Time evolution of a wavepacket I do not understand why if $H\psi = E\psi$, then the time-evolution of the wavefunction is given by $e^{-iEt/h}\psi(x)$.
 A: The equation you've given is the time-independent Schrodinger equation. The time-dependent Schrodinger equation is
$$i \hbar \dot{\psi} = H \psi $$
which you can readily solve to obtain the time-dependence (when $\psi$ is an eigenstate) that you asked for in the question. More generally, you can solve it to show that $\exp\left(-i\hat{H}t/\hbar\right)$ is the generator of time evolution, i.e. the operator that takes $\psi(t=0)\mapsto\psi(t)$.
A: Solving time-dependent SE as danielsmw mentioned above good starting point.
$$
i\hbar\frac{\partial\psi}{\partial t}=H\psi
$$
$$
\frac{d\psi}{\psi}=\frac{H}{i\hbar}dt
$$
$$
log(\psi)\mid^{\psi}_{\psi_{0}}=-\frac{iHt}{\hbar}\mid^{t}_{t_0}
$$
suppose $\psi=|\alpha,t>$ and $\psi_{0}=|\alpha, t_{0}>$
$$
\psi=e^{-\frac{iH(t-t_{0})}{\hbar}}\psi_0
$$
Under infinitesimal changes of time (dt), time evolution operator,
$$
 lim_{dt\to 0}U(t_{0}+dt, t_{0})=1
$$
$$
U(t_{0}+dt, t_{0})=1-iGdt
$$
for $G=\frac{H}{\hbar}$ ,
$$
U(t_{0}+dt, t_{0})=1-i\frac{H}{\hbar}dt \sim exp(-i\frac{H}{\hbar}dt )
$$
For a good guidance I highly recommend J.J.Sakurai Chapter 2.
