Timeline for Is possible to write a separable state as a finite or countable infinite sum of product states?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 20, 2023 at 11:13 | vote | accept | raskolnikov | ||
| Nov 20, 2023 at 9:45 | comment | added | raskolnikov | @forky40 thanks for the reference. I think that this counterexample works only in infinite dimensional systems. Here instead I'm assuming that the Hilbert spaces have finite dimension. But it is interesting to know that for infinite dimensional the answer is no | |
| Nov 19, 2023 at 16:23 | comment | added | forky40 | Possibly related: Holevo (arxiv.org/abs/quant-ph/0504204) defines countably decomposable states as those having an integral representation with "purely atomic" measure. He then provides an example of a separable state that is not countably decomposable in Sec. 4. I don't yet have enough grasp of measure theory to tell if that's what you're looking for. | |
| Nov 18, 2023 at 15:24 | answer | added | John Watrous | timeline score: 5 | |
| Nov 18, 2023 at 10:16 | comment | added | raskolnikov | @Rammus yes, but my question I think still apply, since in the proof of the theorem in the Wikipedia page, they assume that a point in the convex hull is written as a finite sum. So again: is the convex hull just the sum of a finite number of elements in the set? Can I prove a Carathéodory's theorem if my convex hull also consist of continuous sum? I think the question can be also rephrased: there are elements in the convex hull that must be written only as continuous sum, i.e. there is no finite sum that approximate them with arbitrary precision (in some norm)? I hope you see the problem | |
| Nov 17, 2023 at 18:11 | comment | added | Rammus | See Carathéodory's theorem | |
| Nov 17, 2023 at 14:49 | comment | added | raskolnikov | @glS yes but they say it by using a definition of separable state that has a finite amount of element in the convex combination. So the result, at leat how it is proved there, is true if we use the definition of separable states as a finite convex combination of product states. Here, I don't see how to use their argument if the separable state has a continuous sum of product states. | |
| Nov 17, 2023 at 12:36 | comment | added | glS♦ | in that same book, they also discuss that any state can be written as such a convex decomposition of up to $\operatorname{rank}(\rho)^2$ elements. This is also mentioned in quantumcomputing.stackexchange.com/q/13031/55. Is this what you're asking? | |
| Nov 17, 2023 at 11:49 | history | asked | raskolnikov | CC BY-SA 4.0 |