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I am reading about hilbert spaces ( in relation to quantum mechanics ). The book I am reading ( link is not available ) tries to tell how logical relations are defined in hilbert space. I am confused by the following line

If E is an orthogonal projection the set theoretic complement of E(H) is not a subspace.

Here $H$ is the hilbert space. In hilbert space the logical NOT of $E(H)$ is defined as the orthogonal projection space $E^{\dagger}(H)$ but the book says although $E^{\dagger}(H)$ is a natural candidate to define NOT it's not the set theoretic complement/NOT , I don't understand why is it so ?

Asaf Karagila
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advocateofnone
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2 Answers2

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I think you misunderstood it: the "set theoretic complement" is $H \setminus E(H)$, not $E^{\dagger}(H)$

supinf
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I presume you've seen how, in set theory, taking the complement of a subset $A^c = \{x \in X : x \notin A \}$, corresponds naturally to logical NOT. The text you are reading from tries to extend this to Hilbert spaces but then points out that this doesn't work: $A^c$ is not a subspace so is not a useful subset to consider in the context of Hilbert spaces.

So now the book is trying to find a way of defining a complement to a subspace that can act like NOT. We require that complement to a) be a subspace and b) arise naturally from the structure of the Hilbert space. The book chooses the orthogonal complement $A^\perp$, which is a subspace and is therefore not the set-theoretic complement.

You may decide whether you think this choice is suitable or not (hint: it is suitable). Hopefully the book or the class you are taking will have some exercises that will clarify this for you.

Chessanator
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