Why is Gaussian Packet the lower limit?

1. Jan 15, 2012

kokolovehuh

Dear all:
When I was just reading about how Gaussian wave packets allow us to derive(because it has) the lowest limit on uncertainty. So what would be the (mathmetical?) derivation to just obtain that limit before proving/identifying that it's true for Gaussian packet?

2. Jan 15, 2012

jfy4

You really just need the Schwarz Inequality and some basic linear algebra over $\mathbb{C}$. For a derivation of the uncertainty relations see Sakurai for instance.

3. Jan 16, 2012

kokolovehuh

I guess it is easier said than done ;|
Could you type it up or put on a link for the proof? I'm lost.

4. Jan 16, 2012

Staff: Mentor

The momentum and position wave functions are basically Fourier transforms of each other, with a factor of $1 / \sqrt{2 \pi \hbar}$ thrown in:

$$\phi(p) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} {\psi(x) e^{-ipx/\hbar} dx}$$

$$\psi(x) = \frac{1}{\sqrt{2 \pi \hbar}} \int_{-\infty}^{+\infty} {\phi(p) e^{+ipx/\hbar} dp}$$

There is a general uncertainty relation that applies to functions that are Fourier transforms of each other, see the third equation of the section "Uncertainty Principle" here:

http://en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle

You can presumably find the proof in a textbook or perhaps a Web site about Fourier transforms.

http://en.wikipedia.org/wiki/Uncertainty_principle_for_the_short-time_Fourier_transform#Proof_of_the_uncertainty_principle [Broken]

Unfortunately the notation is different on the two pages. I leave it as an exercise to the student to make them consistent, and perform the change of variables to x and p.

Last edited by a moderator: May 5, 2017
5. Jan 16, 2012

jfy4

Well, I wont do it for you, but I'll certainly set it up for you. The Schwarz Inequality $(a|a)(b|b)\geq |(a|b)|^2$. Then let $\Delta A=A-(|A|)$ and set $|a)=\Delta A|)$ and let $|b)=\Delta B|)$. Then note that the expectation values of hermitian operators are real and the expectation values of anti-hermitian operators are purely imaginary. Give this a shot.

Last edited: Jan 16, 2012
6. Jan 16, 2012

kokolovehuh

thanks, you guys!