Consider this Phys.SE question which spells out the energy of a vibrating string as a variant of Hooke's law, and thereby explains why it's proportional to the square of the amplitude (here: the transversal speed).
To what extent can one make the analogy to Born's rule where, this time, it's the probability that's the square of the amplitude?
I.e., could one consider that
quantum mechanical probability $\equiv$ classical energy, and amplitude of wave function $\equiv$ transversal velocity?Ok, if your problem is "why the probability density is given by $|\psi|^2$?", one can show that the integral of this probability density is conserved for all time.
Starting from the Schrodinger equation (I take $\hbar=1$)
$\frac{\partial \psi}{\partial t} = -iH\psi$
$\frac{\partial \psi^*}{\partial t} = +iH\psi^*$
$\begin{align} \frac{\partial |\psi|^2}{\partial t} &= \psi\frac{\partial \psi^*}{\partial t}+\psi^*\frac{\partial \psi}{\partial t} \\ &=i\psi H \psi^* -i \psi^* H \psi \end{align}$
Taking here a 1D case and and Hamiltonian given by $H = -\frac{\partial^2}{\partial x^2}+V(x)$ and integrating the equation above over x, we get
$\begin{align} \frac{\partial}{\partial t}\int\limits_{-\infty}^{+\infty}|\psi(x)|^2 \, dx &= i[\int dx \, \psi(x)(-\psi''^*(x)+V(x)\psi^*(x))-\int dx \, \psi^*(x)(-\psi''(x)+V(x)\psi(x))] \\ &= i[-\int dx \, \psi''(x)\psi^*(x)+\int dx \, \psi^*(x)\psi''(x)] \\ &= 0\end{align}$
Where I used 2 integration by parts in the first integral and used the fact that $\psi(\pm \infty)=0$
Now this does not prove that it is the only possibility for a probability density. If you can show that $\frac{\partial}{\partial t}\int\limits_{-\infty}^{+\infty}|\psi(x)| \, dx=0$ for example then my demonstration is useless.
