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.


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:

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

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 $C^\dagger = (iAB - iBA)^\dagger = -i BA + i AB = C$.


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