Recently, we talked about a homework assignment, which seemed rather odd to me.
Let the Fourier transformed $ F(\xi) $ of a function $f(x)$ be defined via:
$$F(\xi) = \int_{-\infty}^{\infty}f(x) e^{-\mathrm{i} x\xi}\ \mathrm{d} x $$
Now, the assignment is similar to:
Let
$$
S(\lambda) = A \mathrm{e}^{-\frac{1}{2} \left(\frac{\lambda-\lambda_0}{\Delta\lambda}\right)^2}
$$
be the spectrum of a light pulse. Calculate the Forier transformed of $S(\lambda)$ ($\rightarrow$ time space).
It seems to imply that I should plug $S(\lambda)$ into the definition of the Fourier transformed (or the reverse transformation), so that in the above defintion:
$$x = \lambda $$
$$\xi = t$$
I noticed some problems with this approach. For example, the exponent $\mathrm{e}^{-\mathrm{i}\lambda t}$ is not dimensionless.
As a response to the question, whether this is allowed, the response was simply "Why shouldn't it"?
My question is: Is this approach valid?
EDIT
We finally tried to solve the following integral:
$$F(t) = \int_{-\infty}^{\infty}S(\lambda)\ e^{-\mathrm{i} \lambda t}\ \mathrm{d} \lambda $$
To rephrase the question: Is this the correct time space function corresponding to $S(\lambda)$? Or do you need $\omega$ to get the time space function?
Best Answer
Your approach is vaguely valid, but you should have a big red flashing light that says "dimensional analysis" on it: $\lambda t$ has dimensions $[\lambda t]=[LT]$ and it has no business being exponentiated as is.
What should be on that exponent is something like $\omega t$, and you need to transform your spectrum from wavelength to frequency space via $\omega = 2\pi c/\lambda$, but what this really brings up is a flaw in your assignment: normally, when we talk about pulses having a gaussian spectrum, we make them gaussian in frequency space: they're gaussians in the wavevector $k$, $$ S(k) = B \exp\left(-\frac12\left(\frac{k-k_0}{\Delta k}\right)^2\right), \tag1 $$ not over its inverse, the wavelength.
As such, if you want to plow on with the pulse you've been given, it's definitely something you can attempt (once you fix that non-dimensionless exponent), but it won't be particularly pretty.
What does normally get the time on the spotlight, on the other hand, is the $k$-space spectrum above, which does have a nice relationship with the corresponding 1D waveform $$ F(x,t) = \int S(k) e^{ik(x-ct)} \mathrm dk, \tag2 $$ which you can integrate explicitly, and whose relationship with $S(k)$ (and the equivalent $S(\omega)$) is important to keep in mind in many contexts.
Edit: perhaps it's worth going into a little bit more depth into the relationship between the two spectra at hand: the $k$-space gaussian $$ S_k(k) = B \exp\left(-\frac12\left(\frac{k-k_0}{\Delta k}\right)^2\right), \tag1 $$ from above, and the $\lambda$-space gaussian $$ S_\lambda(\lambda) = A \exp\left(-\frac12\left(\frac{\lambda-\lambda_0}{\Delta \lambda}\right)^2\right), \tag3 $$ which when translated to $k$-space via $k=2\pi/\lambda$ gives $$ S_\lambda(k) = A \frac{\sqrt{2\pi}}{k}\exp\left(-\frac{1}{2(\Delta \lambda)^2}\left(\frac{1}{k}-\frac{1}{k_0}\right)^2\right).\tag4 $$
The spectrum in $(4)$ can in principle be used as in $(2)$, and it will give you some pulse, but you're highly unlikely to be able to integrate it effectively. In principle, it is a different function to the $k$-space gaussian $(1)$, but in some parameter regimes it can look similar. To emphasize how this works, consider the following plots: here we have $S_\lambda(k)$ on the left and $S_k(k)$ on the right, for different values of $\Delta k$ - a broadband pulse on the top, with $\Delta k = \frac{1}{5}k_0$, and one with a narrower bandwidth on the bottom, with $\Delta k = \frac{1}{15}k_0$.
Mathematica graphics through Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/Ad7vo.png"]
As you can see, for broadband pulses the functions are obviously different, but as the bandwidth becomes narrower, you just don't have that much bandwidth to explore the nonlinearities of the $k\propto 1/\lambda$ transformation.
So, what is the take-home lesson here? I want to emphasize that gaussian pulses are ultimately just models, and they are useful models in as much as they are both realistic and solvable. In the real world, you are unlikely to encounter a pulse that has a truly gaussian spectrum (unless you work very hard at it) with the kind of broad bandwidth in the top row.
It helps to put in some numbers: say that you have a pulse at optical frequencies that lasts about a nanosecond, which sounds really short. In this case, the bandwidth will be in the GHz regime, but the frequency will be in the order of $10^{14}\: \mathrm{Hz}$, and the bandwidth will look something like $\delta k \sim 10^{-5}k_0$; as far as the plots above go, they would both be indistinguishable delta-like spikes at that resolution. To get spectra that are as broad as the top row you would need an octave-spanning pulse that was shorter than a few tens of femtoseconds; this is nowadays doable but requires lots of dedicated effort.
Ultimately, then, what I'm saying is this: it can be worthwhile to consider what would happen with the time-domain waveform of $S_\lambda(\lambda)$, but if you really want to do it then first you need to be answering some hard questions about what bandwidth you're actually interested in, whether you're not served equally well by a standard $k$-space gaussian, and (if the two spectra are actually different at your bandwidths of interest) just how realistic a model $S_\lambda(\lambda)$ is for the phenomenon of interest, and whether there isn't some better model that you should be calculating.