How does a state in quantum mechanics evolve? I have a question about the time evolution of a state in quantum mechanics. The time-dependent Schrodinger equation is given as
$$
  i\hbar\frac{d}{dt}|\psi(t)\rangle = H|\psi(t)\rangle
$$
I am self-learning the quantum mechanics so there are few conceptual questions confused me. I am reading a book, in which the author emphasizes that any state could be expanded in terms of the eigenstates and eigenvalues of Hamiltonian. So does it mean in above equation, it is not requiring $\psi(t)$ to be an eigenstate (i.e. it could be anything)?
I am wondering it is only true when we are trying to expanding a steady state in terms of the eigenstates, that is
$$
  c_n(0) = \langle E_n|\psi(0)\rangle
$$
But if the state is time-dependent, the expanding coefficient $c_n$ is not longer stationary but evolving in time, the book gives
$$
  c_n(t) = c_n(0)\exp(-iE_nt)
$$
where $E_n$ is the eigenvalue. So the time evolution of the state is 
$$
  |\psi(t)\rangle = \sum_n c_n(t)|E_n\rangle
$$
is it correct? I understand the math. But physically, how do we understand the initial coefficient $c_n(0)$ evolves in the exponential form? and why the exponential has eigenvalue $E_n$? It looks to me that different eigenvalue plays different role in time evolution, why is that? If we look into the math, seems that the contribution of the eigenstate with higher 'energy' (eigenvalue) is decreasing faster in time, what's the physical reason for that?
The last question is even confusing. The text said the system could has many different states. But when you measure it, only one of the eigenstate will be shown and the measured value will be the corresponding eigenvalue. I don't understand this statement but just assume it is true. So if the state of the system is initially $|\psi(0)\rangle$ and it evolves in time as $|\psi(t)\rangle$. I measure the system at time $t=\tau$, so at this time, is it still true that the outcome must be one of the eigenvalues even the state is time-dependent?
 A: First, you asked if wavefunctions need not be eigenfunctions of the Hamiltonian.  That is correct; you can view this through a Fourier decomposition.  Each eigenfunction of the Hamiltonian has a different energy, and hence a different frequency (the two are essentially equivalent through $E = \hbar \omega$), and these frequencies determine how each eigenfunction evolves in time.  Still, it's perfectly legal to add up linear combinations of eigenfunctions together to construct an arbitrary wavefunction--and we often do this.
I think a good way to visualize it is to look at each eigenfunction of the Hamiltonian as precessing on the complex plane.  It is the eigenvalue (the energy, or the frequency) that determines how quickly that eigenfunction precesses over time.
At one point, you mention you thought states with higher energies were decreasing faster in time.  Complex exponentials don't go like that.  They're like sine and cosine waves, or you can imagine them (as I have) as simply tracing out circles on the complex plane.  Higher energy states simply trace out these circles faster.
Your final question gets at a matter that's more one of convention.  Remember that a complex multiple of an eigenstate is still an eigenstate, and as such, even if an eigenstate has evolved in time, the evolved state is still an eigenstate of the Hamiltonian.  So if an evolved wavefunction is collapsed to an evolved eigenstate, the overall idea that the result is an eigenvalue times an eigenstate still applies.
