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The Hamiltonian for a three-state system is, in some basis $|1\rangle ,|2\rangle,|3\rangle$

$$\hat{H}= \left( \begin{array}{ccc} E_0 & 0 & A \\ 0 & E_1 & 0 \\ A & 0 & E_0 \end{array} \right) $$

Let $$|1\rangle =\left( \begin{array}{ccc}1 \\ 0 \\ 0 \end{array} \right)\,, $$ $$|2\rangle =\left( \begin{array}{ccc}0 \\ 1 \\ 0 \end{array} \right) \,,$$ and $$|3\rangle =\left( \begin{array}{ccc}0 \\ 0 \\ 1 \end{array} \right)\,. $$ $E_0, E_1,$ and $A$ are real constants.

If at $t=0$ the system is $|\psi(0)\rangle=|2\rangle,$ what is $|\psi(t)\rangle\;?$ I would like to express this answer in the $|1\rangle ,|2\rangle,|3\rangle$ basis.

So far, I have considered that states 1, 2, and 3 are the eigenstates of the Hamiltonian. I ended up finding their respective eigenvalues to be $E_0+A$, $E_1,$, and $E_0+A$. I'm not sure if this is correct, but I am not sure how to get from knowing $|\psi(0)\rangle$ to knowing $|\psi(t)\rangle\;.$ Could anyone point me in the right direction?

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closed as off-topic by Norbert Schuch, Sebastian Riese, Kyle Kanos, Qmechanic Feb 22 '16 at 19:12

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So, first states 1,2 and 3 are not the eigenstates of the Hamiltonian. To see this, try carrying out this operation

$$ \hat{H} |1\rangle = E |1\rangle $$ What you will find is that this is NOT true. What you should find is that you get a combination of states 1 and 3 when you do this. So the question is what are the eigenstates? The best way to determine this is through the equation $$ (\hat{H} - \hat{1}E_i)|k\rangle = 0 $$ where $k$ goes from 1 to 3 and each $k$ is some 3 component vector.

After solving this equation, you should have a set of eigenstates defined in terms of your initial states. Something like this $$ |k_i\rangle = \sum_{i=1}^3c_{k,i} |i\rangle $$ Or, we can write the initial vectors in terms of the eigenvectors as $$ |i\rangle = \sum_{k=1}^3c^\dagger_{k,i} |k\rangle $$ You can now apply your time evolution operator to your initial state $$ e^{i\hat{H}t/\hbar}|\psi(0)\rangle = e^{i\hat{H}t/\hbar}|2\rangle = \sum_{k=1}^3c^\dagger_{k,2} e^{i\hat{H}t/\hbar}|k\rangle = \sum_{k=1}^3c^\dagger_{k,2}e^{iE_kt/\hbar}|k\rangle $$

So, why was it necessary to find the eigenstates? Specifically because $$ e^{i\hat{H}t/\hbar}|k\rangle = e^{iEt/\hbar}|k\rangle $$ This relationship only applies to eigenstates, thus if you want to determine the time evolution of any initial configuration, you must first write it in terms of eigenstates.

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