^{2}|c_{3}|^{2}+|c_{2}|^{2}|c_{1}|^{2}+2|c_{2}|^{4}$
$-2c_{0}^{*}c_{2}^{*}c_{1}^{2}-c_{0}^{*}c_{1}c_{2}c_{3}^{*}-2(c_{1}^{*})^{2}c_{0}c_{2}-2c_{1}^{*}c_{2}^{2}c_{3}^{*}-c_{0}c_{1}^{*}c_{2}^{*}c_{3}-2c_{1}c_{3}(c_{2}^{*})^{2}.$

By replacing the expressions of $X_{2}$, $X_{3}$, $X_{4}$ in (22) in the right
hand side of (28) one verifies that it coincides with the above expression of
$\det(\rho_{A})$. ∎

From (29) we obtain

$\frac{\tau_{AB}}{2}\geq
2|X_{2}|^{2}+2|X_{4}|^{2}-4|X_{2}X_{4}|=2\left(|X_{2}|-|X_{4}|\right)^{2}\geq
0.$

To have equality, that is the pairwise entanglement equal to zero, both
inequalities have to hold with the equal sign. We must have

$|X_{2}|=|X_{4}|,\quad|X_{3}^{2}-4X_{2}X_{4}|=|X_{3}|^{2}+4|X_{2}||X_{4}|=|X_{3}|^{2}+4|X_{2}|^{2}=|X_{3}|^{2}+4|X_{4}|^{2}.$

The following also follows from Proposition 4.1.

######  Proposition 4.4.

The only states in the symmetric sector that have both distributed and
pairwise entanglements equal to zero are the separable states.

Pairwise entanglement $\tau_{AB}$ and distributed entanglement $\tau$ are
local invariant, that is, they are invariant under local unitary
transformations, which, in the symmetric case are taken symmetric, i.e., of
the form $X\otimes X\otimes X$, with $X\in U(2)$. A complete set of local
invariant for general three qubits states and general local unitary
transformations, is known. For symmetric qubit states a complete set of
invariant can be obtained using the Majorana polynomial representation of
symmetric states [16]. We briefly review this.555We only discuss the Majorana
polynomial representation in the three qubits case. For a general treatment,
we refer to [14] and references therein. Given a general (not necessarily
symmetric) product state $\psi_{1}\otimes\psi_{2}\otimes\psi_{3},$ with
$\psi_{j}:=\alpha_{j}|0\rangle+\beta_{j}|1\rangle$, $j=1,2,3$, one can obtain
a symmetric state of the form (4) as
$A\Pi\psi_{1}\otimes\psi_{2}\otimes\psi_{3}$, where

$\Pi:=\frac{1}{3!}\sum_{P\in S_{3}}P$ (31)

is the total symmetrizer, and $A$ is a normalization factor. In particular,
direct calculation shows, with the definitions (3),

$\Pi(\psi_{1}\otimes\psi_{2}\otimes\psi_{3})=\alpha_{1}\alpha_{2}\alpha_{3}\phi_{0}+\left(\frac{\alpha_{1}\alpha_{2}\beta_{3}+\alpha_{1}\beta_{2}\alpha_{3}+\beta_{1}\alpha_{2}\alpha_{3}}{3}\right)\phi_{1}+\left(\frac{\alpha_{1}\beta_{2}\beta_{3}+\beta_{1}\alpha_{2}\beta_{3}+\beta_{1}\beta_{2}\beta_{3}}{3}\right)\phi_{2}+\beta_{1}\beta_{2}\beta_{3}\phi_{3}.$
(32)

Viceversa given a symmetric state (4) one considers the associate Majorana
polynomial

$p_{M}(x)=c_{0}x^{3}+3c_{1}x^{2}+3c_{2}x+c_{3},$

which by calculating the zeros and up to a common factor can be written as

$p_{M}(x)=(\alpha_{1}x+\beta_{1})(\alpha_{2}x+\beta_{2})(\alpha_{3}x+\beta_{3}).$

By choosing $\psi_{j}:=\alpha_{j}|0\rangle+\beta_{j}|1\rangle$ and using (32),
we see that the resulting symmetric state
$\Pi\psi_{1}\otimes\psi_{2}\otimes\psi_{3}$ is given by (4). Therefore every
symmetric state is in correspondence with a not ordered triple of one qubit
states $\psi_{1}$, $\psi_{2}$, $\psi_{3}$. Since each qubit is in
correspondence with a point on the Bloch sphere (see, e.g., [15]) a symmetric
state is in correspondence with three not ordered vectors from the origin to
the Bloch sphere in $R^{3}$. Furthermore, since for $X\in U(2)$, we have

$X\otimes X\otimes
X\Pi(\psi_{1}\otimes\psi_{2}\otimes\psi_{3})=\Pi(X\psi_{1}\otimes
X\psi_{2}\otimes X\psi_{3}),$

and applying a symmetric local unitary operation corresponds to a simultaneous
rotation of the three Bloch vectors of the three one qubit states $\psi_{1}$,
$\psi_{2}$ and $\psi_{3}$. Therefore the angles between the Bloch vectors give
a complete set of invariants under local symmetric unitary operations.

We remark here that using this representation of symmetric states it is
possible to assume that the states (4) can be written, after local symmetric
unitary operations, in special forms. In particular, after a common rotation,
it is possible to assume that one of the Bloch vectors corresponding to
$\\{\psi_{1},\psi_{2},\psi_{3}\\}$ is in a special position, for example along
the $z$-axis, while the remaining two can be rotated arbitrarily around the
the first one. One possible special form to write the state (4) after a local
unitary transformation is the one with $c_{3}=0$ and $c_{0}$ and $c_{2}$ real
(or have the same phase, recall that states are defined up to a phase factor).
In order to achieve this, take $\psi_{1}\otimes\psi_{2}\otimes\psi_{3}$ and
choose $X\in U(2)$ so that $X\psi_{1}=|0\rangle$ (up to a phase factor).
Therefore we have

$X\otimes X\otimes
X\psi_{1}\otimes\psi_{2}\otimes\psi_{3}=|0\rangle\otimes\left(\cos(\theta_{1})|0\rangle+\sin(\theta_{1})e^{i\chi_{1}}|1\rangle\right)\otimes\left(\cos(\theta_{2})|0\rangle+\sin(\theta_{2})e^{i\chi_{2}}|1\rangle\right),$

for real parameters $\theta_{1},$ $\theta_{2}$, $\chi_{1}$, $\chi_{2}$. Now we
can apply $Y\otimes Y\otimes Y$, with $Y=\begin{pmatrix}e^{i\chi}&0\cr
0&e^{{-i\chi}}\end{pmatrix}$. We obtain

$(Y\otimes Y\otimes Y)X\otimes X\otimes
X\psi_{1}\otimes\psi_{2}\otimes\psi_{3}=$
$e^{i\chi}|0\rangle\otimes\left(\cos(\theta_{1})e^{i\chi}|0\rangle+\sin(\theta_{1})e^{i(\chi_{1}-\chi)}|1\rangle\right)\otimes\left(\cos(\theta_{2})e^{i\chi}|0\rangle+\sin(\theta_{2})e^{i(\chi_{2}-\chi)}|1\rangle\right)=$
$|0\rangle\otimes\left(\cos(\theta_{1})|0\rangle+\sin(\theta_{1})e^{i(\chi_{1}-2\chi)}|1\rangle\right)\otimes\left(\cos(\theta_{2})|0\rangle+\sin(\theta_{2})e^{i(\chi_{2}-2\chi)}|1\rangle\right).$

