Fourier Transform *

To allow wave functions to extend to infinity, we will expand the interval used

$$\bgroup\color{black}L\rightarrow \infty .\egroup$$

As we do this we will use the wave number

$$\bgroup\color{black}k={n\pi\over L}.\egroup$$

As $L\rightarrow \infty .$, $k$ can take on any value, implying we will have a continuous distribution of $k$. Our sum over $n$ becomes an integral over $k$.

$$\bgroup\color{black}dk={\pi\over L}dn.\egroup$$

If we define $A(k)=\sqrt{2\over\pi}La_n$, we can make the transform come out with the constants we want.

$f(x)=\sum\limits_{n=-\infty}^\infty a_ne^{in\pi x\over L}$	Standard Fourier Series
$a_n={1\over 2L} \int\limits_{-L}^L f(x)e^{-in\pi x\over L}dx$	Standard Fourier Series
$A_n=\sqrt{2\over\pi}L{1\over 2L} \int\limits_{-L}^L f(x)e^{-in\pi x\over L}dx \qquad$	redefine coefficient
$A_n={1\over \sqrt{2\pi}} \int\limits_{-L}^L f(x)e^{-in\pi x\over L}dx$
$f(x)=\sqrt{\pi\over 2}{1\over L}\sum\limits_{n=-\infty}^\infty A_ne^{in\pi x\over L}$	f stays the same
$f(x)={\sqrt{\pi}\over\sqrt{2}L} \int\limits_{-\infty}^\infty A(k) e^{ikx} {L\over\pi} dk$	but is rewritten in new A and dk
$f(x)={1\over\sqrt{2\pi}} \int\limits_{-\infty}^\infty A(k) e^{ikx} dk$	result
$A(k)={1\over\sqrt{2\pi}} \int\limits_{-\infty}^\infty f(x) e^{-ikx} dx$	result

This is just the extension of the Fourier series to all $x$.

If $f(x)$ is normalized, then $A(k)$ will also be normalized with this (symmetric) form of the Fourier Transform. Thus, if $f(x)$ is a probability amplitude in position-space, $A(k)$ can be a probability amplitude (in k-space).