Hermitian Conjugate of a Constant Operator

If we have the operator \bgroup\color{black}$O=a+ib$\egroup where \bgroup\color{black}$a$\egroup and \bgroup\color{black}$b$\egroup are real, what is its Hermitian conjugate? By the definition of the Hermitian conjugate

\begin{displaymath}\bgroup\color{black} \langle\phi\vert O\psi\rangle=\langle O^\dagger\phi\vert\psi\rangle .\egroup\end{displaymath}

It is easy to see from the integral that

\begin{displaymath}\bgroup\color{black} \langle (a-ib)\phi\vert\psi\rangle=\lang...
...vert(a+ib)\psi\rangle=(a+ib)\langle\phi\vert\psi\rangle \egroup\end{displaymath}

So the Hermitian conjugate of a constant operator is its complex conjugate.



Jim Branson 2013-04-22