Confusion about interpretation of expectation values in quantum mechanics Given a state $|\psi \rangle$ one can form the expectation value of an observable $O$ as:
$$
\langle \psi|O|\psi \rangle.
$$
For the case $O = H$, where $H$ is the Hamiltonian of the quantum system, the expectation value above gives the expected energy of the state. Similarly, the quantum evolution of a state can be written as a map:
$$
|\psi\rangle \to \mathrm{e}^{-iHt} |\psi \rangle = |\tilde\psi \rangle.
$$
The expectation value
$$
 \langle \psi |\mathrm{e}^{-iHt} |\psi \rangle
$$
thus gives the transition probability of $|\psi \rangle$ to $|\tilde\psi \rangle$. My question is: what is the interpretation of:
$$
 \langle \psi |O\mathrm{e}^{-iHt} |\psi \rangle?
$$
 A: Despite the apparent similarity, the expectation value and the transition probability are not the same things. It becomes clearer when you express these quantities in terms of density operators $\hat{\rho}=|\psi\rangle\langle\psi|$. The expectation value is then
$$ \langle \hat{O}\rangle = \langle\psi| \hat{O}|\psi\rangle = \text{tr}\{ \hat{O}\hat{\rho}\} . $$
The unitary evolution of the state is now represented by
$$ \hat{\rho}(t) = \hat{U}(t)\hat{\rho}(0)\hat{U}^{\dagger}(t) , $$
where (using your convention) $\hat{U}(t)=\exp(-i\hat{H}t)$. So the transition probability becomes
$$ \text{tr}\{\hat{\rho}(0)\hat{\rho}(t)\} 
= \text{tr}\{\hat{\rho}(0)\hat{U}(t)\hat{\rho}(0)\hat{U}^{\dagger}(t)\} , $$
which is the modulus square of the transition amplitude that you computed. To compute the expectation value for an observable with the evolving state, we need
$$ \text{tr}\{\hat{O}\hat{\rho}(t)\}
= \text{tr}\{\hat{O}\hat{U}(t)\hat{\rho}(0)\hat{U}^{\dagger}(t)\} . $$
Note that this represents the Schroedinger picture. The same expression can be interpreted in the Heisenberg picture by incorporating the unitary operators in the observable, so that
$$ \hat{O}(t) = \hat{U}^{\dagger}(t)\hat{O}\hat{U}(t) . $$
