/*! 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 2 A free particle moving in one di... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 40: Problem 2

A free particle moving in one dimension has wave function $$ \Psi(x, t)=A\left[e^{i(k x-\omega t)}-e^{i(2 k x-4 \omega t)}\right] $$ where \(k\) and \(\omega\) are positive real constants. (a) At \(t=0\) what are the two smallest positive values of \(x\) for which the probability function \(|\Psi(x, t)|^{2}\) is a maximum? (b) Repeat part (a) for time \(t=2 \pi / \omega .\) (c) Calculate \(v_{\text {av }}\) as the distance the maxima have moved divided by the elapsed time. Compare your result to the expression \(v_{\mathrm{av}}=\left(\omega_{2}-\omega_{1}\right) /\left(k_{2}-k_{1}\right)\) from Example 40.1

Short Answer

Expert verified
The smallest two positive values of \(x\) where the function is maximized are \(\pi/k\) and \(2\pi/k\) at \(t=0\) and are \(3\pi/k\) and \(4\pi/k\) at \(t=2\pi/\omega\). The calculated average speed of \(\omega/k\) matches the given expression.

Step by step solution

01

Finding the Maximum of the Probability Function at t=0

Firstly, we need to find the maximum of the function \(|\Psi(x, 0)|^{2}\). The wave function at t=0 is \( \Psi(x, 0)=A\left[e^{i(k x)}-e^{i(2 k x)}\right]\). The square of the absolute value of this function is \(|\Psi(x, 0)|^{2}=|A|^2\left[2 - 2\cos(kx)\right]\). We can solve for maximum by setting the derivative of this function to 0, so we get \(kx = n\pi\), where \(n\) is an integer. The smallest two positive values of \(x\) are thus \(x = \pi/k\) and \(x = 2\pi/k\).
02

Finding the Maximum of the Probability Function at t=2蟺/蠅

Now we need to find the maximum of the function \(|\Psi(x, 2\pi/\omega)|^{2}\). Substituting \(t = 2\pi/\omega\) into the wave function, we get \( \Psi(x, 2\pi/\omega)=A\left[e^{i(k x-\omega t)}-e^{i(2 k x-4 \omega t)}\right]=A\left[e^{i(k x-2\pi)}-e^{i(2 k x-4\pi)}\right]=A\left[e^{ikx}e^{-2i\pi}-e^{2ikx}e^{-4i\pi}\right]\). Again the square of the absolute value of this function is \(|\Psi(x, 2\pi/\omega)|^{2}=|A|^2\left[2 - 2\cos(kx)\right]\). By using the same method as in the previous step, we find that the smallest two positive values of \(x\) are \(x = 3\pi/k\) and \(x = 4\pi/k\).
03

Calculating the Average Speed and Comparing it

The average speed is the distance the maxima have moved divided by the elapsed time. In this case, it is \([3\pi/k - \pi/k] / [2\pi/\omega] = \omega/k\). This is the same as the given expression \(\left(\omega_{2}-\omega_{1}\right) /\left(k_{2}-k_{1}\right)\) since for our problem, \(\omega_{2}\) - \(\omega_{1}\) is equal to \(\omega\) and \(k_{2}\) - \(k_{1}\) is equal to \(k\). Therefore, it matches and is consistent with the given result from Example 40.1.

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.

Probability Density Function
Understanding the probability density function is crucial when dealing with quantum mechanics and the behavior of particles at the quantum level. It helps us predict where a particle is likely to be found. In quantum mechanics, the probability density function is given by the square of the absolute value of the wave function, denoted as \(\left|\Psi(x, t)\right|^2\). This value tells us the likelihood of finding the particle at position \(x\) at time \(t\).

For example, in our exercise, the probability function reaches maximum values at certain points. These points are of great interest because they indicate the positions where the particle is most likely to be detected. By determining the locations of these maximum points, we can better understand the motion and behavior of the particle over time. The values found in the solution (such as \(x = \frac{\pi}{k}\) and \(x = \frac{2\pi}{k}\)) are significant because they pinpoint these high-probability locations.
Wave Function Analysis
Wave function analysis involves exploring the properties and behaviors of the wave function, \(\Psi(x, t)\), which encapsulates the quantum state of a particle. Analyzing a wave function provides insights into various physical quantities such as momentum and energy, and also influences the probability density function. The mathematical form of the wave function encodes all the information about the quantum system.

