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.

  • $\begingroup$ As an aside, note that \rangle in the MathJax produced $\rangle$. Looks nicer than $>$, IMO. $\endgroup$ – Kyle Kanos Mar 5 '17 at 13:52
  • $\begingroup$ Expressing your answer in terms of $|1\rangle$ and $|3\rangle$ is a perfectly valid response; those two vectors form a perfectly good basis for a particular subspace of the Hilbert space. Of course, if you preferred, you also could express your response in terms of any other basis of that same subspace; it would just be another way of saying the same thing. The vector $|\psi(t)\rangle$ is unique up to phase, but there are different ways to express it. $\endgroup$ – Michael Seifert Mar 6 '17 at 20:49

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 !



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.

  • $\begingroup$ Additionally, the representation for |2> IS $(0,1,0)^T$ since this state is being represented in its own basis $\endgroup$ – john morrison Mar 4 '17 at 22:07

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