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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.

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  • $\begingroup$ I don't understand your edit. $\rho$ is your quantum state, you want the entanglement negativity of $\rho$, not the entanglement negativity of $\rho^{T_A}$. $\endgroup$ – Mateus Araújo Dec 10 '20 at 12:45
  • $\begingroup$ 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 $A\neq A'$. $\endgroup$ – FriendlyLagrangian Dec 10 '20 at 12:51
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    $\begingroup$ 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 $\rho$. $\endgroup$ – Mateus Araújo Dec 10 '20 at 12:55
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    $\begingroup$ 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. $\endgroup$ – Rammus Dec 10 '20 at 15:49
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    $\begingroup$ @FriendlyLagrangian How do you take the partial transpose without knowing what partition you have chosen? $\endgroup$ – Rammus Dec 10 '20 at 17:03
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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.

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  • $\begingroup$ 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 $B$. $\endgroup$ – FriendlyLagrangian Dec 10 '20 at 13:58
  • $\begingroup$ 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) $\endgroup$ – FriendlyLagrangian Dec 10 '20 at 14:03
  • $\begingroup$ @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). $\endgroup$ – Markus Heinrich Dec 10 '20 at 15:17
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    $\begingroup$ 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. $\endgroup$ – keisuke.akira Dec 11 '20 at 6:35
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    $\begingroup$ @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. $\endgroup$ – Mateus Araújo Dec 11 '20 at 9:13

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