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Question

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?

Unperturbed Energy of a Particle in a 2D Infinite Well

The energy eigenvalues for a particle of mass $m$ in a 2D infinite square well of side length $L$ are given by:

$ E_{n_x,n_y} = \frac{\pi^2\hbar^2}{2mL^2}(n_x^2 + n_y^2) $

where $n_x$ and $n_y$ are positive integers ($n_x, n_y = 1, 2, 3, \dots$).

Identifying the First Excited State

The ground state corresponds to $(n_x, n_y) = (1, 1)$, with energy $E_{1,1} = \frac{\pi^2\hbar^2}{2mL^2}(1^2+1^2) = \frac{2\pi^2\hbar^2}{2mL^2}$.

The first excited state occurs for the next lowest energy values. These correspond to the quantum numbers $(n_x, n_y) = (1, 2)$ and $(2, 1)$.

The energy for these states is:

$ E_{1,2} = E_{2,1} = \frac{\pi^2\hbar^2}{2mL^2}(1^2 + 2^2) = \frac{\pi^2\hbar^2}{2mL^2}(1 + 4) = \frac{5\pi^2\hbar^2}{2mL^2} $

Degenerate Perturbation Theory for the First Excited State

The first excited state is doubly degenerate, meaning the states $\Psi_{1,2}$ and $\Psi_{2,1}$ have the same unperturbed energy $E_0 = \frac{5\pi^2\hbar^2}{2mL^2}$.

A perturbation $V = Cxy$ is applied. For degenerate states, the first-order energy corrections are found by diagonalizing the perturbation matrix within the subspace of degenerate states.

The relevant basis states are $\Psi_{1,2}$ and $\Psi_{2,1}$. The perturbation Hamiltonian matrix elements are:

  • Diagonal elements: $a = \langle \Psi_{1,2} | Cxy | \Psi_{1,2} \rangle = \langle \Psi_{2,1} | Cxy | \Psi_{2,1} \rangle$.
  • Off-diagonal elements: $b = \langle \Psi_{1,2} | Cxy | \Psi_{2,1} \rangle = \langle \Psi_{2,1} | Cxy | \Psi_{1,2} \rangle$.

Due to the symmetry of the wavefunctions and the perturbation, the diagonal elements are equal, and the off-diagonal elements are equal (and real, given $C$ is real).

Secular Equation for Energy Shifts

The energy shifts $\delta$ are the eigenvalues of the matrix representing the perturbation in the degenerate subspace. This leads to the secular equation:

$ \begin{vmatrix} a-\delta & b \\ b & a-\delta \end{vmatrix} = 0 $

Here, $a$ represents the expectation value of the perturbation in each degenerate state, and $b$ represents the coupling between the degenerate states via the perturbation. If $C \ne 0$, both $a$ and $b$ are real and non-zero constants. This equation correctly determines the energy shifts.

The energy shifts are not generally zero, as determined by the eigenvalues of this matrix ($a \pm b$).

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Important Questions from Perturbation Theory Time Independent Degenerate

  1. 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).

  2. 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

  3. 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

  4. 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 ________.

  5. An electric field $\vec{E} = E_0 \hat{z}$ is applied to a Hydrogen atom in $n = 2$ excited state. Ignoring spin, the $n = 2$ state is fourfold degenerate, which in the $|l,m\rangle$ basis are given by $|0,0\rangle, |1,1\rangle, |1,0\rangle$ and $|1,-1\rangle$. If $H'$ is the interaction Hamiltonian corresponding to the applied electric field, which of the following matrix elements is nonzero?
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