# 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 $e^{-iEt/h}\psi(x)$.

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

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