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?
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1$\begingroup$ What does it mean to "express" an operator in terms of a basis? Operators are defined in all bases by their action upon that basis. Do you perhaps mean that it is diagonal in that basis? (That's linear algebra, basically) $\endgroup$– ACuriousMind ♦Commented Oct 31, 2014 at 13:49
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1$\begingroup$ Sorry, Those are my understanding. The original words are: a convenient way to exploit the fact that an operator $\hat{O}$ commutes with an observable $H^0$, here $H^0$ is free hamiltonian(hamiltonian without potential), is to use a basis of eigenvectors of the obsarvable. $\endgroup$– blue skyCommented Oct 31, 2014 at 13:58
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$\begingroup$ Related: math.stackexchange.com/q/6258/11127 Related: physics.stackexchange.com/q/142548/2451 $\endgroup$– Qmechanic ♦Commented Oct 31, 2014 at 14:02
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1 Answer
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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