As observed in the solved problem, the wave function can be complex, involving terms with exponential expressions of imaginary numbers, reflecting the wave-like properties of particles. By examining these terms at various times, such as \(t = 0\) or \(t = \frac{2\pi}{\omega}\), we can track how the probability of finding the particle at certain positions changes with time. Wave function analysis thus forms the backbone of predicting and explaining the probabilistic nature of quantum systems.
Quantum Mechanics Principles
Quantum mechanics principles are fundamental rules that govern the behavior of particles on a quantum scale. These principles challenge our classical understanding of physics, introducing concepts such as wave-particle duality, quantization, and uncertainty. One of the key principles is that physical quantities are often quantized, meaning they can only take on certain discrete values. Another principle is Heisenberg's uncertainty principle, which states that certain pairs of physical properties, like position and momentum, cannot be simultaneously measured to arbitrary precision.

In the context of our exercise, these principles manifest through the wave function and its evolution over time. The analysis showing that the maxima of the probability density function move in a quantized manner with certain values of \(x\) being more probable than others is a direct result of these quantum mechanics principles. This draws a clear distinction from classical physics, where we can normally predict exact outcomes rather than probabilities.

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

The WKB Approximation. It can be a challenge to solve the Schr枚dinger equation for the bound-state energy levels of an arbitrary potential well. An alternative approach that can yield good approximate results for the energy levels is the \(W K B\) approximation (named for the physicists Gregor Wentzel, Hendrik Kramers, and L茅on Brillouin, who pioneered its application to quantum mechanics). The WKB approximation begins from three physical statements: (i) According to de Broglie, the magnitude of momentum \(p\) of a quantum-mechanical particle is \(p=h / \lambda\). (ii) The magnitude of momentum is related to the kinetic energy \(K\) by the relationship \(K=p^{2} / 2 m .\) (iii) If there are no nonconservative forces, then in Newtonian mechanics the energy \(E\) for a particle is constant and equal at each point to the sum of the kinetic and potential energies at that point: \(E=K+U(x),\) where \(x\) is the coordinate. (a) Combine these three relationships to show that the wavelength of the particle at a coordinate \(x\) can be written as $$ \lambda(x)=\frac{h}{\sqrt{2 m[E-U(x)]}} $$ Thus we envision a quantum- mechanical particle in a potential well \(U(x)\) as being like a free particle, but with a wavelength \(\lambda(x)\) that is a function of position. (b) When the particle moves into a region of increasing potential energy, what happens to its wavelength? (c) At a point where \(E=U(x),\) Newtonian mechanics says that the particle has zero kinetic energy and must be instantaneously at rest. Such a point is called a classical turning point, since this is where a Newtonian particle must stop its motion and reverse direction. As an example, an object oscillating in simple harmonic motion with amplitude \(A\) moves back and forth between the points \(x=-A\) and \(x=+A ;\) each of these is a classical turning point, since there the potential energy \(\frac{1}{2} k^{\prime} x^{2}\) equals the total energy \(\frac{1}{2} k^{\prime} A^{2}\). In the WKB expression for \(\lambda(x),\) what is the wavelength at a classical turning point? (d) For a particle in a box with length \(L,\) the walls of the box are classical turning points (see Fig. 40.8\()\) Furthermore, the number of wavelengths that fit within the box must be a half-integer (see Fig. 40.10 ), so that \(L=(n / 2) \lambda\) and hence \(L / \lambda=n / 2,\) where \(n=1,2,3, \ldots\) [Note that this is a restatement of Eq. (40.29).] The WKB scheme for finding the allowed bound-state energy levels of an arbitrary potential well is an extension of these observations. It demands that for an allowed energy \(E\), there must be a half-integer number of wavelengths between the classical turning points for that energy. Since the wavelength in the WKB approximation is not a constant but depends on \(x\), the number of wavelengths between the classical turning points \(a\) and \(b\) for a given value of the energy is the integral of \(1 / \lambda(x)\) between those points: $$ \int_{a}^{b} \frac{d x}{\lambda(x)}=\frac{n}{2} \quad(n=1,2,3, \ldots) $$ Using the expression for \(\lambda(x)\) you found in part (a), show that the \(W K B\) condition for an allowed bound-state energy can be written as $$ \int_{a}^{b} \sqrt{2 m[E-U(x)]} d x=\frac{n h}{2} \quad(n=1,2,3, \ldots) $$ (e) As a check on the expression in part (d), apply it to a particle in a box with walls at \(x=0\) and \(x=L\). Evaluate the integral and show that the allowed energy levels according to the WKB approximation are the same as those given by Eq. (40.31). (Hint: since the walls of the box are infinitely high, the points \(x=0\) and \(x=L\) are classical turning points for any energy \(E .\) Inside the box, the potential energy is zero.) (f) For the finite square well shown in Fig. \(40.13,\) show that the \(\mathrm{WKB}\) expression given in part (d) predicts the same bound-state energies as for an infinite square well of the same width. (Hint: Assume \(E

