I am or rather was interested in the following problem: while we cannot hope to measure with infinite precision in the real world, in the mathematical world can I measure a continuous-spectrum self-adjoint operator given a fixed state? That is measure it with infinite precision?

Let A=:C(\mathbb{X}) be a unital \mathrm{C}^*-algebra, and \varphi\in\mathcal{S}(C(\mathbb{X})) a state on it… actually I will use the Gelfand–Birkhoff picture:

\varphi\in \mathbb{X} \iff \varphi\in\mathcal{S}(C(\mathbb{X}))

Note that \varphi has an extension to a \sigma-weakly continuous state \omega_\varphi on the bidual, C(\mathbb{X})^{**}. The algebra C(\mathbb{X}) sits isometrically in the bidual: the bidual as a von Neumann algebra contains the spectral projections of any self-adjoint f\in C(\mathbb{X}). We use the notation \mathbf{1}_E(f) for Borel E\subseteq\sigma(f).

Let \varepsilon> 0. Define:

p_\varepsilon(f)=\mathbf{1}_{(\lambda-\varepsilon,\lambda+\varepsilon)}(f).

Suppose for a state \varphi\in\mathbb{X} that for all \varepsilon>0:

\omega_\varphi(p_\varepsilon(f))>0.

(the situation where \omega_\varphi(p_\varepsilon(f))=0 will be moot).

Then we want to consider the entity:

\displaystyle \varphi_\lambda(g)=\lim_{\varepsilon\to0^+}\frac{\omega_\varphi(p_\varepsilon(f)gp_\varepsilon(f))}{\omega_\varphi(p_\varepsilon(f)}\qquad (g\in C(\mathbb{X})).

If this limit exists then it is a state.

Alas, this limit does not exist in general. There is a commutative counterexample by Nik Weaver on MO, which we will share here.

Let A=L^{\infty}([0,1]) and f\in A given by f(x)=x and \varphi integration against Lesbesgue measure. Let \lambda=0 and p_{\varepsilon}=\mathbf{1}_{[0,\varepsilon)}.

Define

\displaystyle S=\bigcup_{n=0}^{\infty} [10^{-(2n+1)},10^{-2n}],

and g=\mathbf{1}_S, an element of L^{\infty}([0,1]).

Consider \varphi_0(g) and let \varepsilon =10^{-2m}, with m\in \mathbb{N}. It is possible to show that in this case:

\displaystyle \frac{\varphi(p_\varepsilon gp_\varepsilon)}{\varphi(p_\varepsilon)}=\frac{90}{99}.

However, at \varepsilon=10^{-(2m+1)}, we get 1/99. This means that the limit \varphi_0 above does not exist.

Now, without proof we could expect that for A=C([0,1]), and the same f and \varphi, we could expect that in fact \varphi_0 exists, and \varphi_0(g)=g(0).

The problem here is that for L^{\infty}([0,1]), functions that disagree on a set of measure zero are identified and in general g(0) does not make sense for g\in L^\infty([0,1]).

The best we can do is measure f\in C(\mathbb{X}) up to tolerance \varepsilon>0. Say we measure f\in(\lambda-\varepsilon,\lambda+\varepsilon) with a state \varphi\in \mathbb{X}. Then we get conditioning of \varphi to a state:

\displaystyle \varphi_{\lambda,\varepsilon}(g)=\frac{\varphi(p_{\varepsilon}gp_{\varepsilon})}{\varphi(p_{\varepsilon})}\qquad (g\in C(\mathbb{X}))

In the commutative case, this is giving the average of g on the interval (\lambda-\varepsilon,\lambda+\varepsilon).

I had hoped to use \varphi_\lambda to explain why classical spheres don’t admit quantum symmetry. Alas the above means my argument probably cannot work (well, maybe I can use the \varepsilon of room?)

Perhaps we could try and understand in which \mathrm{C}^*-algebras the state \varphi_\lambda is well-defined… but we can say at least for today that we cannot measure continuous observables with infinite precision… even mathematically.