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.
Hamiltonian Ground State Eigenvector Calculation
The problem asks for the ground state eigenvector of the given unperturbed Hamiltonian $H_0$. The details of the perturbation are irrelevant for finding the eigenvector of the unperturbed system.
Unperturbed Hamiltonian Definition
The unperturbed Hamiltonian matrix is:
$ H_0 = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 5 & 0 \\ 0 & 0 & 2 \end{bmatrix} $Eigenvalue Calculation
To determine the eigenvectors, we first find the eigenvalues $ \lambda $ by solving the characteristic equation $ \det(H_0 - \lambda I) = 0 $, where $ I $ is the identity matrix.
$ \det \begin{bmatrix} 5-\lambda & 2 & 0 \\ 2 & 5-\lambda & 0 \\ 0 & 0 & 2-\lambda \end{bmatrix} = 0 $Expanding the determinant along the third row gives:
$ (2-\lambda) \det \begin{bmatrix} 5-\lambda & 2 \\ 2 & 5-\lambda \end{bmatrix} = 0 $Solving the remaining $ 2 \times 2 $ determinant:
$ (2-\lambda) \left[ (5-\lambda)(5-\lambda) - (2)(2) \right] = 0 $ $ (2-\lambda) \left[ (5-\lambda)^2 - 4 \right] = 0 $ $ (2-\lambda) \left[ 25 - 10\lambda + \lambda^2 - 4 \right] = 0 $ $ (2-\lambda) (\lambda^2 - 10\lambda + 21) = 0 $Factoring the quadratic term:
$ (2-\lambda) (\lambda - 3) (\lambda - 7) = 0 $The eigenvalues are $ \lambda_1 = 2 $, $ \lambda_2 = 3 $, and $ \lambda_3 = 7 $.
Ground State Identification
The ground state corresponds to the eigenvalue with the lowest energy. Comparing the eigenvalues $ \{2, 3, 7\} $, the minimum eigenvalue is $ \lambda_{ground} = 2 $.
Eigenvector Determination
Now, we find the eigenvector $ v = \begin{bmatrix} x \\ y \\ z \end{bmatrix} $ corresponding to the ground state eigenvalue $ \lambda = 2 $ by solving the equation $ (H_0 - 2I)v = 0 $.
$ \begin{bmatrix} 5-2 & 2 & 0 \\ 2 & 5-2 & 0 \\ 0 & 0 & 2-2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $ $ \begin{bmatrix} 3 & 2 & 0 \\ 2 & 3 & 0 \\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $This matrix equation yields the following system of linear equations:
- $ 3x + 2y = 0 $
- $ 2x + 3y = 0 $
- $ 0z = 0 $
From the first two equations, the only solution is $ x = 0 $ and $ y = 0 $. The third equation, $ 0z = 0 $, indicates that $ z $ is arbitrary (any value is possible).
Therefore, the eigenvector has the form $ \begin{bmatrix} 0 \\ 0 \\ z \end{bmatrix} $.
Selected Eigenvector
To obtain a specific eigenvector, we normalize this vector. Choosing $ z = 1 $ (any non-zero value would suffice), we get the normalized eigenvector $ \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} $. This corresponds to the ground state eigenvector.
Important Questions from Perturbation Theory Time Independent Degenerate
- If the perturbation $V = \lambda x^3$ is added to the Hamiltonian of a one-dimensional harmonic oscillator, the matrix element $\langle m|V|0 \rangle$ is/are non-zero for which of the following states? Here, the eigenstates of the harmonic oscillator are denoted by $|n\rangle$.
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
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).The ground state energy of a particle of mass $m$ in an infinite potential well is $E_0$. It changes to $E_0(1 + \alpha \times 10^{-3})$, when there is a small potential bump of height $V_0 = \frac{\pi^2 \hbar^2}{50mL^2}$ and width $a = L/100$, as shown in the figure. The value of $\alpha$ is ________ (up to two decimal places).
- 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?