The choice $\chi:=\frac{\chi_{1}+\chi_{2}}{4}$, gives with
$\eta=\frac{\chi_{1}-\chi_{2}}{2}$, the form

$\psi_{prodcan}:=|0\rangle\otimes\left(\cos(\theta_{1})|0\rangle+\sin(\theta_{1})e^{i\eta}\right)|1\rangle\otimes\left(\cos(\theta_{2})|0\rangle+\sin(\theta_{2})e^{-i\eta}|1\rangle\right).$
(33)

Applying the total symmetrizer $\Pi$ in (31) to $\psi_{prodcan}$ in (33), one
obtains a symmetric state (4) with $c_{3}=0$ and $c_{0}$ and $c_{2}$ real.  We
remark that $3$ is the minimum number of parameters necessary to identify
equivalence classes of (unitary) locally equivalent states since states are
identified up to a common phase factor and therefore (in the symmetric sector)
by $6$ parameters and $SU(2)$ has dimension $3$.

##  5 States in the Two Dimensional Invariant Subspaces and their Entanglement

We now consider the invariant subspaces of dimension two: $V_{1}$ and $V_{2}$
described in section 2. Since the orthogonal basis of $V_{2}$, given in
equation (16), can be obtained from the orthogonal basis of $V_{1}$, given in
equation (15), exchanging $a\leftrightarrow-a$, $b\leftrightarrow-b$,
$c\leftrightarrow-c$, and $(a,b,c)$ are free parameters (not all zero) we will
consider without loss of generality only the subspace $V_{1}$. We shall
calculate the pairwise entanglements and the distributed entanglement. We
remark that since these states are in general not invariant under permutation
(as opposed to states in the symmetric sector treated in the previous section)
there is no a priori reason why $\tau_{AB}$ should be equal to $\tau_{AC}$.

In order to simplify the calculation of the entanglement measures, it is
convenient to anticipate a result on dynamics (treatment of dynamics will be
done in the next section). We recall that we call local (special) unitary
symmetric operations, operations of the type $X\otimes X\otimes X$ with $X\in
U(2)$ ($X\in SU(2)$).

######  Proposition 5.1.

Given two states $|\psi_{1}\rangle$, $|\psi_{2}\rangle$ in the subspaces
$V_{1}$ (same for $V_{2}$) it is always possible to go from $|\psi_{1}\rangle$
to $|\psi_{2}\rangle$, using local operations.

###### Proof.

The Lie algebra corresponding to the Lie group of local symmetric special
unitary matrices is spanned by the matrices $iH_{x,y,z}$ defined in (5). This
Lie algebra leaves $V_{1}$ invariant. It is in fact the standard
representation of $su(2)$. This can be verified directly. Explicit computation
using (15) shows that $H_{z}|v_{1}\rangle=|v_{1}\rangle$,
$H_{z}|w_{1}\rangle=-|w_{1}\rangle$. Calculating $iH_{x}|v_{1}\rangle$ we get

$iH_{x}|v_{1}\rangle=-ix_{3}|101\rangle-
ix_{2}|110\rangle+i(x_{2}+x_{3})|011\rangle.$

This is proportional to $|w_{1}\rangle$ if there exists a non zero $k$ such
that $x_{3}=kx_{6}$ and $x_{2}=kx_{7}$. Now we apply the considerations of the
remark 2.1. If $x_{6}=0$ then we cannot have $x_{7}=0$ but we have $x_{3}=0$.
Therefore we choose $k=\frac{x_{2}}{x_{7}}\not=0$. If $x_{7}=0$, we cannot
have $x_{6}=0$ but we have $x_{2}=0$ and we choose
$k=\frac{x_{3}}{x_{6}}\not=0$. If both $x_{6}$ and $x_{7}$ are different from
zero a direct calculation using the definitions of $x_{2,3,6,7}$ shows that
$\frac{x_{2}}{x_{7}}=\frac{x_{3}}{x_{6}}$. Therefore, defining
$k:=\frac{x_{2}}{x_{7}}=\frac{x_{3}}{x_{6}}\not=0$, we have the result.  In
all cases, we have

$H_{x}|v_{1}\rangle=-k|w_{1}\rangle.$

It can be verified, using the fact that $x_{2}+x_{3}+x_{5}=0$, that
$H_{x}^{2}|v_{1}\rangle=|v_{1}\rangle$. Thus applying $H_{x}$ to the previous
equation $H_{x}|w_{1}\rangle=-\frac{1}{k}|v_{1}\rangle$. Therefore we have
$H_{x}k|w_{1}\rangle=-|v_{1}\rangle$, $H_{x}|v_{1}\rangle=-k|w_{1}\rangle$.

Thus on the orthogonal basis $\\{v_{1},k|w_{1}\rangle\\}$ $H_{z}$ and $-H_{x}$
act as $\sigma_{z}$ and $\sigma_{x}$ on the basis
$\\{|0\rangle,|1\rangle\\}$666Notice also that because of how $k$ is defined
the vectors $|v_{1}\rangle$ and $|w_{1}\rangle$ have the same length and
therefore can be normalized simultaneously. and therefore $iH_{z}$ and
$-iH_{x}$ generate the Lie algebra $su(2)$. Since the corresponding Lie group,
$SU(2)$ is transitive on the complex sphere, the symmetric local
transformations $X\otimes X\otimes X$, $X\in SU(2)$, are able to transfer any
state to any other state (cf. [12]).

∎

We will now compute the entanglement measures for the states in the invariant
subspaces of dimension $2$. Since these quantities remain unchanged by using
local operations, and all the states in $V_{1}$ (or $V_{2}$) can be reached
using local operation starting from an arbitrary state, as proved in the
previous proposition, it is enough to compute the measures for a particular
state.

First we will see that the distributed entanglement $\tau$ is always zero.

######  Proposition 5.2.

Let $|\psi\rangle\in V_{1}$ (or $|\psi\rangle\in V_{2}$) then $\tau=0$

###### Proof.

Direct calculation using (20) shows that $\tau=0$ for the first basis vector
$|v\rangle=x_{2}|001\rangle+x_{3}|010\rangle+x_{5}|100\rangle$ in the
definition (15). Therefore the result follows using Proposition 5.1.

