Time Dependence of a Gaussian Wave Packet *

Assume we start with our Gaussian (minimum uncertainty) wavepacket $A(k)=e^{-\alpha (k-k_0)^2}$ at $t=0$. We are not interested in careful normalization here so we will drop constants.

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

We write explicitly that $w$ depends on $k$. For our free particle, this just means that the energy depends on the momentum. To cover the general case, lets expand $\omega(k)$ around the center of the wave packet in k-space.

$$\bgroup\color{black}\omega(k)=\omega(k_0)+{d\omega\over dk}\v... ..._0)+{1\over 2}{d^2\omega\over dk^2}\vert _{k_0}(k-k_0)^2\egroup$$

We anticipate the outcome a bit and name the coefficients.

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

We still need to do the integral as before. Make the substitution $k'=k-k_0$ giving $A(k')=e^{-\alpha k'^2}$. Factor out the constant exponential that has no $k'$ dependence.

$$\begin{eqnarray*} \psi(x,t)&=&e^{i(k_0x-w_0t)}\; \int\limits_{-\infty}^\infty\; ... ...y}^\infty\; e^{-[\alpha-i\beta t] k'^2} e^{i(k'x-v_gt)}\; dk'\\ \end{eqnarray*}$$

We now compare this integral to the one we did earlier (so we can avoid the work of completing the square again). Dropping the constants, we had

$$\bgroup\color{black}f(x)=e^{ik_0x}\int\limits_{-\infty}^\inft... ...pha k'^2} e^{ik'x} dk'=e^{ik_0x} e^{-{x^2\over 4\alpha}}\egroup$$

Our new integral is the same with the substitutions $k_0x\rightarrow k_0x-\omega_0t$, $k'x\rightarrow k'(x-v_gt)$, and $\alpha\rightarrow (\alpha+i\beta t)$. We can then write down the answer

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