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According to Hudson’s theorem, any pure quantum state with a positive Wigner function is necessarily a Gaussian state. In cases, in which the existing well-known Hudson theorem immediately tells that the output state, which is non-Gaussian, must show negativity in phase space. Is the reasoning flawed?

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closed as unclear what you're asking by ACuriousMind, Danu, JamalS, BMS, Emilio Pisanty Apr 2 '15 at 13:42

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't understand your question. It sounds like this: "The theorem says X. Does X hold?". $\endgroup$ – ACuriousMind Mar 25 '15 at 21:32
  • $\begingroup$ You seem to be asking if "P implies G" does "not G imply not P"? Yes, this is simply the contrapostive statement $\endgroup$ – WetSavannaAnimal Mar 25 '15 at 21:33
  • $\begingroup$ @ ACuriousMind If we consider all the states are pure. In such a case, the existing, well-known, Hudson theorem immediately tells that the output state, which is non-Gaussian, must show negativity in phase space? $\endgroup$ – user0322 Mar 25 '15 at 21:35
  • $\begingroup$ @ ACuriousMind is that always true? $\endgroup$ – user0322 Mar 25 '15 at 21:37
  • $\begingroup$ It still sounds to me as if you are simply asking whether the theorem is true (see Rod Vance's comment). Yes, yes it is. (Also, you must not leave a space after the @ if you want to notify me). $\endgroup$ – ACuriousMind Mar 25 '15 at 21:39
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See also my answer here: Are negativity of the Wigner function and quantum behaviour equivalent?

It seems that it is not clear whether the theorem can be extended to all kinds of mixed states. As you say, given a pure state, if we know that it is not Gaussian, Hudson's theorem implies that its Wigner function must be negative somewhere.

For states not covered by the theorem, nothing of this sort can be said. As far as I know, the question whether Hudson't theorem holds for all mixed states has not been settled yet.

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