I studied in my class, that a plane progressive wave cannot be used to represent the wave nature of a particle as it is not square integrable. Also, the phase velocity can get above the value of $c$, the speed of light. A simplistic argument can be-
$$v_{phase}=\frac{\omega}{k}=\frac{E}{p}=\frac{mc^2}{mv_{particle}}=\frac{c^2}{v_{particle}}$$
(using de Broglie's formula and Planck's formula)
Since $v_{particle}<c$, hence $v_{phase}>c$. Thus we use not a simple wave, but instead a superposition of multiple waves, which represents a wave packet. Wave packets, as far as I know, should have a varying amplitude.
While deriving Schrodinger equation, why don't we treat the amplitude $A$ as variable quantity? Am I missing something?
Best Answer
A monochromatic wave has a fixed amplitude :
$\psi(x,t) = A ~e^{i(k.x-wt)}$
However, a wave packet (WP) is a combination of monochromatic waves :
$\psi_{WP}(x,t) = \int d\tilde k ~A(k) ~e^{i(k.x-wt)}$.
where $A(k)$ is a complex quantity.
So, $\psi_{WP}(x,t)$ is a complex quantity, which of course, can always be written :
$\psi_{WP}(x,t) = A_{WP}(x,t) e^{i\phi_{WP}(x,t)}$, where $\phi_{WP}(x,t)$ is a real phase, and $A_{WP}(x,t)$ is a real quantity, which is variable.
Now, the Schrodinger equation, is considering the whole wave packet $\psi_{WP}(x,t)$ as a probability amplitude (PA) $\psi_{PA}(x,t)$ (so, it is no more a real wave function).
With the decomposition $\psi_{PA}(x,t) = A_{PA}(x,t) e^{i\phi_{PA}(x,t)}$, the probability to find the particle at position $x$, at time $t$, is then just $|\psi_{PA}(x,t)|^2 = (A_{PA}(x,t))^2$.