Understanding the Evolution of a Gaussian Wave Packet

[wany]

Homework Statement

Calculate [itex]\Psi(x, t)[/itex] for the gaussian wave packet according to the amplitude distribution function a(k)=C*[itex]\alpha*e^{-\alpha^2k^2}[/itex]/ [itex]\sqrt{\pi}[/itex]and describe its evolution.

Homework Equations

[itex]\Psi(x, t)=\int_{-\infty}^{\infty} a(k)e^{i\{kx-w(k)t\}}dk[/itex]

The Attempt at a Solution

know that C and [itex]\alpha[/itex] are constants:

So by plugging in for a(k) we get:
[itex]=\frac{c\alpha e^{-iwt}}{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-\alpha^2k^2}e^{i\{kx-w(k)t\}}dk[/itex]

Now we complete the square: [itex]ikx-\alpha^2k^2=-(\alpha*k-ix/(2\alpha}^2)-x^2/4\alpha^2}[/itex]

let [itex]z=\alpha*k-\frac{ix}{2\alpha}[/itex]

so we have now [itex]\Psi(x, t)=\frac{C*\alpha e^{-iwt}}{\alpha*\sqrt{\pi}}e^{-x^2/4*\alpha^2}\int_{-\infty}^{\infty} e^{-z^2}dz[/itex]

which we know the integral equals [itex]\sqrt{\pi}[/itex]
so by plugging that in and canceling we get [itex]\Psi(x, t)=Ce^{-(iwt+x^2/4*\alpha^2)}[/itex]

First of all I do not know if this is right and second of all how do I describe the evolution.
Thank you in advance.

 

[fzero]

You should indicate that the integrals are over k. You also didn't specify if we're dealing with a massive particle or not, since

[tex] w(k) = \frac{\hbar^2k^2}{2m}, ~~m>0, [/tex]
[tex]w(k) = kc, ~~m=0.[/tex]

Once you settle that, you will need to complete the square in the exponent of the integral to obtain a true Gaussian integral.

 

[wany]

sorry I was in the process of trying to get equations to show properly. The latex function in this forum is not working properly for me for some reason, but yes I completed the square as you can see from my updated.
Also we are not told if its a massive particle or not, but its under the section Wavefunction For A Free Particle.

 

[fzero]

w(k) is a function of k, you can't just pull out the time dependence. The time-dependence of the wavefunction will be slightly more complicated than just a periodic phase. Presumably you can use the dispersion relation for a massive particle.

 

[wany]

ok so plugging in [tex]w(k) = \frac{\hbar^2k^2}{2m}[/tex]
will give us: [itex]
=\frac{c\alpha }{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-\alpha^2k^2}e^{i\{kx-\hbar^2k^2t/(2m)\}}dk
[/itex]

So now we need to complete the square of : [itex]
ikx-\alpha^2k^2-\hbar^2k^2t/(2m)= ikx-k^2(\alpha^2-\hbar^2t/(2m))
[/itex]

I am stuck on how to do this now though since we are stuck with three different terms.
Any help would be appreciated.

 

[fzero]

The term is still of the form [tex]ak^2 + b k[/tex], you can complete that square easily and then substitute back for a and b.

 

[wany]

ok so completing the square I get:
[itex]ikx-\alpha^2k^2-\hbar^2k^2t/(2m)= ikx-k^2(\alpha^2-\hbar^2t/(2m))=-(\sqrt{\alpha^2-\hbar^2t/(2m)}k - \frac{ix}{2\sqrt{\alpha^2-\hbar^2t/(2m)}})^2 - \frac{x^2}{4(\alpha^2-\hbar^2t/(2m))}[/itex]

So we let [itex]z=\sqrt{\alpha^2-\hbar^2t/(2m)}k - \frac{ix}{2\sqrt{\alpha^2-\hbar^2t/(2m)}}[/itex]

Now plugging this back in we get that:
[itex]\Psi(x, t)=\frac{c\alpha }{\sqrt{\pi}}*\frac{1}{\sqrt{\alpha^2-\hbar^2t/(2m)}}*e^{\frac{x^2}{4(\alpha^2-\hbar^2t/(2m))}}\int_{-\infty}^{\infty} e^{-z^2}dz=\frac{c\alpha}{\sqrt{\alpha^2-\hbar^2t/(2m)}}*e^{\frac{x^2}{4(\alpha^2-\hbar^2t/(2m))}}[/itex]

Is this correct? And if so how do I then describe the evolution?
Thank you for your help this far.

 

[fzero]

I didn't check every step of the math, but it looks reasonable. A Gaussian function is parametrized by it's amplitude and width, try to figure out how these are varying with time. Maybe make a rough sketch for short and long times. A good question to ask is what sets the scale of short and long times.

 

[wany]

Ok thank you very much. After some more work, I found the evolution of the wave.
Also for anyone referencing this, I forgot an i and made a mistake on a negative sign:
the completing the square should actually be:
[itex]ikx-\alpha^2k^2-i\hbar^2k^2t/(2m)= ikx-k^2(\alpha^2+i\hbar^2t/(2m))=-(\sqrt{\alpha^2+i\hbar^2t/(2m)}k - \frac{ix}{2\sqrt{\alpha^2+i\hbar^2t/(2m)}})^2 - \frac{x^2}{4(\alpha^2+i\hbar^2t/(2m))}
[/itex]
So the rest will change based on this.

Thanks again.