∎

To calculate $\tau_{A(BC)}$, $\tau_{AB}$ and $\tau_{AC}$, we recall that since
$\tau=0$ from (20)

$\tau_{A(BC)}=\tau_{AB}+\tau_{AC}.$ (34)

Since these quantities are constant on $V_{1}$, let us calculate them at
$|v_{1}\rangle$ in (15). We only need to compute $\tau_{A(BC)}$ and
$\tau_{AB}$ since $\tau_{AC}$ will follow from (34). Computation of $\rho_{A}$
and $\rho_{AB}$ for the state $|v_{1}\rangle$ gives

We have:

$\rho_{A}=\left(\begin{array}[]{cc}|x_{2}|^{2}+|x_{3}|^{2}&0\\\
0&|x_{5}|^{2}\end{array}\right),$ (35)

$\rho_{AB}=\left(\begin{array}[]{cccc}|x_{2}|^{2}&0&0&0\\\
0&|x_{3}|^{2}&x_{3}{x}^{*}_{5}&0\\\ 0&x_{5}x^{*}_{3}&|x_{5}|^{2}&0\\\
0&0&0&0\end{array}\right).$ (36)

Using (19), we obtain

$\tau_{A(BC)}=4\left(|x_{2}|^{2}+|x_{3}|^{2}\right)|x_{5}|^{2}=4(|x_{2}|^{2}+|x_{3}|^{2})|x_{2}+x_{3}|^{2}.$
(37)

To compute $\tau_{AB}$ one has to calculate the eigenvalues of
$\rho_{AB}\sigma_{y}\otimes\sigma_{y}\rho^{*}_{AB}\sigma_{y}\otimes\sigma_{y}$
which using formula (36) can be seen to be: zero with multiplicity two and the
eigenvalues of the $2\times 2$ matrix

$\begin{pmatrix}2|x_{3}|^{2}|x_{5}|^{2}&2|x_{3}|^{2}Re(x_{5}x^{*}_{3})\cr
2|x_{5}|^{2}Re(x_{5}x^{*}_{3})&2|x_{5}|^{2}|x_{3}|^{2}\end{pmatrix},$

which are $\lambda_{1}^{2}=2|\mu|(|\mu|+|Re(\mu)|)$,
$\lambda_{2}^{2}=2|\mu|(|\mu|-|Re(\mu)|)$, with $\mu:=x_{5}x_{3}^{*}$. Using
formula (17) we have since $\lambda_{3}=\lambda_{4}=0$,

$\tau_{AB}=(\lambda_{1}-\lambda_{2})^{2}=2|\mu|\left(\sqrt{|\mu|+|Re(\mu)|}-\sqrt{|\mu|-|Re(\mu)|}\right)^{2}=4|\mu|\left(|\mu|-|Im(\mu)|\right).$
(38)

We also have $\tau_{AC}=\tau_{A(BC)}-\tau_{AB}$.

######  Remark 5.3.

From formula (37) it follows that $\tau_{A(BC)}$ is zero if and only if
$x_{5}=0$ ($x_{2}$ and $x_{3}$ cannot be simultaneously zero because this
would imply the vector $|v_{1}\rangle$ to be zero). In this case,
$|v_{1}\rangle$ would be a product state (of the form
$|0\rangle\otimes\tilde{\psi}_{BC}$ for a state $\tilde{\psi}_{BC}$ on the
subsystem $B-C$). Since the local symmetric unitary group is transitive on the
subspace $V_{1}$ every state in this subspace is a product state as expected
when the entanglement is zero. The condition on the entanglement $\tau_{AB}$
is less intuitive. It can be state by saying that $\mu:=x_{5}x^{*}_{3}$ is
purely imaginary.

##  6 Symmetric Dynamics on the Invariant Subspaces

We now study how the Lie group $U^{S_{3}}(8)$ of symmetric dynamics, i.e.,
unitary transformations which commute with the permutation group of three
objects, $S_{3}$, acts on its invariant subspaces. On the two dimensional
invariant subspaces $V_{1}$ and $V_{2}$ this group acts as $U(2)$ and its
induced dynamics is not more rich than the one of the group generated by the
symmetric local transformations $X\otimes X\otimes X$ and multiples of the
$8\times 8$ identity. These transformations do not modify the entanglement
measures and in particular the distributed entanglement which is zero for any
subspace as we have seen in the previous section.

More interesting is the dynamics of $U^{S_{3}}(8)$ on the four dimansional
symmetric sector which can be proven (see, e.g., [1]) to be given by all
possible unitary $4\times 4$ matrices. The local symmetric unitary
transformations are a Lie subgroup of $U^{S_{3}}(8)$ whose Lie algebra is
spanned by $iH_{x,y,z}$ and multiples of the identity. If we consider the
orthonormal basis
$\\{\phi_{0},\frac{\phi_{1}}{\sqrt{3}},\frac{\phi_{2}}{\sqrt{3}},\phi_{3}\\}$,
the matrices $\frac{i}{2}H_{x,y,z}$ give the four dimensional irreducible
representation of $su(2)$. They can be computed using the Clebsch-Gordan
coefficients [9] (or directly by computing their action on the given basis).
They are

$S_{x}:=\frac{1}{2}\begin{pmatrix}0&\sqrt{3}i&0&0\cr\sqrt{3}i&0&2i&0\cr
0&2i&0&\sqrt{3}i\cr 0&0&\sqrt{3}i&0\end{pmatrix},$ (39)
$S_{y}:=\frac{1}{2}\begin{pmatrix}0&-\sqrt{3}&0&0\cr\sqrt{3}&0&-2&0\cr
0&2&0&-\sqrt{3}\cr 0&0&\sqrt{3}&0\end{pmatrix},$ (40)
$S_{z}:=\frac{1}{2}\begin{pmatrix}3i&0&0&0\cr 0&i&0&0\cr 0&0&-i&0\cr
0&0&0&-3i\end{pmatrix},$ (41)

and satisfy the standard commutation relations for $su(2)$,

$[S_{x},S_{y}]=S_{z},\qquad[S_{y},S_{z}]=S_{x},\qquad[S_{z},S_{x}]=S_{y}.$
(42)

The Lie subgroup of $SU(4)$ corresponding to the Lie algebra spanned by
$S_{x,y,z}$ ( corresponding to local symmetric transformations) leaves the
measures of entanglement unchanged. In fact, it is a maximal Lie subgroup
having this property as shown in the following two propositions whose proofs
are postponed to the Appendix.

######  Proposition 6.1.

The local Lie group corresponding to the Lie algebra spanned by the $4\times
4$ $i\times$ identity and $S_{x,y,z}$ is maximal among the Lie groups leaving
the distributed tangle $\tau$ unchanged on the $4$-dimensional symmetric
sector. That is, there is no Lie group leaving such measure invariant which is
larger than the local Lie group.

######  Proposition 6.2.

The local Lie group corresponding to the Lie algebra spanned by the $4\times
4$ identity and $S_{x,y,z}$ is a maximal Lie group leaving the pairwise tangle
$\tau_{AB}=\tau_{AC}=\tau_{BC}$ unchanged on the $4$-dimensional symmetric
sector.

