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10.5 Wave packets (states of minimal uncertainty)

One of the questions of some physical interest is ``how can we create a qunatum-mechanical state that behaves as much as a classical particle as possible?'' From the uncertainty principle,

$\displaystyle \Delta$x$\displaystyle \Delta$p$\displaystyle \ge$$ {\frac{1}{2}}$$\displaystyle \hbar$, (10.13)

this must be a state where $ \Delta$x and $ \Delta$p are both as small as possible. Such a state is known as a ``wavepacket''. We shall see below (and by using a computer demo) that its behavior depends on the Hamiltonian governing the system that we are studying!

Let us start with the uncertainty in x. A state with width $ \Delta$x = $ \sigma$ should probably be a Gaussian, of the form

$\displaystyle \psi$(x, t) = exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$A(x). (10.14)

In order for $ \psi$ to be normalised, we need to require

| A(x)|2 = $\displaystyle \sqrt{\frac{1}{\sigma^2\pi}}$. (10.15)

Actually, I shall show below that with

A(x) = $\displaystyle \sqrt[4]{\frac{1}{\sigma^{2}\pi}}$eip0x/$\scriptstyle \hbar$, (10.16)

we have

$\displaystyle \langle$$\displaystyle \hat{x}$$\displaystyle \rangle$ = x0,    $\displaystyle \langle$$\displaystyle \hat{p}$$\displaystyle \rangle$ = p0,    $\displaystyle \Delta$x = $\displaystyle \sigma$,    $\displaystyle \Delta$p = $\displaystyle \hbar$/$\displaystyle \sigma$ (10.17)

The algebra behind this is relatively straightforward, but I shall just assume the first two, and only do the last two in all gory details.

$\displaystyle \hat{p}$exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$A(x) = (p0 + i$\displaystyle \hbar$$\displaystyle {\frac{(x-x_{0})}{\sigma^{2}}}$)exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$A(x). (10.18)

Thus

$\displaystyle \langle$p$\displaystyle \rangle$ = $\displaystyle \sqrt{\frac{1}{\sigma^2\pi}}$$\displaystyle \int_{-\infty}^{\infty}$(p0 + i$\displaystyle \hbar$$\displaystyle {\frac{(x-x_{0})}{\sigma^{2}}}$)exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$ = p0. (10.19)

Let $ \hat{p}$ act twice,
$\displaystyle \hat{p}^{2}_{}$exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$A(x) = $\displaystyle \left(\vphantom{p_{0}^{2} + 2i\hbar p_{0}\frac{(x-x_{0})}{\sigma^...
...r^{2} \left[\frac{(x-x_{0})^{2}}{\sigma^{4}}
-\frac{1}{\sigma^2}\right]}\right.$p02 + 2i$\displaystyle \hbar$p0$\displaystyle {\frac{(x-x_{0})}{\sigma^{2}}}$ - $\displaystyle \hbar^{2}_{}$$\displaystyle \left[\vphantom{\frac{(x-x_{0})^{2}}{\sigma^{4}}
-\frac{1}{\sigma^2}}\right.$$\displaystyle {\frac{(x-x_{0})^{2}}{\sigma^{4}}}$ - $\displaystyle {\frac{1}{\sigma^2}}$ $\displaystyle \left.\vphantom{\frac{(x-x_{0})^{2}}{\sigma^{4}}
-\frac{1}{\sigma^2}}\right]$ $\displaystyle \left.\vphantom{p_{0}^{2} + 2i\hbar p_{0}\frac{(x-x_{0})}{\sigma^...
...r^{2} \left[\frac{(x-x_{0})^{2}}{\sigma^{4}}
-\frac{1}{\sigma^2}\right]}\right)$ x  
    exp$\displaystyle \left(\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right.$ - $\displaystyle {\frac{(x-x_0)^2}{2\sigma^2}}$ $\displaystyle \left.\vphantom{ - \frac{(x-x_0)^2}{2\sigma^2} }\right)$A(x). (10.20)

Doing all the integrals we conclude that

$\displaystyle \langle$p2$\displaystyle \rangle$ = p02 + $\displaystyle {\frac{\hbar^{2}}{2\sigma^{2}}}$. (10.21)

Thus, finally,

$\displaystyle \Delta$p = $\displaystyle \sqrt{\langle p^{2} \rangle -\langle p \rangle ^{2}}$ = $\displaystyle {\frac{\hbar}{\sigma}}$ (10.22)

This is just the initial state, which clearly has minimal uncertainty. We shall now investigate how the state evolves in time by usin a numerical simulation. What we need to do is to decompose our state of minimal uncertainty in a sum over eigenstates of the Hamiltonian which describes our system!


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Next: 10.6 computer demonstration Up: 10. Time dependent wave Previous: 10.4 Simple example

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk