Can disjoint states be relevant for the same quantum system?

[leo.]

In the algebraic approach, a quantum system has associated to it one $\ast$-algebra $\mathscr{A}$ generated by its observables and a state is a positive and normalized linear functional $\omega : \mathscr{A}\to \mathbb{C}$.

Given the state $\omega$ we can consider the GNS construction to yield the GNS triple $(\mathscr{H}_\omega,\pi_\omega,\Omega_\omega)$ and thus we obtain the folium of $\omega$ as the set $\mathfrak{F}(\omega)$ containing all the algebraic states that can be realized as density matrices on $\mathscr{H}_\omega.$

Now, I believe (and of course I could be wrong), that if we know somehow that the system attains the state $\omega$ (or that it can be prepared on such state), then we have reduced the attention just to the folium of $\omega$.

Actually, in the Heisenberg picture the state is fixed and the observables evolve, so the system ends up always being described by $\omega$, whereas on the Schrödinger picture, I believe the time evolution of the state will lead to another state on its folium. On the other hand, the observables acting on the Hilbert space are $\pi_\omega(a)$, which have eigenstates on $\mathscr{H}_\omega$ which are in turn algebraic states on the folium of $\omega$, so it seems any measurement, because of collapse, yields the system in another state of said folium.

It seems, thus, that for one quantum system, only one folium is relevant. Or in other words, it seems that disjoint states corresponds to states of distinct quantum systems whose observables happen to satisfy the same algebraic properties.

One example that came to my mind is: all angular momenta are described by the same algebraic relations, i.e., they satisfy $[S_i,S_j]=i\hbar \epsilon_{ijk}S_k$. So if we were to build the $\ast$-algebra, I believe we would take the universal enveloping algebra of $\mathfrak{su}(2)$.

Now, this algebra works for all angular momenta. In particular, if we want to describe spin $j$ I think this singles out states $\omega_j$ such that $\pi_{\omega_j}(S^2)= j(j+1)\hbar^2\mathbf{1}$ which I believe to belong to the same folium. So for $j_1\neq j_2$ I believe we have disjoint states $\omega_{j_1},\omega_{j_2}$. The representations seem to be fundamentally different, because in the $j_i$ case $\pi_{\omega_{j_i}}(S_z)$ would have $2j_i+1$ eigenvalues and eigenvectors.

So is this correct? Is this true in general: disjoint states (states which are not in the folium of each other) in truth describe different quantum systems? In other words, are they states for distinct quantum system whose observables satisfy the same algebraic relations? Or it does happen that disjoint states are relevant for the description of a single quantum system?

I'm interested in the general situation, but mainly in QFT where the issue of disjoint states seems to be more relevant.

 

[dextercioby]

Pinging people who could answer this question: @rubi,@A. Neumaier

 

[rubi]

Yes, the quantum physics of a closed system always happens within one particular Hilbert space (up to unitary equivalence) and can thus be characterized by a single algebraic state $\omega$. However, there is Fell's theorem which tells you that the folium of $\omega$ is in a certain sense dense in the set of all algebraic states: If only finitely many measurements are made, states outside the folim can be approximated arbitrarily well by states in the folium if you allow for an arbitrarily small, but finite measurement precision. Hence, experiments can never tell the difference between states in the folium and states outside the folium.
Mathematically: Given $A_1,\ldots,A_n\in\mathfrak A$ and $\epsilon_1,\ldots,\epsilon_n>0$ and some arbitrary state $\eta$, there exists $\omega^\prime\in\mathfrak F(\omega)$ such that $\forall i \leq n: \left|\omega^\prime(A_i)-\eta(A_i)\right| < \epsilon_i$

 

[leo.]

Thanks @rubi. I've seem this theorem explained on nLab as follows:

Fell’s theorem is about a property of vector states of a C-star algebra, it says that if the kernels of two representations of the algebra coincide, then the vector states are mutually weak-* dense. This has a profound consequence for the AQFT interpretation: A state represents the physical state of a physical system. Since one can always only perform a finite number of measurements, with a finite precision, it is only possible to determine a weak-* neigborhood of a given state. This means that it is not possible - not even in principle - to distinguish representations with coinciding kernels by measurements.

For this reason representations with coinciding kernels are sometimes called physically equivalent in the AQFT literature.
​

So it seems that through measurement we can define only one equivalence class of states, whose representations have coinciding kernels. So that physically these states are descriptions of the same quantum system. Is that right?

But what about those states without coinciding kernels? So if I have two states whose representations don't have coinciding kernels, can they still be relevant for the same quantum system? Or in that case they are actually states of distinct quantum systems?

 

[rubi]

Thanks @rubi. I've seem this theorem explained on nLab as follows:

Fell’s theorem is about a property of vector states of a C-star algebra, it says that if the kernels of two representations of the algebra coincide, then the vector states are mutually weak-* dense. This has a profound consequence for the AQFT interpretation: A state represents the physical state of a physical system. Since one can always only perform a finite number of measurements, with a finite precision, it is only possible to determine a weak-* neigborhood of a given state. This means that it is not possible - not even in principle - to distinguish representations with coinciding kernels by measurements.

For this reason representations with coinciding kernels are sometimes called physically equivalent in the AQFT literature.
​

So it seems that through measurement we can define only one equivalence class of states, whose representations have coinciding kernels. So that physically these states are descriptions of the same quantum system. Is that right?

But what about those states without coinciding kernels? So if I have two states whose representations don't have coinciding kernels, can they still be relevant for the same quantum system? Or in that case they are actually states of distinct quantum systems?

As always, the entry in nLab is unnecessarily abstract. The physical content of Fell's theorem can be read off much more easily from the version I have given above: Every algebraic state can be approximated arbitrarily well by a state in the folium in the sense that the state in the folium produces the same physics for an arbitrarily large, but finite number of measurements with fixed finite precision. That means that the states outside the folium don't produce any new physics. The states in the folium suffice for all practical purposes.