# Arbitrarily using Sin and Cos as eigenfunctions of a Hamiltonian?

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 :-).

## 1 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.

• Thanks, can you explain why all of the orthonormal bases have this property or point me to some reference? Also, how would I systematically determine $\theta$? Commented Aug 13, 2013 at 12:58
• When $\theta$ ranges over $[0,2\pi)$, $\vec{v}_1=(\cos\theta,\sin\theta)$ ranges over all the vectors of length $1$. With $\vec{v}_1$ chosen, the other eigenvector has to be orthogonal to it, so it has to be $\pm\vec{v}_2$. Commented Aug 13, 2013 at 13:02
• To find the value of $\theta$ just look at the equation $$A\pmatrix{\cos\theta\cr\sin\theta\cr}=\lambda\pmatrix{\cos\theta\cr\sin\theta},$$ where $A$ is your matrix and $\lambda$ is (one of) the eigenvalue. You get a homogeneous linear equation in the "unknowns" $\cos\theta$ and $\sin\theta$, and can solve for their ratio, i.e. $\tan\theta$. Commented Aug 13, 2013 at 13:05
• Replacing $\theta$ by $\theta+\pi/2$ then gives you the other eigenvector. Those two angles give the same result for $\tan2\theta$, so apparently that helps. I haven't thought about the details. Commented Aug 13, 2013 at 13:09
• I don't think there's any harm in leaving up alternate answers - might help if it's easier for someone who thinks differently. Commented Aug 13, 2013 at 13:30