Perturbation using matrices [closed]

Question

A two-level system is governed by $\mathcal{H_0} = E_0 \left( {\begin{array}{cc} 2 & 0 \\ 0 & 4 \\ \end{array} } \right)$. A small perturbation $\mathcal{H^{'}} = \epsilon \left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$ is applied. So what is the first order correction to the lowest unperturbed energy.

I have tried the following. The eigenvalues $\mathcal{H_0}$ will give me the unperturbed energy, $2$ and $4$. So the eigenvalues of the matrix, $\mathcal{H_0} + \mathcal{H_{'}}$ should give me the perturbed energy. Is that right?

Answer 1

Additional Comment

The nice thing about small matrices is you can easily check the first computation using the pertubative formula $\langle \psi_n | \Delta \hat{H} | \psi_n \rangle$ with a direct expansion, and they should match!

Hint:

compute the eigenvalues directly using

$$\det [\left( {\begin{array}{cc} 2+ \epsilon a & \epsilon b \\ \epsilon c & 4+ \epsilon d \\ \end{array} } \right) - \lambda I] = 0$$

This should give you an exact quadratic for $\lambda$ which you can solve (exactly), and expand to first order in $\epsilon$. These values of $\lambda$ corrected to first order should match whatever you computed previously.

Answer 2

In theory you can obtain exactly the eigenvalues of any Hamiltonian - for a $2\times 2$ Hamiltonian, this is equivalent to solving a quadratic equation - but in practice this is computationally impossible. What perturbation theory does is to write the eigenvalues (and also eigenvectors) of a Hamiltonian of the form $\mathcal{H}_0+\epsilon\Delta\mathcal{H}$ with $\epsilon$ small as a power series in $\epsilon$.

Suppose $v$ is an eigenvector of the unperturbed Hamiltonian; then the first-order correction to the associated eigenenergy is $v\cdot\epsilon\Delta\mathcal{H}v$ (or, in bra-ket formalism, $\epsilon\left\langle v\right|\mathcal{\Delta\mathcal{H}}\left| v\right\rangle$). In this case the unperturbed energy of interest is $2E_0$ so $v=\left(\begin{array}{c} 1\\ 0 \end{array}\right)$. The first-order correction to the unperturbed energy is then $\epsilon a$.