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I do not understand why if $H\psi = E\psi$, then the time-evolution of the wavefunction is given by $e^{-iEt/h}\psi(x)$.

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    $\begingroup$ That formula for the time dependence is only valid for eigenstates, and only for time-independent Hamiltonians. $\endgroup$ – zeldredge Jun 2 '15 at 20:56
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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.

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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 $\psi(t=0)\mapsto\psi(t)$.

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