Quantum mechanics, operator commutes with Hamiltonian My textbook said, if an operator $\hat{O}$ commutes with the Hamiltonian, then we can use the eigen vectors of the Hamiltonian as a basis of the Hilbert space, then express the operator $\hat{O}$ in term of these basis. Is there any proof of this?  
 A: Yes, if $\hat{O}$ commutes with $\hat{H}_0$, $\hat{O}$ is diagonalized in the base of eigenvectors of $\hat{H}_0$. 
Let's show this. If the operator $\hat{O}$ commutes with $\hat{H}_0$, we have
$$\hat{O} \hat{H}_0 = \hat{H}_0 \hat{O}. \qquad (1)$$
Let's assume for simplicity that the spectrum of $\hat{H}_0$ is non-degenerate.
Now, let $V_1$ be an eigenvector of $\hat{H}_0$ with eigenvalue $E_1$, i.e. 
$$\hat{H}_0 V_1 = E_1 V_1 . \qquad (2)$$
Let's multiply the equality (1) on both sides with $V_1$, and apply (2)
$$E_1 \hat{O} V_1 = \hat{H}_0 \hat{O} V_1 . \qquad (3)$$
Let $V_2$ be another eigenvector of $\hat{H}_0$, s.t.
$$\hat{H}_0 V_2 = E_2 V_2 . \qquad (4)$$
Multiplying the equality (3) on both sides with $(V_2)^\dagger$ and applying (4),
$$E_1 (V_2)^\dagger \hat{O} V_1 = E_2 (V_2)^\dagger \hat{O} V_1 . \qquad (5)$$
But, since we assumed a non-degenerate spectrum of $\hat{H}_0$, $E_1 \neq E_2$.
Therefore, in the basis $\{V_i\}$, the off-diagonal matrix-elements of $\hat{O}$ are zero.
Of course, if the spectrum of $\hat{H}_0$ is degenerate but the spectrum of $\hat{O}$ is non-degenerate, one can choose as a basis the eigenvectors of $\hat{O}$. The proof is the same.
Good luck
