# Chemical Dynamics/Electrostatics/Fourier Transforms

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

## Definition

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:

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

## The translation theorem

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

${\displaystyle {\hat {g}}(\mathbf {k} )=e^{-i\mathbf {k} \cdot \mathbf {R} _{0}}{\hat {f}}(\mathbf {k} ).}$

Proof
${\displaystyle {\hat {g}}(\mathbf {k} )={\frac {1}{(2\pi )^{3/2}}}\int g(\mathbf {r} )e^{-i\mathbf {k} \cdot \mathbf {r} }\,d^{3}\mathbf {r} }$
${\displaystyle ={\frac {1}{(2\pi )^{3/2}}}\int f(\mathbf {r} -\mathbf {R} _{0})e^{-i\mathbf {k} \cdot \mathbf {r} }\,d^{3}\mathbf {r} }$
${\displaystyle ={\frac {1}{(2\pi )^{3/2}}}\int f(\mathbf {r} -\mathbf {R} _{0})e^{-i\mathbf {k} \cdot \mathbf {r} }\,d^{3}\mathbf {r} }$

Now, change r to a new variable by: ${\displaystyle \mathbf {r'} =\mathbf {r} -\mathbf {R} _{0}}$

${\displaystyle {\hat {g}}(\mathbf {k} )={\frac {1}{(2\pi )^{3/2}}}\int f(\mathbf {r'} )e^{-i\mathbf {k} \cdot (\mathbf {r'} +\mathbf {R} _{0})}\,d^{3}\mathbf {r'} }$
${\displaystyle ={\frac {1}{(2\pi )^{3/2}}}e^{-i\mathbf {k} \cdot \mathbf {R} _{0}}\int f(\mathbf {r'} )e^{-i\mathbf {k} \cdot \mathbf {r'} }\,d^{3}\mathbf {r'} }$
${\displaystyle ={\frac {1}{(2\pi )^{3/2}}}e^{-i\mathbf {k} \cdot \mathbf {R} _{0}}\int f(\mathbf {r} )e^{-i\mathbf {k} \cdot \mathbf {r} }\,d^{3}\mathbf {r} }$
${\displaystyle =e^{-i\mathbf {k} \cdot \mathbf {R} _{0}}{\hat {f}}(\mathbf {k} ).}$

## The convolution theorem

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:

${\displaystyle (f*g)(t){\stackrel {\mathrm {def} }{=}}\ \int _{-\infty }^{\infty }f(\tau )\,g(t-\tau )\,d\tau }$

The convolution theorem for the Fourier transform says:

If

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

then

${\displaystyle {\hat {h}}(\mathbf {k} )={\hat {f}}(\mathbf {k} )\cdot {\hat {g}}(\mathbf {k} )}$.
Proof
${\displaystyle {\hat {h}}(\mathbf {k} )=\int {e^{-i\mathbf {k} \cdot \mathbf {r} }h(\mathbf {r} )}d\mathbf {r} }$
${\displaystyle =\int {e^{-i\mathbf {k} \cdot \mathbf {r} }\int f(\mathbf {r} )g(\mathbf {r'} -\mathbf {r} )}d^{3}\mathbf {r'} d^{3}\mathbf {r} }$