# How do you determine the degree of localization of a wavefunction?

Suppose that there is a wavefunction $\Psi (x,0)$ where 0 is referring to $t$. Let us also say that $a(k) = \frac{C\alpha}{\sqrt{\pi}}\exp(-\alpha^2k^2)$ is the spectral contents (spectral amplitudes) where $k$ is defined as wavenumber $k$. $\alpha$ and $C$ are constants.

My question is, why do we calculate $\Delta x$ by looking at where the value of $\Psi (x)$ diminish by $1/e$ from the maximum possible value of $\Psi (x)$?

Also, although the width of the $\Psi (x)$ packet is $4\alpha$, we define $\Delta x$ as $\alpha$. Why is it like this?

By the way, $\Delta x$ is used as in uncertainty principle.

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$\Delta x$ is actually the standard deviation of position. It can be calculated from the formula

$$\Delta x^2 = \int\Psi^*(x)\bigl(x - \langle x\rangle\bigr)^2\Psi(x)\mathrm{d}x$$

where $\langle x\rangle = \int \Psi^*(x)x\Psi(x)\mathrm{d}x$ is the expectation value of position. Or you could calculate it in the momentum representation (I guess this would actually be the wavenumber representation), using

$$\Delta x^2 = \int a^*(k)\bigl(x - \langle x\rangle\bigr)^2 a(k)\mathrm{d}k$$

but in this case $x$ is now the position operator, $x = i\frac{\partial}{\partial k}$. All these formulas appear in any introductory quantum mechanics reference.

If you plug in the formula you have for $a(k)$, using the appropriate normalization, you'll find that $\Delta x = \alpha$.

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I am slightly getting confused; why is my textbook then showing me that $\Delta x$ is related to the width of the wave packet (x-axis of wave function)? They use it to calculate $\Delta x \Delta k$.... Edit: I think I can cross out the comment before; so is this only approximation to the equation presented above? – Susan Daisy Aug 2 '12 at 4:09
Well, how do you define the width of the wavepacket? Normally the width is defined as the standard deviation, calculated using a formula like the one in my answer. But if you are using a different definition, naturally it might not agree. A Gaussian wavefunction never actually drops to zero, so there isn't a unique way to define the width. – David Z Aug 2 '12 at 4:30