Quantum Hermiticity Bra-Ket notation please 
If $A$ and $B$ are Hermitian operators, show that $$C~:=~i[A,B]$$ is Hermitian too.

My work:
$$\begin{gather}
C=i(AB-BA) \\
\langle\psi\rvert C\lvert\phi\rangle = i\langle\psi\rvert AB\lvert\phi\rangle-i\langle\psi\rvert BA\lvert\phi\rangle
\end{gather}$$
$A$ and $B$ are Hermitian such that:
$$\begin{align}
\langle\psi\rvert A\lvert\phi\rangle &= \langle\phi\rvert A\lvert\psi\rangle^* \\
\langle\psi\rvert B\lvert\phi\rangle &= \langle\phi\rvert B\lvert\psi\rangle^*
\end{align}$$
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? ie can you use bra-ket notation.
 A: The bra-ket notation is very formal, but well, let's go through it.
We must prove that $\langle \psi | C | \phi \rangle = \langle \phi | C | \psi \rangle ^*$, where the $^*$ is complex conjugation.
First we prove that the adjoint of the composite operator $AB$ is $BA$, that is, that $\langle \psi | AB | \phi \rangle = \langle \phi | BA | \psi \rangle^*$.
For that, notice that 
$\langle \psi | AB | \phi \rangle = 
 \langle \psi | A \, 1 \, B | \phi \rangle =
 \sum_n \langle \psi | A | n \rangle\langle n | B | \phi \rangle =
 \sum_n \langle n | A | \psi \rangle^* \langle \phi | B | n \rangle^* =
 (\sum_n \langle \phi | B | n \rangle \langle n | A | \psi \rangle ) ^* =
 (\langle \phi | B \, 1 \, A | \psi \rangle ) ^* =
 \langle \phi | BA | \psi \rangle^*$.
By the linearity of the inner product, one has then
$\langle \psi | (iAB) | \phi \rangle = 
 i(\langle \psi | AB | \phi \rangle) = 
 (-i)^* (\langle \phi | BA | \psi \rangle)^* = 
 (\langle \phi | (-iBA) | \psi \rangle)^*$.
Switching $AB$ to $BA$ then gives you:


*

*$\langle \psi | (iAB) | \phi \rangle = (\langle \phi | (-iBA) | \psi \rangle)^*$.

*$\langle \psi | (iBA) | \phi \rangle = (\langle \phi | (-iAB) | \psi \rangle)^*$.


Now one has 
$\langle \psi | C | \phi \rangle
 = \langle \psi | (iAB) | \phi \rangle - \langle \psi | (iBA) | \phi \rangle
 = (\langle \phi | (-iAB) | \psi \rangle)^* - (\langle \phi | (-iBA) | \psi \rangle)^*
 = (\langle \phi | C | \psi \rangle)^*$.
That concludes the proof.
--
If you want to know what is going on, I recommend you find the definition of the adjoint of an operator. Mathematicians use the star $A^*$, and physicists use the dagger $A^\dagger$. If you know that $A\mapsto A^\dagger$ is antilinear and satisfies the rule $(AB)^\dagger = B^\dagger A^\dagger$, then your exercise can be solved by checking $C^\dagger = (iAB - iBA)^\dagger = -i BA + i AB = C$. 
