quantum mechanics - Density operator as a function of time

Given the density operator $\rho = \sum_iw_i | \alpha^{i} \rangle \langle \alpha^{i}|$, how does the density operator change with time. Apparently I should get $$i \hbar \frac{\partial \rho}{\partial t} = \sum_{i}w_i(H| \alpha^{i}(t) \rangle \langle \alpha^{i}(t)| - | \alpha^{i}(t) \rangle \langle \alpha^{i}(t)|H).$$ I am having difficulty getting this, it seems that I have to use Shrodingers equation on the $(| \alpha \rangle \langle \alpha|)$ since the intial population $w_i$ is constant in time, but I'm not sure how to differentiate this since $\langle \alpha|$ as I understand is a bra which is a functional in a sense, how does the product rule for differentiation apply then in this case?

Thanks.

Answer

Start with \begin{align} i\hbar \frac{d}{dt}\vert{\alpha}\rangle &=H\vert\alpha\rangle \end{align} and take the adjoint $$ -i\hbar \frac{d}{dt}\langle {\alpha}\vert =\langle \alpha\vert H $$ where $H^\dagger=H$ has been used. Then simply use the product rule: \begin{align} i\hbar\frac{d}{dt} \left[\vert \alpha\rangle\langle\alpha\vert \right]&= \left[i\hbar\frac{d}{dt}\vert \alpha\rangle\right]\langle\alpha\vert + \vert\alpha\rangle\left[i\hbar\frac{d}{dt}\langle \alpha\vert\right]\, ,\\ &=\left[H\vert\alpha\rangle\right]\langle\alpha\vert - \vert\alpha\rangle \left[ \langle\alpha\vert H\right]\, . \end{align}