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To make the partial transpose a complete positive and therefore physical map, one has to mix it with enough of the maximally mixed state to offset the negative eigenvalues.

The most negative eigenvalue is obtained when partial transpose is applied on the maximally entangled state which is $-\frac{1}{2}$.

Therefore, $$\widetilde{I \otimes \Lambda} = (1-p)(I \otimes \Lambda) \rho + p\frac{I \otimes I}{4}$$ where $p = \frac{2}{3}$ will sufficiently offset the values. However, in this paper by Horodecki and Ekert, they say $p$ needs to be greater than $\frac{8}{9}$, which I can't understand why.

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In the paper that you refer to, they are essentially asking "when can we implement the partial transpose map $\Theta=I_2\otimes\Lambda$?". So, that means the SPA of this map must be positive. What you have calculated, by comparison, is to ask when the SPA of the transpose map $\Lambda$ can be made positive. It might sound like these ought to be the same answer, but adding identity to the transpose map and then extending the Hilbert space is different from extending the Hilbert space and adding identity.

So, you need to be finding the value $p$ such that $$ (1-p)(I_4\otimes \Theta)\rho+p\frac{I_4\otimes I_4}{16} $$ is positive. As you did, you need to consider the maximally entangled state, but of course it's of larger dimension now. You'll find there's an eigenvalue $(9p-8)/16$.

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  • $\begingroup$ How do you know what level to stop at? If I want it to be completely positive, the map should be positive however much I extend the space, and we can see that the further I go, the greater the value of p is going to be. $\endgroup$ – Mahathi Vempati May 22 at 8:42
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    $\begingroup$ I guess that's probably true. Presumably it depends on the context of why you want to perform the calculation. (This is not material I know particularly well. I can do the maths, but haven't thought about the philosophy much.) $\endgroup$ – DaftWullie May 22 at 9:51
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    $\begingroup$ @MahathiVempati It is always sufficient to consider an additional system of the same dimension. This follows from the Choi-Jamiolkowksi isomorphism. $\endgroup$ – Norbert Schuch May 31 at 14:35

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