In a finite dimensional commutative $\mathrm{C}^*$-algebra $A\subset B(\mathbb{C}^N)$, if a vector state $\varphi_x(f)=\langle x,f(x)\rangle$ given by $x\in P(\mathbb{C}^N)$ is pure, then there is no projection $p\in A$ that can disturb the state: that is if $p$ is a projection, then $\varphi_x(p)=\langle x,p(x)\rangle$ is zero or one, so measurement (in the sense of Hilbert space quantum mechanics) of the state $x$ with $p$ gives a value of 1/0 with probability one, if one then $x\in \text{ran }p$ so the wave function collapse is $x\mapsto p(x)=x$… if zero then $x\perp \text{ran }p$ and similarly the wave function collapse is trivial in the same way.
I want to show that a noncommutative $\mathrm{C}^*$-algebra $A$ (separable, infinite dimensional, generated by projections), there are pure states $\varphi$ that do not have this property (hopefully without passing the the enveloping vNa and using Borel functional calculus), that there are projections $p\in A$ such that it is not the case that:
$$\varphi(p)=1\text{ or }0.$$
It seems so obviously true… what I think I would like to do is consider the GNS representation $\pi$… as $A$ is noncommutative at least one of the summands give a Hilbert space $\mathsf{H}_\rho$ with dimension greater than one. Let $\xi$ be the associated vector/cyclic vector… now take $y\in (\mathbb{C}\xi)^\perp$, and form the rank one projection $q$ onto the subspace defined by $(\xi+y)/\sqrt{2}$.
Then $$\displaystyle \langle \xi,q(\xi)\rangle=\frac{1}{2}$$… and so we would be done… but there is nothing to say that there is a projection $q_0\in A$ such that $\pi(q_0)=q$…
Best Answer
If $A$ is non-commutative and it is generated by projections then we may choose two projections $p$ and $q$ in $A$ which do not commute.
Let $A_0$ be the closed *-subalgebra of $A$ generated by $p$ and $q$, and notice that, since $pq-qp\neq 0$, there exists an irreducible representation $\pi $ of $A_0$ such that $\pi (pq-qp)\neq 0$.
Considering the universal C*-algebra $B$ generated by two projections $e$ and $f$, there exists a *-homomorphism $\rho :B\to A_0$ sending $e$ to $p$, and $f$ to $q$. The composition $\pi \circ \rho $ is then an irreducible representation of $B$, whose spectrum is fortunately well known!
Observing that $\pi \circ \rho $ cannot be one-dimensional, since $\pi (\rho (e))$ and $\pi (\rho (f))$ do not commute, it necessarily has to be two-dimensional and, up to a unitary conjugation, we must have $$ \pi (\rho (e)) = \left[\matrix{1 & 0\cr 0 & 0}\right],\quad\text{and}\quad \pi (\rho (f)) = \left[\matrix{t & \sqrt{t-t^2}\cr \sqrt{t-t^2} & 1-t}\right], $$ where $t\in (0,1)$.
The vector state $\varphi $ on $A_0$ associated to the first canonical basis vector $e_1$ of $\mathbb C^2$, is therefore pure and satisfies $$ \varphi (q) = \langle \pi (q)e_1,e_1\rangle = \langle \pi (\rho (f))e_1,e_1\rangle = t, $$ so the desired state may be obtained by extending $\varphi $ to any pure state of $A$.