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

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.

• I don't understand your edit. $\rho$ is your quantum state, you want the entanglement negativity of $\rho$, not the entanglement negativity of $\rho^{T_A}$. Dec 10, 2020 at 12:45
• Well, I just wanted to make explicit that $\mathcal{N}$ depends on the choice of over which subsytem the partial transpose is being performed on. In general, $\mathcal{N}(\rho^{T_A})\neq \mathcal{N}(\rho^{T_{A'}})$ when $A\neq A'$. Dec 10, 2020 at 12:51
• This is not true, $\mathcal N$ is the same independently of whether you transpose on A or B, because $\rho^{T_A}$ has the same eigenvalues of $\rho^{T_B}$. And even if it were true, the best way would be to write something like $\mathcal{N}_A(\rho)$, because your variable is still $\rho$. Dec 10, 2020 at 12:55
• If $\mathcal{N}_A(\rho) \neq 0$ then it tells you that for that particular partition, the state is entangled. It does not say anything about whether the state is entangled under other partitions. Note that `entanglement' is defined with respect to a chosen partition. You ask is my state separable with respect to this chosen partition, if not then it's entangled. I'm not really understanding what the problem is. Dec 10, 2020 at 15:49
• @FriendlyLagrangian How do you take the partial transpose without knowing what partition you have chosen? Dec 10, 2020 at 17:03

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.

• Thank you for your answer, but I still don’t understand why if $\mathcal{N}_A(\rho) \neq 0$ then subsystem $A$ is entangled with $B$. Dec 10, 2020 at 13:58
• As a side comment, is it possible to come up with a definition of entanglement so useful and universal that renders the rest of entanglement definitions pointless? Or will we always have to wisely choose our weapons? (This question is purposely vague) Dec 10, 2020 at 14:03
• @FriendlyLagrangian It's basic logic: ($\rho$ is separable $\Rightarrow$ $\mathcal{N}_A(\rho) = 0$) is equivalent to ($\mathcal{N}_A(\rho) \neq 0$ $\Rightarrow$ $\rho$ is not separable, i.e. entangled). Dec 10, 2020 at 15:17
• Just a side comment: negativity is a "nice" measure of entanglement because it is computable and not so much for its operational interpretation; we already have measures of entanglement that have clear operational meaning (in the asymptotic scenario) such as distillable entanglement, entanglement of formation, etc. Dec 11, 2020 at 6:35
• @keisuke.akira: you mean easily computable. All measures of entanglement that I'm aware of are computable in the sense of there existing an algorithm that can provide an $\epsilon$ approximation to them. It's just that usually the complexity is horrifying, but for negativity it is polynomial in the dimension of the state. Dec 11, 2020 at 9:13