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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?

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

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@By Symmetry is right, the integral you need to compute is gaussian (with complex coefficients). First of all, it is integrable since the real part of the exponent gives the gaussian of the initial condition.

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.

Hope it helps, please tell me if you find any mistakes.

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