Spectral widths with increasing central wavelength

Question

One can define an electromagnetic pulse numerically through a field with a Gaussian temporal profile, $$E(t) = E_0\cos(\omega t)\exp(-t^2/t_p^2)$$

Here $E_0$ is the peak field, $\omega$ is the angular frequency, $t$ is time, and $t_p$ is the pulse duration. You can change the central wavelength, $\lambda$, by adjusting $\omega = 2\pi c/\lambda$ accordingly ($c$ is the speed of light).

Doing this for $\lambda = 4\:\rm \mu m$ and $10 \:\rm \mu m$, with $t_p = 60\:\rm fs$, I notice that the spectral widths are different: longer central wavelengths always have a broader spectral width. For example, I can define the wavelength range where the spectral intensity is at the $10^{-2}$ (normalized) intensity level. In that case, my range for $\lambda = 4\:\rm \mu m$ is $3.55\:\rm \mu m$ - $4.58\:\rm \mu m$, while for $\lambda = 10\:\rm \mu m$, my window is $\lambda = 7.6\:\rm \mu m$ - $\lambda = 14.6\:\rm \mu m$.

Is there any insight about why longer central wavelengths yield broader spectra?

Answer 1

Is there any insight about why longer central wavelengths yield broader spectra?

This is exclusively an effect (I would call it an artifact) of the fact that you're looking at the wavelength spectral decomposition instead of the frequency domain. The width in the frequency domain doesn't depend on the carrier frequency, but a constant frequency width translates to a different span of wavelengths depending on where you are in the spectrum.

To start with, there's no need to fiddle with numerics - the frequency-domain spectral representation of your pulse is easy enough to calculate, at $$ \tilde E(\Omega) = \int E(t) e^{i\Omega t} = \sqrt{\pi } t_p \frac{ e^{-\frac{1}{4} t_p^2 (\omega-\Omega )^2} + e^{-\frac{1}{4} t_p^2 (\omega+\Omega )^2}}{2} , $$ and it is just two gaussians of $\omega$-independent width $2/t_p$ centered at $\pm\omega$. So long as those two gaussians are far enough that they don't mess with each other (i.e. so long as $\omega t_p\gtrsim \pi$), then the $\Omega>0$ part of the spectrum is just a single gaussian of constant width.

For the examples you mention, a $t_p = 60 \:\rm fs$ pulse with a carrier wavelength of $4\:\rm \mu m$ and $10 \:\rm \mu m$, the widths are identical:

($4\:\rm \mu m$ in blue and $10 \:\rm \mu m$ in yellow; the horizontal axis is $\omega$ in $\rm rad/fs$).

As you can see, the Full Width at 1% Max is identical for both, at about $0.5\:\rm rad/fs$. However, if you then fold this over into frequency using $$ \lambda = \frac{2\pi c}{\omega} $$ then an equal frequency width will indeed produce a shorter span at shorter wavelengths, simply because of the way the frequency axis maps to the wavelength representation.