Reading one paper, the GNS construction is mentioned as follows:
It is important to recall that a result (theorem) due to Gel'fand, Naimark and Segal (GNS) establishes that for any $\omega$ on $\mathcal{A}$ there always exists a representation $(f_\omega, \mathfrak{h}_\omega)$ of $\mathcal{A}$ and $\Phi_\omega\in \mathfrak{h}_\omega$ (usually called a cyclic vector) such that $f_\omega(\mathcal{A})\Phi_\omega$ is dense in $\mathfrak{h}_\omega$ and $\omega(A)=\langle \Phi_\omega | f_{\omega}(\mathcal{A})|\Phi_\omega\rangle$. Moreover the GNS result warrants that up to unitary equivalence, $(f_\omega,\mathfrak{h}_\omega)$ is the unique cyclic representation of $\mathcal{A}$.
Now, considering the math there is a theorem and a corresponding proof. My point here is not to discuss these. My point here is to discuss the intuition about this construction from the Physics point of view.
So the first thing that makes me confused: in the $C^\ast$-algebra approach, I thought each state $\omega : \mathcal{A}\to \mathbb{R}$ was the counterpart of a ket $|\phi\rangle$ in the traditional approach.
We see in the GNS construction, though, that each state $\omega$ induces one representation. In other words, instead of having for each $\omega$ one ket, we have for each $\omega$ one whole Hilbert space.
More than that, we have that cyclic vector condition, which physically I don't understand.
So my question is: what is the intuition on the GNS construction from the Physics point of view? How does states $\omega$ from the algebraic approach relates to kets $|\psi\rangle$ (state vectors) in the traditional approach? What is that cyclic vector condition about from a physical perspective?
Best Answer
The basic idea of the GNS construction is that you use a single state (often this will be the vacuum, if we're working on flat space) to recreate the entire Hilbert space. This is indeed related to the cyclicity : the set of all vectors generated by the action of the algebra on the vacuum is dense in the resulting Hilbert space. So to generate the full Hilbert space, just apply every member of the $C^*$-algebra to generate a dense subset of the Hilbert space, then do the Cauchy completion of those to generate the full Hilbert space.
A simple way to get back the usual representation as a Hilbert space is to consider the product of three members of the algebra, then their representation $\pi$ as Hilbert space operators becomes
$$\omega(ABC) = \langle \omega, \pi(ABC) \omega \rangle$$
Then you can just define the states $\vert \psi \rangle = \pi(C) \vert \omega \rangle$ and $\vert \phi \rangle = \pi(A) \vert \omega \rangle$, then your state becomes
$$\omega(ABC) = \langle \phi, \pi(B) \psi \rangle$$
This becomes then the usual transition between two states.
A simple example of this would be for instance to consider the creation and annihilation operators on the vacuum. They do form a $C^*$ algebra, and they can act on the vacuum state to create any number of states that will form a Hilbert space. On the other hand, no amount of applying creation operators on the vacuum will give you the state defined by the Fock state
$$\vert 1,1,1,1,1,.... \rangle$$
If we had used this state as the basic $\omega$, we would have a unitarily inequivalent theory.