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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?

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2 Answers 2

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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).

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Given that $(AB)^\dagger =B^\dagger A^\dagger$(*)

You have: $$C^\dagger=[i(AB-BA)]^\dagger =-i(B^\dagger A^\dagger- A^\dagger B^\dagger)= i(AB-BA)$$

(*)$<\phi|A(B|\phi>)^\dagger=(<\phi|B)A|\phi>^*=<\phi|BA|\phi>^*=<\phi|BA|\phi> = \implies (AB)^\dagger=B^\dagger A^\dagger$

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  • $\begingroup$ 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? $\endgroup$
    – user31436
    Oct 20, 2013 at 22:06
  • $\begingroup$ @user31436 : there are some well-known properties of Hermitian conjugation (en.wikipedia.org/wiki/Hermitian_adjoint ), you may wish to check them. $\endgroup$
    – akhmeteli
    Oct 20, 2013 at 22:13

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