Quantum mechanics, operator commutes with Hamiltonian

Question

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?

Answer 1

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