A particle is confined to a one dimensional potential box with the potential V(x)=0, 0 < x < a $= \infty, \text{ otherwise}$ If the particle is subjected to a perturbation, within the box, $W = \beta x$, where $ \beta $ is a small constant, the first order correction to the ground state energy is
Potential Box Perturbation: First Order Energy Correction
This solution calculates the first-order correction to the ground state energy for a particle confined in a one-dimensional potential box subjected to a perturbation.
Problem Setup: Particle in a 1D Box with Perturbation
The system is a particle in a 1D potential box defined by:
- $V(x) = 0$ for $0 < x < a$
- $V(x) = \infty$ otherwise
A perturbation $W(x) = \beta x$ is applied within the box, where $ \beta $ is a small constant.
First Order Energy Correction Formula
The first-order energy correction, $ E_n^{(1)} $, for a state $ n $ is given by the expectation value of the perturbation Hamiltonian $ W $ in the unperturbed state $ \psi_n^{(0)} $:
$E_n^{(1)} = \langle \psi_n^{(0)} | W | \psi_n^{(0)} \rangle = \int \psi_n^{(0)*} (x) W(x) \psi_n^{(0)} (x) dx$Ground State Wave Function
For the 1D potential box, the unperturbed ground state ($n=1$) wave function is:
$ \psi_1^{(0)}(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) $The integration limits are from $ 0 $ to $ a $.
Calculation of First Order Correction
We calculate $ E_1^{(1)} $ using the ground state wave function ($n=1$) and the perturbation $W(x) = \beta x$:
$E_1^{(1)} = \int_0^a \left( \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \right) (\beta x) \left( \sqrt{\frac{2}{a}} \sin\left(\frac{\pi x}{a}\right) \right) dx$Simplify the expression:
$E_1^{(1)} = \frac{2}{a} \beta \int_0^a x \sin^2\left(\frac{\pi x}{a}\right) dx$Use the identity $ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} $:
$E_1^{(1)} = \frac{2\beta}{a} \int_0^a x \left( \frac{1 - \cos\left(\frac{2\pi x}{a}\right)}{2} \right) dx$ $E_1^{(1)} = \frac{\beta}{a} \int_0^a \left( x - x \cos\left(\frac{2\pi x}{a}\right) \right) dx$Split the integral:
$E_1^{(1)} = \frac{\beta}{a} \left[ \int_0^a x dx - \int_0^a x \cos\left(\frac{2\pi x}{a}\right) dx \right]$Evaluate the first integral:
$ \int_0^a x dx = \left[ \frac{x^2}{2} \right]_0^a = \frac{a^2}{2} $Evaluate the second integral using integration by parts ($ \int u dv = uv - \int v du $).
Let $ u = x $ and $ dv = \cos\left(\frac{2\pi x}{a}\right) dx $. Then $ du = dx $ and $ v = \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) $.
$ \int_0^a x \cos\left(\frac{2\pi x}{a}\right) dx = \left[ x \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) \right]_0^a - \int_0^a \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) dx $The boundary term evaluates to zero:
$ \left[ x \frac{a}{2\pi} \sin\left(\frac{2\pi x}{a}\right) \right]_0^a = a \frac{a}{2\pi} \sin(2\pi) - 0 = 0 $The integral term also evaluates to zero:
$ -\frac{a}{2\pi} \int_0^a \sin\left(\frac{2\pi x}{a}\right) dx = -\frac{a}{2\pi} \left[ -\frac{a}{2\pi} \cos\left(\frac{2\pi x}{a}\right) \right]_0^a = \frac{a^2}{4\pi^2} [\cos(2\pi) - \cos(0)] = \frac{a^2}{4\pi^2} [1 - 1] = 0 $Thus, the second integral is $ 0 $.
Substitute the results back:
$E_1^{(1)} = \frac{\beta}{a} \left[ \frac{a^2}{2} - 0 \right]$ $E_1^{(1)} = \frac{\beta a}{2}$Final Result
The first-order correction to the ground state energy is $ \frac{\beta a}{2} $.
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?
