To the given unperturbed Hamiltonian
$\begin{bmatrix} 5 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix}$
we add a small perturbation given by
$\epsilon \begin{bmatrix} 1 & 1 & 1 \\ \epsilon & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix}$
where $\epsilon$ is a small quantity.
Unperturbed Hamiltonian Eigenvalues
First, find the eigenvalues of the unperturbed Hamiltonian $ H_0 = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix} $. Solve the characteristic equation $ \det(H_0 - \lambda I) = 0 $.
$ \begin{vmatrix} 5-\lambda & 2 & 0 \\ 2 & 5-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{vmatrix} = (2-\lambda)[(5-\lambda)^2 - 4] = 0 $
The unperturbed eigenvalues are $ \lambda_1^{(0)} = 2 $, $ \lambda_2^{(0)} = 3 $, and $ \lambda_3^{(0)} = 7 $.
Unperturbed Eigenvectors
Find the normalized eigenvectors corresponding to the eigenvalues $ \lambda=3 $ and $ \lambda=7 $.
- For $ \lambda_2^{(0)} = 3 $, the eigenvector is $ |v_2^{(0)}\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} $.
- For $ \lambda_3^{(0)} = 7 $, the eigenvector is $ |v_3^{(0)}\rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} $.
- For $ \lambda_1^{(0)} = 2 $, the eigenvector is $ |v_1^{(0)}\rangle = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $.
First-Order Perturbation Calculation
The perturbation is given as $ H' = \epsilon \begin{bmatrix} 1 & 1 & 1 \\ \epsilon & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} $. This can be expanded as $ H' = \epsilon M_0 + \epsilon^2 M_1 $, where $ M_0 = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} $ contains terms linear in $ \epsilon $ and $ M_1 = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ contains terms quadratic in $ \epsilon $.
First-order perturbation theory considers the Hamiltonian perturbation to first order in $ \epsilon $, which is $ H'_{pert} = \epsilon M_0 $. The first-order correction to the eigenvalues is $ \lambda_n^{(1)} = \langle v_n^{(0)} | H'_{pert} | v_n^{(0)} \rangle $.
- Correction for $ \lambda_2^{(0)} = 3 $: $ \lambda_2^{(1)} = \epsilon \langle v_2^{(0)} | M_0 | v_2^{(0)} \rangle = \epsilon \frac{1}{2} \begin{bmatrix} 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ -1 \\ 0 \end{bmatrix} = 0 $
- Correction for $ \lambda_3^{(0)} = 7 $: $ \lambda_3^{(1)} = \epsilon \langle v_3^{(0)} | M_0 | v_3^{(0)} \rangle = \epsilon \frac{1}{2} \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = 2\epsilon $
- Correction for $ \lambda_1^{(0)} = 2 $: $ \lambda_1^{(1)} = \epsilon \langle v_1^{(0)} | M_0 | v_1^{(0)} \rangle = \epsilon \begin{bmatrix} 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & -1 \\ 1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \epsilon $
Perturbed Eigenvalues
The perturbed eigenvalues are calculated as $ \lambda_n = \lambda_n^{(0)} + \lambda_n^{(1)} $.
- $ \lambda_2 = 3 + 0 = 3 $.
- $ \lambda_3 = 7 + 2\epsilon $.
- $ \lambda_1 = 2 + \epsilon $.
The pair of eigenvalues $ 3 $ and $ 7+2\epsilon $ is obtained.
Important Questions from Perturbation Theory Time Independent Degenerate
A particle of mass $m$ in an infinite potential well of width $a$ is subjected to a perturbation, $V' = \frac{h^2}{40ma^2}$ as shown in figure, where $h$ is Planck's constant.
The first order energy shift of the fourth energy eigenstate due to this perturbation is
$(\frac{h^2}{Nma^2})$
The value of $N$ is ____________ (in integer).- A particle of mass $m$ in the x-y plane is confined in an infinite two-dimensional well with vertices at $(0, 0)$, $(0, L)$, $(L, L)$, $(L, 0)$. The eigenfunctions of this particle are $\Psi_{n_x,n_y} = \sin(\frac{n_x\pi x}{L}) \sin(\frac{n_y\pi y}{L})$. If perturbation of the form $V = Cxy$, where $C$ is a real constant, is applied, then which of the following statements are correct for the first excited state?
A two-level quantum system has energy eigenvalues $E_1$ and $E_2$. A perturbing potential $H' = \lambda \Delta \sigma_x$ is introduced, where $\Delta$ is a constant having dimensions of energy, $\lambda$ is a small dimensionless parameter, and $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$. The magnitudes of the first and the second order corrections to $E_1$ due to $H'$, respectively, are
Consider a particle in a one-dimensional infinite potential well with its walls at $x = 0$ and $x = L$. The system is perturbed as shown in the figure
The first order correction to the energy eigenvalue is
Consider the Hamiltonian $\hat{H} = \hat{H}_0 + \hat{H}'$ where
\[\hat{H}_0 = \begin{pmatrix} E & 0 & 0 \\ 0 & E & 0 \\ 0 & 0 & E \end{pmatrix}\]and $\hat{H}$ is the time independent perturbation given by
\[\hat{H}' = \begin{pmatrix} 0 & k & 0 \\ k & 0 & k \\ 0 & k & 0 \end{pmatrix}\]where $k>0$. If, the maximum energy eigenvalue of $\hat{H}$ is 3 eV corresponding to $E=2$ eV, the value of $k$ (rounded off to three decimal places) in eV is ________.
