# Can every bipartite state be written as $\rho_{AB} = \sum_{ij} c_{ij}\sigma_A^i\otimes \omega_B^j$?

Can every bipartite quantum state (including entangled ones) be written in the following way

$$\rho_{AB} = \sum_{ij} c_{ij}\sigma_A^i\otimes \omega_B^j$$

where $$\sigma_A^i$$ and $$\omega_B^j$$ are density matrices? And if $$\rho_{AB}$$ is a quantum state, must it also be the case that $$c_{ij}$$ are nonnegative? If the answers are yes, how can one show this?

I ask this because of a claim that the set of product states $$\rho_A\otimes\rho_B$$ span the vector space of all operators acting on $$H_A\otimes H_B$$ in an answer here - https://quantumcomputing.stackexchange.com/a/5066/4831.

• Indeed, product states span the (affine) space of Hermitian trace-one operators - it is actually their affine span. However, if you restrict the cofficients to be non-negative, you actually get the convex hull of these states which are the separable ones, see the answer of DaftWullie below. – Markus Heinrich Oct 20 '20 at 8:50
• @MarkusHeinrich, for the set of product states $S$, the affine span is the set $A = \{\sum_i\alpha_i S_i | S_i\in S, \alpha_i\in \mathbb{R}, \sum_i\alpha_i = 1\}$, correct? I can see that any element of $A$ is Hermitian and unit trace but the reverse inclusion is still a bit unclear to me. – Alice Munroe Oct 20 '20 at 14:21

If all $$c_{ij}$$ are non-negative, this is the definition of a separable state. But you can argue this by giving a protocol that lets you construct, using LOCC, the state $$\rho_{AB}$$.

You probably could use this description to cover entangled states if you let the $$c_{ij}$$ be negative. Here's a simple argument for bipartite states of qubits: the Pauli matrices form a basis, meaning that $$\rho_{AB}=\sum q_{ij}\sigma_i\otimes\sigma_j$$ (I'm using $$\sigma$$ here for Pauli matrices, NOT density matrices). Since $$\rho_{AB}$$ is Hermitian, the $$q_{ij}$$ are real. Now, I can always rewrite this as $$\rho_{AB}=\left(q_{00}-\sum_{(i,j)\neq (0,0)}|q_{ij}|\right)I+ \sum_{(i,j)\neq (0,0)} |q_{ij}|(I+\text{sgn}(q_{ij})\sigma_i\otimes\sigma_j).$$ Now note that every term $$I\pm\sigma_i\otimes\sigma_j$$ is non-negative, with a separable basis, i.e. is a separable state. Indeed, the coefficient in front of every term is positive except, perhaps, that of the identity term.

(I should perhaps clarify that the term in front of the identity being positive is sufficient for the state to be separable, but is not necessary. In some cases there are better separable decompositions than the one I've given.)

There's a similar basis of Hermitian matrices for any Hilbert space dimension, so the result holds just as well for any bipartite state, not just qubits.

• Thank you for the answer. I did not follow what you mean when you say $I \pm \sigma_i\otimes\sigma_j$ is nonnegative with a separable basis. I see that it is Hermitian and has unit trace but why is it a separable state? – Alice Munroe Oct 20 '20 at 14:25
• Consider, for example, $I+\sigma_1\otimes \sigma_3$. The eigenstates are those for the form $|\pm\rangle|0\rangle$ and $|\pm\rangle|1\rangle$. So if you've got a term that has these as eigenvectors with non-negative eigenvalues, they're separable states, with a separable decomposition being given by that basis. – DaftWullie Oct 20 '20 at 15:09

I know the question is already answered, but there was some question on my comment and I wanted to elaborate on that.

First, let us consider one system only. The $$\mathbb{R}$$-span of all states $$\rho$$ is the space of Hermitian operators. Indeed, by the spectral decomposition, already the set of pure states is enough. This also implies that the $$\mathbb{C}$$-span contains all linear operators, since we can decompose any operator into a Hermitian and anti-Hermitian part and expand those in states.

For two (or more) systems, the dimension of $$L(H_A\otimes H_B)=L(H_A)\otimes L(H_B)$$ is the product of the individual dimensions, $$\dim L(H_A) \times \dim L(H_B)$$. This implies that taking tensor products of generating sets of $$L(H_A)$$ and $$L(H_B)$$ has to yield a generating set, too. Just reduce the generating set to a basis of $$L(H_A)$$ and $$L(H_B)$$ and you'll get a product basis which lies in the product of the generating sets.

Note that this also shows that this is not as special property of states. Any generating set will do.

What is special is that states are Hermitian and trace-one. Thus, they all lie in the according affine space in the real vector space of Hermitian operators. Hence, any other Hermitian trace-one operator is not any arbitrary (real) linear combination of states, but an affine one (take the trace on both sides) $$A = \sum_{i,j} c_{i,j} \rho_i \otimes \rho_{j} \quad \Rightarrow \quad \sum_{ij} c_{ij} = 1.$$ As stated before, if we restrict to coefficients to be non-negative, we get the convex hull of product states, which is the set of separable states.

BTW, this is the geometric background for the robustness of entanglement which is an entanglement monotone defined as the minimum negativity in an affine decomposition of a state $$\rho$$ into product states.

• Very nice answer. I guess one cannot enforce positive semidefiniteness of the states through further constraints on $c_{ij}$ without explicitly knowing the $\rho_i$ and $\rho_j$? – user1936752 Oct 21 '20 at 18:52
• Nope. And there's basically no hope even if you know them. This stuff is generally very nasty. The psd constraints are highly nonlinear, thus the intersection of the positive cone with the affine subspace of trace-one matrices is an ugly beast. AFAIK, nobody knows how to express these conditions nicely in terms of coefficients in a basis, except for a single qubit. There you exactly get the Bloch ball. This hasn't prevented some people from trying, though (keyword: higher-dimensional Bloch sphere/ball). – Markus Heinrich Oct 22 '20 at 8:00