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