I recently started reading Algebraic quantum mechanics. So I have no knowledge of the subject. In the GNS construction we construct the Hilbert space of states as follows,

  1. We endow the algebra of observables $\mathfrak{A}$ with an inner product using the state $\omega$ which is a linear functional on the space of observables. This inner product may be degenerate. (non zero element might have zero norm in this inner product)
  2. Remove these null vectors by quotienting the null space $\mathfrak{N}$ hence giving a positive definite inner product on $\mathfrak{A}/\mathfrak{N}$.
  3. Completing this space we get a Hilbert space. The algebra of observables acts naturally on this Hilbert space.

How is the state used to give an inner product? In this case how does the operators corresponding to the observables operate on this Hilbert space? How is it related to the standard Hilbert space state formulism?

  • $\begingroup$ By the algebra mutliplication. $\endgroup$ – ACuriousMind Mar 7 '16 at 16:03
  • $\begingroup$ will you elaborate $\endgroup$ – Boltzee Mar 7 '16 at 16:04
  • $\begingroup$ I don't really see what there is to elaborate. The algebra of observables acts naturally on itself by its defined multiplication, so it also acts naturally by that on any quotient of itself. $\endgroup$ – ACuriousMind Mar 7 '16 at 16:05
  • $\begingroup$ How does it correspond to observation? $\endgroup$ – Boltzee Mar 7 '16 at 16:07

Each element of the Hilbert space is a Cauchy sequence of equivalence classes of operators.

So $\vec v=([a_1],[a_2],\dots)$ where $[a]=\{A\in\mathfrak A: \omega(A-a)=0\}$ and where $(\mathcal C_1,\mathcal C_2,\dots)$ is the specific function (sequence) that maps $n\mapsto \mathcal C_n$ and where the sequence is Cauchy.

So now you have an operator $B$ and a vector $\vec v=([a_1],[a_2],\dots)$ and the obvious operation is $B\vec v=([Ba_1],[Ba_2],\dots)$ but you need to show it is well defined. Firstly that it didn't depend on the representative of the equivalence class, that $\omega(a-b)=0$ implies $\omega(Ba-Bb)=0$ and secondly that$([Ba_1],[Ba_2],\dots)$ is Cauchy. Though if it isn't, then you could just say that $\vec v=([a_1],[a_2],\dots)$ isn't in the domain of the unbounded operator.

How does it correspond to observation?

The same as always, the measurement sends a vector to its orthogonal projection onto an eigenspace. The relative frequency of getting a particular eigenspace is the ratio of the squared norm before and after the projection.

Technically the space of Cauchy sequences still won't be a Hilbert space becasue we didn't finish the completion. Given two Cauchy sequences $([a_1],[a_2],\dots)$ and $([b_1],[b_2],\dots)$ we identify them with the same vector in the Hilbert space if $([a_1-b_1],[a_2-b_2],\dots)$ has zero as a limit (and we have to show that definition is well defined).

So a vector in the Hilbert space is a set of Cauchy sequences. Each Cauchy sequence has values which themselves are sets of operators.

So $\vec v=[([a_1],[a_2],\dots)]$ where the outer $[\,]$ identified two Cauchy sequences if the difference has zero as a limit. And the inner $[\,]$ identifies two operators is their difference has zero as the resukt of $\omega$ and the $(\,)$ just denotes a sequence by listing the values of the sequence in order (and I might be using the axiom of choice in my choice of notation by denoting each equivalence class by a representative).

This means the operator also has to be shown to be well defined on two Cauchy sequence that are identified.

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  • $\begingroup$ I am new to the subject so my statements can be too wrong, but aren't the vectors of the constructed Hilbert space the operators or the elements of the $C^*$ algebra? Why did you write the vector to be a Cauchy sequence of operators? $\endgroup$ – Boltzee Mar 7 '16 at 16:45
  • $\begingroup$ @bgr95 Your post was a bit vague (using a linear functional $\omega$ to make a bilinear form) but you started with an algebra and put something on it that might be zero and then quotient (in what category, as an algebra, as a vector space, again you were vague). Then you complete it. Complete it means consider Cauchy sequences. Quotient means take equivalence classes. So you have a Cauchy sequence of equivalence classes. And the things in your equivalence class are the original objects. In this case, elements of an algebra so you have a multiplication in that algebra. $\endgroup$ – Timaeus Mar 7 '16 at 16:50
  • $\begingroup$ The quotienting is in the category of vector spaces, the Cauchy sequnces that we consider to complete the new vector space (After quotienting) with some inner product I believe was only to make the inner product space a Hilbert space. So the vectors in this algebra should still be the operators in that operator algebra right? $\endgroup$ – Boltzee Mar 7 '16 at 16:52
  • $\begingroup$ @bgr95 You start with operators, they have an addition and a scalar multiplication so they are a vector space. Then you quotient them. So now you have an equivalence class of operators. It also has a natural addition and scalar multiplication and it inherits any bilinear form on the operators, so you can complete it by making the space of Cauchy sequences of these equivalence classes. $\endgroup$ – Timaeus Mar 7 '16 at 16:56
  • $\begingroup$ @bgr95: No, Timaeus is completely correct that "complection" is formally the space of equivalence classes of Cauchy sequences (which again inherits all natural actions from the space in which these sequences take value). Perhaps you should not ask about "algebras of observables", but first try to understand the notions of quotients and completions in general (that is, in the abstract mathematical sense). $\endgroup$ – ACuriousMind Mar 7 '16 at 16:57

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