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

$$\begin{eqnarray*} f(x)&=&{1\over \sqrt{2\pi}}\int\limits_{-\infty}^\infty A(k) e... ...ver \sqrt{2\pi}}\int\limits_{-\infty}^\infty f(x) e^{-ikx} dx\\ \end{eqnarray*}$$

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).