# Why Gaussian wave packet?

1. Mar 10, 2009

### TheDestroyer

Hello,

I have a question about basic Quantum Mechanics.

In free particle study of Quantum Mechanics, why do we use the Gaussian wave packet always? I know that the Fourier Transform of it is also a Gaussian wave packet, but I was thinking, is that what we have selected? or is that what nature has selected? I saw that in Schwabl and Cohen Quantum mechanics books.

I know also that the uncertainty principle is obtained very easily using the standard deviation through Gaussian distributions.

My friend has told me that if we used any other wave packet i.e. Lorentzian or anything else, we will obtain also the uncertainty principle, is that true?

Thanks

2. Mar 10, 2009

### alxm

Not always, but it's usually the most convenient, for reasons you seem aquainted with.

Yes. Try it!
You can also show that a gaussian will minimize $$\Delta x\Delta p$$.

As an unrelated sidenote, Pople got the Nobel for his innovations in quantum chemistry, one of the more noteworthy ones being to use gaussians (his program Gaussian remains the most used QC software) as a basis set to approximate the wave function, which is computationally convenient, but kind of stupid from a mathematical who-cares-about-convenience point of view.
Brain teaser for the thread: Why is that?

3. Mar 10, 2009

### Staff: Mentor

With a wave packet whose shape is different from Gaussian, you can obtain an uncertainty principle, of the form

$$\Delta x \Delta p = K \hbar$$

where K is a numerical constant that depends on the precise shape of the packet (Lorentzian or whatever). With a Gaussian packet, you get the smallest possible value of K, which is 1/2.

Allowing for all possible shapes gives you an uncertainty which is the HUP:

$$\Delta x \Delta p \ge \frac{1}{2} \hbar$$

Last edited: Mar 10, 2009
4. Mar 10, 2009

### TheDestroyer

Thank you people, the thing is clearer now.

But I still have 1 more question:

Why does Gaussian give the minimum uncertainty? how can we understand that physically or mathematically?

Thanks again :)

5. Mar 10, 2009

### alxm

Well, typically you'd first derive the inequality: $$\Delta x\Delta p\ge\frac{\hbar}{2}$$
Then you can show that for a gaussian function $$\Delta x\Delta p=\frac{\hbar}{2}$$, and so minimizes the uncertainty.
Neither calculation is that hard to do.

Showing that a gaussian is the only form to do so requires a bit more work. But I'm pretty sure you can. Intuitively, given that a sharp peak in one representation is a broad one in the other, it stands to reason that there should be some function which minimizes the product.

I've never looked for any deeper physical 'meaning' in the fact that it's a gaussian that does so. I'm not sure there is one, beyond the fact that it had to be something.

6. Mar 10, 2009

### TheDestroyer

Thank you guys, I'm grateful :)

7. Mar 10, 2009

### Staff: Mentor

My understanding is that the proof comes from Fourier analysis, where the uncertainty relation takes the form

$$\Delta x \Delta k \ge \frac{1}{2}$$

but I've never looked up the details.

8. Mar 10, 2009