Building a Localized Single-Particle Wave Packet

We now have a wave function for a free particle with a definite momentum \bgroup\color{black}$p$\egroup

\begin{displaymath}\bgroup\color{black} \psi(x,t)=e^{i(px-Et)/\hbar}=e^{i(kx-\omega t)} \egroup\end{displaymath}

where the wave number \bgroup\color{black}$k$\egroup is defined by \bgroup\color{black}$p=\hbar k$\egroup and the angular frequency \bgroup\color{black}$\omega$\egroup satisfies \bgroup\color{black}$E=\hbar\omega$\egroup. It is not localized since \bgroup\color{black}$P(x,t)=\vert\psi(x,t)\vert^2=1$\egroup everywhere.

We would like a state which is localized and normalized to one particle.

\begin{displaymath}\bgroup\color{black} \int\limits_{-\infty}^\infty\psi^*(x,t)\psi(x,t) dx=1 \egroup\end{displaymath}

To make a wave packet which is localized in space, we must add components of different wave number. Recall that we can use a Fourier Series to compose any function \bgroup\color{black}$f(x)$\egroup when we limit the range to \bgroup\color{black}$-L<x<L$\egroup. We do not want to limit our states in \bgroup\color{black}$x$\egroup, so we will take the limit that \bgroup\color{black}$L\rightarrow \infty$\egroup. In that limit, every wave number is allowed so the sum turns into an integral. The result is the very closely related Fourier Transform

\begin{displaymath}\bgroup\color{black}f(x)={1\over \sqrt{2\pi}}\int\limits_{-\infty}^\infty A(k) e^{ikx} dk\egroup\end{displaymath}

with coefficients which are computable,

\begin{displaymath}\bgroup\color{black}A(k)={1\over \sqrt{2\pi}}\int\limits_{-\infty}^\infty f(x) e^{-ikx} dx.\egroup\end{displaymath}

The normalizations of \bgroup\color{black}$f(x)$\egroup and \bgroup\color{black}$A(k)$\egroup are the same (with this symmetric form) and both can represent probability amplitudes.

\begin{displaymath}\bgroup\color{black}\int\limits_{-\infty}^\infty f^*(x)f(x)dx=\int\limits_{-\infty}^\infty A^*(k)A(k)dk\egroup\end{displaymath}

We understand \bgroup\color{black}$f(x)$\egroup as a wave packet made up of definite momentum terms \bgroup\color{black}$e^{ikx}$\egroup. The coefficient of each term is \bgroup\color{black}$A(k)$\egroup. The probability for a particle to be found in a region \bgroup\color{black}$dx$\egroup around some value of \bgroup\color{black}$x$\egroup is \bgroup\color{black}$\vert f(x)\vert^2dx$\egroup. The probability for a particle to have wave number in region \bgroup\color{black}$dk$\egroup around some value of \bgroup\color{black}$k$\egroup is \bgroup\color{black}$\vert A(k)\vert^2dk$\egroup. (Remember that \bgroup\color{black}$p=\hbar k$\egroup so the momentum distribution is very closely related. We work with \bgroup\color{black}$k$\egroup for a while for economy of notation.)

Jim Branson 2013-04-22