International Journal of Theoretical Physics

, Volume 53, Issue 2, pp 612–621

The Stabilizer Dimension of n-Qubit Symmetric States

Article

DOI: 10.1007/s10773-013-1847-1

Cite this article as:
Li, B., Li, J. & Wang, Z. Int J Theor Phys (2014) 53: 612. doi:10.1007/s10773-013-1847-1
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Abstract

We consider local unitary transformations acting on a multiparty symmetric pure state and determine the stabilizer dimension of any pure symmetric state.

Keywords

Pure symmetric state Stabilizer dimension Local unitary transformation 

1 Introduction

Entanglement is considered a key resource in quantum information. It is a fundamental theoretical problem to classify and quantify entanglement in multiparty quantum systems. Despite its importance for the field of quantum information theory, the properties of entangled states are not fully explored yet [1]. One of the first and most natural proposals for classifying entanglement is to consider two states to have the same entanglement type if they can be transformed to each other by a local unitary transformation [2, 3]. Mathematically, the local unitary Lie group acts on the manifold of quantum states, partitioning it into orbits, each orbit represents a type of entanglement. Generally speaking, the number of orbits is infinite, so we consider the dimension of the orbits.

The study of stabilizer dimension was initiated in Ref. [4], much progress has been made in understanding orbits and orbit dimension these years. Sudbery [5] gave six polynomial invariants for 3-qubit pure states; Makhlin [6] gave 18 polynomial invariants that separate orbits for 2-qubit mixed states; recently, Lyons and Walck changed the problem of orbits into a study the stabilizer dimension of pure multipartite state [7, 8, 9, 10, 11]. In [7, 8], the authors obtain minimum orbit dimension and its classification for n-qubit pure states. In [9], they find the maximum stabilizer dimension of nonproduct pure state is n−1, and obtain that for n≥3,n≠4 only generalized n-qubit GHZ states and their local unitary equivalents have maximum stabilizer dimension [10].

While it is an elementary computation to find the stabilizer dimension for a particular given state, it is very difficult to determine stabilizer dimension for general classes of states. Very recently, Zhang et al. in [12] gave a satisfactory answer for graph states. In this paper, we will determine the stabilizer dimension of general n-qubit symmetric pure states.

2 Symmetric States and Lie Algebra Action

In this section we first introduce the preliminaries which will be used in this paper. Let \(H = (\mathbb{C}^{2})^{\otimes n}\) be the n-qubit Hilbert space. The local unitary group
$$G=U(1)\times\mathit{SU}(2)^n $$
acts on the Hilbert space \(H=(\mathbb{C}^{2})^{\otimes n}\) [7, 8, 9, 10, 11]. For any state |ψ〉∈H, there is an equivalence class of |ψ〉, which constitutes an orbit \(\mathcal{O}_{\psi}= \{g|\psi \rangle \,|\, g\in G\}\). As we know in group theory, there is a natural diffeomorphism
$$\mathcal{O}_\psi\longleftrightarrow G/\mathit{Stab}_\psi, $$
where G/Stabψ is the cosets of the stabilizer group
$$\mathit{Stab}_\psi=\bigl\{ g\in G:g|\psi \rangle =|\psi \rangle \bigr\} . $$
The above correspondence between orbits and stabilizers provides us a way of analyzing stabilizer dimension instead of orbits.
We recall the unnormalized symmetric Dicke state \(|\psi_{n}^{(m)}\rangle \) [14, 15] is defined as
$$ \bigl|\psi_n^{(m)}\bigr\rangle =\sum_{P(0, 1)} |\underbrace{0\cdots0}_{n-m}\underbrace{1\cdots1}_m\rangle , \quad m = 0, 1, \ldots, n , $$
(1)
where the sum runs over all \(C_{n}^{m}\) permutations of 0 and 1 appearing nm and m times, respectively. Then a pure symmetric state |ψ〉 may be written by the unnormalized symmetric Dicke state basis:
$$ |\psi \rangle =\sum_{m=0}^nd_m\bigl|\psi_n^{(m)}\bigr\rangle . $$
(2)
We remind to the reader that any permutation pure symmetric state can be written in this form.
The Lie algebra su(2) of SU(2) is the set of traceless skew Hermitian matrices
$$\mathit{su}(2) = \left \{\left [ \begin{array}{c@{\quad}c} it & u\\ -\overline{u} & -it \end{array} \right ] \colon \ |\ t\in\mathbb{R},\ u\in\mathbb{C} \right \}. $$
Corresponding to the local unitary transformation g=(g1,g2,…,gn) in SU(2)n is the infinitesimal, or the local Lie algebra transformation X=(X1,X2,…,Xn)∈su(2)n. For any product state vector |I〉=|i1〉⊗|i2〉⊗⋯⊗|in〉, g=(g1,g2,…,gn) acts on \(H=({\mathbb{C}}^{2})^{\otimes n}\) by
$$g\cdot|I\rangle =g_1|i_1\rangle \otimes g_2|i_2 \rangle \otimes\cdots\otimes g_n|i_n\rangle . $$
This induces the action of X on |I〉 given by
$$ X\cdot|I\rangle =\sum_{k=1}^n|i_1 \rangle \otimes\cdots\otimes|i_{k-1}\rangle \otimes X_k|i_k \rangle \otimes|i_{k+1}\rangle \otimes\cdots\otimes|i_n\rangle . $$
(3)
Equation (3) extends linearly to the whole Hilbert space; more precisely, let |ψ〉=∑IcI|I〉, then X⋅|ψ〉=∑IcIX⋅|I〉. We use the following formula to compute the dimension of symmetric states [7, 12]:
$$ X\cdot|\psi \rangle =i\theta|\psi \rangle , $$
(4)
where Xsu(2)n, θ is a real parameter.

