eigenenergies when Hamiltonian is $\hat{H}^2$ − $\hat{H}$

Question

If the eigenenergies of the Hamiltonian $\hat{H}$ are $E_n$ and the eigenfunctions are $\psi_n(r)$ , what are the eigenvalues and eigenfunctions of the operator $\hat{H}^2$ − $\hat{H}$ ?

Attempted solution

$$\hat{H}^2 − \hat{H} \psi_n(r) = \hat{H}^2 \psi_n(r) - \hat{H} \psi_n(r) $$

$$\hat{H}^2 \psi_n(r) = \hat{H}[\hat{H}\psi_n(r)]$$

$$\hat{H}[\hat{H}\psi_n(r)] = E_n \hat{H}\psi_n(r)$$

$$\hat{H}^2 − \hat{H}\psi_n(r) = E_n^2\psi_n(r) - E_n\psi_n(r)$$

So Eigenenergies: ($E_n^2 - E_n$)

Eigenfunctions: $\psi_n(r)$

Tell me if I'm completely off track here. Thanks

Answer 1

Correct. The Hamiltonian is (at least, should be) self-adjoint since it is an observable. The spectral theorem says that the eigenfunctions $\psi_n$ form an orthonormal basis, and so $H^2$ can have no more eigenfunctions that are linearly independent of those of $H$. $H\psi_n=E_n\psi_n$, and we have $f(H) \psi_n = f(E_n) \psi_n$, certainly for any polynomial $f$, by iterating $$H^k\psi_n = H^{k-1}(E_n\psi_n) = E_n (H^{k-1}\psi_n) = \dotsb = E_n^k \psi_n. $$