If $A$ and $B$ are Hermitian operators, show that $$C~:=~i[A,B]$$ is Hermitian too.
My work:
$$\begin{gather}
C=i(AB-BA) \\
\langle\psi\rvert C\lvert\phi\rangle = i\langle\psi\rvert AB\lvert\phi\rangle-i\langle\psi\rvert BA\lvert\phi\rangle
\end{gather}$$
I guess I need to split up the operators somehow to use:
$$\begin{align}
\langle\psi\rvert A\lvert\phi\rangle &= \langle\phi\rvert A\lvert\psi\rangle^* \\
\langle\psi\rvert B\lvert\phi\rangle &= \langle\phi\rvert B\lvert\psi\rangle^*
\end{align}$$
I know a little about the identy operator, which I've seen used to do a similar trick, but I'm not that clear on its exact meaning hmm… $$1=\sum_n\lvert n\rangle\langle n\rvert$$
The definition of Hermiticity I learnt from lectures is the one I stated above for A, can you prove it in this way for C?
Best Answer
Just apply Hermitian conjugation to both sides of the equality: $C^+=(i[A,B])^+=-i[AB-BC]^+=-i[B^+A^+-A^+B^+]$= ... (sorry, had to use + sign instead of the dagger).