3 Stabilizer Dimension of Dicke States

In this section, we consider the stabilizer dimension of Dicke states. Note that the unnormalized symmetric Dicke state can be written as
$$ \bigl|\psi_n^{(m)}\bigr\rangle =\sum_{P(0, 1)}|i_1 \rangle \otimes|i_2\rangle \otimes\cdots\otimes|i_n\rangle . $$
(5)
In fact (5) is just an alternative literary style of (1), where P(0,1) denote the sum runs over \(C_{n}^{m}\) permutations of 0 and 1, there are m times 1 and nm times 0 in each term |i1〉⊗|i2〉⊗⋯⊗|in〉. Therefore, from (3) and (5) we obtain
$$ X\cdot\bigl|\psi_n^{(m)}\bigr\rangle =\sum _{P(0, 1)}\sum_{k=1}^n|i_1 \rangle \otimes\cdots\otimes|i_{k-1}\rangle \otimes X_k|i_k \rangle \otimes|i_{k+1}\rangle \otimes\cdots\otimes|i_n\rangle . $$
(6)
Let
$$X_k=\left [ \begin{array}{c@{\quad }c} it_k & {u_k}\\ -\overline{u}_k & -it_k \end{array} \right ], $$
then
$$ X_k|0\rangle =it_k|0\rangle -\overline{u}_k|1 \rangle , \qquad X_k|1\rangle =u_k|0\rangle -it_k|1\rangle . $$
(7)
Applying (7) to (6), we obtain that terms of (6) have only three cases: m−1 times 1, m times 1 and m+1 times 1. Furthermore, the coefficient of each term can be determined easily; as follows.
Case 1: Terms with m−1 times 1 in \(X\cdot|\psi_{n}^{(m)}\rangle \). Since there are m times 1 in each term of \(|\psi_{n}^{(m)}\rangle \), to obtain m−1 times 1, one must change a “1” to “0”, so we select those terms which Xk acts on |1〉 and obtain |0〉, it follows from (7) that the coefficients of these terms are uk. For example, the coefficient of the term \(|\underbrace{0\cdots0}_{n-m+1}\underbrace{1\cdots1}_{m-1}\rangle \) in (6) is u1+u2+⋯+unm+1. In general, if the term |I〉 has m−1 times 1 in (6), suppose “0” is located in the i1-th, i2-th, …, inm+1-th position, “1” is located in the inm+2-th, inm+3-th, …, in-th position, then the coefficient of |I〉 is given by
$$ u_{i_1}+u_{i_2}+\cdots+u_{i_{n-m+1}}. $$
(8)
Case 2: Terms with m times 1 in \(X\cdot|\psi_{n}^{(m)}\rangle \). Similarly, to obtain m times 1, one must select those terms which Xk acts on change a “0” to “0” or a “1” to “1”. It is clear from (7) that the coefficients of terms which Xk acts on change a “0” to “0” is itk, and the coefficients of terms which Xk change a “1” to “1” is −itk. For example, the coefficient of the term \(|\underbrace{0\cdots0}_{n-m}\underbrace{1\cdots1}_{m}\rangle \) in (6) is it1+it2+⋯+itnmitnm+1−⋯−itn. In general, if the term |I〉 has m times 1 in (6). Suppose “0” is located in the i1-th, i2-th, …, inm-th position, “1” is located in the inm+1-th, inm+2-th, …, in-th position, then the coefficient of |I〉 is given by
$$ it_{i_1}+it_{i_2}+\cdots+it_{i_{n-m}}-it_{i_{n-m+1}}- \cdots-it_{i_n}. $$
(9)
Case 3: Terms with m+1 times 1 in \(X\cdot|\psi_{n}^{(m)}\rangle \). There are m times 1 in each term of \(|\psi_{n}^{(m)}\rangle \), to obtain m+1 times 1, one must change a “0” to “1”. From (7) we see the coefficients of these terms is \(-\overline{u}_{k}\). For example, the coefficient of the term \(|\underbrace{0\cdots0}_{n-m-1}\underbrace{1\cdots1}_{m+1}\rangle \) in (6) is \(-\overline{u}_{n-m}-\overline {u}_{n-m+1}-\cdots-\overline{u}_{n}\). In general, if the term |I〉 has m+1 times 1 in (6), suppose “0” is located in the i1-th, i2-th, …, inm−1-th position, “1” is located in the inm-th, inm+1-th, …, in-th position, then the coefficient of |I〉 is given by
$$ -\overline{u}_{i_{n-m}}-\overline{u}_{i_{n-m+1}}-\cdots- \overline{u}_{i_{n}}. $$
(10)

