# Interaction picture Sakurai

I’m going through Sakurai and got stuck with the following in the interaction picture subsection

$$i \hbar \frac{\partial}{\partial t}\left|\alpha, t_{0} ; t\right\rangle_{I}=i \hbar \frac{\partial}{\partial t}\left(e^{i H_{0} t / \hbar}\left|\alpha, t_{0} ; t\right\rangle_{S}\right)$$

$$=-H_{0} e^{i H_{0} t / \hbar}\left|\alpha, t_{0} ; t\right\rangle_{S}+e^{i H_{0} t / \hbar}\left(H_{0}+V\right)\left|\alpha, t_{0} ; t\right\rangle_{S}$$

$$= e^{i H_{0} t / \hbar} V e^{-i H_{0} t / \hbar} e^{i H_{0} t / \hbar}\left|\alpha, t_{0} ; t\right\rangle_{S}$$

What is the justification for going from line 1 to 2? It honestly just looks like they added and subtracted the $$H_0$$ term, but that’s the term that would come from the derivative. Any help greatly appreciated

The first term in the second line comes from $$i\hbar\frac{\partial}{\partial t}e^{iH_0t} = -H_0e^{iH_0t}$$ The second term comes from the time dependent Schrodinger equation $$i\hbar\frac{\partial}{\partial t}|\alpha, t\rangle= (H_0+V)|\alpha, t\rangle.$$