The symmetric Lie group $U^{S_{3}}(8)$ acts on the symmetric sector spanned by
the vectors $\\{\phi_{0},\phi_{1},\phi_{2},\phi_{3}\\}$ as the unitary group
$U(4)$ and the associated Lie algebra acts like $u(4)$ (see, e.g., [1]). The
Lie algebra spanned by $S_{x,y,z}$ in (39), (40), (41) is a subalgebra of
$u(4)$ corresponding to local symmetric operations and isomorphic to $su(2)$,
the symmetric sector giving an irreducible representation of $su(2)$. Since
$U(4)$ is transitive on the complex unit sphere, it is possible, using
elements of the Lie group $U^{S_{3}}(8)$, to transfer the state from a product
state $\phi\otimes\phi\otimes\phi$ to any arbitrary state in the symmetric
sector, independently of the entanglement value of the target state. In order
to analyze how the elements of $U^{S_{3}}(8)$ modify the entanglement on the
symmetric sector, we analyze the structure of the Lie algebra $u(4)$ starting
with describing how the Lie algebra spanned by $\\{S_{x},S_{y},S_{z}\\}$ which
leaves such measures unchanged ‘sits’ in $u(4)$.  Our goal is to arrive at a
factorization of elements of $U(4)$ which separates factors which modify the
entanglement measures from the symmetric local transformations that do not,
trying to maximize the latter ones.

The Lie algebra $u(4)$ admits, up to conjugacy, a Cartan decomposition (see,
e.g., [10])

$u(4)=sp(2)\oplus sp^{\perp}(2),$ (43)

where $sp(2)$ is the symplectic Lie algebra and $sp^{\perp}(2)$ is its
orthogonal complement.777The inner product considered is the inner product
$\langle A,B\rangle=kTr(AB^{\dagger})$, for an appropriate positive constant
$k$. They satisfy the basic (Cartan-like) commutation relations

$\left[sp(2),sp(2)\right]\subseteq
sp(2),\qquad\left[sp^{\perp}(2),sp(2)\right]\subseteq
sp^{\perp}(2),\qquad\left[sp^{\perp}(2),sp^{\perp}(2)\right]\subseteq sp(2).$
(44)

As it is customary, we denote by $Sp(n)$ the connected Lie group associated
with $sp(n)$. According to Cartan decomposition theorem, every unitary
$4\times 4$ matrix $U$ can be written as

$U:=K_{1}AK_{2},$ (45)

where $K_{1,2}$ are in $Sp(2)$ and $A$ is the exponential of an element in a
maximal Abelian subalgebra in $sp^{\perp}(2)$ which in this case has dimension
$2$ since we are including multiples of the identity as well in
$sp^{\perp}(2)$.888In general the maximal Abelian subalgebra for the Cartan
decomposition $su(n)=sp(\frac{n}{2})\oplus sp^{\perp}(\frac{n}{2})$ has
dimension $\frac{n}{2}-1$ which would give $1$ in this case. However, we have
included multiples of the identity since we looked at the decomposition for
$u(4)$. A full treatment of the decompositions for $u(n)$ can be found in [10]
and a summary with applications for quantum systems can be found in [5].

The symplectic Lie algebra $sp(n)$ (which has dimension $n(2n+1)$) and its
associated Lie group $Sp(n)$ have several important properties that are of
interest for the study of quantum dynamics. In particular, $sp(n)$ is a
maximal Lie subalgebra of $su(2n)$ which means that $sp(n)$ along with any
nonzero element $X\notin sp(n)$ of $su(2n)$ generates all of $su(2n)$. Every
Lie subalgebra of $su(2n)$ which is isomorphic to $sp(n)$ is actually
conjugate to $sp(n)$. Furthermore, $Sp(n)$ is transitive on the complex unit
sphere $S^{2n-1}$ representing quantum states. This means that, for any two
(normalized) quantum states $|\psi_{1}\rangle$ and $|\psi_{2}\rangle$, there
exists a matrix in $Sp(n)$ such that $|\psi_{2}\rangle=X|\psi_{1}\rangle$.
This means, in particular, that any possible value of the entanglement in the
symmetric sector can be achieved by only using the transformations $K_{1}$ and
$K_{2}$ in (45).

For our purposes we consider a Lie subalgebra ${\cal S}$ conjugate to $sp(2)$
in $u(4)$. We consider the Lie algebra of $4\times 4$ matrices of the form

$F:=\begin{pmatrix}ir&\alpha&\beta&\gamma\cr-\alpha^{*}&is&\delta&-\beta\cr-\beta^{*}&-\delta^{*}&-is&\alpha\cr-\gamma^{*}&\beta^{*}&-\alpha^{*}&-ir\end{pmatrix},$

with $r$ and $s$ arbitrary real numbers and $\alpha,$ $\beta$, $\gamma$,
$\delta$ arbitrary complex numbers. Matrices in this Lie algebra ${\cal S}$
satisfy,

$FJ+JF^{T}=0,$

where $J$ is the matrix

$J:=\begin{pmatrix}0&0&0&1\cr 0&0&-1&0\cr 0&1&0&0\cr-1&0&0&0\end{pmatrix},$
(46)

and formulas (43)-(45) hold with $sp(2)$ replaced by ${\cal S}$. The reason
for this choice is that the matrices $S_{x,y,z}$ (39), (40), (41), giving the
$4$-dimensional irreducible representation of $su(2)$, belong to ${\cal S}$
(see, e.g., [7] for a treatment of how irreducible representations of $su(2)$
fit in the corresponding unitary Lie algebra). The decomposition (45)
therefore holds with $K_{1}$ and $K_{2}$ belonging to the connected Lie group
$e^{\cal S}$ conjugate to $Sp(2)$ and associated with the Lie algebra ${\cal
S}.$999Here and in the following we use the convention of denoting by $e^{\cal
S}$ the connected Lie group associated with a Lie algebra ${\cal S}$. On the
other hand, the for the matrix $A$ in (45), we can take the product of the
exponentials of two elements in a Cartan subalgebra in ${\cal S}^{\perp}$. For
such a Cartan subalgebra, we take $\texttt{span}\\{i{\bf 1}_{4},iH_{zz}\\}$
where on $(C^{2})^{\otimes 3}$, $H_{zz}$ is defined as $H_{zz}$ in (6). In the
symmetric sector, in the basis $\\{\phi_{0},\phi_{1},\phi_{2},\phi_{3}\\}$ it
is given by the matrix

$iH_{zz}=\begin{pmatrix}3i&0&0&0\cr 0&-i&0&0\cr 0&0&-i&0\cr
0&0&0&3i\end{pmatrix}.$ (47)

With this choice formula (45) can be written as

$U=K_{1}e^{i{\bf 1}_{4}z}e^{iH_{zz}w}K_{2},$ (48)

for real parameters $z$ and $w$, with $K_{1}$ and $K_{2}$ in $e^{\cal S}$
(isomorphic to $Sp(2)$). We now turn to the factorization of $K_{1}$ and
$K_{2}$ in (48).