Lemma 1

For any symmetric Dicke state\(|\psi_{n}^{(m)}\rangle \), Xsu(2)n, \(\theta\in\mathbb {R}\), if (4) is satisfied, thenu1=u2=⋯=un=0.

Proof

Notice that \(i\theta|\psi_{n}^{(m)}\rangle \) is also a symmetric state which has terms with m times “1”, and the coefficients of the terms with m−1 times 1 and m+1 times 1 are all zero. By (8) and (10) we obtain the following \(C_{n}^{m-1}\) and \(C_{n}^{m+1}\) equations respectively
$$\begin{gathered} u_{i_1}+u_{i_2}+\cdots+u_{i_{n-m+1}}=0, \end{gathered}$$
(11)
$$\begin{gathered} -\overline{u}_{i_{n-m}}-\overline{u}_{i_{n-m+1}}-\cdots- \overline{u}_{i_{n}}=0. \end{gathered}$$
(12)
It is easy for us to obtain u1=u2=⋯=un=0 by solving the system of homogeneous linear equations. □

Theorem 1

The stabilizer dimension of\(|\psi_{n}^{(0)}\rangle \), \(|\psi_{n}^{(n)}\rangle \)and its LU equivalents is 1; the stabilizer dimension of\(|\psi_{n}^{(m)}\rangle \)and its LU equivalents isn, where 1≤mn−1.

