Review of Traveling Waves

A normal traveling wave may be given by

$$\bgroup\color{black}\cos(kx-\omega t).\egroup$$

The phase of the wave goes through $2\pi$ in one wavelength in $x$. So the wavelength $\lambda$ satisfies

$$\bgroup\color{black}k\lambda=2\pi .\egroup$$

Similarly the phase goes through $2\pi$ in one period $\tau$ in time.

$$\bgroup\color{black}\omega\tau=2\pi\egroup$$

$\omega$ is the angular frequency. It changes by $2\pi$ every cycle. The frequency $\nu$ increases by 1 every cycle so

$$\bgroup\color{black}\omega=2\pi\nu .\egroup$$

There is no reason to memorize these equations. They should be obvious.

Lets see how fast one of the peaks of the wave moves. This is called the phase velocity. At time $t=0$, there is a peak at $x=0$. This is the peak for which the argument of cosine is 0. At time $t=1$, the argument is zero when $kx=\omega t$ or at $x={\omega\over k}$. If we compute the phase velocity by taking ${\Delta x\over\Delta t}$, we get

$$\bgroup\color{black}v_{\mathrm{phase}}={\omega\over k}.\egroup$$

That is, one of the peaks of this wave travels with a velocity of $\omega\over k$.

$$\bgroup\color{black}v={\omega\over k}={2\pi\nu\over {2\pi\over\lambda}}=\nu\lambda\egroup$$

In non-relativistic QM, we have $\hbar k=p$, $E=\hbar\omega$, and $E={p^2\over 2m}$, so

$$\bgroup\color{black}\omega(k)={E\over\hbar}={\hbar^2k^2\over 2m\hbar}={\hbar k^2\over 2m}\egroup$$

You may remember that a pulse will move at the group velocity which is given by

$$\bgroup\color{black}v_g=\left({d\omega\over dk}\right)={2\hbar k\over 2m}={\hbar k\over m}={p\over m}.\egroup$$

(The phase velocity for the non-relativistic case is $v_p={p\over 2m}$.)