Hermitian Conjugate of \bgroup\color{black}${\partial\over\partial x}$\egroup

We wish to compute the Hermitian conjugate of the operator \bgroup\color{black}${\partial\over\partial x}$\egroup. We will use the integral to derive the result.

\begin{displaymath}\bgroup\color{black} \left<\phi\left\vert{\partial\over\parti...
...nfty}^\infty\phi^*(x){\partial\psi(x)\over\partial x}dx \egroup\end{displaymath}

We can integrate this by parts, differentiating the \bgroup\color{black}$\phi$\egroup and integrating to get \bgroup\color{black}$\psi$\egroup.

\begin{displaymath}\bgroup\color{black} \left<\phi\left\vert{\partial\over\parti...
...ft.{-\partial\over\partial x}\phi\right\vert\psi\right> \egroup\end{displaymath}

So the Hermitian conjugate of \bgroup\color{black}${\partial\over\partial x}$\egroup is \bgroup\color{black}$-{\partial\over\partial x}$\egroup.

Note that the Hermitian conjugate of the momentum operator is \bgroup\color{black}${\hbar\over -i}{-\partial\over\partial x}$\egroup which is the same as the original operator. So the momentum operator is Hermitian.



Jim Branson 2013-04-22