Bohr frequency of an expectation value?

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vector
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Homework Statement



Consider a two-state system with a Hamiltonian defined as

\begin{bmatrix}
E_1 &0 \\
0 & E_2
\end{bmatrix}

Another observable, ##A##, is given (in the same basis) by

\begin{bmatrix}
0 &a \\
a & 0
\end{bmatrix}

where ##a\in\mathbb{R}^+##.

The initial state of the system is ##\lvert\psi(0)\rangle = \lvert a_1\rangle##, where ##\lvert a_1\rangle## is the eigenstate corresponding to the larger of the two possible eigenvalues of ##A##. What is the frequency of oscillation (the Bohr frequency) of the expectation value of ##A##?

Homework Equations



Equations for finding an expectation value?

The Attempt at a Solution



I expressed ##\lvert\psi(0)\rangle = \alpha_1 \lvert E_1\rangle + \alpha_2 \lvert E_2\rangle##, and so ##\lvert\psi(t)\rangle = \alpha_1 e^{-iE_1 t/\hslash} \lvert E_1\rangle + \alpha_2 e^{-iE_2t/\hslash}\lvert E_2\rangle##.

Do I now need to find the expectation value of ##A## and then see what is in an exponent defined in terms of the difference of ##E_1## and ##E_2##? But what is the use of the fact that ##a_1## is the larger eigenvalues of the two?

I'm lost here, as I don't understand what this question actually means. I'd appreciate if someone could please clarify, preferably in detail, what one is supposed to do to solve this problem, and the exact meaning of the problem.
 
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First find the eigenvalues and the corresponding eigenvectors of ##A##. Then choose the eigenvector of the larger eigenvalue of ##A## to be ##|\psi(0)\rangle## and calculate its time evolution.
 
I've calculated the eigenvector corresponding to ##a_1## to be ##1/\sqrt{2} (1, 1)##, so I think ##\lvert \psi(0) \rangle = 1/\sqrt{2} ( \lvert E_1 \rangle + \lvert E_2 \rangle)##. So the expectation value appears to be ##1/2 (E_1+E_2)##. But how can we read the Bohr frequency from here?
 
Thanks, I managed to do the question. The Bohr frequency turned out to be ##\frac{E_2-E_1}{\hbar}##, if I was correct.
 
vector said:
Thanks, I managed to do the question. The Bohr frequency turned out to be ##\frac{E_2-E_1}{\hbar}##, if I was correct.
Yes.