Free particle has a Gaussian wave packet wave function.

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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)=Ae^{-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
 
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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
$$-\left(\frac{\beta}{4a}k^2 - ikx\right)$$
 
Ah ha! Thanks a lot Vela.
 
To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.
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