# Proof for $\langle i[A,B]\rangle$ [closed]

I have to prove the following equation: $$\langle i[A,B]\rangle = 2\mathfrak{Im}\left[\int dV(\overline{B\psi)}(A\psi)\right]\,,$$ where A,B are hermitian operators. Here is my calculation, but I don't get the right result. Can you help me to find my mistake?

\begin{aligned} \langle i[A,B]\rangle &=(\psi,i[A,B]\psi) \\&=(\psi,i(AB-BA)\psi) \\&=(\psi,iAB\psi-iBA\psi) \\&=(\psi,iAB\psi)-(\psi,iBA\psi) \\&=i(A^\dagger\psi,B\psi)-i(B^\dagger\psi,A\psi) \\&=i(A\psi,B\psi)-i(B\psi,A\psi) \\&=-i[(B\psi,A\psi)-(A\psi,B\psi)] \\&=-i[(B\psi,A\psi)-\overline{(B\psi,A\psi)}] \\&=-2i\,\mathfrak{Im}[(B\psi,A\psi)] \end{aligned}

## closed as off-topic by Emilio Pisanty, Kyle Kanos, GiorgioP, Carl Brannen, YashasApr 25 at 4:51

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## 1 Answer

I think the mistake is in the last line. Using the following identity helps to get the result what you are seeking for. $$(a+ib)-(a-ib)=2ib=2i Im(a+ib)$$