Proof

Since the stabilizer dimension is LU invariant, it is sufficient for us to consider only with \(|\psi_{n}^{(m)}\rangle \), for any 0≤mn, inserting (9) into (4) and using Lemma 1 we have the following \(C_{n}^{m}\) equations
$$ it_{i_1}+it_{i_2}+\cdots+it_{i_{n-m}}-it_{i_{n-m+1}}- \cdots-it_{i_n}=i\theta, $$
(13)
where {i1,i2,…,in}={1,2,…,n}.
If 1≤m<n, it is easy to see that
$$t_1=t_2=\cdots=t_n=t, $$
thus
$$ (n-2m)t=\theta. $$
(14)
In case of n=2m, then θ=0, t is the free unknown, so the stabilizer dimension is 1. Otherwise, n≠2m, there is only one free unknown, the stabilizer dimension is 1.
If m=0, then \(|\psi_{n}^{(m)}\rangle =|0\cdots0\rangle \). Equation (13) turns into only one equation
$$ t_1+t_2+\cdots+t_n=\theta. $$
(15)
So the stabilizer dimension is n. In fact, there is only one equation, but n+1 unknowns.
If m=n, then \(|\psi_{n}^{(m)}\rangle =|1\cdots1\rangle \). Similarly, it follows that
$$-t_1-t_2+\cdots-t_n=\theta, $$
thus the stabilizer dimension is n. □

4 Stabilizer Dimension of Pure Symmetric States

In this section we discuss an arbitrary n-qubit pure symmetric state which has the form in (2). Suppose at least two dm≠0 (in case only one dm≠0, then the state is just a Dicke state), X acts on |ψ〉 is given by
$$ X\cdot|\psi \rangle =\sum_{m=0}^nd_mX \cdot\bigl|\psi_n^{(m)}\bigr\rangle . $$
(16)
For simplicity, we discuss the coefficient of \(|\underbrace{0\cdots0}_{n-m}\underbrace{1\cdots1}_{m}\rangle \) in (16). (It is very similar to discuss the coefficient of |I〉, where “0” is located in the i1-th, i2-th, …, inm-th position, “1” is located in the inm+1-th, inm+2-th, …, in-th position.) The coefficient of \(|\underbrace{0\cdots0}_{n-m}\underbrace{1\cdots1}_{m}\rangle \) consists of three parts: the contribution from \(d_{m-1}X\cdot|\psi_{n}^{(m-1)}\rangle \) is given by
$$d_{m-1}(-\overline{u}_{n-m+1}-\overline{u}_{n-m+2}-\cdots- \overline{u}_{n}); $$
the contribution from \(d_{m}X\cdot|\psi_{n}^{(m)}\rangle \) is given by
$$d_m(it_1+it_2+\cdots+it_{n-m}-it_{n-m+1}- \cdots-it_n); $$
the contribution from \(d_{m+1}X\cdot|\psi_{n}^{(m+1)}\rangle \) is given by
$$d_{m+1}(u_1+u_2+\cdots+u_{n-m}). $$
Substituting these coefficients into (16), and we agree on d−1=dn+1=0. Then (4) can be written in the following equations
$$\begin{aligned} &d_{m-1}(-\overline{u}_{n-m+1}-\overline{u}_{n-m+2}- \cdots-\overline{u}_{n}) +d_m(it_1+it_2+ \cdots+it_{n-m}-it_{n-m+1} \\ &\quad {} -\cdots-it_n) +d_{m+1}(u_1+u_2+ \cdots+u_{n-m})=i\theta d_m, \end{aligned}$$
(17)
where 0≤mn.

We first determine the stabilizer dimension of a class of particular states.

Proposition 1

Let\(|\varPhi_{0}\rangle =\sum_{0 \leq m\leq \frac{n}{2}}(-1)^{m}e^{-im\delta}|\psi_{n}^{(2m)}\rangle \), \(|\varPhi_{1}\rangle =\sum_{0 \leq m\leq \frac{n-1}{2}}(-1)^{m}e^{-im\delta}| \psi_{n}^{(2m+1)}\rangle \)be unnormalized symmetric states, whereδ∈[0,2π] is a parameter, then the stabilizer dimension of |Φ0〉, |Φ1and their LU equivalents aren−1.

