Where is noncommutativity in the state-effect formalism of quantum mechanics?

Question

In quantum information theory, one can adopt the basic formalism where every system is given by an operator algebra, state preparation procedures correspond to linear functionals on that algebra (macroscopic systems preparing quantum systems), yes-no measurements correspond to projections in that algebra (quantum system affecting a macroscopic measurement device), and finally evaluating a state at a given projection is interpreted as the relative frequency of obtaining `yes' for the given projection.

More general observables are given by projection-valued measures on a set of measurement outcomes, or equivalently, self-adjoint operators. The need for an operator algebra (in the notes above, for example) is apparently that the spectral projections of a self-adjoint operator must be found in the observable algebra.

The trouble I have here is that it is not clear where, in the above formalism, one is to multiply linear operators representing observables? If an observable corresponds to a measurement, hence giving a `reading' on a macroscopic device (which is stipulated in the definition of measurement), how is it possible to perform a second measurement once a first has been made? (I'm thinking of a detector, here...like in Stern-Gerlach or double-slit.)

Does the need for noncommutativity come from an observed statistical uncertainty relation? Does one prepare a system in a specific way, say one of the beams of a Stern-Gerlach apparatus, and then make a measurement of two different quantities on the single prepared beam, and then try to observe a statistical uncertainty relation on the measured quantities? (i.e. does one look for an uncertainty relation in the standard deviations of the resulting measurements?) Is this how we'd determine that the observables assigned to the two measurements taken do not commute?

Note: "composed" Stern-Gerlach apparati and composition of quantum operations are not what I'm interested in. I want to know about noncommutativity of observables (which are viewed as measurements).

EDIT: It should be noted that a positive answer to the above question essentially protects uncertainty principle from weak measurement objections...

Thanks for your help in advance!

Further Edit:

Please tell me if the following is flawed. In classical mechanics, every observable is given by a real-valued function on the phase space. Each such function has a Fourier decomposition. In the case of a classical radiating atom under Maxwell's equations, it is predicted that the Fourier transform of the algebra of functions (observables) yields (and is isomorphic to) the convolution algebra of an additive abelian subgroup of the real numbers. However experimentally, the Rydberg-Ritz combination formula holds. Heisenberg drew a parallel between the convolution algebra of the group and that of the "Rydberg-Ritz rules" which yields a matrix algebra.

My question originally asked, essentially, how one is to get a noncommutative algebra by looking only at measurement results. This seems to be precisely what Heisenberg did.

Regarding the observables of this algebra (hermitian elements): at first glance it seems strange that Heisenberg assigns a noncommutative algebra to rules derived from spacing of frequencies of spectral lines, when Quantum Mechanics tells us to use measurement outcomes to label the eigenspaces of a hermitian operator in an operator algebra. I was under the impression that the spectral projections of the Hermitian operator, should generate the observable algebra. This sometimes holds in the commutative (classical) case by coincidence...since all the operators commute, a hermitian operator can determine a common eigenbasis for all observables, therefore the spectral projections generate the algebra. What an observable seems to do in the general (perhaps noncommutative matrix algebra) case is select an orthonormal basis with respect to which the matrices of the algebra are represented. If the observable is not invertible (e.g. a projection) this requires the additional choice of a basis for the kernel. Of course, everywhere I'm restricting my attention to finite dimensions.

The latter role for the Hermitian operator is not incompatible with the view that "measurement is final". We never need to compose two observables, but can effectively look at every operator in the algebra representing the system in terms of this basis. The noncommutativity came before this fact. (von Neumann) measurement, as orthogonal projection into eigenspaces of the given observable seems to make more sense from this viewpoint, as well.

Would a genuine physicist tell me if I'm off the mark here?

Answer 1

Let us formulate the question in mathematical terms. We have a system in a general state $\rho$, and two observables $A$ and $B$, which do not commute each other ($AB\not= BA$). To prove this non-commutativity, you have just to measure the observable $AB$ (which means measure first $B$ and then $A$, and take as result the product of the two single measurements) several times, and estimate the value $\langle AB\rangle_{\rho}$, then make the same for the observable $BA$, estimating $\langle BA\rangle_{\rho}$. If $\langle AB\rangle_{\rho}\not=\langle BA\rangle_{\rho}$ for one $\rho$, then $A$ and $B$ cannot commute. If you get an equality for all possible initial states $\rho$, it means that the two observables commute each other.

Basically what I have written comes out from a possible mathematical definition of commutativity of two observables: $[A,B]=0$ if and only if $$ \|AB-BA\|_\infty\equiv\sup_{\rho} \left|\text{Tr}[\rho(AB-BA)]\right|=0. $$

Answer 2

The paper that you link uses ordinary quantum mechanics (e.g. check sections 2.1.1--2.1.4). The state is given by the state operator $\rho$ and averages are obtained by the usual trace $\langle A \rangle = \mathrm{Tr}(A \rho)$. The algebra of operators can be found in many (all?) textbooks in quantum mechanics. The non-commutativity is present in that the algebra includes a non-commutative product. For some operators $AB \neq BA$. For instance, this is true for P and X operators as shown in ordinary textbooks in quantum mechanics.