Time Development of a Gaussian Wave Packet *

So far, we have performed our Fourier Transforms at \bgroup\color{black}$t=0$\egroup and looked at the result only at \bgroup\color{black}$t=0$\egroup. We will now put time back into the wave function and look at the wave packet at later times. We will see that the behavior of photons and non-relativistic electrons is quite different.

Assume we start with our Gaussian (minimum uncertainty) wavepacket \bgroup\color{black}$A(k)=e^{-\alpha (k-k_0)^2}$\egroup at \bgroup\color{black}$t=0$\egroup. We can do the Fourier Transform to position space, including the time dependence.

\begin{displaymath}\bgroup\color{black}\psi(x,t)=\int\limits_{-\infty}^\infty\; A(k) e^{i(kx-\omega(k)t)}\; dk\egroup\end{displaymath}

We write explicitly that \bgroup\color{black}$\omega$\egroup depends on \bgroup\color{black}$k$\egroup. For our free particle, this just means that the energy depends on the momentum. For a photon, \bgroup\color{black}$E=pc$\egroup, so \bgroup\color{black}$\hbar\omega=\hbar kc$\egroup, and hence \bgroup\color{black}$\omega=kc$\egroup. For an non-relativistic electron, \bgroup\color{black}$E={p^2\over 2m}$\egroup, so \bgroup\color{black}$\hbar\omega={\hbar^2 k^2\over 2m}$\egroup, and hence \bgroup\color{black}$\omega={\hbar k^2\over 2m}$\egroup.

To cover the general case, lets expand \bgroup\color{black}$\omega(k)$\egroup around the center of the wave packet in k-space.

...2}\left.{d^2\omega\over dk^2}\right\vert _{k_0}(k-k_0)^2\egroup\end{displaymath}

We anticipate the outcome a bit and name the coefficients.

\begin{displaymath}\bgroup\color{black}\omega(k)=\omega_0+v_g (k-k_0)+\beta (k-k_0)^2\egroup\end{displaymath}

For the photon, \bgroup\color{black}$v_g=c$\egroup and \bgroup\color{black}$\beta=0$\egroup. For the NR electron, \bgroup\color{black}$v_g={\hbar k_0\over m}$\egroup and \bgroup\color{black}$\beta={\hbar\over 2m}$\egroup.

Performing the Fourier Transform, we get

\psi(x,t)&=&\sqrt{\pi\over\alpha+i\beta t}\; e^{i(k_0x-\omega_...
...2 t^2}}\; e^{-\alpha(x-v_gt)^2\over 2(\alpha^2+\beta^2 t^2)}.\\

We see that the photon will move with the velocity of light and that the wave packet will not disperse, because \bgroup\color{black}$\beta=0$\egroup.

For the NR electron, the wave packet moves with the correct group velocity, \bgroup\color{black}$v_g={p\over m}$\egroup, but the wave packet spreads with time. The RMS width is \bgroup\color{black}$\sigma=\sqrt{\alpha^2+\left({\hbar t\over 2m}\right)^2}$\egroup.

A wave packet naturally spreads because it contains waves of different momenta and hence different velocities. Wave packets that are very localized in space spread rapidly.

Jim Branson 2013-04-22