Towards an Einsteinian Quantum Mechanics, II: Psi from Trajectory

Here I argued for $$\omega=\frac{v}{x}.$$  

(For simplicity I work in 1+1 dimensions)

In this post I show how one can define the wavefunction of a particle from its trajectory and how it's consistent with the orthodox definitions of quantum mechanical operator. 

Let

$$\boxed{\psi:=x^{-i}}$$

Then $$\frac{d\psi}{dt}=\frac{\partial\psi}{\partial t}=-ix^{-i-1}\frac{dx}{dt}=-i\frac{x^{-i}}{x}v=-i\psi \frac{v}{x},$$

but from the above definition $$\frac{v}{x}=\omega,$$

so finally

$$\frac{\partial\psi}{\partial t}=-i\psi\omega,$$

which is 

$$\hat{\omega}=i\frac{\partial}{\partial t},$$

when expressed as an operator/eigenvalue problem.