In a bipartite system $AB$, why does the entanglement negativity $\mathcal{N}(\rho^{T_A})$ measure the entanglement between $A$ and $B$?

Question

Consider a system composed of two subsystems $A$ and $B$ living in $\mathcal{H}=\mathcal{H}_A \otimes \mathcal{H}_B$. The density matrix of the system $AB$ is defined to be $\rho$. The entanglement negativity of $\rho$, defined as $$\mathcal{N}_A(\rho) := \frac12(\|\rho^{T_A}\|_1 -1),$$ where $\rho^{T_A}$ is the partial transposition of $\rho$ and $\|\cdot\|_1$ is the trace norm, measures by how much $\rho^{T_A}$ fails to be positive semidefinite. This is useful since would $\rho$ be separable, $\rho^{T_A}$ would be positive semidefinite, hence $\mathcal{N}_A(\rho)=0$. This, along with some other nice properties makes $\mathcal{N}$ a nice entanglement measure.

I have read that if $\mathcal{N}_A(\rho)\neq 0$ then one can claim $A$ is entangled with $B$. This is what I don’t understand. By definition, $\mathcal{N}_A(\rho)$ measures by how much $\rho^{T_A}$ fails to be positive semidefinite, an essential property of a separable and hence a non-entangled system. Great, we know whether $\rho$ is entangled or not. However, just because we are told $\rho$ is entangled it doesn’t necessarily mean that the degrees of freedom in $A$ are entangled with those in $B$ right? I guess my problem could steem from the fact that I don’t understand the physical consequences of taking a partial transpose of $\rho$ w.r.t. some subsystem (i.e. what is the physical significance of $\rho^{T_A}$?).

Edit: First of all for your all your comments and generous patience. I edited the question to better address my last issue with understanding entanglement negativity.

Answer 1

There is no good definition of what is an "amount of entanglement". We have some requirements, such as saying that a measure of entanglement must be convex and cannot increase under local operations, but beyond that it is really a matter of taste.

There is a nice interpretation of entanglement negativity, though, in the case that $\rho^{T_A}$ only has a single negative eigenvalue. Let it be $-\lambda$. Then by construction $\mathcal N(\rho) = \lambda$, and $d\mathcal N(\rho)$ almost coincides with the amount of white noise you must add to $\rho$ before it becomes separable.

This is another measure of entanglement, called random robustness, defined more precisely as $R(\rho)$ being the minimal $s \ge 0$ such that the state $$\rho' = \frac1{1+s}(\rho + s I/d)$$ is separable.

I'm saying almost because $\rho^{T_A} \ge 0$ in general does not imply that $\rho$ is separable. But in the cases when it does, $R(\rho)$ is the minimal $s$ such that $${\rho'}^{T_A} = \frac1{1+s}(\rho^{T_A} + s I/d) \ge 0,$$ which is precisely $d\lambda$.

More generally, I don't know any nice interpretation for entanglement negativity.