Even $sp(2)$ has a Cartan decomposition $sp(2)={\cal L}\oplus{\cal L}^{\perp}$
that can be chosen (up to conjugacy) between two possibilities denoted by CI
and CII (cf. Chapter X in [10]). Given such a decomposition, the matrices
$K_{1}$ and $K_{2}$ in (45) can be written as $K$ in the following formula

$K=L_{1}\hat{A}L_{2},$ (49)

where $\hat{A}$ is the exponential of a matrix in a maximal Abelian subalgebra
inside ${\cal L}^{\perp}$. We choose the decomposition CI because this allows
us to separate $S_{x}$, $S_{y}$ and $S_{z}$ in ${\cal L}$ and ${\cal
L}^{\perp}$. In particular we have ${\cal L}={\cal S}\cap so(4)$ which is
given by the matrices of the form

$\hat{L}:=\begin{pmatrix}0&a&k&r\cr-a&0&f&-k\cr-k&-f&0&a\cr-r&k&-a&0\end{pmatrix},$
(50)

for four real parameter $a,k,r,f$.101010The Lie algebra ${\cal S}$ is only
conjugate to $sp(2)$, therefore, the Lie algebra characterizing the
decomposition CI is different but conjugate with respect to the one for
$sp(2)$. The matrix $M$ which gives the conjugacy is
$M:=\begin{pmatrix}U_{1}&0\cr 0&U_{1}^{T}\end{pmatrix}$ with
$U_{1}=\frac{1}{\sqrt{2}}\begin{pmatrix}1&1\cr-1&1\end{pmatrix}$. We have
$Msp(2)M^{T}={\cal S}$. The matrix $S_{y}$ in (40) belongs to the Lie
subalgebra ${\cal L}$ (choose $a=-\frac{\sqrt{3}}{2}$, $f=-1$ and the other
parameters equal to zero) while $S_{x}$ and $S_{z}$ belong to ${\cal
L}^{\perp}$. The dimension of the Cartan subalgebra associated to this
decomposition (the rank of the decomposition) is $2$. We choose as Cartan
subalgebra the one spanned by $S_{z}$ and $H$, where
$H=\texttt{diag}(0,\frac{i}{2},-\frac{i}{2},0)$. Therefore $\hat{A}$ in
formula (49) can be written as

$\hat{A}=e^{S_{z}x}e^{Hy},$ (51)

for real values $x$ and $y$. The Lie algebra ${\cal L}$ is isomorphic to
$u(2)$ with the isomorphism given by

$i{\bf 1}_{2}\leftrightarrow J,$ (52)

with $J$ in (46),

$\frac{i}{2}\begin{pmatrix}0&1\cr
1&0\end{pmatrix}\leftrightarrow\frac{1}{2}\begin{pmatrix}0&0&0&1\cr 0&0&1&0\cr
0&-1&0&0\cr-1&0&0&0\end{pmatrix}$ (53)
$\frac{1}{2}\begin{pmatrix}0&1\cr-1&0\end{pmatrix}\leftrightarrow\frac{1}{2}\begin{pmatrix}0&1&0&0\cr-1&0&0&0\cr
0&0&0&1\cr 0&0&-1&0\end{pmatrix},$ (54) $\frac{i}{2}\begin{pmatrix}1&0\cr
0&-1\end{pmatrix}\leftrightarrow\frac{1}{2}\begin{pmatrix}0&0&1&0\cr
0&0&0&-1\cr-1&0&0&0\cr 0&1&0&0\end{pmatrix}.$ (55)

Consider the matrix in ${\cal L}$

$R:=\begin{pmatrix}0&1&0&0\cr-1&0&-\sqrt{3}&0\cr 0&\sqrt{3}&0&1\cr
0&0&-1&0\end{pmatrix},$ (56)

which is orthogonal to $S_{y}$. Performing an Euler-like decomposition on
$e^{\cal L}$, we can write any element in $e^{\cal L}$, such as $L_{1}$ and
$L_{2}$ in (49), as $L:=e^{S_{y}t_{1}}e^{Rt_{2}}e^{S_{y}t_{3}}e^{Jt_{4}}$.
Combining this with $\hat{A}$ in (51) we obtain that every element $K$ in
$e^{\cal S}$ can be written as

$K:=e^{S_{y}t_{1}}e^{Rt_{2}}e^{S_{y}t_{3}}e^{Jt_{4}}e^{S_{z}t_{5}}e^{Ht_{6}}e^{Jt_{7}}e^{S_{y}t_{8}}e^{Rt_{9}}e^{S_{y}t_{10}}.$
(57)

In this decomposition, the only elements that change the entanglement measures
are factors of the type $e^{Rt}$, $e^{Jt}$, and $e^{Ht}$. In the resulting
decomposition for $U\in U(4)$ (48) to these needs to be added the evolution
$e^{iH_{zz}w}$ in (48). The full decomposition of unitary transformations
which combines (48) with (57) separates factors which do not change the
entanglement on the symmetric sector from factors that do.

##  7 Control of a Three Spin Symmetric Network

In this section, we briefly consider a possible physical scenario where the
analysis carried out in the previous sections applies. We consider in
particular a network of three identical spin $\frac{1}{2}$ particles subject
to a controlling electromagnetic field which simultaneously interacts with the
three spins. The three spins interact with each other via Ising interaction.
The time varying Hamiltonian for this system is

$H_{S}:=H_{S}(t)=H_{zz}+H_{x}u_{x}(t)+H_{y}u_{y}(t)+H_{z}u_{z}(t),$ (58)

with $H_{zz}$ defined in (6) and $H_{x,y,z}$ defined in (5). The functions
$u_{x,y,z}=u_{x,y,z}(t)$ represent (spatially uniform) components of an
electromagnetic field in the $x,y,z$ direction. The dynamics of the state is
given by the Schrödinger equation

$\dot{\psi}=-iH_{S}(t)\psi,\qquad\psi(0)=\psi_{0}.$ (59)

The controllability analysis of a quantum system (see, e.g., [5]) describes
the set of states that can be reached starting from $\psi_{0}$. In this case,
such a set is [1]

${\cal R}_{\psi_{0}}:=\\{X\psi_{0}\,|\,X\in U^{S_{3}}(8)\\}.$ (60)

In particular, if $\psi_{0}$ belongs to one of the invariant subspaces of
$U^{S_{3}}(8)$ (or $u^{S_{3}}(8)$), then, for every control function, the
state will remain in that subspace. In fact, every state in that subspace can
be reached from $\psi_{0}$ with an appropriate control (transitivity of the
unitary group [1]). The invariant subspaces for the group $U^{S_{3}}(8)$ were
described in section 2.

Let us restrict ourselves to the symmetric sector where the dynamics was
described in section 6 and let us pose the problem of reaching from a
symmetric product state a state with maximum distributed entanglement
$\tau=1$, using an appropriate control function. The problem will be solved if
we prove that there exists a symmetric local state
$\hat{\psi}_{0}:=\phi\otimes\phi\otimes\phi$ and a time $\bar{t}$ such that
$e^{-iH_{zz}\bar{t}}\hat{\psi}_{0}$ has maximum distributed entanglement. This
property is referred to, in the case of (pairwise) entanglement of two qubits,
as $e^{-iH_{zz}\bar{t}}$ being a perfect entangler [21] and we shall adopt
this terminology here, mutatis mutandis. If there exists such a product state
$\hat{\psi}_{0}$, we can use in (59) high amplitude short time pulses (so that
we can neglect in the dynamics the effect of $H_{zz}$) which will produce a
local symmetric transformation from $\psi_{0}$ to the state $\hat{\psi}_{0}$.
Then we can set the controls equal to zero and allow the evolution go as
$\hat{\psi}_{0}\rightarrow e^{-iH_{zz}t}\hat{\psi}_{0}$ for time $\bar{t}$. We
are left with proving that $e^{-iH_{zz}\bar{t}}$ is a perfect entangler.

