# Duality and Fourier Transforms [closed]

I read that

$(FF(f))(x)=2\pi f(-x)$, where $F$ is the Fourier transform

and $F(f(x-a))(k)=\exp(-ika) X(k)$ where $X(k)=F(f(x))$

implies $F(\exp(iax)f(x))(k)=X(k-a)$.

But I don't see how that is done... I am quite happy with getting $F^{-1}X(k-a)=\exp(iax)f(x)$ by brute force calculation. I would like to see how to use duality though.

## closed as off topic by Colin K, Qmechanic♦, David Z♦Nov 15 '11 at 21:02

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## 2 Answers

You need to know the basic Fourier transform delta-function identity

$$\int_{-\infty}^{\infty} e^{ikx} {dk\over 2\pi} = \delta(x)$$

Which implies Fourier inversion. Proving this identity is slightly subtle, because the right hand side is a distribution, but you can do the integral explicitly over a long interval from -M to M to get an object which has a unit integral and is shrinking in size with M as 1/M, so it must be a delta-function in any reasonable sense of limits.

The double fourier-transform is

$$FF(f) (x') = \int dk e^{ix'k} (\int dk e^{ikx} f(x))$$

And you can do the k integral using the identity to get the result.

There are quite a few good dualities involving Fourier transforms and they are all really good to know. The one you are dealing with is quite descriptive. First consider an wavefunction in the time domain. Multiplying it by an exponential is like modulating an AM radio signal: it adds two sidebands to the frequency spectrum which are images of the original signal, displaced by plus/minus the multiplying frequency. This is how it works for sine wave modulation: for exponential modulation, its only one sideband, so you get simple displacement.

This is one half of the duality. To bring it full circle, you now have to do the complimentary problem: take a signal in the time domain, delay it by a fixed time T, and ask what this does to the spectrum in the frequency domain. Obviously, it has to contain all the same frequencies, and to the exact same amplitude. Multiplication by an exponential...any exponential...obviously does this. So far so good. But that is far from proving the duality. An argument of phase delays can actually tie things together in a fairly convincing way. That's pretty much how I understand it.