# Relationship between phase velocity and group velocity using the de Broglie postulates

If I have to show that the group velocity of a free particle is twice the phase velocity, is the following argument correct (avoiding to use the wave function and the momentum operator):

For a particle with energy $$E$$ and momentum $$p$$ we have the circular frequency $$\omega = \frac{E}{\hbar}$$ and the wave length $$\lambda = \frac{2 \pi \hbar}{p}\, .$$

As usual for waves we get the phase velocity by the formula $$v_{p} = \frac{\omega \lambda}{2\pi} \, .$$ This gives us $$v_{p} = \frac{\omega \lambda}{2\pi} = \frac{E \cdot 2\pi \hbar }{\hbar p \cdot 2\pi} = \frac{E }{p} = \frac{ p^2 }{2 m p} = \frac{ p}{2 m} = \frac{v}{2}.$$ But the speed of the particle, $$v$$, is nothing but the group velocity $$v_g$$ of the corresponding wave function. Therefore $$v_g = 2 v_p$$.

• – Qmechanic Jun 10 '15 at 15:24

If you want to do it more formally, you can also start from the usual definitions of the group velocity and phase velocity, $$v_g = \frac{d \omega}{dk}, \quad v_p = \frac{\omega}{k}.$$ The simplest form of the de Broglie relations are $$E = \hbar \omega, \quad p = \hbar k.$$ Your forms are perfectly right too, but a little more complicated because there are $$2\pi$$'s all over the place. Next, we know that for a free particle, the energy is $$E = \frac12 mv^2 = \frac{p^2}{2m}.$$ Combining this with the de Broglie relations, we have $$\omega = \frac{\hbar k^2}{2m}.$$ Using the definitions of the group and phase velocity $$v_g = \frac{\hbar k}{m}, \quad v_p = \frac{\hbar k}{2m}.$$ Then $$v_g = 2 v_p$$ as desired. Incidentally, another way to that the classical velocity is $$v = v_g$$ is to note $$p = \hbar k = m v$$ and comparing with our previous expression gives $$v = v_g$$.