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I like to Fourier transform the following product of functions:

$$g(\vec{r})f(\vec{r}).$$

So I like to calculate the following:

$$\int g(\vec{r})f(\vec{r}) e^{-i\vec{k}\cdot\vec{r}}d^3r.$$

$\vec{k}$ is a discrete wave vector. I also know how the solution has to look like

$$\sum_\vec{q} g_{\vec{k}-\vec{q}} \,\,f_{\vec{q}}.$$

I figured that I somehow have to use the convolution theorem. But I am not sure how I can correctly use it. I find that

$$FT(gf) = FT(g) * FT(f)$$

where $FT$ is the Fourier transform and $*$ is the convolution. I applied this as follows

$$\int g(\vec{r})f(\vec{r}) e^{-i\vec{k}\cdot\vec{r}}d^3r=\underbrace{FT(g(\vec{r}))}_{g_\vec{k}} * \underbrace{FT(f(\vec{r}))}_{f_\vec{k}}= \int f_\vec{q}g_\vec{k-q} d^3q .$$

The last step is the definition of $*$. This almost looks like where I need to go but instead of a sum I have an integral. Do I just have to use the definition of the discrete convolution in the last step? Then I think I have found the solution but I am not 100% sure if the way to get there is correct.

DaPhil
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  • when you say that $k$ is a discrete wave vector, do you mean it takes values in a discrete set? That is only possible on $\mathbb R^3$ if your functions $f$ and $g$ are triply periodic. In that case the Fourier transform is supported (as delta functions) on discrete points so that the integral turns into a sum. It may be more natural in that setting to work with Fourier series. – doetoe Jun 27 '14 at 09:39
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    Although the Fourier transform is frequently used in physics, this question appears to be off-topic because it is about the transform solely, and not any application or reference to physics is featured. – JPhy Jun 27 '14 at 09:43
  • yes, $\vec{k}$ only can take values from a discrete set. But the Fourier transform from r to k is an integral since r is continous right? – DaPhil Jun 27 '14 at 09:43
  • That is right. To evaluate the Fourier coefficients of a periodic function you integrate over their fundamental domain (a parallelepiped on which the function is s.t. if it would be tiled it would give you back the whole function). An integral over the whole $\mathbb R^3$ wouldn't converge. – doetoe Jun 27 '14 at 10:15

1 Answers1

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"Do I just have to use the definition of the discrete convolution in the last step?"

Yes, so the result is just $\sum_\vec{q} g_{\vec{q}} \,\,f_{\vec{k} - \vec{q}}$, or alternatively $\sum_\vec{q} g_{\vec{k} -\vec{q}} \,\,f_{ \vec{q}}$.

But it is not $\sum_\vec{q} g_{\vec{k}-\vec{q}} \,\,f_{\vec{k}}$ (third expression in your question)

Trimok
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