Wave packets, integration

1. Aug 26, 2009

AntiStrange

I'm reading up on Quantum Mechanics and I don't follow an integration they use.

$$\psi(x,t) = \int^{\infty}_{-\infty} dk A(k) e^{i(kx-\omega t)}$$

They begin by considering the wave packet at time t=0:
$$\psi(x,0) = \int^{\infty}_{-\infty} dk A(k) e^{ikx}$$

"and illustrate it by considering a special form, called the gaussian form":
$$A(k) = e^{-\alpha (k-k_{0})^{2} / 2}$$

I'm ok with all of this so far, although not entirely sure why they chose (or what even the purpose is of) the "gaussian form". I do know a very little bit about the gaussian distribution and I see that this "gaussian form" looks a little similar to a portion of the noramal distribution but why just that part, and why the differences. But, at any rate I can live with all of that so far, however I get completely lost once they do the next step.
They make a change of variables to q' = k-k₀, and end up with:

$$\psi(x,0) = e^{ik_{0}x}e^{-x^{2} / 2\alpha} \int^{\infty}_{-\infty} dq' e^{-\alpha q'^{2} / 2}$$
$$= \sqrt{\frac{2\pi}{\alpha}}e^{ik_{0}x}e^{-x^{2} / 2\alpha}$$

Any help, advice, insights, or even just a point in the right direction would really help me right now. Thanks.

2. Aug 27, 2009

Feldoh

Well for starters that looks like the wave function for a free particle. In contrast to say an infinite well there are no stationary states, so instead of a discrete number of coefficients for each stationary state of the wave function (as seen in the infinite well) we have a continuous distribution of coefficients for a free particle.

They choose a Gaussian distribution because it is a probability distribution which shows up a lot in quantum mechanics. You thought it looked like a normal distribution and that's because it essentially is just that. Although in this case A(k) by itself is not normalized.

So looking at the Gaussian distribution we can get a few properties from it:
$$A(k) = e^{-\alpha (k-k_{0})^{2} / 2}$$

The average value is $$k_0$$ and the standard deviation is $$\sqrt{\frac{1}{\alpha}}$$

As for the integral, I'm not sure how they are rearranging the exponents but Mathematica says that it's correct, sorry...

3. Aug 27, 2009

gabbagabbahey

Are you okay with the change of variables itself? Just straight substitution should lead you to the integral,

$$\psi(x,0) =e^{i k_0 x} \int^{\infty}_{-\infty} dq' e^{-\alpha (q')^{2} / 2} e^{iq' x}$$

From here, you could recognize the above integral as an inverse Fourier Transform and basically look up the final result in a table of said transforms.

Alternatively, just complete the square on the exponent:

$$-\frac{\alpha}{2}q'^2+iq'x=\frac{-\alpha}{2}\left(q'^2+\frac{2ix}{\alpha}q'\right)=\frac{-\alpha}{2}\left((q'+\frac{i}{\alpha}x)^2+\frac{x^2}{\alpha^2}\right)$$

then take out the constant (w.r.t q') factor of $e^{x^2/2\alpha}$ and perform another change of variables $q=q'+\frac{i}{\alpha}x$.

And the final solution is obtained by applying the well-known result $\int_{-\infty}^{\infty}e^{-y^2}dy=\sqrt{\pi}$.

4. Aug 28, 2009

AntiStrange

I'm sorry but I don't follow the first step for the variable change.
So if we have q' = k-k₀
then:
dq' = 1*dk
correct? If it is right, then we now have:

$$\psi(x,0) = \int^{\infty}_{-\infty} dq' e^{-\alpha q'^{2} / 2} e^{ikx}$$

which doesn't look right.

5. Aug 28, 2009

AntiStrange

Does it just require more simplifying? I see how I could get it like this:

$$\psi(x,0) = e^{-ixk_{0}}\int^{\infty}_{-\infty} dq' e^{-\alpha q'^{2} / 2} e^{ixq'}$$

but the final form doesn't have the power of e in front of the integral as negative like I do.

6. Aug 28, 2009

gabbagabbahey

You should have a positive sign...

$$q'=k-k_0\implies k=q'+k_0\implies e^{ikx}=e^{iq'x}e^{ik_0 x}$$

7. Aug 28, 2009

AntiStrange

Ah, ok that is the way you wrote it too :)
Nice ty very much.