/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none}

91Ó°ÊÓ

Solve the time-independent Schrödinger equation for a centered infinite square well with a delta-function barrier in the middle: $$ V(x)=\left\\{\begin{array}{ll} \alpha \delta(x), & \text { for }-a

Short Answer

Expert verified
Odd solutions energy remains unchanged; even solutions are shifted. For \(\alpha=0\), it's the standard well; \(\alpha=\infty\), even states disappear.

Step by step solution

01

Identify the Region and Boundary Conditions

The potential is zero for \(-a < x < a\) and infinite elsewhere. The boundary conditions require that the wave function vanish at \(x = \pm a\), ensuring an infinite potential outside is a wall.
02

Set Up the Schrödinger Equation Inside the Well

Inside the well, the time-independent Schrödinger equation is given by: \(-\frac{\hbar^2}{2m} \frac{d^2 \psi}{dx^2} = E \psi\), where \(0 < x < a\). This simplifies to \(\frac{d^2 \psi}{dx^2} = -k^2 \psi\), with \(k^2 = \frac{2mE}{\hbar^2}\). This equation has solutions \(\psi(x) = A \cos(kx) + B \sin(kx)\).
03

Apply the Even Condition

For even wave functions, \(\psi(-x) = \psi(x)\), thus \(\psi(x) = A \cos(kx)\). Applying boundary conditions at \(x = a\), \(\psi(a) = A \cos(ka) = 0\). This implies \(\cos(ka) = 0\), leading to \(ka = (2n + 1)\frac{\pi}{2}, n \in \mathbb{Z}\).
04

Apply the Odd Condition

Odd wave functions satisfy \(\psi(-x) = -\psi(x)\), implying \(\psi(x) = B \sin(kx)\). Again bounding at \(x = a\), \(\psi(a) = B \sin(ka) = 0\), leading to \(ka = n\pi, n \in \mathbb{Z}\). These solutions are unaffected by the delta barrier.
05

Delta Function Discontinuity Condition

At \(x=0\), \(\psi'\) must have a discontinuity. Integrating the Schrödinger equation around \(x=0\), \(\psi'(0^+) - \psi'(0^-) = \frac{2m\alpha}{\hbar^2} \psi(0)\). For even states, \(\psi(0) = A eq 0\); for odd states, \(\psi(0) = 0\) results no effect from the delta.
06

Solve for Even State Energies

The energy equation for even states becomes \(k \tan(ka) = \frac{m\alpha}{\hbar^2}\). Solve graphically or numerically, resulting in modified energies due to the delta function.
07

Compare with Delta Function Absent

Without the delta barrier, energies satisfy \(ka = n\pi\), for both even (\(n>0\)) and odd solutions, leading to unaffected odd solutions and shifted even solutions.
08

Analyze Limiting Cases

For \(\alpha \to 0\), the system reverts to a normal infinite square well, while ·\(\alpha \to \infty\) diverges, forcing even solutions to nullify around the barrier.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Infinite Square Well
The infinite square well is a fundamental quantum mechanics problem. Imagine a particle trapped in a one-dimensional box, but unlike a normal box, it can't escape past the walls. The walls are impenetrable and represented by infinite potential energy. Inside the well, the potential energy is zero, allowing a wave function to describe the particle’s position.

For the quantum particle, its allowed positions are determined by boundary conditions—specifically that the wave function must vanish at the walls of the well. These boundary conditions arise because particles can't exist in regions of infinite potential. These constraints lead to quantized solutions. The particle inside the well can only have energies and shapes that fit perfectly within this box.

The wave functions are sinusoidal (sine or cosine waves) because they satisfy the Schrödinger equation under these boundary conditions. The continuity and smoothness of the wave function lead to quantization of energy levels.
Delta Function Barrier
A delta function barrier is a unique quantum potential representing an "infinitely thin" and "infinitely high" obstacle at a specific point, affecting the particle’s wave function at that point. In practical terms, it acts as a sharp spike or bump in potential energy that disrupts the continuity of the wave function’s derivative.

In the problem of an infinite square well with a delta function barrier right in the middle, this disruption imposes an additional condition. The wave function remains continuous, but its derivative experiences a discontinuity. This is mathematically described by the "jump" condition in the derivative of the wave function at the location of the delta function.

The impact on the wave function varies based on its symmetry (even or odd). In particular, for even wave functions, this barrier modifies the energy levels because the wave function's value is non-zero at the barrier, causing an impact. Odd wave functions, however, do not get affected by the barrier since they inherently have a nodal zero at the center—the same as the position of the barrier.
Even and Odd Wave Functions
In quantum mechanics, wave functions can be classified based on symmetry as even or odd. An even wave function remains unchanged when flipped about an axis, while an odd wave function acquires a sign change.

For a particle in an infinite square well bisected by a delta barrier, even wave functions have the form of cosine waves. They are symmetric about the origin. As a result, any barriers placed symmetrically in the well, such as at the center, directly affect them. Since the barrier is centered, these wave functions encounter the delta at their maximum, directly altering their energy states.

In contrast, odd wave functions follow the sine wave pattern and are antisymmetric. They naturally have a node at the center of the well. This point of zero amplitude results in no change or interaction with the delta barrier, as the wave function value there is zero, making these odd states immune to the barrier’s effects.
Energy Eigenvalues
In the infinite square well problem, energy eigenvalues are the specific energy levels a particle can occupy. They are solutions to the Schrödinger equation under the given boundary conditions.

