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Chapter 2: Problem 11

An electron is confined to a one-dimensional box of length \(L .\) What should be the length of the box in order for its zero-point energy to be equal to its rest mass energy \(\left(m_{\mathrm{e}} c^{2}\right) ?\) Express the result in terms of the Compton wavelength, \(\lambda_{\mathrm{C}}=h / m_{\mathrm{e}} c\)

Short Answer

Expert verified
The length of the box for the zero-point energy to be equal to its rest mass energy is \(\frac{1}{\sqrt{2}}\) times the Compton wavelength of the electron, i.e. \(L = \dfrac{1}{\sqrt{2}} \lambda_C\).

Step by step solution

01

Write down Given Information

We know that \(E_0 = m_ec^2\) and \( \lambda_C=\frac{h}{m_ec}\).
02

Express Zero-Point Energy

The zero-point energy (ground state energy) for a particle in a box of length \(L\) is \(E_0 = \dfrac{\pi^2 \hbar^2}{2mL^2}\). Equate this to \(m_ec^2\) to get \(\dfrac{\pi^2 \hbar^2}{2mL^2} = m_ec^2\).
03

Solve for Length L

On rearranging the equation we get, \(L = \dfrac{\pi \hbar}{\sqrt{2}m_ec}\). Replace \(\hbar = \dfrac{h}{2\pi}\) and rearrange the terms to get \(L\).
04

Express in terms of Compton wavelength

We are asked for the result in terms of the Compton wavelength. Replace \(h/m_ec\) by \(\lambda_C\) in our derived expression to get \(L = \dfrac{1}{\sqrt{2}} \lambda_C\).

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Key Concepts

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

Zero-point Energy
Zero-point energy is an intriguing concept in quantum mechanics. Imagine a particle trapped in a box as if it were a marble in a shoebox. Even in its lowest energy state, the particle doesn't just sit still. It constantly jiggles around, a consequence of quantum physics known as the "Zero-Point Energy". In simple terms, it's like the minimum energy that the particle can have, even at absolute zero temperature.
In the solution provided, zero-point energy for an electron in a box is given by the equation:
  • \(E_0 = \dfrac{\pi^2 \hbar^2}{2mL^2}\).
Here, \(E_0\) signifies the zero-point energy, \(m\) is the mass of the electron, and \(L\) represents the length of the box. This formula derives from solving the Schr枚dinger equation for a particle in a one-dimensional box.
It's important to grasp that zero-point energy is different from other energy levels because it does not depend on the temperature but on other system's properties, like the box's length.
Compton Wavelength
The Compton wavelength is a fundamental concept in quantum mechanics. It represents a quantum limit to measuring the position of a particle. If you try to pinpoint the location of a particle more precisely than its Compton wavelength, bizarre quantum effects dominate, and new particles can pop up as pairs created from pure energy.
In terms of formula, Compton wavelength \(\lambda_C\) is expressed as:
  • \(\lambda_C = \dfrac{h}{m_ec}\)
where \(h\) is Planck鈥檚 constant, \(m_e\) is the electron mass, and \(c\) denotes the speed of light. This wavelength sets the fundamental scale for quantum mechanical effects and ultimately links the zero-point energy of an electron in the box to its rest mass energy.
The problem presented asks us to depict the length of the box in terms of the Compton wavelength, illustrating the bridge between quantum theory and the physical attributes of particles.
Particle in a Box
The particle in a box is a cornerstone problem in quantum mechanics. It provides insights into how quantum particles, such as electrons, behave in confined spaces. Picture it like a boxing ring where the boxer (the particle) can bounce back and forth but not leave.
As the problem describes, the electron is confined to a box where it exhibits certain quantifiable behaviors, leading to quantized energy levels. The simpler case is for a one-dimensional box. The ground state energy, or zero-point energy, as calculated in this scenario, arises from the particle being confined, which inherently comes with an uncertainty in position and consequently in momentum.
Through deriving the expression \(L = \dfrac{1}{\sqrt{2}} \lambda_C\), we connect the particle's confinement (the variable \(L\)) with the Compton wavelength. This linkage supports the understanding of the particle's physical behavior within the limitations of quantum mechanics, demonstrating that quantum systems behave differently from classical predictions.

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Most popular questions from this chapter

Confirm that the completeness relation, eqn 1.25, may be expressed in terms of wavefunctions as \\[ \sum_{n} \psi_{n}(r) \psi_{n}^{*}\left(r^{\prime}\right)=\delta\left(r-r^{\prime}\right) \\] where \(\delta\left(r-r^{\prime}\right)\) is the Dirac \(\delta\) -function described in Section 2.1

A particle of mass \(m\) is confined to a one-dimensional box of length \(L\). Calculate the probability of finding it in the following regions: (a) \(0 \leq x \leq \frac{1}{2} L,\) (b) \(0 \leq x \leq \frac{1}{4} L\) (c) \(\frac{1}{2} L-\delta x \leq x \leq \frac{1}{2} L+\delta x .\) Derive expressions for a general value of \(n\). Then evaluate the probabilities (i) for \(n=1\) (ii) in the limit \(n \rightarrow \infty\). Compare the latter to the classical expectations.

Determine the probability of finding the ground-state harmonic oscillator stretched to a displacement beyond the classical turning point. Hint. Relate the expression for the probability to the error function, erf \(z,\) defined as \\[ \operatorname{erf} z=1-\frac{2}{\pi^{1 / 2}} \int_{z}^{\infty} \mathrm{e}^{-y^{2}} \mathrm{d} y \\] and evaluate it using erf \(1=0.8427 .\) The error function is incorporated into most mathematical software packages.

For a particle in a box, the mean value and mean square value of the linear momentum are given by \(\int_{0}^{L} \psi^{*} p \psi \mathrm{d} x\) and \(\int_{0}^{L} \psi^{*} p^{2} \psi \mathrm{d} x,\) respectively. Evaluate these quantities. Form the root mean square deviation \(\Delta p=\left\\{\left\langle p^{2}\right\rangle-\langle p\rangle^{2}\right\\}^{1 / 2}\) and investigate the consistency of the outcome with the uncertainty principle. Hint. Use \(p=(\hbar / \mathrm{i}) \mathrm{d} / \mathrm{d} x .\) For \(\left\langle p^{2}\right\rangle\) notice that \(E=p^{2} / 2 m\) and we already know \(E\) for each \(n\). For the last part, form \(\Delta x \Delta p\) and show that \(\Delta x \Delta p \geq \frac{1}{2} \hbar,\) the precise form of the principle, for all \(n\) evaluate \(\Delta x \Delta p\) for \(n=1.\)

Calculate the energies and wavefunctions for a particle in a one-dimensional square well in which the potential energy rises to a finite value \(V\) at each end, and is zero inside the well; that is \\[ \begin{array}{ll} V(x)=V & x \leq 0 \text { and } x \geq L \\ V(x)=0 & 0
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