Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 :-).

share|improve this question
1  
Cross-posted to physics.stackexchange.com/q/74061/2451 –  Qmechanic Aug 13 '13 at 13:07
add comment

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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 $\theta$? –  TCTopCat Aug 13 '13 at 12:58
1  
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 $\pm\vec{v}_2$. –  Jyrki Lahtonen Aug 13 '13 at 13:02
1  
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. $\tan\theta$. –  Jyrki Lahtonen Aug 13 '13 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. –  Jyrki Lahtonen Aug 13 '13 at 13:09
2  
I don't think there's any harm in leaving up alternate answers - might help if it's easier for someone who thinks differently. –  TCTopCat Aug 13 '13 at 13:30
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.