# Time evolution of Gaussian packet

From Shankar's QM book pg. 154:

Consider the Gaussian wave packet at time $$t=0:$$ $$\psi(x',0)=e^{ip_0x'/\hbar} \frac{e^{-x'^2/2\Delta^2}}{(\pi\Delta^2)^{1/4}}.$$

Using the propagator $$U(t)$$ in the position $$X$$ basis, which is (eqn. 5.1.10)

$$\langle x|U(t)| x'\rangle=(\frac{m}{2\pi\hbar it})^{1/2} \space e^{im(x-x')^2/2\hbar t},$$ we can find the time evolution of the Gaussian wave packet by

\begin{aligned} \psi(x,t) &= \int \langle x| U(t) |x'\rangle \psi(x',0)dx' \\ &=\int(\frac{m}{2\pi\hbar it})^{1/2} \space e^{im(x-x')^2/2\hbar t} \space e^{ip_0x'/\hbar} \space\frac{e^{-x'^2/2\Delta^2}}{(\pi\Delta^2)^{1/4}} dx'.\end{aligned}

How can this complicated looking integral be evaluated?

• Combining the exponentials, unless there is a term I have missed, this is just a gaussian Apr 21, 2022 at 7:39

Generally, for $$a,b\in\mathbb C^2$$ and $$a$$ having a (strictly) positive real part (for integrability issues), the formula generalizes to: $$\int_{-\infty}^{+\infty}e^{-ax^2/2}dx = \sqrt{\frac{2\pi}{a}}$$ with the square root here sending the right half of the complex plane inside itself. By completing the square and contour integration (or recognizing a Fourier transform), you also have $$\int_{-\infty}^{+\infty}e^{-ax^2/2+bx}dx = \sqrt{\frac{2\pi}{a}}e^{b^2/2a}$$ Back to the problem, setting $$a=\frac{1}{\Delta^2}-im/\hbar t$$ and $$b=i(p_0t/m-x)m/\hbar t$$ which satisfy the necessary conditions, you’ll get: $$\psi(x,t)=\frac{1}{(\pi\Delta^2)^{1/4}\sqrt{1+i\frac{\hbar t}{\Delta^2m}}}e^{imx^2/2\hbar t-i\frac{m/\hbar t}{2(1+i\frac{\hbar t}{\Delta^2m})}(x-p_0t/m)^2}$$ So you get a wave packet moving uniformly, and spreading due to dispersion.