In the context of quantum optics, the rotating wave Hamiltonian can be written:
$\hbar\begin{pmatrix}
-\Delta & \Omega/2\\
\Omega/2 & 0
\end{pmatrix}$
The eigenvalues can then be calculated in the conventional way, yielding:
$\hbar\frac{1}{2} \left(-\Delta -\sqrt{\Delta ^2+\Omega ^2}\right)$ and
$\hbar\frac{1}{2} \left(-\Delta +\sqrt{\Delta ^2+\Omega ^2}\right)$.
So far so good.
My question is how do we get from here to the eigenvectors
$\{\sin(\theta),\cos(\theta)\}$ and $\{\cos(\theta),-\sin(\theta)\}$ with:
$\tan(2\theta)=-\Omega/\Delta$?
I've seen this result used many times without proof(we're only Physicists after all), but I can't find a proper justification anywhere. I assume it has something to do with sin an cos being orthonormal, but clarity would be welcome :-).
Best Answer
The matrix is real and symmetric, so it has an orthonormal set of eigenvectors. In two dimensions all the orthonormal bases (up multiple $\pm1$) are of the form $\vec{v}_1=(\cos\theta,\sin\theta)$, $\vec{v}_2=(-\sin\theta,\cos\theta)$ for some choice of $\theta$. After all, we can normalize them to have unit length, so the question is just about their direction, and as $\theta$ goes around, we get all the combos.
You just need to study the eigenvalue-equation to figure out which value of $\theta$ works for this particular matrix.