# Diagonalizing a matrix using perturbation theory.

1. Oct 9, 2011

### PsychoDash

1. The problem statement, all variables and given/known data

Consider the following Hamiltonian.

H=$\begin{pmatrix} 20 & 1 & 0 \\1 & 20 & 2 \\0 & 2 & 30 \end{pmatrix}$
Diagonalize this matrix using perturbation theory. Obtain eigenvectors (to first order) and eigenvalues (to second order).

Ho=$\begin{pmatrix} 20 & 0 & 0 \\0 & 20 & 0 \\0 & 0 & 30 \end{pmatrix}$
H'=$\begin{pmatrix} 0 & 1 & 0 \\1 & 0 & 2 \\0 & 2 & 0 \end{pmatrix}$

2. Relevant equations

3. The attempt at a solution

In general, diagonalizing a matrix involves finding its eigenvalues and then writing the eigenvalues on the diagonal with zeros elsewhere. Despite that, I'm just not sure how to approach this question.

2. Oct 9, 2011

### vela

Staff Emeritus
How do you calculate corrections in perturbation theory? What equations do you have? Do you have degenerate states?

It's really just a matter of figuring out which formulas you need to use and plugging everything in. You can get the needed matrix elements by inspection.

3. Oct 9, 2011

### PsychoDash

That's what's throwing me off. There is no "physical system" as I have come to understand it. All I'm given is what I wrote above. I understand that to calculate perturbations in general, you use <Psi|H'|Psi>, but that gets me back to needing a wavefunction to operate on. All I have is this set of matrices.

4. Oct 9, 2011

### vela

Staff Emeritus
You don't need a wave function for the system. You need the eigenstates of the unperturbed Hamiltonian. Let's call those $\vert 1 \rangle$, $\vert 2 \rangle$, and $\vert 3 \rangle$. The matrix you've been given is the representation of $\hat{H}$ relative to that basis. In other words,
$$\begin{pmatrix} \langle 1 \lvert \hat{H} \rvert 1 \rangle & \langle 1 \lvert \hat{H} \rvert 2 \rangle & \langle 1 \lvert \hat{H} \rvert 3 \rangle \\ \langle 2 \lvert \hat{H} \rvert 1 \rangle & \langle 2 \lvert \hat{H} \rvert 2 \rangle & \langle 2 \lvert \hat{H} \rvert 3 \rangle \\ \langle 3 \lvert \hat{H} \rvert 1 \rangle & \langle 3 \lvert \hat{H} \rvert 2 \rangle & \langle 3 \lvert \hat{H} \rvert 3 \rangle \end{pmatrix} = \begin{pmatrix} 20 & 1 & 0 \\1 & 20 & 2 \\0 & 2 & 30 \end{pmatrix}$$
Does that clear things up?

5. Oct 9, 2011

### PsychoDash

How does that help diagonalize H?