Show that [hat{q}left(t_{0}+Delta t ight)=e^{frac{i}{hbar} hat{H} Delta t} hat{q}left(t_{0} ight) e^{-frac{i}{hbar} hat{H} Delta t}] where we define

Show that

\[\hat{q}\left(t_{0}+\Delta t\right)=e^{\frac{i}{\hbar} \hat{H} \Delta t} \hat{q}\left(t_{0}\right) e^{-\frac{i}{\hbar} \hat{H} \Delta t}\]

where we define

\[e^{-\frac{i}{\hbar} \hat{H} \Delta t}=\sum_{n=0}^{\infty} \frac{1}{n !}\left(\frac{-i \hat{H}}{\hbar}\right)^{n}\]

implies

\[\hat{q}\left(t_{0}+\Delta t\right)=\hat{q}\left(t_{0}\right)+(-i) \frac{\Delta t}{\hbar}[\hat{q}, \hat{H}]+(-i)^{2} \frac{1}{2 !} \frac{\Delta t^{2}}{\hbar^{2}}[[\hat{q}, \hat{H}], \hat{H}]+\cdots\]

Do this by showing the pattern for the first few terms only.