######  Proposition 7.1.

There exists a $\bar{t}$ such that $e^{-iH_{zz}\bar{t}}$ is a perfect
(distributed) entangler transferring the local symmetric state

$\hat{\psi}_{0}=\left(\frac{1}{\sqrt{2}}|0\rangle+\frac{1}{\sqrt{2}}|1\rangle\right)^{\otimes
3},$ (61)

to a state of maximum distributed entanglement $\tau=1$.

###### Proof.

The state $\hat{\psi}_{0}$ in (61) can be written in the form (4) with
$c_{0}=c_{1}=c_{2}=c_{3}=\frac{1}{2\sqrt{2}}$. Using the explicit expression
of $H_{zz}$ in (47) we have that $c_{0,1,2,3}$ vary with time as

$c_{0}(t)=c_{3}(t)=\frac{e^{3it}}{2\sqrt{2}},\qquad
c_{1}(t)=c_{2}(t)=\frac{e^{-it}}{2\sqrt{2}}.$ (62)

Using the formula for the distributed entanglement $\tau$ in (25) together
with the expressions of $X_{2}$, $X_{3}$, $X_{4}$ in (22) we get
$X_{2}=X_{4}=\frac{1}{8}(e^{2it}-e^{-2it})$,
$X_{3}=\frac{1}{8}(e^{6it}-e^{-2it})$, and

$\tau(t)=4\left|X_{3}^{2}-4X_{2}X_{4}\right|=\frac{1}{16}\left|(e^{6it}-e^{-2it})^{2}-4(e^{2it}-e^{-2it})^{2}\right|=\frac{1}{16}\left|(e^{6it}-e^{-2it})^{2}+16\sin^{2}(2t)\right|.$

The function on the right hand side has a maximum equal to $1$ at
$t=\frac{\pi}{4}$. Therefore the proposition holds with
$\bar{t}=\frac{\pi}{4}$. ∎

##  8 Conclusions

In this paper, we have given an analysis of the states of a three qubit
quantum system under the action of the Lie group $U^{S_{3}}(8)$ of unitary
matrices which commute with the symmetric group of three objects. This is
motivated by the controlled dynamics of symmetric spin networks with three
spin $\frac{1}{2}$ particles, as described in section 7. The Hilbert space of
three qubits splits into subspaces of dimension $4$, $2$ and $2$, which are
invariant under the action of the Lie group $U^{S_{3}}(8)$. The subspace of
dimension $4$ is uniquely determined and corresponds to the so called
symmetric sector $W$ of states which are invariant under permutation
(symmetric states). The subspaces of dimension $2$ are not uniquely determined
although they are orthogonal to $W$ and orthogonal to each other. We have
provided the following results:

  * We have parametrized all the possible decompositions of the state space in invariant subspaces under the Lie group $U^{S_{3}}(8)$ (Theorem 1).

  * For states in the symmetric sector $W$, we have introduced three quantities $X_{2},$ $X_{3}$, $X_{4}$ in (22) which are easily calculated from the expression of the state and give a simple criterion of separability (Proposition 4.1).

  * We have calculated and expressions of the distributed entanglement and the pairwise entanglement in terms of the quantities $X_{2}$, $X_{3}$, $X_{4}$ Propositions 4.2 , 4.3 and concluded that the only states on the symmetric sector which have both entanglements equal to zero are the separable states (Proposition 4.4).

  * For states in the two dimensional invariant subspaces, we have proven that the distributed entanglement is always equal to zero (Proposition 5.2) while the pairwise entanglement will depend on the subspace considered. We provided a simple formula for it (formula (38)).

  * We have proven that it does not exists any connected Lie subgroup of $U^{S_{3}}(8)$ which properly contains the Lie subgroup of local symmetric transformations and leaves unchanged the distributed entanglement (Proposition 6.1) or the pairwise entanglement (Proposition 6.2) on the symmetric sector.

  * We have given a decomposition of any evolution in $U^{S_{3}}(8)$ on the symmetric sector into (local) elements which do not modify the entanglement and factors which modify it (formulas (48) and (57)).

  * We have proven that the free evolution given by a pairwise Ising interaction gives a perfect entangler for distributed entanglement on a symmetric network of three spin (Proposition 7.1) and used this to prescribe a control law to drive a separable symmetric state to a state of maximal distributed entanglement.

In future research, it will be of interest to extend the results presented
here to symmetric states for more than three qubits. Such extensions however
will depend on a better understanding of multipartite entanglement beyond the
three qubits case.

##  9 Acknowledgement

D. D’Alessandro’s research was supported by NSF under Grant EECS-1710558

## Data Availability Statement

The data that support this study are available from the corresponding author
upon reasonable request.

## References

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## Appendix: Proofs of Propositions 6.1 and 6.2

We consider a general vector in the symmetric sector

$\psi=c_{0}\phi_{0}+\hat{c}_{1}\frac{1}{\sqrt{3}}{\phi_{1}}+\hat{c}_{2}\frac{1}{\sqrt{3}}{\phi_{2}}+c_{3}\phi_{3},$

where, comparing with (4), we have $\hat{c}_{1}=\sqrt{3}c_{1}$,
$\hat{c}_{2}=\sqrt{3}c_{2}$.

We also follow the convention of denoting by $y_{R}$ and $y_{I}$ the real and
imaginary part of a quantity $y$.

### Proof of Proposition 6.1

###### Proof.

Recall, see (25), that the distributed tangle $\tau$ is given by
$\tau=4|X_{3}^{2}-4X_{2}X_{4}|$. Consider the quantity $X_{3}^{2}-4X_{2}X_{4}$
written separating its real and imaginary parts as
$X_{3}^{2}-4X_{2}X_{4}:=R+iI$ Then invariance of $\tau$ is equivalent to
invariance of the function $f:=\frac{1}{2}(R^{2}+I^{2})$. $R$ and $I$ are
functions of the complex vector
$\vec{v}:=(c_{0},\hat{c}_{1},\hat{c}_{2},c_{3})^{T}:=\vec{v}_{R}+i\vec{v}_{I}$,
for real vectors $\vec{v}_{R}$ and $\vec{v}_{I}$, where
$\vec{v}_{R}:=(c_{0,R},\hat{c}_{1,R},\hat{c}_{2,R},c_{3,R})^{T}$,
$\vec{v}_{I}:=(c_{0,I},\hat{c}_{1,I},\hat{c}_{2,I},c_{3,I})^{T}$. If $F$ is an
element of the Lie algebra associated to a given Lie subgroup of $U(4)$, under
the action of $e^{Ft}$, $\vec{v}$ changes as $\vec{v}\rightarrow
e^{Ft}\vec{v}$ and therefore $f(t)$ varies as

$f(t)=\frac{1}{2}\left(R^{2}(e^{Ft}\vec{v})+I^{2}(e^{Ft}\vec{v})\right).$ (63)

Since $F$ is skew-Hermitian, it can be written as $F=A+iB$ with $A$ skew
symmetric and $B$ symmetric (with both of them real). Invariance of $f=f(t)$
implies

$0=\frac{df}{dt}|_{t=0}=R\nabla R+I\nabla
I\begin{pmatrix}A\vec{v}_{R}-B\vec{v}_{I}\cr
A\vec{v}_{I}+B\vec{v}_{R}\end{pmatrix}.$ (64)

The idea of the proof is to show that if the matrix $F:=A+iB$ satisfies (64)
for all possible vectors $\vec{v}$, then it must be in the Lie algebra spanned
by $S_{x,y,z}$, plus the $i\times$ identity. To show this, we will first
compute (64) for special vectors.

