- #1

Lambda96

- 189

- 65

- Homework Statement
- Calculate the eigenvalues and eigenvectors of the Hamilton operator

- Relevant Equations
- none

Hi,

unfortunately, I am not sure if I have calculated the task a correctly.

I calculated the eigenvalues with the usual formula ##\vec{0}=(H-\lambda I) \psi## and got the following results

$$\lambda_1=E_1=-\sqrt{B^2+\nabla^2}$$

$$\lambda_2=E_2=\sqrt{B^2+\nabla^2}$$

I'm just not sure about the normalized eigenvectors, I got the following as eigenvectors

$$\vec{\psi}_1= \left(\begin{array}{c} \frac{B- \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$

$$\vec{\psi}_2= \left(\begin{array}{c} \frac{B+ \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$

For the normalization, I must divide the vectors by their norm. Since writing this vector via Latex is very time-consuming, I have calculated this with Mathematica, instead of the triangle, I have simply written an A. For the eigenvector ##\vec{\psi_1}## I got the following:

Since this eigenvector looks very messy, I am not sure if I have calculated the task correctly now.

unfortunately, I am not sure if I have calculated the task a correctly.

I calculated the eigenvalues with the usual formula ##\vec{0}=(H-\lambda I) \psi## and got the following results

$$\lambda_1=E_1=-\sqrt{B^2+\nabla^2}$$

$$\lambda_2=E_2=\sqrt{B^2+\nabla^2}$$

I'm just not sure about the normalized eigenvectors, I got the following as eigenvectors

$$\vec{\psi}_1= \left(\begin{array}{c} \frac{B- \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$

$$\vec{\psi}_2= \left(\begin{array}{c} \frac{B+ \sqrt{B^2+\triangle^2}}{\triangle} \\ 1 \end{array}\right)$$

For the normalization, I must divide the vectors by their norm. Since writing this vector via Latex is very time-consuming, I have calculated this with Mathematica, instead of the triangle, I have simply written an A. For the eigenvector ##\vec{\psi_1}## I got the following:

Since this eigenvector looks very messy, I am not sure if I have calculated the task correctly now.