Timeline for In a bipartite system $AB$, why does the entanglement negativity $\mathcal{N}(\rho^{T_A})$ measure the entanglement between $A$ and $B$?
Current License: CC BY-SA 4.0
14 events
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| Dec 13, 2020 at 13:07 | comment | added | Mateus Araújo | That's a good point, the asymptotic measures might very well be uncomputable, I completely forgot about them. | |
| Dec 12, 2020 at 1:53 | comment | added | Norbert Schuch | See e.g. arxiv.org/abs/1111.5425 | |
| Dec 12, 2020 at 1:47 | comment | added | Norbert Schuch | @Mateus "All measures of entanglement that I'm aware of are computable in the sense of there existing an algorithm that can provide an ϵ approximation to them." -- This is not true (afaik) for quantities defined in the limit of many copies (distillable entanglement, entanglement cost,...), which arguably from the conceptual point are the best definitions. Neither for e.g. sqashed entanglement. | |
| Dec 11, 2020 at 13:25 | comment | added | FriendlyLagrangian | I have edited my question and I believe it now addresses my issue better. | |
| Dec 11, 2020 at 10:58 | comment | added | keisuke.akira | @MateusAraújo Yes, that's what I meant :) | |
| Dec 11, 2020 at 9:13 | comment | added | Mateus Araújo | @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 8:53 | comment | added | glS♦ | @FriendlyLagrangian if I understand your last comment: yes, you can equivalently define it using $\mathcal N_B$ rather than $\mathcal N_A$. Entanglement is a property of the bipartite system, not of one of its components | |
| Dec 11, 2020 at 6:35 | comment | added | keisuke.akira | 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 10, 2020 at 15:59 | history | edited | Mateus Araújo | CC BY-SA 4.0 |
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| Dec 10, 2020 at 15:40 | comment | added | FriendlyLagrangian | @MarkusHeinrich I see that, $\rho$ is entangled. But I cannot see that within $\rho$ it is the degrees of freedom in $A$ the ones entangled with those in $B$. Cause we could find upon partial trace to $A$ that $A$ is unentangled whereas $B$ is entangled, right? | |
| Dec 10, 2020 at 15:17 | comment | added | Markus Heinrich | @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 14:03 | comment | added | FriendlyLagrangian | 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 13:58 | comment | added | FriendlyLagrangian | 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 13:14 | history | answered | Mateus Araújo | CC BY-SA 4.0 |