On page 96 of his book, Griffiths explains that determinate states of some observable $Q$ are eigenfunctions of that operator. So if a particle starts out in that state it will continue to be in that state as long as a measurement of an observable is not being made. This is all well and good and seemed to make sense as I went through the book but then I encountered an example (not in Griffiths) of a spin $1$ particle which starts out in the spin state $(1,1)$ which is an eigenstate of $S_z$ and evolved out of that state (with the Hamiltonian being $H = kS_x$ where $k$ is a constant) when $t > 0$. But how can that be? we know that if particle is in a determinate state it should remain in that state for all time unless a measurement is made.
More generally I considered the following scenario: Suppose you have a definite state of Angular momentum, call it $\psi(0)$ [we will consider it the initial condition which we need to evolve in time]. Then (suppose for simpicity) you can expand this definite state in terms of two eigenstates of the Hamiltonian so:
$\psi(0) = aE_1 + bE_2$
But then to obtain the state at $t>0$ we just tack on the wiggle factor corresponding to each energy eigenstate and we can easily see that the state will evolve out of our initial angular momentum eigenstate! So again: How come that be given that it was a determinate state of an observable.
The conclusion I came to was that given any observable $Q$, it will have determinate states only if they are also energy eigenstates, i.e. only if $Q$ and $H$ are compatible observables in which case they will have a common set of eigenfunctions. But there's not even a hint of that in Griffiths who explicitly defines determinate states as eigenfunctions of observables regardless of whether or not they commute with $H$. So given an observable $Q$, get the eigenfunction and you're done: you got the determinate states. But that contradicts what I've stated above so I must be missing something.