# Quantum Mechanics - Finding expectation value

by Kyle91
Tags: expectation, mechanics, quantum
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 11,774 What happened to the factor of $\sin\left[\frac{(E_1-E_2)t}{\hbar}\right]$? After you multiply the integrand out, you should have \begin{align*} \langle x \rangle &= A^2\int_0^L \left(x\psi_1^2(x) + x\psi_2^2(x) - 2x\psi_1(x)\psi_2(x) \sin [(E_1-E_2)t/\hbar]\right)\,dx \\ &= A^2\left(\int_0^L x\psi_1^2(x)\,dx + \int_0^L x\psi_2^2(x)\,dx - 2\sin [(E_1-E_2)t/\hbar]\int_0^L x\psi_1(x)\psi_2(x)\,dx\right) \end{align*} As long as that last integral isn't 0, the expectation value will vary with time.