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Find the eigenvalues and eigenvectors of the $2\times2$ hermitian matrix.

$$\pmatrix{\epsilon_1&|V|e^{i\alpha}\\ |V|e^{-i\alpha}&\epsilon_2}$$

I know to find eigenvalues, you use $|A-\lambda I|$, but this is giving me difficult results to find an exact value for $\lambda$.

$V$, $\epsilon_1$, $\epsilon_2$, $\alpha$ are all constants.

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  • $\begingroup$ What is V? What is epsilon? Constants? $\endgroup$ – Vladhagen Dec 17 '13 at 8:32
  • $\begingroup$ V, epsilon, and alpha are all constants. $\endgroup$ – anonymous Dec 17 '13 at 8:39
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We can start off by solving the more general case system in order to simplify matters:

$$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$

This produces the eigenvalue / eigenvector pairs:

  • $\lambda_1 = \dfrac{1}{2} \left(-\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$
  • $v_1 = \left(\dfrac{-(-a+d+\sqrt{a^2+4 b c-2 a d+d^2})}{2 c}, 1\right)$
  • $\lambda_2 = \dfrac{1}{2} \left(\sqrt{a^2-2 a d+4 b c+d^2}+a+d\right)$
  • $v_2 = \left(\dfrac{-(-a+d-\sqrt{a^2+4 b c-2 a d+d^2})}{2 c}, 1\right)$

We can now use this result to write the eigenvalues and eigenvectors of the original system:

$$\begin{bmatrix}\epsilon_1&|V|e^{i\alpha} \\ |V|e^{-i\alpha}&\epsilon_2 \end{bmatrix}$$

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  • $\lambda_1 = \dfrac{1}{2} \left(-\sqrt{\epsilon_1^2-2 \epsilon_1 \epsilon_2+4 |V|^2 +\epsilon_2^2}+\epsilon_1+\epsilon_2\right)$
  • $v_1 = \left(\dfrac{-e^{i~\alpha}(-\epsilon_1+\epsilon_2+\sqrt{\epsilon_1^2+4 |V|^2 -2 \epsilon_1 \epsilon_2+\epsilon_2^2})}{2 |V|}, 1\right)$
  • $\lambda_2 = \dfrac{1}{2} \left(\sqrt{\epsilon_1^2-2 \epsilon_1 \epsilon_2+4 |V|^2+\epsilon_2^2}+\epsilon_1+\epsilon_2\right)$
  • $v_2 = \left(\dfrac{-e^{i~\alpha}(-\epsilon_1+\epsilon_2-\sqrt{\epsilon_1^2+4 |V|^2 -2 \epsilon_1 \epsilon_2+\epsilon_2^2})}{2 |V|}, 1\right)$
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  • $\begingroup$ All this great work...Needs an UV! +1 $\endgroup$ – Namaste Dec 18 '13 at 0:07
  • $\begingroup$ @Amzoti Did you miss a leftmost $($ in the two $v_1$ and in the two $v_2$? $\endgroup$ – Alessandro Jacopson Dec 22 '13 at 8:58

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