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Question

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 ________.

Hamiltonian Eigenvalue Calculation

The total Hamiltonian $\hat{H}$ is the sum of the unperturbed Hamiltonian $\hat{H}_0$ and the perturbation $\hat{H}'$.

The given matrices are:

\[\hat{H}_0 = \begin{pmatrix} E & 0 & 0 \\ 0 & E & 0 \\ 0 & 0 & E \end{pmatrix}, \quad \hat{H}' = \begin{pmatrix} 0 & k & 0 \\ k & 0 & k \\ 0 & k & 0 \end{pmatrix}\]

Adding these matrices gives the total Hamiltonian:

\[\hat{H} = \hat{H}_0 + \hat{H}' = \begin{pmatrix} E & k & 0 \\ k & E & k \\ 0 & k & E \end{pmatrix}\]

Eigenvalue Determination

The energy eigenvalues $\lambda$ are found by solving the characteristic equation $\det(\hat{H} - \lambda I) = 0$, where $I$ is the identity matrix.

\[ \det \begin{pmatrix} E-\lambda & k & 0 \\ k & E-\lambda & k \\ 0 & k & E-\lambda \end{pmatrix} = 0 \]

Expanding the determinant:

\[ (E-\lambda) \left| \begin{matrix} E-\lambda & k \\ k & E-\lambda \end{matrix} \right| - k \left| \begin{matrix} k & k \\ 0 & E-\lambda \end{matrix} \right| = 0 \] \[ (E-\lambda) [(E-\lambda)^2 - k^2] - k [k(E-\lambda)] = 0 \] \[ (E-\lambda) [(E-\lambda)^2 - k^2 - k^2] = 0 \] \[ (E-\lambda) [(E-\lambda)^2 - 2k^2] = 0 \]

This equation yields three eigenvalues:

  • The first eigenvalue is found from \( E - \lambda_1 = 0 \), so \( \lambda_1 = E \).
  • The other two eigenvalues are found from \( (E-\lambda)^2 = 2k^2 \), which gives \( E - \lambda = \pm \sqrt{2}k \).
  • Thus, \( \lambda_2 = E - \sqrt{2}k \) and \( \lambda_3 = E + \sqrt{2}k \).

Perturbation Parameter k Solution

We are given that $E = 2$ eV and the maximum energy eigenvalue of $\hat{H}$ is 3 eV.

Since $k > 0$, the largest eigenvalue is $\lambda_{max} = E + \sqrt{2}k$. Setting this to the given maximum energy:

\[ 2 \text{ eV} + \sqrt{2}k = 3 \text{ eV} \]

Solving for $k$:

\[ \sqrt{2}k = 3 \text{ eV} - 2 \text{ eV} \] \[ \sqrt{2}k = 1 \text{ eV} \] \[ k = \frac{1}{\sqrt{2}} \text{ eV} \]

Calculating the numerical value and rounding to three decimal places:

\[ k = \frac{\sqrt{2}}{2} \text{ eV} \approx 0.70710678 \text{ eV} \]

Therefore, $k \approx 0.707$ eV.

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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 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?
  3. 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

  4. 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

  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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