Timeline for Can any channel be written as $\Phi(X)=\operatorname{Tr}_{\mathcal Z}[U(X\otimes \sigma)U^\dagger]$ for any state $\sigma$?
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| Aug 20, 2020 at 12:16 | vote | accept | glS♦ | ||
| Jul 19, 2020 at 15:25 | comment | added | Norbert Schuch | @glS My argument (or generalizations thereof) is to some extent quantitative. I suspect it can still be phrased for a general state $\sigma$ (basically you want to know what kind of state $\mathrm{tr}_B(I\otimes \sigma)$ can be). This was the intuition which led me to the argument: I thought that the mixedness of sigma should be reflected in the mixedness of the output. Here, the "mixedness" is measured by the distance to the maximally mixed states, but I'm sure other arguments are possible. E.g., the partial trace should not decrease the rank. | |
| Jul 19, 2020 at 14:05 | comment | added | glS♦ | do you know if this can be made more precise? Like if there is a relation between the rank of $\sigma$ and some property of $\Phi$ (intuitively, I would say something that quantifies the number of extremal maps required to express $\Phi$). It looks like there might be some nice relation of this sort (I might ask this as a separate question if the answer is nontrivial) | |
| Jul 19, 2020 at 14:03 | comment | added | glS♦ | nice. I guess an equivalent way to state this is to observe this $\operatorname{Tr}_{\mathcal Z}[U(\frac{I}{d}\otimes \frac{I}{d'})U^\dagger]=\frac{I}{d}$, which means that any map representable as $\Phi(X)=\operatorname{Tr}_{\mathcal Z}[U(X\otimes I/d')U^\dagger]$ must preserve the identity: $\Phi(I/d)=I/d$. This is clearly not the case for all maps (e.g. for the replacement map you use). | |
| Jul 19, 2020 at 1:53 | comment | added | keisuke.akira | +1 for the (sanity-check) list! | |
| Jul 19, 2020 at 1:14 | history | edited | Norbert Schuch | CC BY-SA 4.0 |
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| Jul 18, 2020 at 23:44 | history | edited | Norbert Schuch | CC BY-SA 4.0 |
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| Jul 18, 2020 at 21:37 | history | answered | Norbert Schuch | CC BY-SA 4.0 |