Free particle has a Gaussian wave packet wave function.

[Agent M27]

Homework Statement

This is problem 2.22 from D.J. Griffiths Introduction to Quantum Mechanics

A free particle has the initial wave function:

$\Psi(x,0)$=A$e^{-ax^{2}}$

Find $\Psi(x,t)$. Hint Integrals of the form:
$\int_{-\infty}^{\infty}$$e^{-(ax^{2}+bx)}dx$

can be handled by completing the square: Let $y\equiv \sqrt{a}[x+(b/2a)]$, and note that $(ax^{2}+bx)=y^{2}-(b^{2}/4a)$.

Homework Equations

$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \phi(k)e^{i(kx-\omega t)}dk$

$\phi(k)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} \Psi(x,0)e^{-ikx}dx$

$\omega=\frac{\hbar k^{2}}{2m}$

The Attempt at a Solution

Homework Statement

So I found $\phi(k)=\left(\frac{1}{2\pi a}\right)^{1/4}e^{-k^{2}/4a}$.

Plugging this into my eq for $\Psi(x,t)$ I get the following:

$\Psi(x,t)=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty} e^{-k^{2}/4a}e^{i(kx-(\hbar k^{2}/2m)t)}dk$

$=\frac{1}{\sqrt{2\pi}}\left(\frac{1}{2\pi a}\right)^{1/4}\int_{-\infty}^{\infty}exp[-\left(\left(\frac{i\hbar t}{2m}+\frac{1}{4a}\right)k^{2}-ikx\right)]dk$

Now here is where I get stuck. I feel like I need to do another completing the square manipulation to argument of the exponential,but I am having trouble seeing how the obtained the following solution:

$\Psi(x,t)=\left(\frac{2a}{\pi}\right)^{(1/4)}\frac{e^{-ax^{2}}/[1+(i2\hbar at/m}{\sqrt{1+(i2\hbar at/m)}}$

Any help would be greatly appreciated. Seems as though Professor Griffiths has some real cute tricks up his sleeve. Thanks in advance.

Joe

 

[vela]

If you pull out a factor of 1/4a in the coefficient of the quadratic term, the exponent will be equal to
$$-\left[\left(1+\frac{i2\hbar a t}{m}\right) \frac{k^2}{4a} - ikx\right]$$ To save you some writing, it would convenient to define $\beta = 1+ i2\hbar at/m$ since that quantity appears in the expression you're trying to derive. So you want to complete the square on
$$-\left(\frac{\beta}{4a}k^2 - ikx\right)$$

 

[Agent M27]

Ah ha! Thanks a lot Vela.