Chemical Dynamics/Electrostatics/Fourier Transforms

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The Fourier transform is a useful mathematical transformation often utilized in many scientific and engineering fields. Here we extract useful concepts of Fourier transformation and logically arrange them to form a foundation for the Ewald summation and other related methods in electrostatics. Readers could check out other more mathematically formal introduction of Fourier transform


We use the following convention in which the Fourier transform is a unitary transformation on the 3-D Cartesian space R3, the Fourier transform and its inverse transform are symmetric:

 \hat{f}(\mathbf{k}) = \frac{1}{(2\pi)^{3/2}} \int f(\mathbf{r}) e^{- i\mathbf{k}\cdot \mathbf{r}}\,d^{3}\mathbf{r}
f(\mathbf{r}) = \frac{1}{(2\pi)^{3/2}} \int \hat{f}(\mathbf{k}) e^{ i\mathbf{k} \cdot \mathbf{r}}\,d^{3}\mathbf{k}

The translation theorem[edit]

Given a fixed position vector R0, if g(r) = ƒ(r − R0), then  

\hat{g}(\mathbf{k})= e^{- i \mathbf{k}\cdot \mathbf{R}_0 }\hat{f}(\mathbf{k}).

\hat{g}(\mathbf{k})= \frac{1}{(2\pi)^{3/2}} \int g(\mathbf{r}) e^{- i\mathbf{k}\cdot \mathbf{r}}\,d^{3}\mathbf{r}
 = \frac{1}{(2\pi)^{3/2}} \int f(\mathbf{r}-\mathbf{R}_0) e^{- i\mathbf{k}\cdot \mathbf{r}}\,d^{3}\mathbf{r}
 = \frac{1}{(2\pi)^{3/2}} \int f(\mathbf{r}-\mathbf{R}_0) e^{- i\mathbf{k}\cdot \mathbf{r}}\,d^{3}\mathbf{r}

Now, change r to a new variable by:  \mathbf{r'}= \mathbf{r}-\mathbf{R}_0

 \hat{g}(\mathbf{k}) = \frac{1}{(2\pi)^{3/2}} \int f(\mathbf{r'}) e^{- i\mathbf{k}\cdot (\mathbf{r'}+\mathbf{R}_0)}\,d^{3}\mathbf{r'}
  = \frac{1}{(2\pi)^{3/2}} e^{-i \mathbf{k}\cdot \mathbf{R}_0} \int f(\mathbf{r'}) e^{- i\mathbf{k}\cdot \mathbf{r'}}\,d^{3}\mathbf{r'}
  = \frac{1}{(2\pi)^{3/2}} e^{-i \mathbf{k}\cdot \mathbf{R}_0} \int f(\mathbf{r}) e^{- i\mathbf{k}\cdot \mathbf{r}}\,d^{3}\mathbf{r}
 = e^{- i \mathbf{k}\cdot \mathbf{R}_0 } \hat{f}(\mathbf{k}).

The convolution theorem[edit]

The convolution of f and g is usually denoted as fg, using an asterisk or star. It is defined as the integral of the product of the two functions after one is reversed and shifted:

(f * g )(t)
\stackrel{\mathrm{def}}{=}\ \int_{-\infty}^\infty f(\tau)\, g(t - \tau)\, d\tau

The convolution theorem for the Fourier transform says:


 h(\mathbf{r}) = (f * g )(\mathbf{r})


 \hat{h}(\mathbf{k})= \hat{f}(\mathbf{k})\cdot \hat{g}(\mathbf{k}).
 \hat{h}(\mathbf{k}) = \int {e^{-i\mathbf{k} \cdot \mathbf{r}} h(\mathbf{r})} d\mathbf{r}
  = \int {e^{-i\mathbf{k} \cdot \mathbf{r}} \int f(\mathbf{r}) g(\mathbf{r'}-\mathbf{r})}