Origin of the de Broglie Equation

Of course, there are many ways to motivate/justify the de Broglie relation. Along the lines of reasoning you are playing with, there is a simple way to prove the following statement which one can regard as a version of the de Broglie relation:

The momentum $p$ of a quantum particle is related to the wavenumber $k$ (that characterizes the wave-like properties of a quantum particle) by the relation $p=hbar k$.

Let's consider the momentum operator in the position basis and find out the eigenstates of the momentum operator (i.e., the only states for which a well-defined momentum can be said to exist). This eigenvalue equation reads begin{align} hat{p}psi_p&=ppsi_p implies -ihbarfrac{partial}{partial x}psi_p(x)&=ppsi_p(x) implies frac{1}{psi_p(x)}frac{partial psi_p(x)}{partial x}&=ip/hbar implies psi_p(x)&=C e^{ipx/hbar} end{align} where $C$ is some overall normalization (which doesn't matter because the momentum eigenstates cannot be normalized anyway). So, we already see that the particle states with a definite momentum $p$ have a wave-like nature and their wavelength is $frac{2pihbar}{p}$. Or, in other words, their wavenumber is $k=p/hbar$.

Notice that we didn't have to assume anything except the axioms of the modern formulation of quantum mechanics. Of course, de Broglie derived his relation before the advent of the modern formulation of quantum mechanics and he needed to take ingenious leaps to posit such a relation. Most peculiarly, we took the fact that the state of a particle is described by some wave-function for granted. de Broglie, on the other hand, actually guessed that a particle also has some wavelike character and he then arrived at the conclusion that the wavenumber that would characterize the wavelike character of a particle would be related to its momentum by the relation $p=hbar k$. We don't need to do all this guess-work because the insights from such guess-works have already been built into the modern formulation of quantum mechanics.