# 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.

• And why would you think that $\Delta x$ represents the half-width at half-maximum? It's the standard deviation, not the HWHM! – ACuriousMind Nov 20 '14 at 23:12
• The QM book I'm using, by N. Zettili, says "It is convenient to define the half widths $\Delta x$ and $\Delta k$ as corresponding to the half-maxima of $|\psi(x)|^{2}$ and $|\phi(k)|^{2}$. In this way, when x varies from 0 to $\pm \Delta x$ and k from $k_{0}$ to $k_{0} \pm \Delta k$, the functions $|\psi(x)|^{2}$ and $|\phi(k)|^{2}$ drop to $e^{-1/2}$". So, is the book wrong then? – Zack Nov 20 '14 at 23:29
• @Zack Maybe the book isn't wrong - they are defining $\Delta x$ and $\Delta k$ to be the HWHM. You haven't defined them here: the functions have the property you state (dropping to $e^{-1/2}$ times their peak magnitudes) when $x=\pm a/\sqrt{2}$ and $k=k_0\pm \sqrt{2}/a$. So I think maybe the book's statement is making a definition (without spelling it out) that $\Delta x=\sqrt{\log 2}\,a$ and $\Delta k = 2\,\sqrt{\log 2}/a$. – WetSavannaAnimal Nov 20 '14 at 23:41
• Thanks, I was thinking the same as well by working backwards from where the half of the Gaussian should occur at, but I was a bit hesitant as the book isn't as explicit as I'd have liked it to be. – Zack Nov 20 '14 at 23:55
• @WetSavannaAnimal aka Rod Vance Now, that I look back at my notes and the book, I'm not sure if this the case as book uses this fact to relate $e^{-2\Delta x^{2}/a^{2}}$ to $e^{-1/2}$ and deduces that $\Delta x = \frac{a}{2}$, from the fact that $\frac{|\psi(\pm \Delta x, 0)|^2}{|\psi(0, 0)|^2} = e^{-1/2}$. But, if $\Delta x$ was already defined, then this argument would be nonsensical, no? – Zack Nov 21 '14 at 0:10

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 !

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 $e^{−1/2}$.