Consider the following relations
$$H_0|\psi_a\rangle = E_a|\psi_a\rangle$$ $$H_0|\psi_b\rangle = E_b|\psi_b\rangle$$
I am struggling then to understand why the following identity holds (its probably straight forward but I just can not see it.) Note that r represents the position operator.
$$ \langle \psi_b| rH_0 - H_0r|\psi_a\rangle = (E_a - E_b)\langle \psi_b| r|\psi_a\rangle$$
The reason is that $H_0$ as an observable is hermitian $H_0 = H_0^\dagger$. That means you can use $$\langle\Psi_b|H_0 r|\Psi_a\rangle = (\langle\Psi_a|r^\dagger H_0^\dagger|\Psi_b\rangle)^\ast = (\langle\Psi_a|r^\dagger H_0|\Psi_b\rangle)^\ast$$
You might also just say that it is part of the definition of the bra-ket notation, that operators $A$ inside act as $A$ to the right and as $A^\dagger$ to the left.
Here’s an alternate derivation that uses the expansion of unity: $$ \mathbb{1}=\sum_c \vert \psi_c\rangle\langle \psi_c\vert $$ rather than explicit hermiticity.
From \begin{align} \langle\psi_b\vert H_0 r\vert\psi_a\rangle= \langle\psi_b\vert H_0\mathbb{1} r\vert\psi_a\rangle =\sum_c\langle\psi_b\vert H_0\vert\psi_c\rangle\langle \psi_c\vert r\vert\psi_a\rangle\, . \end{align} At this stage use $H_0\vert\psi_c\rangle = E_c\vert\psi_c\rangle$ to get \begin{align} \langle \psi_b\vert H_0 r\vert\psi_a\rangle &= \sum_c E_c\langle\psi_b\vert\psi_c\rangle\langle\psi_c\vert r\vert\psi_a\rangle\, ,\\ &=E_b\langle \psi_b\vert r\vert\psi_a\rangle, \end{align} where in the last step $\langle \psi_b\vert\psi_c\rangle=\delta_{bc}$ has been used to eliminate the sum.
