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In the simplest particle-in-a-box experiment, a particle is confined to a potential where $V(x)=0$ on the interval $[0,a]$ and $V(x) = \infty$ otherwise. Then the energy eigenvalue equation, $\hat{H} \psi_n = E_n \psi_n$ is solved, where $\hat{H} = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2}$, and the energy eigenstates are found to to be $\psi_n(x) = \sqrt{\frac{2}{a}} \sin(k_n x)$. This is usually one of the very first exercises in an introductory physics class.

Now, these eigenstates were found using the Hamiltonian, so they are eigenstates of the Hamiltonian. I see a lot of places on the internet that then try to find an expectation value for position, $\langle x \rangle$, using the eigenstates of the Hamiltonian. But this Hamiltonian operator and the position operator don't commute. Therefore, these two operators don't have a common set of eigenstates. Therefore, you shouldn't be able to use the eigenstates of the Hamiltonian to calculate the position expectation value. Is this right, or am I missing something?

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Just because something is not an eigenstate, does not mean you can or should not calculate expectation values of an operator in that state.

It is just that in the case of an eigenstate, the calculation becomes particularly easy. You will also notice that the variance of that expectation value does not vanish, but for an eigenstate it does.

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