- #1
leo.
- 96
- 5
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.
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.