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?


1 Answer 1


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 $|\psi_{PA}(x,t)|^2 = (A_{PA}(x,t))^2$.

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    $\begingroup$ Splitting the wave function into amplitude/phase is also done in the WKB approximation. $\endgroup$ Nov 23, 2013 at 18:42

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