Dispersion Relation of Quantum Mechanics

From the definition of wavenumber, rearrange de Broglie's relation to find:

$p = \hbar k.$

In quantum mechanics, a general particle (in 1D) has the energy:

$E = \frac{p^2}{2m} + V.$

Since the potential $V$ is not momentum-dependent it can be ignored for computation of group and phase velocities. The kinetic term, substituting in from de Broglie's relation, is:

$E = \frac{\hbar^2 k^2}{2m}.$

Since $E = \hbar \omega$ in QM, have:

$\omega = \frac{\hbar k^2}{2m}.$

Differentiating to obtain the group velocity gets an extra factor of two from the power rule compared to the phase velocity.