Consider the Hamiltonian $$ \left[ \begin{matrix} E_1 & -A\\ -A& E_2\\ \end{matrix} \right] $$

where $A$, $E_1,E_2$ are real numbers. I have seen a different formula to calculate the eigenvalues of this matrix, here it is:

$$\lambda = \frac{E_1 +E_2}{2} \pm \sqrt{\frac{(E_1 - E_2)^2}{4} + A^2}.$$

To demonstrate it, I have written:

$$\begin{equation} H = \frac{E_1 +E_2}{2}I + \frac{E_1 - E_2}{2} \sigma_z -A\sigma_x \end{equation}$$ where $I$ is the identity and $\sigma_x, \sigma_z$ are the Pauli matrices.

So, the part $\frac{E_1 +E_2}{2}$ comes from the eigenvalues of the identity,but how do you obtain the second part?


closed as off-topic by Aaron Stevens, Norbert Schuch, Qmechanic Jul 9 at 15:34

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  • 1
    $\begingroup$ There is no easy way to calculate the eigenvalues of a matrix. The straight-forward way is to solve the equation $\text{det}(\lambda I - H)=0$. In your case, this will lead to a quadratic equation for $\lambda$ with the 2 solutions given in your question. $\endgroup$ – Thomas Fritsch Jul 9 at 12:09
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    $\begingroup$ I think it should be $$\lambda = \frac{E_1+E_2}{2}\pm\sqrt{\frac{(E_1-E_2)^2}{4}+A^2}$$ The square of the difference under the root is missing. $\endgroup$ – denklo Jul 9 at 12:15
  • $\begingroup$ I fixed the missing square, thank you. I know that I can solve the problem by calculating det($\lambda I - H$)=0, but I was interested in understanding this different approach with the Pauli matrices. $\endgroup$ – BlackPhoenix Jul 9 at 13:13
  • $\begingroup$ @ThomasFritsch What do you mean there is no easy way to calculate eigenvalues? $\endgroup$ – Aaron Stevens Jul 9 at 13:29
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    $\begingroup$ So, is the question how to get from your 3rd equation (H=...) to your 2nd equation (λ=...) without calculating the characteristic polynomial? $\endgroup$ – Noiralef Jul 9 at 14:32

Eigenvectors of the hamiltonian are null vectors $\psi$ of $H-\lambda \mathbb{I}$ for some eigenvalues $\lambda$.

So $$ (H-\lambda \mathbb{I})\psi =0=\left ( \frac{E_1 +E_2-2\lambda}{2}\mathbb{I} + \frac{E_1 - E_2}{2} \sigma_z -A\sigma_x \right )\psi \qquad \Longrightarrow \\ \left (\frac{E_1 - E_2}{2} \sigma_z -A\sigma_x \right)\psi=-\left ( \frac{E_1 +E_2-2\lambda}{2}\mathbb{I}\right)\psi \qquad \Longrightarrow \\ \left (\frac{E_1 - E_2}{2} \sigma_z -A\sigma_x \right)^2\psi=-\left ( \frac{E_1 +E_2-2\lambda}{2}\mathbb{I} \right ) \left (\frac{E_1 - E_2}{2} \sigma_z -A\sigma_x \right )\psi =\left ( \frac{E_1 +E_2-2\lambda}{2}\mathbb{I}\right)^2\psi \qquad \Longrightarrow \\ \left (A^2 +\left (\frac{E_1 - E_2}{2}\right )^2 \right )\psi =\left ( \frac{E_1 +E_2-2\lambda}{2}\right )^2\psi , $$ since each Pauli matrix squares to the identity and the two anticommute.

You then have your algebraic equation $$ \pm \sqrt{A^2 +\frac{(E_1 - E_2)^2}{4}} = \frac{E_1 +E_2}{2} -\lambda $$ without determinants, etc... Perhaps hardly worth it.


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