# Quantum Mechanics Operator Hermiticity

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?

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).
Given that $(AB)^\dagger =B^\dagger A^\dagger$(*)
You have: $$C^\dagger=[i(AB-BA)]^\dagger =-i(B^\dagger A^\dagger- A^\dagger B^\dagger)= i(AB-BA)$$
(*)$<\phi|A(B|\phi>)^\dagger=(<\phi|B)A|\phi>^*=<\phi|BA|\phi>^*=<\phi|BA|\phi> = \implies (AB)^\dagger=B^\dagger A^\dagger$