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

## 1 Answer

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