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Chapter 3: Problem 33

Derive the equation for the allowed energies of a particle in a one- dimensional box by assuming that the particle is described by standing de Broglie waves within the box.

Short Answer

Expert verified
The allowed energies are given by \( E_n = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \).

Step by step solution

01

Understanding the Problem

We are tasked to derive the equation for the allowed energies of a particle in a 1-D box. The particle is described by standing de Broglie waves within a box of length \( L \). The endpoints of the box are at \( x = 0 \) and \( x = L \).
02

Wave Function Formation

In a one-dimensional box, the wave function \( \psi(x) \) for a standing de Broglie wave must meet boundary conditions: it must be zero at the endpoints of the box. This implies a sinusoidal function, so consider the general form \( \psi(x) = A \sin(kx) \).
03

Boundary Condition Application

Apply the boundary conditions to determine \( k \):1. \( \psi(0) = 0 \) implies no constraint as \( \sin(0) = 0 \).2. \( \psi(L) = 0 \) leads to \( A \sin(kL) = 0 \). For this to be true and \( A eq 0 \), \( \sin(kL) \) must be zero, hence \( kL = n\pi \), where \( n \) is a positive integer.
04

Determine the Wavenumber

From the boundary conditions, we found \( k = \frac{n\pi}{L} \). This represents the wavenumber for different modes, corresponding to different integers \( n \).
05

Relating Wavenumber to Energy

The de Broglie wavelength is \( \lambda = \frac{2\pi}{k} \), and using de Broglie's relation, \( p = \frac{h}{\lambda} \), the momentum \( p = \hbar k \). The energy \( E \) of the particle is given by \( E = \frac{p^2}{2m} = \frac{\hbar^2 k^2}{2m} \).
06

Substitute Wavenumber to Find Energies

Substitute \( k = \frac{n\pi}{L} \) into the expression for the energy: \[ E_n = \frac{\hbar^2}{2m} \left( \frac{n\pi}{L} \right)^2 = \frac{n^2 \pi^2 \hbar^2}{2mL^2} \]. This gives the allowed energy values for the particle in a box, where \( n = 1, 2, 3, \ldots \).

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Key Concepts

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

Particle in a Box
The 'Particle in a Box' is a foundational concept in quantum mechanics. It represents a simplified model where a particle moves freely within an enclosed space with infinitely high walls. In this system, the particle cannot exist outside the box and cannot escape its boundaries. The model helps in understanding the quantization of energy levels and basic wave properties of particles.
Here, a 'box' is a one-dimensional space, with boundaries defined at two points: one at position zero and the other at position L. Since the particle is confined, this setup leads to quantized energy levels. The model assumes no potential within the box, meaning the particle is free to move inside this region until it hits one of these boundaries.
A key takeaway is that the particle cannot have just any energy. Instead, its energy levels are discrete (quantized). This quantization arises from the boundary conditions imposed by the walls of the box.
De Broglie Waves
De Broglie Waves introduce the wave-particle duality concept, suggesting that particles can exhibit both wave and particle characteristics. According to de Broglie, each particle has an associated wavelength given by the equation: \(\lambda = \frac{h}{p}\), where \(h\) is Planck's constant and \(p\) is the particle's momentum.
In the 'Particle in a Box' model, particles such as electrons are considered to form standing waves within the box. A standing wave is a wave that remains in a constant position. The idea is central in deriving the energy states. These waves must fit within the boundary conditions imposed by the finite box size. This means only certain wavelengths (or energy states) are permissible.
  • De Broglie鈥檚 hypothesis bridges classical and quantum worldviews.
  • The concept supports why electrons orbit in specific quantized levels in atoms.
Wave Function
The wave function \(\psi(x)\) is a crucial part of the quantum mechanical description of a particle. It encodes all the information about a particle's state. For a particle in a box, the wave function must form standing waves that satisfy specific boundary conditions.
A typical wave function for a particle in a box takes the sinusoidal form \(\psi(x) = A \sin(kx)\). Here, \(A\) represents the amplitude, and \(k\) is the wave number related to the de Broglie wavelength. The wave function must be zero at the boundaries of the box because the particle cannot exist outside the box.
Standing wave solutions explain how the wave reflects back on itself within this confined space, forming nodes at the walls. The wave number \(k\) determines the frequency and energy of the wave which directly correspond to the quantized energy levels the particle can have. Each mode or energy level corresponds to a specific state that the particle can occupy.
Boundary Conditions
Boundary conditions are essential constraints that define the allowable solutions for a physical system's wave functions. In the context of the 'Particle in a Box', these boundary conditions mean the wave function must be zero at the endpoints of the box. This arises because the particle cannot exist where the wave function does not have a defined value.
For a particle in a box, applying the boundary conditions results in solutions that are \( \psi(0) = 0 \) and \( \psi(L) = 0 \). This results in the quantization condition \( kL = n\pi \) for a non-zero amplitude. Essentially, these conditions ensure that only certain wave numbers \(k\) and corresponding energies \(E\) are possible.
  • Boundary conditions provide the constraints that lead to discrete (quantized) rather than continuous energy levels.
  • This concept underscores the inherently quantized nature of microscopic systems in contrast to classical systems that can take any energy level.
Understanding these constraints helps in comprehending why certain energy levels are "allowed" while others are not.

