Help with proving ($\langle$i[$\hat{A}$, $\hat{B}$]$\rangle$)$^{2}$ = |$\langle$[$\hat{A}$, $\hat{B}$]$\rangle$|$^{2}$

Question

Let $\hat{A}$ and $\hat{B}$ be hermitian operators. Show that ($\langle$i[$\hat{A}$, $\hat{B}$]$\rangle$)$^{2}$ = |$\langle$[$\hat{A}$, $\hat{B}$]$\rangle$|$^{2}$

Answer 1

The appearance of $i$ is important here, otherwise $[A,B]$ is anti-self-adjoint and thus does not (strictly) define an expectation value (though multiplication by $i$ fixes it, so it's not a big deal - it's when you leave symmetric/anti-self-adjoint operators that this becomes an issue). As such, the right hand side is technically an abuse of notation, but no real harm is done. Given appropriate $f$ in the Hilbert space,

$$\langle i[A,B]\rangle_f = \langle f \mid i[A,B]f\rangle.$$

Thus

$$\langle i[A,B]\rangle^2 = \langle f \mid i[A,B]f\rangle^2 = \langle f \mid i[A,B]f\rangle \langle f \mid i[A,B] f\rangle = \langle i[A,B] f\mid f\rangle \langle f\mid i[A,B]f\rangle.$$

At this point the factors of $i$ cancel, leaving us with

$$\langle i[A,B]\rangle^2 = \langle [A,B] f\mid f\rangle \langle f\mid [A,B]f\rangle = \overline{\langle f \mid [A,B] f\rangle}\langle f\mid [A,B]f\rangle.$$

Can you take it from here?