Consider the Hamiltonian $\hat{H} = \hat{H}_0 + \hat{H}'$ where and $\hat{H}$ is the time independent perturbation given by 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 ________.
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:
\[\hat{H} = \hat{H}_0 + \hat{H}' = \begin{pmatrix} E & k & 0 \\ k & E & k \\ 0 & k & E \end{pmatrix}\]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:
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.
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 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