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Most popular questions from this chapter

The quantized energies of a particle in a box result from the boundary conditions, or from the fact that the particle is restricted to a finite region. In this problem, we investigate the quantum-mechanical problem of a free particle, one that is not restricted to a finite region. The potential energy \(V(x)\) is equal to zero and the Schr枚dinger equation is $$\frac{d^{2} \psi}{d x^{2}}+\frac{2 m E}{\hbar^{2}} \psi(x)=0 \quad-\infty

Consider a particle of mass \(m\) in a one-dimensional box of length \(a\). Its average energy is given by $$\langle E\rangle=\frac{1}{2 m}\left\langle p^{2}\right\rangle$$ Because \(\langle p\rangle=0,\left\langle p^{2}\right\rangle=\sigma_{p}^{2},\) where \(\sigma_{p}\) can be called the uncertainty in \(p .\) Using the Uncertainty Principle, show that the energy must be at least as large as \(\hbar^{2} / 8 m a^{2}\) because \(\sigma_{x},\) the uncertainty in \(x,\) cannot be larger than \(a\)

Determine whether the following operators are linear or nonlinear: a. \(\hat{A} f(x)=\operatorname{SQR} f(x) \text { [square } f(x)]\) b. \(\hat{A} f(x)=f^{*}(x)[\text { form the complex conjugate of } f(x)]\) c. \(\hat{A} f(x)=0[\text { multiply } f(x) \text { by zero }]\) d. \(\left.\hat{A} f(x)=[f(x)]^{-1} \text {[take the reciprocal of } f(x)\right]\) e. \(\hat{A} f(x)=f(0) \text { [evaluate } f(x) \text { at } x=0]\) f. \(\hat{A} f(x)=\ln f(x)[\text { take the logarithm of } f(x)]\)

