1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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

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

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

$\endgroup$
  • $\begingroup$ 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 '13 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 '13 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.