Collapse of Wavefunction, and Subsequent Time Evolution To keep it simple, suppose the system is the well-known particle in a 1D infinite potential well.
Suppose the wavefunction is  $ a|1\rangle +  b|2 \rangle +  c|3\rangle$, where the $|i\rangle$ are energy eigenfunctions.
Suppose I measure the energy, and I get $E_1$.  
Does this mean that after the measurement, the wavefunction becomes  $ |1 \rangle$?  This is my understanding of "collapsing the wavefunction."
What happens after the measurement?
The propagator (from Shankar) is the sum over $n$ of:
$$ |n \rangle \langle n| \exp\left(\frac{-i E_n t} {\hbar} \right)  $$
If I apply this to $ |1\rangle$, only the inner product for $ n = 1$ survives, so the answer is $  |1\rangle \exp\left(\frac{-i E_1 t} {\hbar} \right)$.  
So it stays in state $|1\rangle$  forever?
 A: The expression of "collapse of the wavefunction" means that you are taking a measurement which means a change in the interactions and boundary conditions, which means a new wavefunction. 
Let us take the hydrogen atom. How would you check that it is in the ground state? By sending a photon that would change it from the ground state to a higher state,(a different wave function) and getting the photon coming back down to the ground state. The wavefunction changes with each interaction, or decay, into another wavefunction fulfilling different boundary conditions. The original "collapses" , i.e. no longer describes the  probabilities of the system. 
A: In your specific example the state collapses to the energy eigenstate $\vert 1\rangle$.  Since energy eigenstates are stationary, i.e. they evolve as $e^{-i E_n t/\hbar}\vert b\rangle$, and since there is only one
state with energy $E_1$ in your system, the system will remain in 
$\vert 1\rangle$ once it has been measured to have energy $E_1$. 
I interpret "ideal" in a slightly different way, as meaning that your outcome is $E_1$ without possibility of errors.  A non-ideal detector would have some (presumably small) probability of "confusing" $E_2$ with $E_1$, in which case the system would have a (presumably small) probability of being "measured" as having energy $E_1$ but being projected to the $\vert 2\rangle$ eigenstate.  This could happen for instance if the energy resolution of your detector was larger than the energy difference $\vert E_2-E_1\vert$.  Obviously if the energy levels are separated by a few $eV$s and your detector has resolution of $0.01$ eV, your measurements will be close to ideal.
A: To expand on @Geoffrey's comment and answer your question:
Yes, in this particular case the system stays in the state |1> forever (or until some other interaction comes along and the Hamiltonian changes!)
This is because the state |1> is an energy eigenstate. 
If instead you would have made a measurement of a conjugate variable, then your state would have temporarily collapsed into that state. But as you see from the propagator, that will have some expansion in multiple energy eigenstates, so the state will evolve as a superposition of eigenstates with the coefficient for an given eigenstate changing with time.
