# Polynomial entanglement invariants

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# Problem

We say that two bipartite quantum states $\rho$ and $\sigma$ are "equally entangled" if they differ only by a choice of bases in Alice's and Bob's subspaces, i.e., if we can find $U_A$, $U_B$, such that

$\rho=(U_A\otimes U_B) \sigma (U_A\otimes U_B)^*.$

An entanglement invariant is by definition any real valued function on the space of bipartite density operators, which assigns the same value to equally entangled density operators. A polynomial invariant is an entanglement invariant, which can be computed as a polynomial in the matrix elements of $\rho$. Note that because we only consider hermitian operators, allowing polynomials in the matrix elements and their complex conjugates does not enlarge this class.

The basic problem is to decide the following question:

• Are the polynomial entanglement invariants complete?, i.e., if all polynomial invariants of $\rho$ and $\sigma$ agree, can we infer the existence of unitaries $U_A$, $U_B$ satisfying the above equation?

But we may add some further, closely related problems:

• Given the dimensions of Alices's and Bob's Hilbert spaces, name a finite set of invariants which is already complete.
• Do all this for multi-partite states. In this case even the case of pure states is not obvious.
• Decide whether the set of separable states can be described in terms of a polynomial invariant $f$, such that $f(\rho)\ge 0$ is equivalent to separability. There are many weaker versions of

# Background

1. When $d_1$ and $d_2$ are the dimensions of Alice's and Bob's Hilbert space, respectively, the state space is $(d_1\, d_2)^2-1$ dimensional. Since phases for $U_A$, $U_B$ drop out of the transformation equation, we may fix their determinant to be $1$, and hence get $(d_1^2-1)+(d_1^2-1)$ for the dimension of the symmetry group. Subtracting we get an expected manifold dimension of $(d_1^2-1)(d_1^2-1)$ for the quotient manifold, i.e., the manifold of all invariants. Of course, it may happen that no set of this many differentiable invariants is sufficient to pin down each equivalence class uniquely, and more invariants are needed to rule out some discrete choices.
1. It is perhaps useful to recall the "unipartite" version of this problem, i.e., the characterization of density operators up to unitary equivalence. The well-known complete set of invariants in that case is the spectrum of the density operator (including multiplicities). The eigenvalues are not polynomial, but the coefficients of the characteristic polynomial (i.e., the elementary symmetric functions of the eigenvalues) or, equivalently the numbers $a_n=tr(\rho^n)$ are polynomial, and from these the eigenvalues can be determined. Hence a complete set of invariants are the $a_n$ for $n=1,...,\text{dimension}$.
1. A basis for the ring of of invariant polynomials (even in the multi-partite case, and for arbitrary Hilbert space dimensions) was given in  and . Note that any homogeneous polynomial of degree $k$ in $\rho$ can be written as an expectation value of the $k^\text{th}$ tensor power of $\rho$, i.e., as $tr(\rho^{\otimes k} X)$, with a uniquely determined $X$. For an $n$-partite system this is an operator on a tensor product of $n\, k$ Hilbert spaces. Invariance requires that it commutes with all unitaries of the

form $U_1^{\otimes k}\otimes ...\otimes U_n^{\otimes k}$, where $U_m$ is an operator on the Hilbert space of the $m^\text{th}$ type of systems ($m=1,...,n$). Then the commutation theorem of von Neumann algebras, and the corresponding result for $n=1$, imply that $X$ must be a tensor product of $n$ permutation operators, each one permuting the $k$ tensor factors belong to one of the $n$ system types.

# Partial Results

• Y. Makhlin  has shown completeness in the bipartite qubit case. Moreover, he has identified a set of 18 invariants, which is sufficient in that case, and has shown that none of these may be omitted without destroying completeness.
• A. Sudbery  has solved the case of pure three qubit states, finding 8 polynomial invariants (6 being the dimension of the manifold of all invariants).

# Solution

The basic question of principle (are the polynomial entanglement invariants complete?) is answered in Onishchik and Vinberg's book "Lie Groups and Algebraic Groups", which contains the theorem .

The orbits of a compact linear group acting in a real vector space are separated by the polynomial invariants.

In other words (those of quantum information theory), if two states of a multipartite system are not related by local unitary transformations, then they have different values for some polynomial entanglement invariant.

It follows that the space of entanglement types of states, i.e. the space of orbits factored by normalisation, can be identified with the space of polynomial invariants (more precisely, the ring of polynomial functions on this space is isomorphic to the ring of polynomial invariants). The dimension of this space is known in full generality for pure states . For two parties it is one less than the dimension of the smaller state space (a complete set of invariants is the set of Schmidt coefficients, which sum to 1 by normalisation). For $n\gt2$, if the parties have state spaces with dimensions $d_1,\ldots,d_n$ in increasing order, then the space of orbits of normalised states has dimension

$D_{\text{pure}} = 2\prod_{r=1}^n d_r - \sum_{r=1}^n d_r^2 + n - 2 + \Delta^2$

where $\Delta = d_n - d_1 \ldots d_{n-1}$ if this is positive, otherwise $\Delta = 0$. If all the parties are qudits ($d_1 = \cdots = d_n = d$) this becomes

$\textstyle D_{\text{pure}} = 2d^n - nd^2 + n - 2 .$

The corresponding dimension for mixed states is

$\textstyle D_{\text{mixed}} = d^{2n} - nd^2 + n - 1$

which is probably correct, though a careful treatment has never appeared in the literature. The general case for mixed states has not been discussed.

