# 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.