Proof

We establish only with |Φ0〉, the proof is similar to |Φ1〉 and its LU equivalents. Suppose X⋅|Φ0〉=|Φ0〉, since d2m=(−1)meimδ, d2m+1=d2m−1=0, it follows from (17) that
$$d_{2m}(it_1+it_2+\cdots+it_{n-2m}-it_{n-2m+1}- \cdots-it_n)=i\theta d_{2m}; $$
$$-d_{2(m-1)}(\overline{u}_{n-(2m-1)+1}+\overline {u}_{n-(2m-1)+2}+\cdots +\overline{u}_{n}) +d_{2m}(u_1+u_2+ \cdots+u_{n-(2m-1)})=0, $$
i.e., for any \(0\leq m\leq \frac{n}{2}\), we have that
$$ \begin{gathered} t_1+t_2+\cdots+t_{n-2m}-t_{n-2m+1}- \cdots-t_n=\theta \\ e^{i\delta}(\overline{u}_{n-2m}+\overline{u}_{n-2m+1}+ \cdots+\overline{u}_{n}) +(u_1+u_2+ \cdots+u_{n-(2m-1)})=0. \end{gathered} $$
(18)
By permutation of 1,…,n, we obtain
$$t_1=t_2=\cdots=t_n=\theta=0, $$
and subtracting two equations which differ only in that one contains ui where the other contains uj with (18)
$$u_i-u_j=e^{i\delta}(\overline{u}_{i}- \overline{u}_{j}). $$
Thus all the difference uiuj have the same argument \(\frac{\delta}{2}\), and therefore, the ui lie on a line in the complex plane, \(u_{i}=r_{i}e^{i\frac{\delta}{2}}\), where ri is real, substituting into (18), we have that
$$(r_1+\cdots+r_n)e^{i\frac{\delta}{2}}=u_1+ \cdots+u_n=0, $$
thus the stabilizer dimension of |Φ0〉 is n−1.

It is interesting for us to note that the result of Ref. [10], we find that |Φ0〉 and |Φ1〉 (possibly after local unitary transformation) has the maximum stabilizer dimension n−1, therefore, |Φ0〉, |Φ1〉 must LU equivalent to the generalized n-qubit GHZ state. In fact, |Φ0〉 and |Φ1〉 have stabilizers conjugate to the stabilizer of the GHZ state. □

Lemma 2

For any pure symmetric state |ψ〉, Xsu(2)n, \(\theta\in\mathbb{R}\), if |ψsatisfies the following conditions: (i) X⋅|ψ〉=|ψ〉, (ii) |ψis not local unitary equivalent to |Φ0〉, |Φ1〉, thenu1=u2=⋯=un=0.

Proof

We separate the argument into four cases.

