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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I don't understand your question. It sounds like this: "The theorem says X. Does X hold?". – ACuriousMind Mar 25 '15 at 21:32
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You seem to be asking if "P implies G" does "not G imply not P"? Yes, this is simply the contrapostive statement – Selene Routley Mar 25 '15 at 21:33
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@ 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? – user0322 Mar 25 '15 at 21:35
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@ ACuriousMind is that always true? – user0322 Mar 25 '15 at 21:37
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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). – ACuriousMind Mar 25 '15 at 21:39
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@ACuriousMind, I would like to know if there's any exception that non-Gaussianity of states with positive Wigner functions. – user0322 Mar 25 '15 at 21:41
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No, that is precisely what Hudson's theorem tells you: The Wigner function of a pure state is positive if and only if it is a Gaussian state. – ACuriousMind Mar 25 '15 at 21:43
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@ACuriousMind, I'm a bit confused with reading this paper http://journals.aps.org/pra/abstract/10.1103/PhysRevA.79.062302 – user0322 Mar 25 '15 at 21:48
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The question is perfectly clear: Given a PURE quantum state with positive Wigner function, the state must be Gaussian. The obvious question is: What about MIXED quantum states? Can the theorem be extended to this case? – Martin Mar 26 '15 at 00:16
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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.
