How do I get $\omega$ in de Broglie’s formula? Suppose I have the ground state $\psi$ of some Hamiltonian $H$ and I want to get energy eigenvalue of $\psi$ without doing $H \psi = E \psi$.
De Broglie’s formula $$
E = \hbar \omega,
$$says I can get $E$ provided I know $\omega$. How do I calculate this $\omega$ in that case? I imagine the Fourier transform would give me a range of $\omega$. Is de Broglie a bridge between Fourier analysis and quantum mechanics that can be made mathematically precise?
 A: Two quantum mechanical systems which are defined on the same Hilbert space (e.g. the free particle on a line and the harmonic oscillator, which are both defined on $L^2(\mathbb R)$) are distinguished only by which operator is chosen to be the Hamiltonian. It follows immediately that any procedure which gives you information about the set of energy eigenstates must reflect this choice.

I have never seen the relation you quote referred to as the "de Broglie formula."  Nevertheless, this doesn't provide a workaround to dealing with the Hamiltonian operator; it merely relates the energy of an eigenstate to the frequency at which its phase oscillates.  Finding the allowed frequencies simply amounts to finding the allowed energies and then dividing by $\hbar$.
Explicitly, the Schrödinger equation says that if $\psi(t)$ is the state of the system at time $t$, then
$$i\hbar \psi'(t) = H \psi(t)$$
If $\psi(t)$ is an eigenstate of $H$ with eigenvalue $E$, then it follows that
$$i\hbar \psi'(t) = E\psi(t) \implies \psi'(t) = -\frac{iE}{\hbar} \psi(t)$$
$$\implies \psi(t) = \psi_0 e^{-iEt/\hbar} =\psi_0 e^{-i\omega t},\ \ \ \omega\equiv \frac{E}{\hbar}$$
This tells us that eigenstates of the Hamiltonian evolve via a periodic phase factor with frequency $\omega = E/\hbar$.
