The problem asks for the first-order energy eigenvalues of a quantum system subjected to a perturbation. The unperturbed Hamiltonian is degenerate.
Hamiltonian and Perturbation Details
The unperturbed Hamiltonian is given by:
$H_0 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
The eigenvalues of $H_0$ are $E^{(0)}=1$, with a degeneracy of 2. The standard orthonormal eigenvectors are $|1\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$ and $|2\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$.
The perturbation Hamiltonian is:
$H' = \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix}$
where $\delta \ll 1$.
Degenerate Perturbation Theory Application
Since the unperturbed energy level $E^{(0)}=1$ is degenerate, we must use degenerate perturbation theory to find the first-order energy corrections. This involves finding the eigenvalues of an effective Hamiltonian matrix constructed within the degenerate subspace.
Effective Hamiltonian Matrix Construction
The matrix elements of the perturbation $H'$ in the basis of the degenerate subspace (spanned by $|1\rangle$ and $|2\rangle$) are calculated as follows:
- $H'_{11} = \langle 1 | H' | 1 \rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \delta$
- $H'_{12} = \langle 1 | H' | 2 \rangle = \begin{pmatrix} 1 & 0 \end{pmatrix} \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \delta$
- $H'_{21} = \langle 2 | H' | 1 \rangle = \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \delta$
- $H'_{22} = \langle 2 | H' | 2 \rangle = \begin{pmatrix} 0 & 1 \end{pmatrix} \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \delta$
The effective Hamiltonian matrix $H'_{eff}$ in this subspace is:
$H'_{eff} = \begin{pmatrix} \delta & \delta \\ \delta & \delta \end{pmatrix}$
Secular Equation and Energy Corrections
The first-order energy corrections, $E^{(1)}$, are the eigenvalues of $H'_{eff}$. We solve the secular equation $\det(H'_{eff} - E^{(1)} I) = 0$:
$\det \begin{pmatrix} \delta - E^{(1)} & \delta \\ \delta & \delta - E^{(1)} \end{pmatrix} = 0$
$(\delta - E^{(1)})^2 - \delta^2 = 0$
$(\delta - E^{(1)})^2 = \delta^2$
$\delta - E^{(1)} = \pm \delta$
This yields two possible values for the energy correction:
- $E^{(1)}_1 = \delta - \delta = 0$
- $E^{(1)}_2 = \delta - (-\delta) = 2\delta$
Final Perturbed Energy Eigenvalues
The total energy eigenvalues are the sum of the unperturbed energy and the first-order correction: $E = E^{(0)} + E^{(1)}$.
The perturbed energy eigenvalues are:
- $E_1 = 1 + 0 = 1$
- $E_2 = 1 + 2\delta$
Therefore, the energy eigenvalues of the perturbed system, using first-order perturbation approximation, are $1$ and $(1+2\delta)$.
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?
