(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

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

A free particle has the initial wave function:

[itex]\Psi(x,0)[/itex]=A[itex]e^{-ax^{2}}[/itex]

Find [itex]\Psi(x,t)[/itex]. Hint Integrals of the form:

[itex]\int_{-\infty}^{\infty}[/itex][itex]e^{-(ax^{2}+bx)}dx[/itex]

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

2. Relevant equations

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

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

[itex]\omega=\frac{\hbar k^{2}}{2m}[/itex]

3. The attempt at a solution

1. The problem statement, all variables and given/known data

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

Plugging this into my eq for [itex]\Psi(x,t)[/itex] I get the following:

[itex]\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[/itex]

[itex]=\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[/itex]

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:

[itex]\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)}}[/itex]

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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# Homework Help: Free particle has a Gaussian wave packet wave function.

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