This is a small derivation of the De Broglie wavelength formula that I found kinda cute, and don’t want to forget about.

We start with the wave equation of a free particle:

$$ \Psi(x, t) = Ae^{-i(kx-\omega t)} $$

Here $k = \frac{2\pi}{\lambda}$ is the wave number, and $\omega$ the angular frequency.

Now we apply the operator momentum $\hat{p} = -i\hbar\partial_{x}$:

$$ \hat{p}\Psi(x, t) = -i\hbar\partial_x\Psi(x, t)= -i\hbar[-ikAe^{-i(kx-\omega t)}] = \hbar k\Psi(x, t) $$

So we have that the eigenvalue of $\hat{p}$ is $\hbar k=\frac{2\pi\hbar}{\lambda} = \frac{h}{\lambda}$, and from here, we just rearrange to get the original De Broglie postulate for the wavelength of a moving massive object:

$$ \lambda = \frac{h}{p} $$