The Position Operator

What about the position operator, \bgroup\color{black}$x^{(op)}$\egroup? The answer is simply

\bgroup\color{black}$\displaystyle x^{(op)}=x$\egroup
when we are working in position space with \bgroup\color{black}$u_{p0}(x,t)={1\over\sqrt{2\pi\hbar}}e^{i(p_0x-E_0t)/\hbar}$\egroup (as we have been above).



Jim Branson 2013-04-22