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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!

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  • $\begingroup$ @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 $\left[\hbar\,\omega(k)-\hbar^2\frac{k^2}{2m}\right] =0$. $\endgroup$
    – elduge
    May 2, 2018 at 8:38
  • $\begingroup$ Oh, apologies then! But anyway, what does $>$ and $<$ mean here? There is no (total) ordering on $\mathbb C$. $\endgroup$
    – AccidentalFourierTransform
    May 2, 2018 at 13:09
  • $\begingroup$ What do you mean by the symbols $\ngtr$ (not greater than) and $\nless$ (not less than) in the context of complex numbers? $\endgroup$
    – freecharly
    May 2, 2018 at 16:50
  • $\begingroup$ @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? $\endgroup$
    – elduge
    May 3, 2018 at 8:38
  • $\begingroup$ @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. $\endgroup$
    – elduge
    May 3, 2018 at 8:39

1 Answer 1

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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.

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  • $\begingroup$ Why is there no dispersion relation for a wave packet? Shoudn't there be one if a particle is described quantum mechanically by a wave packet? E.g. on wikipedia it is claimed that there exists one. $\endgroup$
    – elduge
    May 2, 2018 at 8:36
  • $\begingroup$ 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. $\endgroup$
    – freecharly
    May 2, 2018 at 13:26
  • $\begingroup$ @elduge - At the indicated link, Wikipedia does not give a dispersion relation for a wave packet. The dispersion relation there corresponds to eq. (1) and is again the relation between frequency and wave vector of a sinusoidal wave. $\endgroup$
    – freecharly
    May 2, 2018 at 14:01
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    $\begingroup$ @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. $\endgroup$
    – freecharly
    May 3, 2018 at 13:38
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    $\begingroup$ @elduge - The nonlinear dependence of the frequencies and wave vectors of the Fourier components of the wave packet leads to the "dispersion" of the wave form. $\endgroup$
    – freecharly
    May 3, 2018 at 16:42

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