# Dispersion relation of wave packet from Schrödinger equation

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.

• @AccidentalFourierTransform Thanks for the edit, but I really wanted to say $\phi(k)\ngtr 0$ and $e^{i(kx-\omega(k)t)}\ngtr 0$ (or equivalently $\phi(k)\nless 0$ and $e^{i(kx-\omega(k)t)}\nless 0$) as there exists the possibility of a zero crossing. Consequently, one cannot conclude that $\left[\hbar\,\omega(k)-\hbar^2\frac{k^2}{2m}\right] =0$. May 2, 2018 at 8:38
• Oh, apologies then! But anyway, what does $>$ and $<$ mean here? There is no (total) ordering on $\mathbb C$. May 2, 2018 at 13:09
• What do you mean by the symbols $\ngtr$ (not greater than) and $\nless$ (not less than) in the context of complex numbers? May 2, 2018 at 16:50
• @AccidentalFourierTransform Sorry for the confusion and sloppy writing! I indeed meant that since there is no ordering in $\mathbb{C}$ one cannot make the statement that either function is positive or negative. That was actually my question: How can one infer the dispersion relation if one cannot make a statement about the integrands? May 3, 2018 at 8:38
• @freecharly My current understanding is that one demands the Schrödinger equation to hold for every $\phi(k) \in \mathbb{C}$ and for every $\omega(k)$ and $k$ such that the inner bracket under the integral has to be identically zero. May 3, 2018 at 8:39

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.
• Dispersion relations give the functional relation $\omega=f(k)$ between the frequency $\omega$ (or energy $E=\hbar \omega)$ and wave vector (momentum $p=\hbar k)$ of a sinusoidal wave. This means quantum-mechanically that the sinusoidal wave is both an eigenfunction of the energy and the momentum operator and thus both the energy and the momentum of the particle are exactly known, while the position is completely undetermined. A wave packet is not a (simultaneous) eigenfunction of the energy and momentum operator and thus energy and momentum of the particle are not exactly defined. May 2, 2018 at 13:26
• @elduge - The wave packet is, in general, not an eigenfunction of the energy and the momentum operator. For energy and momentum you get expectation values and finite uncertainties. The Fourier components $\phi(k) expi(kx-\omega t)$ are eigenfunctions of the energy and momentum operator. The wave packet is composed of many sinusoids with different frequencies and wave vectors. May 3, 2018 at 13:38