/*! 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} Problem 85 Exercises \(78-88\) refer to a p... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 85

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ Calculate the uncertainty in the particle's momentum.

Short Answer

Expert verified
The uncertainty in the particle's momentum can be calculated by finding the difference between mean square momentum and the square of the mean momentum using the wavefunction. The mathematical process involves evaluating certain integrals and differentiating the wavefunction. The exact numerical result depends on the concrete normalization constant \(a\) and mass \(m\) of a particle.

Step by step solution

01

Mean value of square of momentum

The operator for the square of the momentum is \( p^2= -\hbar^2 \frac{d^2}{dx^2} \). So, the mean value is given by the following integral: \[ \langle p^2 \rangle = -\hbar^2 \int_{-\infty}^{\infty}\psi^* (x)\frac{d^2}{dx^2} \psi(x) dx\] Also, consider that \( \psi^* (x) = \psi(x) \) because the wavefunction is real.
02

Square of the mean value of momentum

The operator for the momentum is \( p = -i \hbar \frac{d}{dx} \). So, the square of the mean value of the momentum is given by: \[ \langle p \rangle^2 = \left(-i \hbar \int_{-\infty}^{\infty} \psi^* (x) \frac{d}{dx} \psi(x) dx \right)^2 \]
03

Use the uncertainty relation

The uncertainty of the momentum is given by the following equation according to Heisenberg's uncertainty principle: \[ (\Delta p)^2 = \langle p^2 \rangle - \langle p \rangle^2\] Substitute the result from step 1 and step 2 in this equation to get the desired result.

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 91影视!

Key Concepts

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

Uncertainty Principle
The Heisenberg Uncertainty Principle is a key underpinning of quantum mechanics, highlighting the intrinsic limitations in measuring complementary properties of a particle, such as position and momentum. It states that the more precisely one property is known, the less precisely the other can be known, encapsulated by the formula:
\[ \Delta x \Delta p \geq \frac{\hbar}{2} \]where \(\Delta x\) is the uncertainty in position and \(\Delta p\) is the uncertainty in momentum, and \(\hbar\) is the reduced Planck鈥檚 constant.
This principle emphasizes the probabilistic nature of quantum mechanics, reflecting the idea that particles do not have definite locations or speeds at the atomic scale until they are measured. Instead, these properties are described by probabilities.
Wave Function
A wave function, commonly denoted by \( \psi(x) \), is a mathematical description of the quantum state of a particle or system. It contains all the necessary information about a system's state and its interaction with the environment. In quantum mechanics, the wave function is a solution to the Schr枚dinger equation.
The square of its absolute value, \( |\psi(x)|^2 \), gives the probability density of finding a particle at a specific position. For example, in the given function:
\[ \psi(x)=\left\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-ax} & x>0 \ 0 & x<0 \end{array}\right. \]it shows us that the probability of finding the particle decreases exponentially as \(x\) increases for \(x>0\), and it is zero for \(x<0\).
This function shapes the way we predict a particle's behavior within a quantum system.
Momentum Operator
In quantum mechanics, the momentum operator is a crucial tool that represents the momentum of a particle in the position space representation. It is given by the expression:
\[ \hat{p} = -i \hbar \frac{d}{dx} \]where \( \hat{p} \) is the momentum operator, \( \hbar \) is the reduced Planck's constant, and \( i \) is the imaginary unit.
This operator functions as part of the Schr枚dinger equation, acting on wave functions to yield momentum-related information.
  • In the given exercise, the momentum operator is applied to the wave function \( \psi(x) \) to determine the mean momentum.
  • The momentum operator allows us to calculate expected values when used with the wave functions, highlighting the deep connections between states and measurements in quantum systems.
By working with this operator, physicists can predict how particles behave, offering insights into their dynamic properties.

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

Whereas an infinite well has an infinite number of bound states, a finite well does not. By relating the well height \(U_{0}\) to the kinetic energy and the kinetic energy (through \(\lambda\) ) to \(n\) and \(L\). show that the number of bound states is given roughly by \(\sqrt{8 m L^{2} U_{0} / h^{2}}\). (Assume that the number is Large.)

The term interaction is sometimes used interchangeably with for'e, and (other times interchangeably with potential energy. Although force and potential energy certainly aren't the same thing, what justification is there for using the same term to cover both?

Exercises \(78-88\) refer to a particle of mass \(m\) described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} 2 \sqrt{a^{3}} x e^{-a x} & x>0 \\ 0 & x<0 \end{array}\right. $$ What is the probability that the particle would be found hetween \(x=0\) and \(x=1 / a ?\)

A particle is bound by a potential energy of the form \(U(x)=\left\\{\begin{array}{cl}0 & |r|<\frac{1}{2} a \\ \infty & |x|>\frac{1}{2} a\end{array}\right.\) This differs from the infinite well of Section 5.5 in being symmetric about \(x=0\), which implies that the probability densities must also he symmetric. Noting that either sine or cosine would fit this requirement and could be made continuous with the zero wave function outside the well. determiner the allowed energies and corresponding normalized wave functions for this potential well.

A finite potential energy function \(U(x)\) allows \(\psi(x)\). the solution of the time-independent Schr枚dinger equation. to penetrate the classically forbidden region. Without assuming any particular function for \(U(x)\). show that W. \(x\) ) must have an inflection point at any value of \(x\) where it enters a classically forbidden region.
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Measurements

Read Explanation

Mechanics and Materials

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Modern Physics

Read Explanation

Electricity and Magnetism

Read Explanation
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.