Resolution of the Identity in Quantum Mechanics I have a conceptual question about the resolution of the identity in quantum mechanics.
For a continuous spectrum, a self-adjoint operator $F$ can be written in the following way
$$
F = \int \lambda\,dE_F(\lambda) = \int \lambda\frac{dE}{d\lambda}d\lambda = \int \lambda |\lambda\rangle\langle\lambda|d\lambda
$$
where $E_F(\lambda)$ is the projection operator.
Now suppose we want of find out the resolution of the identity $E_{F^2}(\lambda)$ for $F^2$. 
By definition we have
$$
F^2 = \int \lambda^2\,dE_F(\lambda) = \int \lambda\,dE_{F^2}(\lambda)
$$
And I don't know how to proceed to write out $E_{F^2}$ in terms of $E_F$. I also wonder if there is a way to find the resolution of the identity for any operator $g(F)$ that is a function of $F$?
 A: The two projection-valued measures (PVMs) are related in a straight-forward fashion. For better legibility, let me call the spectral parameter for $F^2$ with the letter $\mu$ instead of $\lambda$. The spectrum of $F^2$ really equals $\bigl ( \sigma(F) \bigr )^2 = \sigma(F^2)$. Moreover, the equality you give tells you that the level set 
\begin{align*}
U_{\mu} := \bigl \{ \lambda \in \sigma(F) \; \; \vert \; \; \mu = \lambda^2 \bigr \}
\end{align*}
is crucial in relating the two PVMs, so that morally speaking you have 
\begin{align*}
\mathrm{d} E_{F^2}(\mu) = \sum_{\lambda \in U_{\mu}} \mathrm{d} E_{F}(\lambda)
. 
\end{align*}
To make this rigorous, you have to consider subsets of the spectra which are related by the function $q(\lambda) = \lambda^2$, considered as a function from $\mathbb{R}$ to $\mathbb{R}$. If $\Omega$ is any Borel set, then the associated PVMs are related by
\begin{align*}
E_{F^2}(\Omega) = E_{F} \bigl ( q^{-1}(\Omega) \bigr )
. 
\end{align*}
This generalizes the “differential” version since the level set $U_{\mu} = q^{-1}(\{ \mu \})$ is just the preimage of the square function $q$. 
