In wave mechanics the dispersion relation between frequency $\omega$ and wave number $k$ is linear: $$\omega_n=c k_n$$

But in quantum mechanics, based on Schrödinger's equation, one can show that we have a quadratic relation between the two :

$$\omega_n=\frac{\hbar}{2m}k_n^2$$

• What is the implication of this difference?
• Does it say something on the wave nature of wave-functions in quantum mechanics compared to e.g. EM waves?

Just looking for intuitive physical interpretations here.

• Note that there are electromagnetic waves which have dispersion relations that are quadratic, cubic, quartic, etc. So this phenomena is not limited to just quantum mechanics. Technically, the EM waves to which you refer are typically called free modes because they do not interact with the background plasma in space (well, even these do… Faraday Rotation). Oct 28 '14 at 14:11
• @honeste_vivere good to know, thanks, I wasn't aware of this. And what does that imply in general? (i.e. having non-linear dispersion relations) Oct 28 '14 at 14:18
• That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen upstream of collisionless magnetized shocks in space. Oct 28 '14 at 14:48
• You can also graph $\omega$ vs. $\kappa$ and show that the slope of the line at any given ($\omega$, $\kappa$) point corresponds to the group velocity and the ratio $\omega$/$\kappa$ corresponds to the phase velocity. Oct 28 '14 at 15:06

The wave mechanics dispersion relation you cite is for EM waves propagating in free space. In other media, the dispersion relation is not necessarily linear (it can be quadratic or have some more complex dependence). So in this context, there's nothing special about quantum mechanics.

More generally, the dispersion relation tells us about the phase speed of the wave and the group velocity:

$$v_\text{phase} = \frac{\omega_n}{k_n}$$

and

$$v_\text{g} \equiv \frac{\partial \omega_n}{\partial k_n}$$

So for example, in contrast to EM waves in free space, the particular quantum dispersion relation you cite will have a group velocity that depends on the wavenumber. The quantum mechanics interpretation of this is that the particle's momentum will depend on its wavenumber ($p = \hbar k$).

• Thanks for this answer, I upvoted earlier. One small question: physically, what does it imply that the group velocity in the QM case depends on momentum $\hbar k$? Last comment if I may: you said"particle's momentum depends on its wavenumber", but isn't that always true? I mean they just differ by constant of proportionality always, i.e. $\hbar.$ Thanks for your help. Oct 17 '14 at 6:28
• @user929304 First, I made a mistake that I just caught given this comment. It's the group velocity, not phase velocity that corresponds to the particle velocity in the quantum sense (it's fixed now). A quantum particle is typically represented as a superposition of many waves that form a packet. It is the group velocity that says how fast this packet is moving. The phase velocity just says how fast one partial wave that comprises the wave function is moving. It's less meaningful in this context. Oct 17 '14 at 14:08

That the phase speed can have a dependence on the wavelength/frequency of the wave. For instance, a whistler mode wave can have a cubic dispersion relation at low frequencies. In this limit, the higher(smaller) frequencies(wavelengths) propagate faster than the converse. It results in a sort of "spreading out" of the wave modes. This if often seen upstream of collisionless magnetized shocks in space.

You can also graph $\omega$ vs. $\kappa$ and show that the slope of the line at any given ($\omega$, $\kappa$) point corresponds to the group velocity and the ratio $\omega$/$\kappa$ corresponds to the phase velocity.

For instance, see the image below, modified from an figure in Krauss-Varban and Omidi, : The $\omega$ vs. $\kappa$ diagram has been Doppler-shifted into a shock rest frame for this specific example, but it should illustrate my point. Where the frequency goes to zero on this plot corresponds to a phase-standing mode (i.e., zero phase velocity) in this frame of reference. In another frame, the same mode would have a finite phase velocity. Where the slope of the blue line goes to zero, the wave's group velocity is zero (or group standing).

Does that help?

References
Krauss-Varban, D., and N. Omidi "Structure of medium Mach number quasi-parallel shocks: Upstream and downstream waves," Journal of Geophysical Research 11, pp. 17,715--17,731, 1991.

• Thanks for writing it as an answer, yes helpful +1. Do you have any comments on the quantum mechanics part of my question? Oct 28 '14 at 15:18
• The relationship you showed is similar, but in quantum the dispersion relation is often associated with a very specific problem (at least for those problems that have analytical/tractable solutions). For your case, the phase speed has a linear dependence upon the wavenumber (or inverse dependence on wavelength) and a group velocity that is twice the phase velocity (in magnitude, ignoring angular effects). Oct 28 '14 at 15:22