Timeline for In a bipartite system $AB$, why does the entanglement negativity $\mathcal{N}(\rho^{T_A})$ measure the entanglement between $A$ and $B$?
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| Mar 26, 2021 at 4:42 | history | edited | Adam Zalcman |
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| Dec 12, 2020 at 16:29 | comment | added | FriendlyLagrangian | @Rammus , When you say “If $\mathcal{N}_A(\rho) \neq 0$ then […] is entangled”, do you mean the partition $AB$ is entangled or the partition $B$ is entangled with the partition $A$? If the latter, why if $\rho^{T_A}$ fails to be positive semidefinite then it follows that $A$ is entangled with $B$? (I don’t understand the physical significance of $\rho \mapsto \rho^{T_A}$) | |
| 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 13:22 | history | edited | FriendlyLagrangian | CC BY-SA 4.0 |
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| Dec 10, 2020 at 17:03 | comment | added | Rammus | @FriendlyLagrangian How do you take the partial transpose without knowing what partition you have chosen? | |
| Dec 10, 2020 at 16:05 | comment | added | FriendlyLagrangian | @Rammus Okay good, so in which step is clear that the mathematics “know” that we have chosen $A$ and $B$ fixed? Because to me, so far, $\mathcal{N}_A(\rho)$ is just a mathematical trick to study the separability (or positive semidefiniteness) of some object “$\rho^{T_A}$”. This is at most interesting, but I fail to see how this object $\rho^{T_A}$ knows anything about the way we have partitioned the system, sure the partial transpose is w.r.t. $A$, so what? | |
| Dec 10, 2020 at 15:49 | comment | added | Rammus | 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 14:35 | comment | added | FriendlyLagrangian | @Rammus Maybe it helps if I put an example of my question. Say $AB$ is some spin chain and $A$ and $B$ are some selection of spins such that “$A+B=AB$”. If I vary the size of $A$ and $B$ but keeping $AB$ fixed, $\mathcal{N}_A(\rho)$ will vary accordingly, from being zero when $A=AB$ and $B= \emptyset$ to perhaps a non zero value for other choices of $A$ and $B$. Suppose for some particular choice of $A$ and $B$ we find $\mathcal{N}_A(\rho)\neq 0$. What is it specific about $\mathcal{N}_A(\rho)$ that tells us that it is $A$ and $B$ and not any other parts that are the ones entangled? | |
| Dec 10, 2020 at 14:18 | comment | added | Rammus | @FriendlyLagrangian Because all separable states for that bipartition would satisfy $\mathcal{N}_A(\rho) = 0$, if we find that it is non-zero then it cannot be the case that $\rho$ is separable, hence it is entangled. Note that there are PPT states that are entangled so this doesn't capture the entanglement of all states. | |
| Dec 10, 2020 at 14:00 | comment | added | FriendlyLagrangian | @Rammus Indeed, but why if $\mathcal{N}_A(\rho)\neq 0$ then $A$ is entangled with $B$? | |
| Dec 10, 2020 at 13:17 | comment | added | Rammus | @FriendlyLagrangian You would expect that any meaningful measure of entanglement would depend on how you partition the whole system into subsystems. | |
| Dec 10, 2020 at 13:14 | answer | added | Mateus Araújo | timeline score: 4 | |
| Dec 10, 2020 at 13:01 | history | edited | FriendlyLagrangian | CC BY-SA 4.0 |
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| Dec 10, 2020 at 12:59 | comment | added | FriendlyLagrangian | Im not saying that. I am saying that the negativity depends on your definition of subsystems. $\rho$ exists without the need of $A$ or $B$ so $\mathcal{N}(\rho)$ to me is ambiguous as it depends on how you partition the system. I also like your notation of $\mathcal{N}_A(\rho)$, I will edit that. | |
| Dec 10, 2020 at 12:55 | comment | added | Mateus Araújo | 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 12:51 | comment | added | FriendlyLagrangian | 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 12:45 | comment | added | Mateus Araújo | 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 12:39 | history | edited | FriendlyLagrangian | CC BY-SA 4.0 |
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| S Dec 10, 2020 at 12:37 | history | suggested | Mateus Araújo | CC BY-SA 4.0 |
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| Dec 10, 2020 at 12:25 | review | Suggested edits | |||
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| Dec 10, 2020 at 11:57 | review | First posts | |||
| Dec 10, 2020 at 12:51 | |||||
| Dec 10, 2020 at 11:57 | history | asked | FriendlyLagrangian | CC BY-SA 4.0 |