How do I find the time evolution of a ket? I have a question which reads:
Let \begin{bmatrix}
    {E_0}       & 0 & A \\
    0       & E_1 &  0 \\
    A       & 0 & E_0 
\end{bmatrix}
be the matrix representation of the Hamiltonian for a three-state system with basis states $|1>, |2> \mbox{and } |3>$.
a.  If the state of the system at time $t$ = $0$ is $|\psi(0)>=|2>$ what is $|\psi(t)>$? 
b. If the state of the system at time $t$ = $0$ is $|\psi(0)>=|3>$ what is $|\psi(t)>$?
$\textbf{My attempt at a solution:}$
a.  For both problems we can use $|\psi(t)> = \hat{U}(t)|\psi(0)>$ where $\hat{U}= e^{\frac{-i\hat{H}t}{\hbar}}$.  Since
$$|2> = \begin{bmatrix}
           0 \\
           1 \\
           0
         \end{bmatrix}$$
is an eigenvector with eigenvalue $E_1$ we can simply replace the Hamiltonian in the time evolution operator by $E_1$, so $$|\psi(t)> = e^{\frac{-iE_1t}{\hbar}}|2> $$
Is this correct? I am finding other solutions online which have a different answer, although I can't see how this could possibly be wrong, unless my representation for $|2>$ is wrong.  
Assuming this is the correct way of doing this, I am having a hard time doing b.  I can find the eigenvalues and eigenvectors of the hamiltonian easily, and can represent |3> = $(0,0,1)^T$ as a linear combination of those vectors, thereby allowing me to operate on it.  However, my final answer is in terms of |1> and |3>, which I feel is incorrect somehow.
 A: Your reasoning is perfectly correct. Here it is in a complete form.
Let us write the Hamiltonian in the following way to make things clearer
$$ \hat{H} = E_0(|1 \rangle \langle 1|+|3 \rangle \langle 3|) + E_1|2 \rangle \langle 2| + A(|1 \rangle \langle 3| + |3 \rangle \langle 1|) $$ 
It is then straightforward to see that :


*

*$|2 \rangle$ is an eigenstate  with eigenvalue $E_1$ as you have already noticed. Hence if the initial state is $|2 \rangle$ then $$|\psi(t)\rangle = e^{-iE_1t/\hbar}|2\rangle$$

*$(|1\rangle+|3\rangle)$ and $(|1\rangle-|3\rangle)$ are eigenstates with respective eigenvalues of $E_0 + A$ and $E_0 - A$. Hence if the initial state is $|3\rangle = \frac{1}{2} [(|1\rangle+|3\rangle) - (|1\rangle-|3\rangle)]$, then $$|\psi(t)\rangle = \frac{1}{2} \left[ e^{-i(E_0+A)t/\hbar}(|1\rangle+|3\rangle) - e^{-i(E_0-A)t/\hbar}(|1\rangle-|3\rangle)\right]$$
I hope my explanation was clear !
Cheers
A: Actually I believe both answers are correct.  I can't seem to find anything wrong with either.  Certainly a. is correct since the hamiltonian in the time operator should just be replaced by the eigenvalue, seen simply if we expand the matrix exponential.
For b, there is nothing wrong with expressing our time dependent state as a linear combination of the initial state and another basis state.
