Projectors of unbounded operators in *-algebra

Question

Let's suppose we have a Hilbert space $\mathcal{H}$ and the C*-algebra of the set of bounded operators $\mathcal{B}(\mathcal{H})$. For what I have understood, unbounded operators as for example position $\hat{x}$ and momentum $\hat{p}$ operators enter in the algebra $\mathcal{B}(\mathcal{H})$ as generators of strongly continuous group of unitary operators. Can we say that the projectors into the eigenstates of $\hat{x}$ and $\hat{p}$ are in $\mathcal{B}(\mathcal{H})$?

Edit

For eigenstates of $\hat{x}$ and $\hat{p}$ I mean here the vectors that approximate as close as you want to eigenstates of the unbounded operators

Answer 1

The correct notion of "projectors" for self-adjoint operators with continuous spectrum is via their spectral measure. For any $A$, there is a projection-valued measure $\mathrm{d}\pi_A$ on $\mathbb{R}$ supported on the spectrum $\sigma(A)$ of $A$. For any measureable $\Omega\subset \sigma(A)$, the projector $P_\Omega = \int_\Omega \mathrm{d}\pi_A$ is a bounded operator.

Note that when the operator has continuous spectrum, then single points in the continuous spectrum have zero measure: There is strictly speaking no projector $\lvert x\rangle\langle x\rvert$ onto an eigenspace with value $x$ for the position operator, but there is the projector $\int_{x_1}^{x_2}\lvert x\rangle\langle x\rvert$ onto states with position between $x_1$ and $x_2$ for any $x_1\neq x_2$.

Note that in order to talk about $C^\ast$-algebras for unbounded operators $A$, the easier way to get bounded operators out of them is to talk about their exponentiated versions $\mathrm{e}^{\mathrm{i}At}$, with which they are in one-to-one correspondence via Stone's theorem.