Fourier Transforms in D Dimensions

Abstract

Let us introduce the abreviated notation

$$ \int\limits_{{ - \infty }}^{\infty } {d{{x}_{1}}} \int\limits_{{ - \infty }}^{\infty } {d{{x}_{2}} \ldots \int\limits_{{ - \infty }}^{\infty } {d{{x}_{d}}f\left( {{{x}_{1}},{{x}_{2}}, \ldots ,{{x}_{d}}} \right) \equiv \int {dxf\left( {{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{x}}} \right)} } } $$

(4-1)

and

$$ {{e}^{{i\left( {{{k}_{1}}{{x}_{1}}{\text{ }} + {\text{ }}{{k}_{2}}{{x}_{2}} + \ldots + {{k}_{d}}{{x}_{d}}} \right)}}} \equiv {{e}^{{i{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{k}}\cdot {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{x}}}}} $$

(4-2)

Then the d-dimensional Fourier transform of the function f(x) is given by

$$ {{e}^{{i({{k}_{1}}{{x}_{1}} + {{k}_{2}}{{x}_{2}} + \ldots + {{k}_{d}}{{x}_{d}})}}} \equiv {{e}^{{i{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{k}} \cdot {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{x}}}}} $$

(4-3)

while the inverse transform is

$$ {{f}^{t}}({\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{k}}) = \tfrac{1}{{(2\pi ) \frac{\hbox{$\scriptstyle 1$}}{\hbox{$\scriptstyle 2$}} d}}{{\int {dx\;e} }^{{i{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{k}} \cdot {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{x}}}}}f({\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle-}$}}{x}}) $$

(4-4)

We would like to show that the scalar product of two functions in direct space is equal to the scalar product of their Fourier transforms in reciprocal space.