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
To solve this problem, we need to find the first-order correction to the energy eigenvalue for a particle in an infinite potential well with a perturbation. The perturbation is shown as a linear potential within the well.
The potential inside the well is given by:
\(V(x) = \frac{V_0}{L}x \quad \text{for} \quad 0 \leq x \leq L\)
In perturbation theory, the first-order correction to the energy is given by:
\(E_n^{(1)} = \int_{0}^{L} \psi_n^*(x) V(x) \psi_n(x) \, dx\)
where \(\psi_n(x)\) is the normalized wave function for the unperturbed system. The wave function for a particle in a 1D box is:
\(\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\)
Substituting \(V(x)\) and \(\psi_n(x)\) into the energy correction formula:
\(E_n^{(1)} = \int_{0}^{L} \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right)^2 \frac{V_0}{L}x \, dx\)
Simplifying, we have:
\(E_n^{(1)} = \frac{2V_0}{L^2} \int_{0}^{L} x \sin^2\left(\frac{n\pi x}{L}\right) \, dx\)
Using the trigonometric identity \(\sin^2\theta = \frac{1 - \cos(2\theta)}{2}\), the integral becomes:
\(E_n^{(1)} = \frac{V_0}{L^2} \int_{0}^{L} x \left(1 - \cos\left(\frac{2n\pi x}{L}\right)\right) \, dx\)
Which is:
\(E_n^{(1)} = \frac{V_0}{L^2} \left[\int_{0}^{L} x \, dx - \int_{0}^{L} x \cos\left(\frac{2n\pi x}{L}\right) \, dx\right]\)
The first integral is straightforward:
\(\int_{0}^{L} x \, dx = \left[\frac{x^2}{2}\right]_{0}^{L} = \frac{L^2}{2}\)
The second integral can be solved using integration by parts, resulting in zero since it involves a sine term evaluated between 0 and \(L\).
Thus, the correction is:
\(E_n^{(1)} = \frac{V_0}{L^2} \cdot \frac{L^2}{2} = \frac{V_0}{2}\)
Therefore, the first-order correction to the energy eigenvalue is \(\frac{V_0}{2}\). Hence, the correct answer is:
Option: \(\frac{V_0}{2}\)
Important Questions from Perturbation Theory Time Independent Degenerate
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 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?
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 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 ________.
- 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?
