# How do I get a complex conjugation of one of the deltas?

1. Mar 11, 2014

### aaaa202

For the attached matrix A I can find the eigenvalues:

λ$\pm$ = $\pm$√($\xi$k2-lΔl2)
But when I try to solve for the eigenvectors I get it wrong:
I solve:
(A-λ$\pm$)(x,y) = (0,0)
gives the eigenvectors:

e$\pm$=(Δ,$\xi$k$\pm$√($\xi$k2-lΔl2))
But these eigenvectors are not orthogonal. How do I get a complex conjugation of one of the deltas?

#### Attached Files:

• ###### Matrix.png
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2. Mar 11, 2014

### BvU

Hello Manya,

Forgive me for asking, but why does the distinction between $\xi_k$ and $\xi_{-k}$ disappear when you state the eigenvalues ?

3. Mar 11, 2014

### Dick

If you ignore the distinction between k and -k that BvU points out, then you should get that the eigenvalues are $\pm \sqrt{\xi_k^2+|\Delta|^2}$ which in turn gives the eigenvectors $\left( \Delta, \xi_k \pm \sqrt{\xi_k^2 + |\Delta^2|} \right)$. So you are making some algebra mistake as well. Show your work. Those ARE orthogonal. The complex conjugation comes from the inner product. The inner product of two complex vectors is one dotted with the conjugate of the other. Now try it paying attention to the difference between +k and -k.