The Dirac Delta Function

The Dirac delta function is zero everywhere except at the point where its argument is zero. At that point, it is just the right kind of infinity so that

\begin{displaymath}\bgroup\color{black}\int\limits_{-\infty}^\infty dx\; f(x)\; \delta(x)=f(0).\egroup\end{displaymath}

This is the definition of the delta function. It picks of the value of the function \bgroup\color{black}$f(x)$\egroup at the point where the argument of the delta function vanishes. A simple extension of the definition gives.

\begin{displaymath}\bgroup\color{black}\int\limits_{-\infty}^\infty dx\; f(x)\; \delta(x-a)=f(a)\egroup\end{displaymath}

The transformation of an integral allows us to compute

\begin{displaymath}\bgroup\color{black}\int\limits_{-\infty}^\infty dx\; f(x)\; ...
...left[{1\over \vert{dg\over dx}\vert}f(x)\right]_{g(x)=0}\egroup\end{displaymath}

the effect of the argument being a function.

If we make a wave packet in p-space using the delta function, and we transform to position space,

\begin{displaymath}\bgroup\color{black}\psi(x)={1\over \sqrt{2\pi\hbar}}\int\lim...
...\hbar} dp
= {1\over \sqrt{2\pi\hbar}}\; e^{ip_0x/\hbar}\egroup\end{displaymath}

we just get the state of definite \bgroup\color{black}$p$\egroup.

This is a state of definite momentum written in momentum space. \bgroup\color{black}$\delta(p-p_0) $\egroup
Its Fourier transform is \bgroup\color{black}$\psi_p(x,t)={1\over \sqrt{2\pi\hbar}}\; e^{i(px-Et)/\hbar}$\egroup
This is a state of definite position written in position space. \bgroup\color{black}$\delta(x-x_0) $\egroup

Jim Branson 2013-04-22