Operators in Momentum Space

If we want to work in momentum space, we need to look at the states of definite position to find our operators. The state (in momentum space) with definite position \bgroup\color{black}$x_0$\egroup is

\begin{displaymath}\bgroup\color{black}v_{x0}(p)={1\over\sqrt{2\pi\hbar}}e^{-ipx_0/\hbar}\egroup\end{displaymath}

The operators are
\bgroup\color{black}$\displaystyle x^{(op)}=i\hbar{\partial\over \partial p}$\egroup
and

\begin{displaymath}\bgroup\color{black}p^{(op)}=p .\egroup\end{displaymath}

The \bgroup\color{black}$(op)$\egroup notation used above is usually dropped. If we see the variable \bgroup\color{black}$p$\egroup, use of the operator is implied (except in state written in terms of \bgroup\color{black}$p$\egroup like \bgroup\color{black}$\phi(p)$\egroup).

Gasiorowicz Chapter 3

Griffiths doesn't cover this.

Cohen-Tannoudji et al. Chapter



Jim Branson 2013-04-22