Let us consider the case of vectors for which $\vec{v}_{I}=0$. A direct
calculation using the definitions (22) shows that $I=0$ so that (64)
simplifies as

$\dot{f}(0)=R\nabla R\begin{pmatrix}A\vec{v}_{R}\cr
B\vec{v}_{R}\end{pmatrix}=0.$ (65)

Using the definitions (22) we get

$R=X_{3R}^{2}-X_{3I}^{2}-4\left(X_{2R}X_{4R}-X_{2I}X_{4I}\right),$ (66)

and an explicit calculation, using the constraint that $\vec{v}_{I}=0$, gives
$X_{2I}=X_{3I}=X_{4I}=0$, so that we have

$\nabla R(\vec{v}_{R},0)=2X_{3R}\nabla X_{3R}-4X_{4R}\nabla
X_{2R}-4X_{2R}\nabla X_{4R}.$ (67)

Now we specialize further the vector $\vec{v}$ in (65).

  1. 1.

Set $\hat{c}_{2}=c_{2}=\hat{c}_{1}=c_{1}=0$. We have $X_{2}=X_{4}=0$ and
$X_{3R}=c_{0R}c_{3R}$. A direct calculation gives

$\nabla
X_{3R}=\nabla\left(c_{0R}c_{3R}-c_{0I}c_{3I}+\frac{\hat{c}_{1I}\hat{c}_{2I}}{3}-\frac{\hat{c}_{1R}\hat{c}_{2R}}{3}\right)=$
$\left(c_{3R},\,-\frac{\hat{c}_{2R}}{3},\,-\frac{\hat{c}_{1R}}{3},\,c_{0R}\,-c_{3I},\,\frac{\hat{c}_{2I}}{3},\,\frac{\hat{c}_{1I}}{3},\,-c_{0I}\right),$

which, using $\hat{c}_{1}=\hat{c}_{2}=0$ along with $c_{0I}=c_{3I}=0$, gives

$\nabla X_{3R}=(c_{3R},\,0,\,0,\,c_{0R},\,0,\,0,\,0,\,0).$ (68)

Placing this and $X_{3R}=c_{0R}c_{3R}$ in (65), we have

$\dot{f}(0)=c_{0R}c_{3R}\begin{pmatrix}c_{3R}&0&0&c_{0R}\end{pmatrix}A\begin{pmatrix}c_{0R}\cr
0\cr 0\cr c_{3R}\end{pmatrix}=0.$ (69)

Assume $c_{0R}c_{3R}\neq 0$. Using the fact that $A$ is skew-symmetric, and
using $c_{0R}^{2}\neq c_{3R}^{2}$, this relation implies $a_{1,4}=a_{4,1}=0$.

  2. 2.

Set $c_{0}=c_{2}=0$. A direct calculation using (22) gives $X_{3}=0$ and (66)
gives $R=4c_{1R}^{3}c_{3R}$. We have using (67) and $X_{3}=0$

$\nabla R=-4X_{4R}\nabla X_{2R}-4X_{2R}\nabla X_{4R}=-4c_{1R}c_{3R}\nabla
X_{2R}+4c_{1R}^{2}\nabla X_{4R}.$ (70)

Now use

$\nabla
X_{2R}=\nabla\left(c_{0R}\frac{\hat{c}_{2R}}{\sqrt{3}}-c_{0I}\frac{\hat{c}_{2I}}{\sqrt{3}}-\frac{\hat{c}_{1R}^{2}}{3}+\frac{\hat{c}_{1I}}{3}\right)=$
(71)
$\left(\frac{\hat{c}_{2R}}{\sqrt{3}},\,-\frac{2}{3}\hat{c}_{1R},\,\frac{c_{0R}}{\sqrt{3}},\,0,\,-\frac{\hat{c}_{2I}}{\sqrt{3}},\,\frac{2\hat{c}_{1I}}{3},\,-\frac{c_{0I}}{\sqrt{3}},\,0\right),$

and

$\nabla
X_{4R}=\nabla\left(\frac{\hat{c}_{1R}}{\sqrt{3}}c_{3R}-\frac{\hat{c}_{1I}}{\sqrt{3}}c_{3I}-\frac{\hat{c}_{2R}^{2}}{3}+\frac{\hat{c}_{2I}^{2}}{3}\right)=$
(72)
$\left(0,\,\frac{c_{3R}}{\sqrt{3}},\,-\frac{2}{3}\hat{c}_{2R},\,\frac{\hat{c}_{1R}}{\sqrt{3}},\,0,\,-\frac{c_{3I}}{\sqrt{3}},\,\frac{2}{3}\hat{c}_{2I},\,-\frac{\hat{c}_{1I}}{\sqrt{3}}\right),$

with $c_{0}=\hat{c}_{2}=0$ and $\hat{c}_{1I}=c_{3I}=0$, in (70). We get

$\nabla
R=\frac{4}{\sqrt{3}}\hat{c}_{1R}^{2}\left(0,\,c_{3R},\,0,\,\frac{\hat{c}_{1R}}{3},\,0,\,0,\,0,\,0\right).$

Placing this in (65) , we get

$\dot{f}(0)=\frac{16}{9}\hat{c}_{1R}^{5}c_{3R}\begin{pmatrix}0&c_{3R}&0&\frac{\hat{c}_{1R}}{3}\end{pmatrix}A\begin{pmatrix}0\cr\hat{c}_{1R}\cr
0\cr c_{3R}\end{pmatrix}=0.$ (73)

This gives using $a_{1,4}=a_{4,1}=0$,
$(c_{3R}^{2}-\frac{\hat{c}_{1R}^{2}}{3})a_{2,4}=0$ which implies $a_{2,4}=0$,
if we choose $\hat{c}_{1R}$ and $c_{3R}$ different from zero.

This shows that every matrix $F=A+iB$ in the Lie algebra corresponding to the
Lie group which leaves $\tau$ unchanged has to be such that
$a_{2,4}=a_{1,4}=a_{4,2}=a_{4,1}=0$. Since we assume that $S_{y}$ in (40) also
belongs to such Lie algebra and $[S_{y},A]$ is real (and skew-symmetric) while
$[S_{y},B]$ is purely imaginary (and symmetric), imposing this condition on
$[S_{y},A]$, shows that we must have also $a_{1,3}=a_{3,1}=0$ and
$a_{3,4}=-a_{4,3}=\frac{\sqrt{3}}{2}a_{2,3}$. Furthermore, imposing the
condition that the $(1,3)$ component is zero to $[S_{y},A]$, we also get
$a_{1,2}=a_{3,4}$. This shows that the real part of $A+iB$ must be a multiple
of $S_{y}$ in (40).

