$\newcommand{\1}{1\negthickspace{\mathrm{I}}}$
Von Neumann ergodic theorem states that, if $U(t)$ is a one-parameter group of unitaries acting on a Hilbert space $\mathcal{H}$, we have
$$ \lim_{T\to \infty} \parallel X(T)(\psi) - P_0(\psi) \parallel = 0 $$
for all $\psi\in\mathcal{H}$ where $\parallel \bullet \parallel $ is the norm in $\mathcal{H}$,
$$ X(T) := \frac{1}{T}\int_0^T dt U(t) \, , $$
and $P_0$ is the orthogonal projector onto the invariant subspace $V_0$ of $U(t)$, $V_0=\{ \phi\in \mathcal{H} |\, U(t)\phi=\phi \ \forall t\}$.
Although reminiscent of quantum mechanics it is hard to apply the above theorem in a quantum mechanical setting.
In quantum mechanics we evolve states or operators with the adjoint action (respectively Schroedinger and Heisenberg evolution). Let us pick the Heisenberg evolution:
$$ \mathcal{U}(t)[A] :=U(t)^\dagger A U(t) \, , $$
where $U(t)=e^{-i t H}$, where $H$ is a self-adjoint (Hamiltonian) operator. Define, analogously as before,
$$ \mathcal{X}(T) := \frac{1}{T}\int_0^T dt \mathcal{U}(t) \, . $$
Let us consider the case where $H$ has purely continuous spectrum (for example the free article on the line).
Question Assume $H$ has purely continuous spectrum, what is
$$ \lim_{T\to \infty} \mathcal{X}(T) = ?$$
Remarks
1) Clearly we have, for all $T$, and trace class $\xi$ $$ \mathrm{Tr} \Big ( \mathcal{X}(T)[\xi] \Big )= \mathrm{Tr} \xi $$
2) It seems that the only invariant operator of $\mathcal{U}(t)$ is the identity $\1$. Unfortunately this operator is not trace class. Hence is not in $B(\mathcal{H})$ seen as an Hilbert space with Hilbert-Schmidt scalar product (which is the setting one would consider when trying to apply the Von Neumann ergodic theorem).
3) In a way the question is: Is there a way to give a meaning to the limit such that it is non-zero?
It seems that this problem should have been well studied but I could not find any reference.