Case 1: If there exist dm−1=dm=0, dm+1≠0 for some m, then (17) turns into
$$ d_{m+1}(u_1+u_2+\cdots+u_{n-m})=0. $$
(19)
Now we consider the coefficient of other terms which have also m times “1”, then we obtain \(C_{n}^{m}\) equations similar to (19). Combine these equations, then we get
$$u_1=u_2=\cdots=u_n=0. $$
Case 2: If there exist dm=0, dm−1≠0, dm+1≠0, and |dm+1|≠|dm−1| for some m, (17) turns into
$$ d_{m-1}(\overline{u}_{n-m+1}+\overline{u}_{n-m+2}+ \cdots+\overline{u}_{n}) =d_{m+1}(u_1+u_2+ \cdots+u_{n-m}). $$
(20)
Similarly, by considering the coefficient of the term \(|0\cdots010\underbrace{1\cdots1}_{m-1}\rangle \), (17) turns into
$$ d_{m-1}(\overline{u}_{n-m}+\overline{u}_{n-m+2}+ \cdots+\overline{u}_{n}) =d_{m+1}(u_1+u_2+ \cdots+u_{n-m-1}+u_{n-m+1}). $$
(21)
Subtracting (20) and (21), we obtain
$$ d_{m-1}(\overline{u}_{n-m+1}-\overline{u}_{n-m}) =d_{m+1}(u_{n-m}-u_{n-m+1}). $$
(22)
Taking norms of both sides of (22), we obtain
$$\bigl|d_{m+1}(u_{n-m}-u_{n-m+1})\bigr|^2=\bigl|d_{m-1}(u_{n-m}-u_{n-m+1})\bigr|^2. $$
Since |dm+1|≠|dm−1|, then unm=unm+1. By considering the coefficient of other terms with m times “1”, we get
$$u_1=u_2=\cdots=u_n. $$
Substituting into (20), then
$$u_1=u_2=\cdots=u_n=0. $$
Case 3: If there exist dm=0, dm−1≠0, dm+1≠0, and |dm+1|=|dm−1| for some m, (17) turns into
$$ e^{i\delta}(\overline{u}_{n-m}+\overline{u}_{n-m+1}+ \cdots+\overline{u}_{n}) +(u_1+u_2+ \cdots+u_{n-(m-1)})=0, $$
(23)
where \(e^{i\delta}=-\frac{d_{m-1}}{d_{m+1}}\). Thus we obtain that ui lie on a line in the complex plane, \(u_{i}=r_{i}e^{i\frac{\delta}{2}}\), where ri is real, furthermore, \(\sum_{i=1}^{n} u_{i}=0\). If there exist m′≠m also satisfy the condition of case 2, i.e., dm=0, dm′−1≠0, dm′+1≠0, and |dm′+1|=|dm′−1|, then from (23), we obtain
$$u_i-u_j=e^{i\delta}(\overline{u}_{i}- \overline{u}_{j})=e^{i\delta '}(\overline{u}_{i}- \overline{u}_{j}), $$
if δδ′, then
$$u_i=u_j, $$
inserting into (23), we have that
$$u_1=u_2=\cdots=u_n=0. $$
So, in the remaining case we need only consider that many three consecutive dm≠0 except for one thing, |ψ〉 is local unitary equivalent to |Φ0〉, |Φ1〉, which we have determined the stabilizer dimension in Proposition 1.
Case 4: If there exit many three consecutive dm≠0, let dm=am+ibm, um=rm+ism, (17) turns into
$$\begin{aligned} &(a_{m-1}+ib_{m-1}) \bigl(-r_{n-m+1}-\cdots-r_n+i(s_{n-m+1}+\cdots+ s_n)\bigr) \\ &\qquad {}+(a_m+ib_m) (it_1+it_2+ \cdots+it_{n-m}-it_{n-m+1}-\cdots-it_n) \\ &\qquad {}+(a_{m+1}+ib_{m+1}) \bigl(r_1+\cdots+r_{n-m}+i(s_1+ \cdots+s_{n-m})\bigr) \\ &\quad {}=i\theta(a_m+ib_m). \end{aligned}$$
(24)
That is
$$\begin{aligned} &{-}a_{m-1}(r_{n-m+1}+ \cdots+r_n)-b_{m-1}(s_{n-m+1}+\cdots+s_n) -b_m(t_1+t_2+\cdots+t_{n-m} \\ &\qquad {}-t_{n-m+1}-\cdots-t_n)+a_{m+1}(r_1+ \cdots+r_{n-m}) -b_{m+1}(s_1+\cdots+s_{n-m}) \\ &\quad {}=-\theta b_m, \end{aligned}$$
(25)
$$\begin{aligned} &{-}b_{m-1}(r_{n-m+1}+ \cdots+r_n)+a_{m-1}(s_{n-m+1}+\cdots+s_n) +a_m(t_1+t_2+\cdots+t_{n-m} \\ &\qquad {}-t_{n-m+1}-\cdots-t_n)+b_{m+1}(r_1+ \cdots+r_{n-m})+a_{m+1}(s_1+\cdots+s_{n-m}) \\ &\quad {}=\theta a_m. \end{aligned}$$
(26)
A simple calculation on (25) and (26) gives
$$\begin{aligned} &{-}(a_{m-1}a_m+b_{m-1}b_m) (r_{n-m+1}+\cdots+r_n)+(a_ma_{m+1}+b_mb_{m+1}) (r_1+\cdots+r_{n-m}) \\ &\qquad {}+(a_{m-1}b_m-b_{m-1}a_m) (s_{n-m+1}+\cdots+s_n)+(b_ma_{m+1}-a_mb_{m+1}) (s_1+\cdots+s_{n-m}) \\ &\quad {}=0. \end{aligned}$$
(27)
For any given state, dm is a constant, so are am, bm, where 0≤mn. So (27) is a homogeneous linear equation in 2n unknowns, by permutations of 1,…,n, we obtain a homogeneous linear system of \(C_{n}^{m}\) equations in 2n unknowns. Suppose |ψ〉 has l three consecutive dm≠0, then we obtain \(lC_{n}^{m}\) equations in 2n unknowns, from (27), we see that the coefficients of ri,si can not be all the same, thus we can select 2n independent equations, the solution of the system of homogeneous linear equations must be 0, i.e.,
$$r_1=\cdots=r_n=0,\qquad s_1= \cdots=s_n=0. $$
That is, u1=⋯=un=0. Having exhausted all these cases, we have proved the argument. □