Now consider the restrictions on the $B$ symmetric matrix. Since, with $S_{z}$
in (41), $[S_{z},iB]$ is real, it must be proportional to $S_{y}$. From this
restriction, it follows that $B$ must be the sum of a diagonal matrix and a
matrix proportional to $S_{x}$ in (39). Then, considering the Lie bracket
$[S_{x},iB]$ which must also be proportional to $S_{y}$ it follows that the
diagonal part of $B$ must be a linear combination of the identity and $S_{z}$.
This concludes the proof.

∎

### Proof of Proposition 6.2

###### Proof.

We use the notations of the proof of Proposition 6.1. Let us first consider
the function $\det(\rho_{A})$ in (28) as we act on the vector $\vec{v}$ as
defined in the previous proof of Proposition 6.1 with $e^{Ft}$. That is,
similarly to (63), we define a function

$g(t)=\det(\rho_{A})\left(e^{Ft}\vec{v}\right)=\left|X_{3}(e^{Ft}\vec{v})\right|^{2}+2\left|X_{2}(e^{Ft}\vec{v})\right|^{2}+2\left|X_{4}(e^{Ft}\vec{v})\right|^{2},$
(74)

and we calculate $\frac{d}{dt}|_{t=0}g(t)$. Notice that this is not set to
zero yet. Similarly to what was done in (64), we have

$\frac{d}{dt}|_{t=0}g(t)=(\nabla|X_{3}|^{2}+2\nabla|X_{2}|^{2}+2\nabla|X_{4}|^{2})\begin{pmatrix}A\vec{v}_{R}-B\vec{v}_{I}\cr
A\vec{v}_{I}+B\vec{v}_{R}\end{pmatrix}.$ (75)

Choose now an initial point so that $\vec{v}_{I}=0$ which implies that the
imaginary parts of $X_{2},$ $X_{3}$ and $X_{4}$ are also zero. Therefore (75)
gives

$\frac{d}{dt}|_{t=0}g(t)=(2X_{3R}\nabla X_{3R}+4X_{2R}\nabla
X_{2R}+4X_{4R}\nabla X_{4R})\begin{pmatrix}A\vec{v}_{R}\cr
B\vec{v}_{R}\end{pmatrix}.$ (76)

We now proceed considering subcases as in Proposition 6.1, in fact, the same
subcases.

  1. 1.

Set $\hat{c}_{1}=\hat{c}_{2}=0$. In this case $X_{2}=X_{4}=0$ and $\nabla
X_{3R}$ was already computed in (68). Using this expression and the expression
of $X_{3R}$, which in this case is $X_{3R}=c_{0R}c_{3R}$, we obtain

$\frac{d}{dt}|_{t=0}g(t)=2c_{0R}c_{3R}K,$ (77)

where (cf. (69))

$K:=\begin{pmatrix}c_{3R}&0&0&c_{0R}\end{pmatrix}A\begin{pmatrix}c_{0R}\cr
0\cr 0\cr c_{3R}\end{pmatrix}.$

With the given definitions, the expression of the pairwise entanglement (29)
as a function of $t$ is

$\tau_{AB}(t)=2\left(g(t)-\sqrt{2f(t)}\right).$

The condition $\frac{d\tau_{AB}}{dt}|_{t=0}$ gives

$\dot{g}(0)-\frac{\dot{f}(0)}{\sqrt{2f(0)}}=0,$ (78)

which using (77) and (64) and (69) and the expression of
$f(0)=\frac{1}{2}c_{0R}^{2}c_{3R}^{2}$, gives

$2c_{0R}c_{3R}|c_{0R}c_{3R}|K-c_{0R}c_{3R}K=0.$

The quantity $c_{0R}c_{3R}$ can be chosen so that this implies $K=0$ and, as
in the proof of Proposition 6.1, this gives $a_{14}=a_{4,1}=0$.

  2. 2.

Set $c_{0}=c_{2}=0$.

This gives $X_{3}=0$. We also have
$X_{2R}=-c_{1R}^{2}=-\frac{\hat{c}_{1R}^{2}}{3}$,
$X_{4R}=c_{1R}c_{3R}=\frac{\hat{c}_{1R}}{\sqrt{3}}c_{3R}$. The quantities
$\nabla X_{2R}$ and $\nabla X_{4R}$ in this case were calculated in (71) (72).
These formulas give, with $c_{0}=c_{2}=0$,

$\nabla X_{2R}=\left(0,-\frac{2}{3}\hat{c}_{1R},0,0,0,0,0,0\right),$ (79)
$\nabla
X_{4R}=\left(0,\frac{c_{3R}}{\sqrt{3}},0,\frac{\hat{c}_{1R}}{\sqrt{3}},0,0,0,0\right).$
(80)

Using these formulas in (76), we obtain

$\frac{d}{dt}|_{t=0}g(t)=\frac{4}{3}\hat{c}_{1R}\left(0,c_{3R}^{2}+\frac{2}{3}\hat{c}_{1R}^{2},0,\hat{c}_{1R}c_{3R},0,0,0,0\right)\begin{pmatrix}A\vec{v}_{R}\cr
B\vec{v}_{R}\end{pmatrix}.$ (81)

In this case,
$X_{3}-4X_{2}X_{4}=4c_{1R}^{3}c_{3R}=\frac{4}{3}\frac{\hat{c}_{1R}^{3}}{\sqrt{3}}c_{3R}$,
and therefore
$f(0)=\frac{1}{2}|X_{3}-4X_{2}X_{4}|^{2}=\frac{8}{27}\hat{c}_{1R}^{6}c_{3R}^{2}$.
Replacing this in (78), together with the expression of $\dot{f}(0)$
calculated in (73), we get

$\frac{4}{3}\hat{c}_{1R}\left(0,c_{3R}^{2}+\frac{2}{3}\hat{c}_{1R}^{2},0,\hat{c}_{1R}c_{3R}\right)A\begin{pmatrix}0\cr\hat{c}_{1R}\cr
0\cr
c_{3R}\end{pmatrix}-\frac{4}{3}\sqrt{3}\frac{\hat{c}_{1R}^{5}c_{3R}}{|c_{1R}^{3}c_{3R}|}(0,c_{3R},0,\frac{\hat{c}_{1R}}{3})A\begin{pmatrix}0\cr\hat{c}_{1R}\cr
0\cr c_{3R}\end{pmatrix}=0.$ (82)

Now choose $c_{3R}=\hat{c}_{1R}$. Using this in (82), we obtain, after
simplifications,

$\begin{pmatrix}0&5-3\sqrt{3}&0&3-\sqrt{3}\end{pmatrix}A\begin{pmatrix}0\cr
1\cr 0\cr 1\end{pmatrix}=0$

This, using $a_{1,4}=a_{4,1}=0$, implies $a_{2,4}=a_{4,2}=0$.

The rest of the proof proceeds as the proof of Proposition 6.1.

∎