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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: $$ \langle \psi |O\mathrm{e}^{-iHt} |\psi \rangle? $$

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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 $$ \hat{O}(t) = \hat{U}^{\dagger}(t)\hat{O}\hat{U}(t) . $$

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