# Gaussian wave packet

At our QM intro our professor said that we derive uncertainty principle using the integral of plane waves $\psi = \psi_0(k) e^{i(kx - \omega t)}$ over wave numbers $k$. We do it at $t=0$ hence $\psi = \psi_0(k) e^{ikx}$

$$\psi = \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k$$

where $\psi_0(k)$ is a $k$-dependent normalisation factor (please correct me if I am wrong). This dependency was said to be a Gaussian function

$$\psi_0(k)= \psi_0 e^{i(k-k_0)^2/4\sigma k^2}$$

where $\psi_0$ is an ordinary normalisation factor (please correct me if I am wrong).

QUESTION 1: Why do we choose $\psi_0(k)$ as a Gaussian function? Why is this function so appropriate in this case?

QUESTION 2: I don't know how did our professor get a Gaussian function with an imagnary number $i$ in it. His Gaussian is nothing like the one on Wikipedia which is

$$f(x) = a e^{-(x-b)^2/2c^2}$$

QUESTION 3: We used the first integral i wrote down to calculate the Heisenberg's uncertainty principle like shown below, but it seems to me that most of the steps are missing and this is the reason i don't understand this. Could anyone explain to me step by step how to do this.

$$\begin{split} \psi &= \int\limits_{-\infty}^{+\infty} \psi_0\!(k) \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \int\limits_{-\infty}^{+\infty} \psi_0 e^{i(k-k_0)^2/4\sigma k^2} \cdot e^{ikx} \, \textrm{d} k\\ \psi &= \psi_0 2 \sqrt{\pi} e^{ik_0x} e^{-x/2 \sigma k^2} \end{split}$$

I think this is connected to a Gaussian integral, but it doesn't look quite like it to me. Well in the end our professor just says that out of the above it follows that

$$\boxed{\delta x \delta k = \frac{1}{2}}$$

I don't understand this neither. It was way too fast or me.

• For question 1): first of all, your teacher was probably thinking about wave packets, which intuitively are more or less Gaussian - it's not a stretch to think that a wavepacket should have some mean wavenumber $k_0$ and some variance $\sigma$. There's also a mathematical reason: the bound you're deriving holds for all possible $L^2$ functions, and the Gaussian is the only one that saturates it. So apart from doing the full proof, it's the most interesting test case you can work with. Feb 24, 2013 at 11:51
• It seems the homework tag applies even if this is not actually a homework problem. Feb 24, 2013 at 17:17
• Gaussian wave function is selected because it minimizes the value of uncertainty to $\hbar/2$. For any other shape the uncertainty is higher. Jun 26, 2014 at 10:02

It is hard to know what exact the professor has done, but from what I have understood made an effort to help the situation.

Q1. A Gaussian function is chosen to represent a freely expanding wave function for several reasons:

(i) The Gaussian function represents a normal probability distribution function. Since $|\psi(x)|^2$ represents a probability distribution function for a huge class of particles moving in a similar way, and having momentum within a certain range, the large numbers theorem points towards a Gaussian function, to represent the wave function of particles within a region of space determined by the width, $\sigma$, of the Gaussian function. The $\sigma$ also happens to be the standard deviation, i.e. the uncertainty in the position of the particle.

(ii) The expectation value of the position of the particle turns out to be the value of $b$ in your Gaussian wave packet.

(iii) The Gaussian wave packet also contains the wavy bit shown by the phase factor

$e^{ikx}$.

This is what makes Gaussian wave packets an excellent representation of particles, which are known to be located within some region of width $w$, but nevertheless they are all moving as plane waves while the packet travels along in space.

(iv) The wave packet is made of an infinitely large number of momentum values in a momentum width which relates to $\sigma$ by a Fourier transformation.

Q2. So to put some order in all these, let us consider the general Gaussian function

$\psi(x)=\psi_0e^{-A(x-x_0)^2+Bx+C}$

which has a Fourier transformation

$\psi(k)=\psi_0\sqrt{\frac{\pi}{A}}e^{\frac{B^2}{4A}+Bx_0+C}$.

The normalisation constant $\psi_0$ is given by an integration in the range [-$\infty, +\infty$] and has the value $\sqrt{\frac{A}{\pi}}$ but leave it as $\psi_0$.

Q3. To make contact with your professor, try to apply these results using the following:

$A=\frac{1}{2\sigma^2}$

$B=ik$

$x_0=b$

$C=0$

• I think that our professor was lecturing by using this document (tjhsst.edu/~2011akessler/notes/hup.pdf) where they use a suare root of a gauss so this is where he got $\psi_0(k)$: $\psi_0(k)= \sqrt{\text{gauss}} = \sqrt{\psi_0 e^{-(k-k_0)^2/2\sigma_k^2}} = \psi_0 e^{-(k-k_0)^2/4\sigma_k^2}$
– 71GA
Feb 25, 2013 at 19:19
• @71GA Thank you for your notes. I don't think the square rooting should make any difference to the form of the uncertainty principle, since it is only itroducing a scale factor of 1/2. Not sure why should one prefer to go through the square root though.
– JKL
Feb 25, 2013 at 21:36