10.1 Dirac delta function

The above allows us to relate the Fourier transform of an Aperture and the resulting E field from diffraction through that aperture. To extend this to anarray of apertures, requires that one introduce a new concept, the Dirac delta function.

The fundamental definition of the Dirac delta function is

$${\displaystyle \begin{array}{c}\delta (x)=0ifx\ne 0\\ f(0)={\int }_{-\infty }^{\infty }f(x)\delta (x)dx\end{array}}$$

As a special case if $f(x)=1$

$${\int }_{-\infty }^{\infty }\delta (x)dx=1$$

This function has some important properties:

$$f(x^{\prime })={\int }_{-\infty }^{\infty }f(x)\delta (x-x^{\prime })dx$$ which follows direction from the definition ie. define a new coordinate $$a=x-x^{\prime }$$ , then $$x=a+x^{\prime }$$ and $$da=dx\text{.}$$ Then

$${\displaystyle \begin{array}{c}{\int }_{-\infty }^{\infty }f(x)\delta (x-x^{\prime })dx={\int }_{-\infty }^{\infty }f(a+x^{\prime })\delta (a)da\\ =f(x^{\prime })\end{array}}$$

We also note that $\delta (x)=\delta (-x)\text{.}$

It gets even more interesting. Recall $$f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })e^{ik(x^{\prime }-x)}d{x}^{\prime }dk$$ which we rearrange suggestively $${\displaystyle \begin{array}{c}f(x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }f(x^{\prime })e^{ik(x^{\prime }-x)}d{x}^{\prime }dk\\ ={\int }_{-\infty }^{\infty }\left[\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{ik(x^{\prime }-x)}dk\right]f(x^{\prime })d{x}^{\prime }\end{array}}$$ we also have $$f(x)={\int }_{-\infty }^{\infty }f(x^{\prime })\delta (x^{\prime }-x)d{x}^{\prime }$$ so we must have $$\delta (x^{\prime }-x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{ik(x^{\prime }-x)}dk$$ or $$\delta (x)=\delta (-x)=\frac{1}{2\pi }{\int }_{-\infty }^{\infty }{e}^{-ikx}dk$$

That is the Dirac delta function is the inverse Fourier transform of 1. This is a very useful property that allows us to do problems like Young's doubleslit. Consider the aperture function: $$A=E_0(\delta (y-a/2)+\delta (y+a/2))$$ where we have represented the slits by Dirac delta functions. Then we obtain $${\displaystyle \begin{array}{c}E={\int }_{-\infty }^{\infty }{E}_0\delta (y-a/2)e^{iy{k}_y}+{\int }_{-\infty }^{\infty }{E}_0\delta (y+a/2)e^{iy{k}_y}\\ =E_0(e^{ia{k}_y/2}+e^{-ia{k}_y/2})\\ =E_{0}2\cos (k_{y}a/2)\\ =E_{0}2\cos \left(\frac{ka\sin \theta }{2}\right)\end{array}}$$ which is exactly what we obtained before. Well that was a lot easier than what we did earlier in the course!