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1. • CommentRowNumber1.
   • CommentAuthorTobyBartels
   • CommentTimeAug 16th 2010
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   Author: TobyBartels
   Format: MarkdownItexThis is mathematically much simpler than the classical Gleason's Theorem, but I added it to [[Gleason's theorem]] anyway.

   This is mathematically much simpler than the classical Gleason’s Theorem, but I added it to Gleason’s theorem anyway.

2. • CommentRowNumber2.
   • CommentAuthorIan_Durham
   • CommentTimeAug 16th 2010
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   Author: Ian_Durham
   Format: MarkdownItexI'm half-thinking outloud here with this question but, if the POVM version of Gleason's theorem is weaker than the classical theorem, does that imply that POVMs are inherently weaker than Hermitian operators in some regard?

   I’m half-thinking outloud here with this question but, if the POVM version of Gleason’s theorem is weaker than the classical theorem, does that imply that POVMs are inherently weaker than Hermitian operators in some regard?

3. • CommentRowNumber3.
   • CommentAuthorTobyBartels
   • CommentTimeAug 17th 2010
   • (edited Aug 17th 2010)
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   Author: TobyBartels
   Format: MarkdownItexPOVMs are inherently more general than Hermitian operators. This is nothing deep; every Hermitian operator defines a POVM, but not every POVM comes from a Hermitian operator. So if the hypothesis of a theorem has to be true for all POVMs, then this is a stronger hypothesis than merely assuming it for all Hermitian operators, and therefore the theorem is weaker. Physically, the point of interest is that, at least according to some quantum information theorists (and possibly other physicsts, I don't know), thinking of observables as Hermitian operators is too restrictive, and we ought to count every POVM as an observable. Really, that's the only reason that one even cares about the theorem for POVMs.

   POVMs are inherently more general than Hermitian operators. This is nothing deep; every Hermitian operator defines a POVM, but not every POVM comes from a Hermitian operator. So if the hypothesis of a theorem has to be true for all POVMs, then this is a stronger hypothesis than merely assuming it for all Hermitian operators, and therefore the theorem is weaker.

   Physically, the point of interest is that, at least according to some quantum information theorists (and possibly other physicsts, I don’t know), thinking of observables as Hermitian operators is too restrictive, and we ought to count every POVM as an observable. Really, that’s the only reason that one even cares about the theorem for POVMs.

4. • CommentRowNumber4.
   • CommentAuthorIan_Durham
   • CommentTimeAug 18th 2010
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   Author: Ian_Durham
   Format: MarkdownItex> at least according to some quantum information theorists Well, _some_ quantum information theorists. There's a whole sub-class of them that don't think that way - in fact, in some sub-circles of QI POVMs have become passé. But that's beside the point. The point is: "duh." I'm an idiot. I knew that. Thank you for reminding me, though. ;)

   at least according to some quantum information theorists

   Well, some quantum information theorists. There’s a whole sub-class of them that don’t think that way - in fact, in some sub-circles of QI POVMs have become passé.

   But that’s beside the point. The point is: “duh.” I’m an idiot. I knew that. Thank you for reminding me, though. ;)