# Evolution of the propagator in the Interaction picture?

The evolution operator in the interaction picture is defined as

$$U_I=e^{iH_0t}e^{-iH_St}e^{-iH_0t}$$

Where $$H_S=H_0+V$$

I am trying to find the evolution of the operator $$U_I$$. In literature it is given by:

$$i\dfrac{dU_I}{dt}=V_IU_I$$.

However when I take the derivative of the operator:

\begin{align}i\dfrac{dU_I}{dt}&=i^2H_0e^{iH_0t}e^{-iH_St}e^{-iH_0t}+ie^{iH_0t}(-iH_s)e^{-iH_St}e^{-iH_0t}+ie^{iH_0t}e^{-iH_St}(-iH_0)e^{-iH_0t}\\ &=-H_0e^{iH_0t}e^{-iH_St}e^{-iH_0t}+e^{iH_0t}(H_0+V)e^{-iH_St}e^{-iH_0t}+e^{iH_0t}e^{-iH_st}H_0e^{-iH_0t}\end{align}

Since $$H_0$$ and $$e^{iH_0t}$$ commute, the first two terms cancel and we are left with

\begin{align} i\frac{dU_I}{dt}&=e^{iH_0t}Ve^{-iH_St}e^{-iH_0t}+e^{iH_0t}e^{-iH_st}H_0e^{-iH_0t}\\&=e^{iH_0t}Ve^{-iH_0t}e^{iH_0t}e^{-iH_St}e^{-iH_0t}+e^{iH_0t}e^{-iH_st}H_0e^{-iH_0t}\\&=V_IU_I+e^{iH_0t}e^{-iH_st}H_0e^{-iH_0t} \end{align}

In the second line I inserted $$1=e^{-iH_0t}e^{iH_0t}$$, so I arrive at the correct term, $$V_IU_I$$ but + some extra term. What is going wrong here?