Multipartite entanglement¶
Pure state¶
For a general multipartite pure state $\vert \psi \rangle \in \mathcal{H}_1 \otimes\mathcal{H}_2\otimes\cdots\otimes\mathcal{H}_n$, we use tensor instead of matrix to describe the multipartite states. Tensor rank is defined as the minimum number $r$ such that there exist states $\vert \phi_i^{(k)}\rangle \in \mathcal{H}_k (1\leq i\leq r, 1\leq k\leq n)$ satisfying $$ \vert \psi \rangle = \sum_{i=1}^r \lambda_i \vert \phi_i^{(1)}\rangle \otimes \vert \phi_i^{(2)}\rangle \otimes \cdots \otimes \vert \phi_i^{(n)}\rangle $$ where the coefficients $\lambda_i$ are positive but the different product terms may not be orthonormal to each other. Mutipartite entanglement means the tensor rank of a pure state is larger than 1. In other words, the pure state cannot decompose into fully product state among different parties.
In the multipartite scenario, the notion of border rank also arises. This concept acknowledges that certain tensors can be approximated with arbitrary precision using other tensors with smaller tensor ranks -- a phenomenon not possible with matrices. Border rank is defined as the minimum number $r$ of product terms that are sufficient to approximate the given tensor with arbitrarily small error (under certain distance definition). In other words, the given tensor is a limit of tensors of rank $r$.
A famous example of the gap between tensor rank and border rank is that the tensor rank of the W state in 3-qubit is 3 while its border rank is 2 since $$ \left|W\right\rangle=\frac{1}{\sqrt{3}}\left[\lim _{t \rightarrow 0} \frac{1}{t}\left(\left(|0\rangle+t |1\rangle\right)^{\otimes 3}-|0\rangle^{\otimes 3}\right) \right]. $$ For any $t \neq 0$, the tensor on the right hand has rank 2 and in the limit $t \rightarrow 0$, it becomes W state. For brevity, in the following, we will denote the tensor and border rank as $R$ and $\underline{R}$ separately.
The gap between these two different ranks is related to the problem of determining a best rank-r approximation for the given tensor $A$, which has no solution in general implying $R(A)> \underline{R}(A)$. However, there exist two notable exceptions: the cases $n=2$ (the given tensor is a matrix) and $r=1$ (approximation by rank-1 tensors). The former implies $R(A)=\underline{R}(A)$ for a matrix $A$ while the latter means there does not exist a tensor $A$ with $R(A)>\underline{R}(A)=1$.
Quantum subspace¶
Given a subspace $\mathcal{S}$ spanned by a set of multipartite states $\{ \vert \psi_1\rangle, \vert \psi_2\rangle,\dots,\vert \psi_m\rangle\}$, we define the minimal rank of $\mathcal{S}$ as the following: $$ \operatorname{r}(\mathcal{S})=\min_{\vert\psi\rangle \in \mathcal{S}}\underline{R}(\vert\psi\rangle). $$ The definition is consistent across the bipartite and multipartite cases since the Schmidt rank is equivalent to the border rank for bipartite states.
Completely entangled subspace¶
Consider a subspace $\mathcal{S} \subset \mathcal{H}_1 \otimes\mathcal{H}_2\otimes\cdots\otimes\mathcal{H}_n$, we call $\mathcal{S}$ is completely entangled iff it contains no fully product state. Due to the existence of best rank-1 approximation as mentioned above, we know it is equivalent to the detection of subspaces with $r(\mathcal{S}) > 1$.
Genuinely entangled subspace¶
A subspace $\mathcal{S}\subset \mathcal{H}_1 \otimes\mathcal{H}_2\otimes\cdots\otimes\mathcal{H}_n$ is genuinely entangled iff it contains no product state for any bipartition $K \vert K^c$, where $K$ is a subset of $\{1,2,\dots,n\}$ and $K^c$ denotes the complementary set. For example, for a tripartite state $\vert \psi \rangle \in \mathcal{H}_A \otimes \mathcal{H}_B \otimes \mathcal{H}_C$, there are three possible bipartition $AB\vert C$, $AC\vert B$ and $BC\vert A$. The genuine entanglement of a subspace is equivalent to $r(\mathcal{S})>1$ for any bipartition $K\vert K^c$.