I have a question regarding the derivation of the dispersion relation of a wave packet from the Schrödinger equation.
The wave packet is given by
$$\psi(x,t)=\int_{-\infty}^{\infty}\frac{dk}{2\pi}\,\phi(k)\,e^{i(kx-\omega(k)t)}$$
where $\phi(k)$ is the Fourier transform of $\psi(x,t=0)$
$$\phi(k)=\int_{-\infty}^{\infty}dx\,\psi(x,0)\,e^{-ikx},$$
i.e. $\phi(k)=|\phi(k)|\,e^{i\,\varphi(k)}$ with $\varphi(k) \in \mathbb{R}$ in general.
Plugging the general form of the wave packet into the time-dependent Schrödinger equation
$$\left[i\hbar \partial_t+\hbar^2\frac{\nabla^2}{2m}\right]\psi(x,t)=0$$
thus yields
$$\int_{-\infty}^{\infty}\frac{dk}{2\pi}\,\phi(k)\,\left[\hbar\,\omega(k)-\hbar^2\frac{k^2}{2m}\right]\,e^{i(kx-\omega(k)t)}=0.$$
My question is:
What is the reasoning that $\omega(k)=\frac{\hbar\,k^2}{2m}$ given that $\phi(k) \in \mathbb{C}$, i.e. $\phi(k)\ngtr 0$ and $e^{i(kx-\omega(k)t)}\ngtr 0$? Since then the vanishing integral cannot yield a vanishing integral kernel.
Many thanks in advance!
Best Answer
There is no specific dispersion equation for a wave packet. The dispersion equation $$\omega (k)=\frac {\hbar k^2}{2m} \tag 1$$ of the Schrödinger equation for a particle with constant (zero) potential energy holds for plane wave solutions $$\psi=\psi_0 \exp i(\vec k·\vec r-\omega t) \tag 2$$ The wave packet is composed of a superposition of such plane waves.