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A one-dimensional potential well has infinite walls at \(x=0\) and \(x=L\). The bottom of the well is not flat, but rather increases linearly from 0 at \(x=0\) to \(V\) at \(x=L\). Find the first-order shift in the energy levels as a function of principal quantum number \(n\).

Short Answer

Expert verified
The energy shift is \( \Delta E_n = \frac{VL}{n^2 \pi^2} \).

Step by step solution

01

Understand the Problem

We have a one-dimensional potential well with infinite walls at positions \(x = 0\) and \(x = L\). The potential inside the well increases linearly from 0 at \(x=0\) to \(V\) at \(x=L\). We need to find the first-order shift in the energy levels, considering this linear potential in the well.
02

Set Up the Perturbation

The linear potential inside the well can be described by the function \( V(x) = \frac{V}{L}x \). This acts as a perturbation to the potential well with infinite walls, which typically has zero potential inside.
03

Use Perturbation Theory

In first-order perturbation theory, the shift in energy \( \Delta E_n \) is given by the integral of the product of the eigenfunction and the perturbing potential over the limits of the well:\[\Delta E_n = \int_{0}^{L} \psi_n^*(x) V(x) \psi_n(x) \ dx,\]where \(\psi_n(x)\) is the unperturbed wave function for state \(n\).
04

Identify Unperturbed Wave Functions

For an infinite potential well, the normalized wave function \( \psi_n(x) \) is:\[\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right).\]
05

Substitute into the Integral

Substitute the expressions for \( \psi_n(x) \) and \( V(x) \) into the integral:\[\Delta E_n = \int_{0}^{L} \left(\sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\right)^2 \left(\frac{V}{L}x\right) \ dx.\]
06

Simplify the Expression and Integrate

Simplify the integral\[\Delta E_n = \frac{2V}{L^3} \int_{0}^{L} x \sin^2\left(\frac{n\pi x}{L}\right) dx.\]Using the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \), further simplify and solve the integral to get:\[\Delta E_n = \frac{2V}{L^3} \left(\frac{L^3}{2n^2\pi^2}\right) = \frac{VL}{n^2\pi^2}.\]
07

Derive the Final Expression

Conclude that the first-order energy shift for state \(n\) is:\[\Delta E_n = \frac{VL}{n^2\pi^2}.\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Well
A potential well in quantum mechanics is a concept used to describe a scenario where a particle is confined in a particular region by potential barriers. In this exercise, we are dealing with a one-dimensional potential well with infinite walls at positions \(x = 0\) and \(x = L\). These walls prevent the particle from escaping, essentially creating a scenario where the particle's energy states can only exist within this bounded region.

The potential inside the well increases linearly from zero at the left wall to some value \(V\) at the right wall. This means the floor of the well is not flat, but slopes upward. This is significant because changes in the potential shape can affect the energy levels of the particles within the well.

In an ideal infinite potential well, the potential inside the well is constant (usually zero), meaning the energy levels are determined solely by the boundary conditions. However, in our problem, the linear potential creates a non-constant environment, adding complexity, as it slightly shifts the energy levels of the system compared to a system with a flat potential floor.
Perturbation Theory
Perturbation theory is a method used in quantum mechanics to find an approximate solution to a problem that cannot be solved exactly. It is especially useful when dealing with complex systems by breaking them into simpler parts. In this exercise, the linear potential inside the well acts as a perturbation to the system.

Originally, the infinite potential well with flat potential allows us to determine basic energy states. Here's where perturbation theory comes into play. We treat the linear potential \(V(x) = \frac{V}{L}x\) as a small correction to the overall potential. First-order perturbation theory aims to provide the first approximation of how these energy levels shift under the new potential.

The method involves calculating the integral of the product of the eigenfunction, the perturbing potential, and the conjugate of the eigenfunction over the limits of the well. The result gives us the energy shift \(\Delta E_n\) which is dependent on the principal quantum number \(n\). This shift reveals how the introduction of the slope in the potential well affects the energy levels.
Eigenfunctions
Eigenfunctions are critical in quantum mechanics as they represent the possible states of a particle within a well-defined boundary. In our potential well example, the eigenfunctions \(\psi_n(x)\) are determined by the original infinite potential well structure before considering perturbations.