With no delta function, energy eigenvalues for even and odd solutions can be expressed by the quantum number, known as "n," and are determined by the principle that the wave function must fit an integer number of half-wavelengths within the well.

The presence of a delta function shifts these energy eigenvalues for even solutions. Because the wave function at the center is non-zero for even states, they experience an effective potential due to the delta barrier. The new quantized energies can be calculated with an additional mathematical condition involving the tangent function and other constants like the mass of the particle, \(m\), and Planck's constant, \(\hbar\).

Odd states, however, experience no change in energy because the delta function coincides with their node at the center, where they inherently possess zero probability amplitude. Therefore, the odd solutions retain their energy as if the delta barrier wasn't present.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In this problem we explore some of the more useful theorems (stated without proof) involving Hermite polynomials. (a) The Rodrigues formula says that $$ H_{n}(\xi)=(-1)^{n} e^{\xi^{2}}\left(\frac{d}{d \xi}\right)^{n} e^{-\xi^{2}} $$ Use it to derive \(H_{3}\) and \(H_{4}\). (b) The following recursion relation gives you \(H_{n+1}\) in terms of the two preceding Hermite polynomials: $$ H_{n+1}(\xi)=2 \xi H_{n}(\xi)-2 n H_{n-1}(\xi) $$ Use it, together with your answer in (a), to obtain \(H_{5}\) and \(H_{6}\).

A particle of mass \(m\) in the infinite square well (of width \(a\) ) starts out in the left half of the well, and is (at \(t=0\) ) equally likely to be found at any point in that region. (a) What is its initial wave function, \(\Psi(x, 0)\) ? (Assume it is real. Don't forget to normalize it.) (b) What is the probability that a measurement of the energy would yield the value \(\pi^{2} \hbar^{2} / 2 m a^{2}\) ?

This problem is designed to guide you through a "proof" of Plancherel's theorem, by starting with the theory of ordinary Fourier series on a finite interval, and allowing that interval to expand to infinity. (a) Dirichlet's theorem says that "any" function \(f(x)\) on the interval \([-a,+a]\) can be expanded as a Fourier series: $$ f(x)=\sum_{n=0}^{\infty}\left[a_{n} \sin (n \pi x / a)+b_{n} \cos (n \pi x / a)\right] $$ Show that this can be written equivalently as $$ f(x)=\sum_{n=-\infty}^{\infty} c_{n} e^{i n \pi x / a} $$ What is \(c_{n}\), in terms of \(a_{n}\) and \(b_{n}\) ? (b) Show (by appropriate modification of Fourier's trick) that $$ c_{n}=\frac{1}{2 a} \int_{-a}^{+a} f(x) e^{-i n \pi x / a} d x $$ (c) Eliminate \(n\) and \(c_{n}\) in favor of the new variables \(k=(n \pi / a)\) and \(F(k)=\) \(\sqrt{2 / \pi} a c_{n} .\) Show that (a) and (b) now become $$ f(x)=\frac{1}{\sqrt{2 \pi}} \sum_{n=-\infty}^{\infty} F(k) e^{i k x} \Delta k ; \quad F(k)=\frac{1}{\sqrt{2 \pi}} \int_{-a}^{+a} f(x) e^{-i k x} d x $$ where \(\Delta k\) is the increment in \(k\) from one \(n\) to the next. (d) Take the limit \(a \rightarrow \infty\) to obtain Plancherel's theorem. Comment: In view of their quite different origins, it is surprising (and delightful) that the two formulas-one for \(F(k)\) in terms of \(f(x)\), the other for \(f(x)\) in terms of \(F(k)\) - have such a similar structure in the limit \(a \rightarrow \infty\).

Show that \(\left[A e^{i k x}+B e^{-i k x}\right]\) and \([C \cos k x+D \sin k x]\) are equivalent ways of writing the same function of \(x\), and determine the constants \(C\) and \(D\) in terms of \(A\) and \(B\), and vice versa. Comment: In quantum mechanics, when \(V=0\), the exponentials represent traveling waves, and are most convenient in discussing the free particle, whereas sines and cosines correspond to standing waves, which arise naturally in the case of the infinite square well.

Delta functions live under integral signs, and two expressions \(\left(D_{1}(x)\right.\) and \(D_{2}(x)\) ) involving delta functions are said to be equal if $$ \int_{-\infty}^{+\infty} f(x) D_{1}(x) d x=\int_{-\infty}^{+\infty} f(x) D_{2}(x) d x $$ for every (ordinary) function \(f(x)\). (a) Show that $$ \delta(c x)=\frac{1}{|c|} \delta(x) $$ where \(c\) is a real constant. (Be sure to check the case where \(c\) is negative.) (b) Let \(\theta(x)\) be the step function: $$ \theta(x) \equiv\left\\{\begin{array}{ll} 1, & \text { if } x>0 \\ 0, & \text { if } x<0 \end{array}\right. $$ (In the rare case where it actually matters, we define \(\theta(0)\) to be \(1 / 2 .\) ) Show that \(d \theta / d x=\delta(x)\).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.