# Intuition on the GNS construction and how it relates to usual quantum mechanics

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?

• For motivation you might look at how Glimm and Jaffe used GNS to construct an interacting theory from "locally correct" dynamics defined by a Hamiltonian with a space cut-off interaction Summers, p. 60. Also, the remark about Haag's theorem on p. 7 (roughly speaking, taking the space cut-off to infinity cannot result in a theory in Fock space). Commented Jun 16, 2017 at 4:10
• This is an old question but still, I am having the same difficulties that you describe in your question ,namely the fact there will be a bunch of Hilbert spaces (one for each state) with other states inside. How did you solve this? I posted a question recently about this, because yours has no accepted answer. Commented Feb 19, 2021 at 16:52
• In the end I didn't study AQFT further, but what made this point clearer for me was the concept of folium. For every state you may take the associated GNS Hilbert space and you may group together in a set all states which can be realized as density operators in that Hilbert space. This is the folium of the state you began with and for them the GNS construction yields essentially equivalent Hilbert spaces. What is new is the fact that you have disjoint folia. These correspond to inequivalent representations. BTW it's been long since I've read about this, so it may be imprecise.
– Gold
Commented Feb 19, 2021 at 17:30

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.

• Thanks for the answer! Some doubts: (i) this idea of the vacuum generating the entire Hilbert space seems reasonable, and it works for QFT. But I believe the GNS construction also works for non-relativistic QM, where AFAIK there's no vacuum in general. So what would be the intuition of a state generating everything in this context? (ii) As I understand, $A,B,C$ are observables right? So what is the point in considering the states $|\phi\rangle=\pi(A)|\omega\rangle$ and $|\psi\rangle=\pi(C)|\omega\rangle$? These are observables acting on the distinguished state $\omega$, but then what?
– Gold
Commented Jun 14, 2017 at 1:56
• As said, the state doesn't have to be the vacuum. Usually people still try to use a state that resembles the vacuum in some way (quasi-free states, as they are called), but technically this work for any state, since any state can generate any other state if there are no superselectors involved. Commented Jun 14, 2017 at 5:51
• Also the algebra members aren't observables. You'll note that $\omega(A)$ isn't necessarily real, only $\omega(A^* A)$. The only big restriction is that they have to be bounded operators. Which means that you can generate states with bounded observables from them. Commented Jun 14, 2017 at 5:56

In reverse order:

1. Cyclicity should be thought of as a kind of irreducibility condition. Observe that every vector of an irreducible representation is cyclic, and that therefore the existence of a non-cyclic vector would indicate reducibility. So there is little significance to the cyclicity beyond the usual idea to study all irreducible representations since these contain together all relevant information about the algebra. One aspect that may be worth to mention is that demanding cyclicity makes the GNS construction unique - there may be many spaces in which any given abstract state is represented by a vector, but all representations in which it is cyclic are unitarily isomorphic.

2. The relation between states and vectors is the following: In one direction, from vectors to states, we have that for every representation $\rho : \mathcal{A}\to \mathrm{B}(H)$ on a Hilbert space $H$ with bounded operators $\mathrm{B}(H)$ and every vector $v\in H$, the map $\mathcal{A}\to\mathbb{C}, A\mapsto \langle v\vert \rho(A)\vert v\rangle$ is a state in the abstract sense. Conversely, it is precisely the point of the GNS construction that to every abstract state one can find a Hilbert space such that the state is given by a vector on that space in that sense.

3. I see nothing intutive about it (and I am a bit puzzled what sort of intuition you expect for abstract $C^\ast$-algebras), but physically, the GNS construction assures us that the abstract $C^\ast$-algebraic perspective and the traditional approach that starts with an algebra of observables on a Hilbert space are equivalent: The direct sum over all the GNS representations associated to (pure) states of the algebra $\mathcal{A}$ is faithful and an isometry, that is, the abstract algebra is isometrically isomorphic to the algebra of bounded operators on that Hilbert space. Therefore, it makes no difference in the outcomes whether we take the "abstract" or the "concrete" point of view. This is the content of the Gel'fand-Naimark theorem.

• Thanks for the answer! With respect to 2, I believe what confuses me the most is: in the traditional approach, every state $|\varphi\rangle$ is an element of the same Hilbert space $\mathcal{H}$. Now in the $C^\ast$-algebra approach, for every state there's a different Hilbert space, with lots of other states? This is what I feel kind of strange.
– Gold
Commented Jun 14, 2017 at 1:59

As a physicist, I understand GNS as follow.

## short version

Given observables, expectation values and symmetries, we can reconstruct the usual QM with its Hilbert space, its definition of expectation values as "sandwiches", and its usual unitary representation of symmetries.

## more formal version

We give ourselves

• an algebra $\mathcal{A}$ stable under $A \mapsto A^*$: those are to be identified with our operators;
• a function $\omega$ associating a complex number to each element of that algebra: those will be the expectation values of the operators;
• a symmetry group $G$ acting on that algebra such that
• any symmetry $s$ satisfies $s(AB)=s(A)s(B)$,
• and it leaves $\omega$ invariant: $\omega(s(A))=\omega(A)$.

Then GNS constructs:

• a Hilbert space $\mathcal{H}$,
• a vacuum vector $\mid 0\rangle$,
• a representation $\phi$ of the algebra $\mathcal{A}$, i.e. a mapping from $\mathcal{A}$ onto $\mathcal{H}$ such that $\phi(AB)=\phi(A)\phi(B)$, which furthermore has the property that the expectation of any element $A \in \mathcal{A}$ is the quantum expectation of $\phi(A)$: $$\omega(A) = \langle 0\mid\phi(A)\mid 0\rangle$$
• a unitary reprentation of the symmetry group which brings the symmetry on the Hilbert space, i.e. to each $s \in G$ is associated a unitary operator $U_s$ on the Hilbert space, such that $$\phi(s(A))=U_s\phi(A)U_s^*$$

## cyclicity of the vacuum

The short version is that by applying all the operator representations to the vacuum, we get almost all elements of $\mathcal{H}$. The rigorous version being that $\left\{ \phi(A) \mid\! 0 \rangle \mid A\in\mathcal{A} \right\}$ is dense in $\mathcal{H}$.