The number of invariants needed to uniquely specify a state up to local unitary transformations is not the same as the dimension $D$ of the space of entanglement types; this is in general a curved space, with complicated geometry. Makhlin's work $M$ shows that the space of entanglement types of mixed states of two qubits is a nine-dimensional manifold in $\mathbb{R}^{18}$ (the ring of polynomial invariants has 18 generators subject to 9 relations). For pure states of three qubits, which have $D=6$ (including the norm), a complete set of invariants  consists of the six independent invariants given in  together with one more found by Grassl. Thus the space of orbits of non-normalised state vectors is a hypersurface in $\mathbb{R}^7$; normalising, the space of entanglement types of pure states of three qubits is a hypersurface in real projective 6-space.

The above theorem was used by Hilary Carteret and myself in our proof  that on an orbit whose dimension is exceptionally low, some entanglement invariant has an extreme value. We classified these exceptional orbits for pure states of three qubits.

The condition that the group should be compact is essential, as is shown by the example of the general linear group $\mathrm{GL}(n,\mathbb{C})$ acting on $n\times n$ complex matrices by the similarity transformation $X\mapsto GXG^{-1}$ where $G\in$ $\mathrm{GL}(n,\mathbb{C})$. The polynomial invariants here are the coefficients in the characteristic equation of $X$, so two matrices have the same values of the invariants if and only if they have the same eigenvalues. But having the same eigenvalues is not sufficient for two matrices to be similar; if some of the eigenvalues are repeated, there are different possible Jordan normal forms which are not related by similarity.

An even simpler example, and one which is relevant to quantum information theory, is the action of $\mathrm{GL} (m,\mathbb{C})\times\mathrm{GL} (n,\mathbb{C})$ on $m\times n$ matrices by $X\mapsto PXQ^T$ where $P\in \mathrm{GL} (m,\mathbb{C})$ and $Q\in \mathrm{GL}(n,\mathbb{C})$. In this case there are no polynomial invariants, but matrices can only be transformed into each other by such a transformation if they have the same rank. (The rank is a non-polynomial invariant.) If we take $X$ to be an element of $\mathbb{C}^m\otimes\mathbb{C}^n$ representing a pure state of a bipartite system, two states are related by this action if there are local operations which will convert them into each other with non-zero probability. This generalises the deterministic (unitary) local operations which define equally entangled states in the statement of the problem. The corresponding orbits for three qubits have been determined by Dür, Vidal and Cirac , and for four qubits by Verstraete, Dehaene, De Moor and Verschelde .

# Literature

1. M. Grassl, M. Rötteler, and T. Beth, »Computing local invariants of qubit systems«, Phys. Rev. A 58, 1833 (1998) and quant-ph/9712040 (1997)
2. E. M. Rains, »Polynomial invariants of quantum codes«, quant-ph/9704042 (1997)
3. Y. Makhlin, »Nonlocal properties of two-qubit gates and mixed states and optimization of quantum computations«, quant-ph/0002045 (2000)
4. A. Sudbery, »On local invariants of pure three-qubit states«, J. Phys. A 34, 643 (2001) and quant-ph/0001116 (2000)
5. A. L. Onishchik and E. B. Vinberg, »Seminar on Lie groups and algebraic groups«, Springer (Berlin) 1990, p.144 (in Russian); English translation »Lie groups and algebraic groups«, Springer (Berlin) 1990, Chap. 3, Paragraph 4, Theorem 3
6. H. A. Carteret, A. Higuchi, and A. Sudbery, »Multipartite generalisation of the Schmidt decomposition«, J. Math. Phys. 41 (2000) and quant-ph/0006125 (2000)
7. A. Acin, A. Andrianov, E. Jane, and R. Tarrach, »Three-qubit pure-state canonical forms«, J. Phys. A 34, 6725 (2001) and quant-ph/0009107 (2000)
8. H. A. Carteret and A. Sudbery, »Local symmetry properties of pure states of three qubits«, J. Phys. A 33, 4981 (2000) and quant-ph/0001091 (2000)
9. W. Dür, G. Vidal, and J. I. Cirac, »Three qubits can be entangled in two inequivalent ways«, Phys. Rev. A 62, 062314 (2000) and quant-ph/0005115 (2000)
10. B. De Moor, F. Verstraete, J. Dehaene, and H. Verschelde, »Four qubits can be entangled in nine inequivalent ways«, quant-ph/0109033 (2001)