0
$\begingroup$

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?

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

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