We are interested in quantum systems defined on a system \(S = A \cup B\), and quantifying the entanglement between the A and B subsystems. Let \(\mathcal{H}_X\) denote the Hilbert space of register \(X\), so \(\mathcal{H}_S = \mathcal{H}_A \otimes \mathcal{H}_B\). A quantum state on \(X\) is a unit-trace positive operator \(\rho\in \mathcal{D}(\mathcal{H}_X)\). A state is pure iff it is rank-one. Owing to the annoyance caused by writing kets and bras without the physics package, I will just write \(\psi\) to denote the density matrix of a pure state.
Let us first set the ground and talk about the simpler case. For pure states, the notion of entanglement is very well defined and well captured by von-Neumann entropy. Specifically, if I define the reduced state of A as \(\rho_A = tr_B(\psi)\), then the entropy \(S(\rho_A) = -tr(\rho_A \log{\rho_A})\) captures the amount of entanglement in the bipartition \(A \cup B\). This is often called the entropy of entanglement. Operationally, in the limit of a large number of copies of the state, this quantity quantifies
Hence, the entropy of entanglement is a fundamental quantity for pure states, and does a good job at characterizing the entanglement.
It is trivial to see that the von-Neumann entropy, or any Renyi entropy for that matter, is not a good object to characterize mixed-state entanglement. A counter example is given by the fully mixed state, exhibiting no correlations of any kind yet scoring high on entropies. To build up the story for mixed state, we must first understand what’s not entangled. Well, a pure state is not entangled if it can be written as a product across that bipartition. It turns out a mixed state is not entangled if it can be written in the following form,
\begin{equation} \rho_{AB} = \sum_i p_i \rho_A^i \otimes \rho_B^i \end{equation}
and any state of the form above is said to be separable (with respect to that bipartition). The reason any such object is not entangled (again, across that bipartition!) is because such a state can always be prepared by LOCC on that bipartition, and hence displays classical correlations. One cannot create entanglement via LOCC. Specifically, there could be correlations in the state, but they are classical and the extent of those correlations is quantified by the Shannon entropy of the distribution \(p\).
Well then, I give you the state \(\rho_{AB}\), how do you tell me if it is entangled? One way, and perhaps the most prominent way, to answer this question was given in [1] by Peres. Specifically, they outline the following: if a state is separable, it’s partial transpose must be positive. Let us unpack this. What is the transpose? Well, it’s a map \(A \mapsto A^{T}\) such that \([A^{T}]_{ij} := [A]_{ji}\). A partial transpose is when we transpose only the \(A(B)\) subsystem, we denote this as \(\rho^{T_A}\).
The proof is trivial, note that if \(\sigma\) is a valid density matrix, then so is \(\sigma^T\). Hence, upon partial transposing a separable state, we get another valid density matrix, which must be positive. Hence, any separable state has a positive partial tranpose—we say that it passes the PPT test.
If a state does not pass PPT, we denote it to be NPT. Then, we can say that a state which is NPT must be entangled. However, the chain of impliciations must be paid attention to. There do exist entangled states which are PPT. Thus, if I give you a state and it passes PPT, a priori we cannot say if it is entangled or not. However, if it violates PPT and is thus NPT, then we say that it is entangled.
\begin{equation} \text{Separable} \implies \text{PPT} \end{equation}
\begin{equation} \text{NPT} \implies \text{Entangled} \end{equation}
It is good to note a stronger statements proven for \(2 \otimes 2\) and \(2 \otimes 3\). In [2], the \(\text{Horodecki}^{\otimes 3}\) proved that for these dimensions, PPT and separability are equivalent.
\begin{equation} (2 \otimes 2) \text{ and } (2 \otimes 3) : \text{PPT} \Leftrightarrow \text{Separable} \end{equation}
We then perform the following task. Take the state \(\rho_{AB}\), partial transpose it \(\rho_{AB}^{T_A}\), and find it’s spectrum \(\Lambda(\rho_{AB}^{T_A}) = \{\lambda_i\}_i\). Then, define,
\begin{equation} \mathcal{N}(\rho_{AB}) := \sum_{\lambda_i : \lambda_i < 0} |\lambda_i| \end{equation}
If this quantity is positive, we have entanglement. This is called negativity. That is \(\mathcal{N}(\rho_{AB}) = 0\) for separable states. A nicer way to write this down is,
\begin{equation} \mathcal{N}(\rho_{AB}) := \frac{||\rho_{AB}^{T_A}||_1 - 1}{2} \end{equation}
where \(\|O\|_1 := tr\sqrt{O^{\dagger}O}\) is the trace norm. Another quantity born out of this discussion is the logarithmic negativity,
\begin{equation} E_N(\rho_{AB}) := \log_2||\rho_{AB}^{T_A}||1 \equiv \log_2(2\mathcal{N}(\rho{AB}) + 1) \end{equation}
These quantities first showed up in [4], and are now commonly used in analyzing mixed state entanglement.
[1] Peres, Asher. “Separability criterion for density matrices.” Physical Review Letters 77.8 (1996): 1413.
[2] Horodecki, Michal, Pawel Horodecki, and Ryszard Horodecki. “On the necessary and sufficient conditions for separability of mixed quantum states.” Phys. Lett. A 223.1 (1996).
[3] Horodecki, Michał, Paweł Horodecki, and Ryszard Horodecki. “Mixed-state entanglement and distillation: Is there a “bound” entanglement in nature?.” Physical Review Letters 80.24 (1998): 5239.
[4] Vidal, Guifré, and Reinhard F. Werner. “Computable measure of entanglement.” Physical Review A 65.3 (2002): 032314.