For such a well, these eigenfunctions are expressed as \(\psi_n(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{n\pi x}{L}\right)\), which are sine functions that satisfy the boundary conditions of the well. These functions describe the probability amplitude of finding the particle at position \(x\) within the well.

When solving the energy shift due to the perturbation, these eigenfunctions are used in the integral to calculate \( \Delta E_n \). The shape and properties of the eigenfunctions significantly influence the result of this calculation.

Each eigenfunction corresponds to a specific energy level, defined by the quantum number \(n\), which implies that higher \(n\) values correlate with higher energy states. By understanding these eigenfunctions, we understand how the particle behaves when confined within an infinite potential well.

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Most popular questions from this chapter

Consider a particle bound to a fixed center by a spherically symmetric potential \(V(r)\). a. Prove $$ |\psi(0)|^{2}=\left(\frac{m}{2 \pi \hbar^{2}}\right)\left\langle\frac{d V}{d r}\right\rangle $$ for all \(s\) states, ground and excited. b. Check this relation for the ground state of a three-dimensional isotropic oscillator, the hydrogen atom, and so on. (Note: This relation has actually been found to be useful in guessing the form of the potential between a quark and an antiquark. See Moxhay and Rosner, J. Math. Phys., 21 (1980) 1688.)

This chapter derived two of the three relativistic corrections to the one- electron atom, namely \(\Delta_{K}^{(1)}\) from "relativistic kinetic energy," and \(\Delta_{L S}^{(1)}\) from the spin-orbit interaction. A third term comes from the spread of the electron wave function in the region of changing electric field. The perturbation for this "Darwin term" is $$ V_{D}=-\frac{1}{8 m^{2} c^{2}} \sum_{i=1}^{3}\left[p_{i},\left[p_{i}, e \phi(r)\right]\right] $$ where \(\phi(r)\) is the Coulomb potential. Find \(\Delta_{D}^{(1)}\) and show that $$ \Delta_{n j}^{(1)} \equiv \Delta_{K}^{(1)}+\Delta_{L S}^{(1)}+\Delta_{D}^{(1)}=\frac{m c^{2}(Z \alpha)^{4}}{2 n^{3}}\left[\frac{3}{4 n}-\frac{1}{j+1 / 2}\right] . $$ In Section \(8.4\) we will compare this expression to the result of solving the Dirac equation in the presence of the Coulomb potential.

A one-electron atom whose ground state is nondegenerate is placed in a uniform electric field in the \(z\)-direction. Obtain an approximate expression for the induced electric dipole moment of the ground state by considering the expectation value of ez with respect to the perturbed state vector computed to first order. Show that the same expression can also be obtained from the energy shift \(\Delta=-\alpha|\mathbf{E}|^{2} / 2\) of the ground state computed to second order. (Note: \(\alpha\) stands for the polarizability.) Ignore spin.

Consider an isotropic harmonic oscillator in two dimensions. The Hamiltonian is $$ H_{0}=\frac{p_{x}^{2}}{2 m}+\frac{p_{y}^{2}}{2 m}+\frac{m \omega^{2}}{2}\left(x^{2}+y^{2}\right) $$ a. What are the energies of the three lowest-lying states? Is there any degeneracy? b. We now apply a perturbation $$ V=\delta m \omega^{2} x y $$ where \(\delta\) is a dimensionless real number much smaller than unity. Find the zerothorder energy eigenket and the corresponding energy to first order [that is, the unperturbed energy obtained in (a) plus the first-order energy shift] for each of the three lowest-lying states. c. Solve the \(H_{0}+V\) problem exactly. Compare with the perturbation results obtained in (b).

A hydrogen atom in its ground state \([(n, l, m)=(1,0,0)]\) is placed between the plates of a capacitor. A time-dependent but spatial uniform electric field (not potential!) is applied as follows: \(\mathbf{E}=\left\\{\begin{array}{ll}0 & \text { for } t<0 \\ \mathbf{E}_{0} e^{-t / \tau} & \text { for } t>0\end{array}\left(\mathbf{E}_{0}\right.\right.\) in the positive \(z\)-direction \()\) Using first-order time-dependent perturbation theory, compute the probability for the atom to be found at \(t \gg \tau\) in each of the three \(2 p\) states: \((n, l, m)=(2,1, \pm 1\) or 0\()\). Repeat the problem for the \(2 s\) state: \((n, l, m)=(2,0,0)\). Consider the limit \(\tau \rightarrow \infty\).

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