Theorem 2

For anyn-qubit pure symmetric state\(|\psi \rangle =\sum_{m=0}^{n}d_{m}|\psi_{n}^{(m)}\rangle \), the stabilizer dimension of |ψcan only be four cases, i.e., 0, 1, n−1 andn.

Proof

By Lemmas 1, 2 and (17), if dm≠0, for some m, it follows that
$$ t_1+t_2+\cdots+t_{n-m}-t_{n-m+1}- \cdots-t_n=\theta $$
(28)
and we have \(C_{n}^{m}\) equations altogether. We now discuss the equations in four different cases:

(1) If |ψ〉 is LU equivalent to \(|\psi_{n}^{(0)}\rangle \) or \(|\psi_{n}^{(n)}\rangle \), by Theorem 1, then the stabilizer dimension is n.

(2) If |ψ〉 is LU equivalent to \(|\psi_{n}^{(m)}\rangle \) where 1≤mn−1, by Theorem 1 the stabilizer dimension is 1.

(3) If |ψ〉 is LU equivalent to the generalized n-qubit Greenberger-Horne-Zeilinger states, there are only two equations which have the form in (28), and it is easy to know that the dimension is n−1. From Proposition 1 we see that the stabilizer dimension of |Φ0〉, |Φ1〉 are also n−1.

(4) The only remaining possibility is that |ψ〉 must has more than one dm≠0, further more, there must be at least one dm≠0 with 1≤mn−1. By (28) we obtain \(C_{n}^{m}\) equations, then we see that
$$ t_1=t_2=\cdots=t_n=t,\quad(n-2m)t= \theta. $$
(29)
Suppose some other dm≠0, we obtain the similar equation, thus, it follows that (n−2m′)t=θ, i.e., t=θ=0, and therefore the dimension of the stabilizer is 0. □
To summarize, for any pure symmetric states |ψ〉 in the form of (2), we showed our result in the following Table 1.
Table 1

Four kinds of possible stabilizer dimension for n-qubit symmetric states

Case

Stabilizer dimension

The LU equivalents of \(|\psi_{n}^{(0)}\rangle \) or \(|\psi_{n}^{(n)}\rangle \)

n

The LU equivalents of \(|\psi_{n}^{(m)}\rangle \), where 1≤mn−1

1

The LU equivalents of the generalized GHZ state

n−1

Other

0

5 Conclusions

The entanglement classification and some other relevant properties for symmetric states have been studied by many authors [13, 16, 17, 18]. In this paper, we analyzed the stabilizer dimension of n-qubit pure symmetric states. We show the stabilizer dimension for any pure symmetric state can only be four different cases. We further provide the states (under local unitary transformation) which have the corresponding dimension. We hope our findings will shed new light on the characterization of entanglement of multiparty pure symmetric state.

Acknowledgements

The authors thank Professor Shaoming Fei for his advice. Bo Li is supported by Natural Science Foundation of China (Grants No. 11305105), the Natural Science Foundation of Jiangxi Province (Grants No. 20132BAB212010). Jiao-jiao Li is supported by Youth Foundation of Henan Normal University (12QK02) and Zhixi Wang is supported by KZ201210028032.

Copyright information

© Springer Science+Business Media New York 2013

Authors and Affiliations

  1. 1.Department of Mathematics and ComputerShangrao Normal UniversityShangraoChina
  2. 2.College of Mathematics and Information ScienceHenan Normal UniversityXinxiangChina
  3. 3.School of Mathematical SciencesCapital Normal UniversityBeijingChina

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