This question came to me while reading Where does de Broglie wavelength $\lambda=h/p$ for massive particles come from? This question has a nice answer that explains that wave number has be proportional to momentum because of Lorentz symmetry. The proportionality constant is Planck's constant $h$.
As I understand it $h$ can be thought of as the charge under translation transformation,
$$U(d) \, f(x) \equiv f(x+d) \cong 1 + \frac{df}{dx} d \, .$$
$U(d)$ is the time translation operator which is defined as $U(d) = \text{e}^{i P/\hbar}$. This definition leads to
$$ P = -i \hbar \frac{\text{d}}{\text{d}x} \, ,$$
from which $\lambda p = h$ can be deduced.
My question is the following: How come there is only one proportionality constant for all particles? Why is is not like with the relation in between spin and magnetic moment where there is a particle dependent gyromagnetic ratio?
Answer
We know from translation symmetry that the expectation value of the wavevector operator is constant -- that is, $$ \langle \mathbf{k} \rangle = \langle \mathbf{k}_1 + \mathbf{k}_2 + \ldots \rangle = \text{const.} $$
In other words, wavevector is a conserved quantity. If the constant of proportionality between $\mathbf{k}$ and momentum were different for different particles, then momentum would not be conserved.
Here's how I think of these things: the wavevector of a particle is the really fundamental quantity. Momentum is simply defined to be proportional to the wavevector because it can be shown using Schrodinger's equation that if a particle of mass $m$ is moving as a wavepacket with an average wavevector $\mathbf{k}$, then $m\mathbf{v} = \hbar\mathbf{k}$, where $\mathbf{v}$ is the particle's group velocity.
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