Wave packets and half-width at half-maximum Suppose we have a Gaussian wave function and amplitude distribution function
$$\psi(x) = (\frac{2}{\pi a^{2}})^{1/4}e^{-x^{2}/a^{2}}e^{ik_{0}x}, \qquad \phi(k) = (\frac{a^{2}}{2\pi})^{1/4}e^{-a^{2} (k-k_{0})^{2}/4}.$$
Now, according to my textbook, when $x$ and $k$ vary from $0$ and $k_{0}$ to $\pm \Delta x$ and $k_{0} \pm \Delta k$, the functions $|\psi(x)|^{2}$ and $|\phi(k)|^{2}$ drop to $e^{-1/2}$. I'm having trouble seeing why that is the case, as $e^{-1/2}$ is clearly not half the amplitude, which I would expect to be 0.5.
 A: Let's calculate $\Delta x$ as the fellow with @ACuriousMind said. Shall we? But, for simplicity of formulas let me introduce the usual notation $2\alpha^2=\sigma^2$, such that the Gaussian takes the form 
$$\frac1{\pi\sigma^2}\int e^{-(1/2)(x/\sigma)^2}{\rm d}x$$
Also for simplicity let's take $k_0=0$. To pass to $k_0\ne0$ is not difficult.
The standard deviation is $\sigma$, and introducing in the Gaussian $x=\sigma$ you get immediately the drop by  $e^{-1/2}$. With the function $\varphi(k)$ you do the same.
Good luck !
A: I was also puzzled with the way Zettili defined $Δx$ and $Δk$ to derive uncertainty relations. His definition is not the same as the well known $\text{FWHM}=2.3548\sigma$. First we need to realize that $|\psi(x)|^2$ and $|\phi(k)|^2$ have the standard Gaussian form 
$$\frac{1}{\sigma\sqrt{2π}} \exp\left[−\frac{x^2}{2\sigma^2}\right]$$ 
with $a^2=2\sigma^2$ - not $|\psi(x)|$ and $|\phi(k)|$. He defines the half maxima (0.6 to be precise) of $|\psi(x)|^2$ and $|\phi(k)|^2$ at their height at standard deviation, $\sigma$, where they drop to $e^{−1/2}$.
