- #1
confused_man
- 16
- 1
I'm trying to wrap my head around the dispersion relation ##\omega(k)##. I understand how you can construct a wavepacket by combining multiple traveling waves of different wavelengths. I can then calculate the phase and group velocities of this wavepacket:
\begin{align*}
v_p &= \frac{\omega}{k} \\
v_g &= \frac{d\omega}{dk} .
\end{align*}
Suppose that I'm looking at a free quantum mechanical particle, the dispersion relation is
\begin{align*}
E &= \hbar\omega =\frac{\hbar^2 k^2}{2m}\\
\omega &= \frac{\hbar k^2}{2m}
\end{align*}
Since ##\omega## depends the wavenumber in a non-linear way, I understand that a wavepacket made of a combination of these plane waves will experience dispersion and will spread out.
If I'm talking about a pure plane wave with frequency ##k##, then I wouldn't use the group velocity to describe how fast it's moving, I'd just use the phase velocity.
My question is if I'm looking at a wave packet built up from a number of different plane waves and I want to calculate its group velocity, what value of ##k## should I use? The group velocity equation only has a single ##k## in it.
\begin{align*}
v_p &= \frac{\omega}{k} \\
v_g &= \frac{d\omega}{dk} .
\end{align*}
Suppose that I'm looking at a free quantum mechanical particle, the dispersion relation is
\begin{align*}
E &= \hbar\omega =\frac{\hbar^2 k^2}{2m}\\
\omega &= \frac{\hbar k^2}{2m}
\end{align*}
Since ##\omega## depends the wavenumber in a non-linear way, I understand that a wavepacket made of a combination of these plane waves will experience dispersion and will spread out.
If I'm talking about a pure plane wave with frequency ##k##, then I wouldn't use the group velocity to describe how fast it's moving, I'd just use the phase velocity.
My question is if I'm looking at a wave packet built up from a number of different plane waves and I want to calculate its group velocity, what value of ##k## should I use? The group velocity equation only has a single ##k## in it.