Transcription
This application of the Fourier transform is much more subtle than it looks. Consider the simplest function of all, a constant. Then what happens when we try to plug it into the Fourier transform? The integral looks pretty intimidating.
So, here's a more intuitive way to understand the answer. One of the most important facts about the Fourier transform is that when you plug in a Gaussian function G, you get back out another Gaussian, but with a key difference. If the original Gaussian had standard deviation sigma, then its Fourier transform will have standard deviation one over sigma. Meaning that if the original curve was very narrow in X space, its Fourier transform will be very broad in K space, and vice versa.
In particular, if we let sigma shrink down to zero size, then the first Gaussian becomes an infinitely tall, infinitesimally narrow spike at the origin. In other words, it approaches the Dirac delta function, delta of X. At the same time, the Fourier transformed curve simply becomes an infinitely broad horizontal line. Meaning that it's just the constant function that we wanted from the beginning, up to a factor of 2 pi.
Thus, the Fourier transform of a constant is a delta function, and vice [clears throat] versa. Again, reflecting the fact that a narrow spike in one space turns into a broad plateau in the other. And reading this relationship in reverse, the Fourier transform of f of x equals one is then the square root of 2 pi times delta of k. Meaning that we found the answer to that intimidating looking integral from earlier.
In fact, we've stumbled upon one of the key representations of the delta function that's used all the time in physics, as an integral over all complex waves. The integral doesn't actually converge though, and so this formula needs to be interpreted in the mathematical language of distributions in order to make rigorous sense of it. And that's why