This problem shows that the intensity of a wave is proportional to the square of its amplitude. Figure \(3.7\) illustrates the geometry of a vibrating string. Because the velocity at any point of the string is \(\partial u / \partial t\), the kinetic energy of the entire string is $$ K=\int_{0}^{l} \frac{1}{2} \rho\left(\frac{\partial u}{\partial t}\right)^{2} d x $$ where \(\rho\) is the linear mass density of the string. The potential energy is found by considering the increase of length of the small arc \(P Q\) of length \(d s\) in Figure 3.7. The segment of the string along that arc has increased its length from \(d x\) to \(d s\). Therefore, the potential energy associated with this increase is $$ V=\int_{0}^{l} T(d s-d x) $$ where \(T\) is the tension in the string. Using the fact that \((d s)^{2}=(d x)^{2}+(d u)^{2}\), show that $$ V=\int_{0}^{l} T\left\\{\left[1+\left(\frac{\partial u}{\partial x}\right)^{2}\right]^{1 / 2}-1\right\\} d x $$ Using the fact that \((1+x)^{1 / 2} \approx 1+(x / 2)\) for small \(x\), show that $$ V=\frac{1}{2} T \int_{0}^{l}\left(\frac{\partial u}{\partial x}\right)^{2} d x $$ for small displacements. The total energy of the vibrating string is the sum of \(K\) and \(V\) and so $$ E=\frac{\rho}{2} \int_{0}^{l}\left(\frac{\partial u}{\partial t}\right)^{2} d x+\frac{T}{2} \int_{0}^{l}\left(\frac{\partial u}{\partial x}\right)^{2} d x $$ We showed in Chapter 2 (Equations \(2.23\) through \(2.25\) ) that the \(n\)th normal mode can be written in the form $$ u_{n}(x, l)=D_{n} \cos \left(\omega_{n} t+\phi_{n}\right) \sin \frac{n \pi x}{l} $$ where \(\omega_{n}=v n \pi / l\). Using this equation, show that $$ K_{n}=\frac{\pi^{2} v^{2} n^{2} \rho}{4 l} D_{n}^{2} \sin ^{2}\left(\omega_{n} t+\phi_{n}\right) $$ and $$ V_{n}=\frac{\pi^{2} n^{2} T}{4 l} D_{n}^{2} \cos ^{2}\left(\omega_{n} t+\phi_{n}\right) $$ Using the fact that \(v=(T / \rho)^{1 / 2}\), show that $$ E_{n}=\frac{\pi^{2} v^{2} n^{2} \rho}{4 l} D_{n}^{2} $$ Note that the total energy, or intensity, is proportional to the square of the amplitude. Although we have shown this proportionality only for the case of a vibrating string, it is a general result and shows that the intensity of a wave is proportional to the square of the amplitude. If we had carried everything through in complex notation instead of sines and cosines, then we would have found that \(E_{n}\) is proportional to \(\left|D_{n}\right|^{2}\) instead of just \(D_{n}^{2}\). Generally, there are many normal modes present at the same time, and the complete solution is (Equation 2.25) $$ u(x, t)=\sum_{n=1}^{\infty} D_{n} \cos \left(\omega_{n} t+\phi_{n}\right) \sin \frac{n \pi x}{l} $$ Using the fact that (see Problem 3-16) $$ \int_{0}^{l} \sin \frac{n \pi x}{l} \sin \frac{m \pi x}{l} d x=0 \quad \text { if } m \neq n $$ show that $$ E_{n}=\frac{\pi^{2} v^{2} \rho}{4 l} \sum_{n=1}^{\infty} n^{2} D_{n}^{2} $$

Using a table of integrals, show that $$ \begin{gathered} \int_{0}^{a} \sin ^{2} \frac{n \pi x}{a} d x=\frac{a}{2} \\ \int_{0}^{a} x \sin ^{2} \frac{n \pi x}{a} d x=\frac{a^{2}}{4} \end{gathered} $$ and $$ \int_{0}^{a} x^{2} \sin ^{2} \frac{n \pi x}{a} d x=\left(\frac{a}{2 \pi n}\right)^{3}\left(\frac{4 \pi^{3} n^{3}}{3}-2 n \pi\right) $$ All these integrals can be evaluated from $$ I(\beta)=\int_{0}^{a} e^{\beta x} \sin ^{2} \frac{n \pi x}{a} d x $$ Show that the above integrals are given by \(I(0), I^{\prime}(0)\), and \(I^{\prime \prime}(0)\), respectively, where the primes denote differentiation with respect to \(\beta\). Using a table of integrals, evaluate \(I(\beta)\). and then the above three integrals by differentiation.
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