Understanding the Evolution of a Gaussian Wave Packet

[wany]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

know that C and $\alpha$ are constants:

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

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

let $z=\alpha*k-\frac{ix}{2\alpha}$

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

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

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

$$w(k) = \frac{\hbar^2k^2}{2m}, ~~m>0,$$
$$w(k) = kc, ~~m=0.$$

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 $$w(k) = \frac{\hbar^2k^2}{2m}$$
will give us: $=\frac{c\alpha }{\sqrt{\pi}}\int_{-\infty}^{\infty} e^{-\alpha^2k^2}e^{i\{kx-\hbar^2k^2t/(2m)\}}dk$

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

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 $$ak^2 + b k$$, you can complete that square easily and then substitute back for a and b.

 

[wany]

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

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

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

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:
$ikx-\alpha^2k^2-i\hbar^2k^2t/(2m)= ikx-k^2(\alpha^2+i\hbar^2t/(2m))=-(\sqrt{\alpha^2+i\hbar^2t/(2m)}k - \frac{ix}{2\sqrt{\alpha^2+i\hbar^2t/(2m)}})^2 - \frac{x^2}{4(\alpha^2+i\hbar^2t/(2m))}$
So the rest will change based on this.

Thanks again.