de Broglie associated a free particle with momentum $p$ to a plane wave with wavelength $\lambda=\frac{h}{p}$, where $h$ is Planck’s constant. This relation only works in the quantum world since it is dependent on $h$, which doesn’t exist in classical theory.

The momentum of the photons $p_{photon}$, is equal to $E/c$, since it is massless. On the other hand, $E_{gamma}$, the energy of the photon, is given by $\frac{h \nu}{c}$. Since the velocity of light is $c$, we have $E_{gamma}=\frac{h}{\lambda}$.

We can define the wave as $\psi(x,t)\in\mathbb{C}$, with $p=\frac{h}{\lambda}=\frac{h}{2\pi}*\frac{2\pi}{\lambda}=\hbar k$, where $k$ is the wavenumber. When we consider wave phenomena, a good question to ask is how a wave looks to different observers.

So we will consider two different observers in two frames of reference $S$ and $S’$. Denote $v$ the velocity which the frame $S’$ is moving with respect to the frame $S$. The frames are arranged in such a way that the origins coincide at zero time. For simplicity, we relate them by a Galilean transformation. Given a point $x’$, we have the relation to its other frame $x$ by $x’=x-vt$, $t’=t$.

We want to check if the two frames agree on the wavelength of the observed wave. If it is equal for both frames, what does it imply? hmm

To do this, we assume a particle is moving at $\tilde v>0$ in the $S$ frame. The momentum of this particle will thus be $p=m\tilde v>0. \\ \lambda=\frac{h}{p}$.

In frame $S’$, $\tilde v’=\frac{dx’}{dt}=\frac{d}{dt}(x-vt)=\tilde v-v$. $p’=m\tilde v'=m(\tilde v-v)=p-mv$. Therefore, we conclude the momentum in the frame $S'$ differs substantially different from the momentum in the frame $S$. This implies the wavelengths also are rather different mathematically.

In our daily lives, ordinary measurable waves like waterwaves or soundwaves don’t transform under different frames of observers. This is partial evidence that these types of waves are not directly measurable, and $\psi$, again is not a property that can be directly observed in an experiment.