The Schr枚dinger equation for the quantum- mechanical harmonic oscillator in Eq. (40.44) may be written as $$ -\frac{\hbar^{2}}{2 m} \psi^{\prime \prime}+\frac{1}{2} m \omega^{2} x^{2} \psi=E \psi $$ where \(\omega=\sqrt{k^{\prime} / m}\) is the angular frequency for classical oscillations. Its solutions may be cleverly addressed using the following observations: The difference in applying two operations \(A\) and \(B,\) in opposite order, called a commutator, is denoted by \([A, B]=A B-B A\). For example, if \(A\) represents differentiation, so that \(A f=\frac{d f}{d x},\) where \(f=f(x)\) is a function, and \(B\) represents multiplication by \(x,\) then \([A, B]=\left[\frac{d}{d x}, x\right] f=\frac{d}{d x}(x f)-x \frac{d f}{d x} .\) Applying the product rule for differentiation on the first term, we determine \(\left[\frac{d}{d x}, x\right] f=f,\) which we summarize by writing \(\left[\frac{d}{d x}, x\right]=1 .\) Similarly, consider the two operators \(a_{+}\) and \(a_{-}\) defined by \(a_{\pm}=1 / \sqrt{2 m \hbar \omega}\left(\mp \hbar \frac{d}{d x}+m \omega x\right)\) (a) Determine the commutator \(\left[a_{-}, a_{+}\right]\). (b) Compute \(a_{+} a_{-\psi}\) in terms of \(\psi\) and \(\psi^{\prime \prime},\) where \(\psi=\psi(x)\) is a wave function. Be mindful that \(\frac{d}{d x}(x \psi)=\psi+x \psi^{\prime} .\) (c) The Schr枚dinger equation above may be written as \(H \psi=E \psi,\) where \(H\) is a differential operator. Use your previous result to write \(H\) in terms of \(a_{+}\) and \(a_{-}\) (d) Determine the commutators \(\left[H, a_{\pm}\right] .\) (e) Consider a wave function \(\psi_{n}\) with definite energy \(E_{n},\) whereby \(H \psi_{n}=E_{n} \psi_{n} .\) If we define a new function \(\psi_{n+1}=a_{+} \psi_{n},\) then we can readily determine its energy by computing \(H \psi_{n+1}=H a_{+} \psi_{n}=\left(a_{+} H+\left[H, a_{+}\right]\right) \psi_{n}\) and comparing the result with \(E_{n+1} \psi_{n+1}\). Finish this calculation by inserting what we have determined to find \(E_{n+1}\) in terms of \(E_{n}\) and \(n .\) (f) We can prove that the lowest-energy wave function \(\psi_{0}\) is annihilated by \(a_{-},\) meaning that \(a_{-} \psi_{0}=0 .\) Using your expression for \(H\) in terms of \(a_{\pm},\) determine the ground-state energy \(E_{0}\). (g) By applying \(a_{+}\) repeatedly, we can generate all higher states. The energy of the \(n\) th excited state can be determined by considering \(\mathrm{Ha}_{+}^{n} \psi_{0}=E_{0} \psi_{0}\). Using the above results, determine \(E_{n}\). (Note: We did not have to solve any differential equations or take any derivatives to determine the spectrum of this theory.)

When low-energy electrons pass through an ionized gas, electrons of certain energies pass through the gas as if the gas atoms weren't there and thus have transmission coefficients (tunneling probabilities) \(T\) equal to unity. The gas ions can be modeled approximately as a rectangular barrier. The value of \(T=1\) occurs when an integral or half-integral number of de Broglie wavelengths of the electron as it passes over the barrier equal the width \(L\) of the barrier. You are planning an experiment to measure this effect. To assist you in designing the necessary apparatus, you estimate the electron energies \(E\) that will result in \(T=1\). You assume a barrier height of \(10 \mathrm{eV}\) and a width of \(1.8 \times 10^{-10} \mathrm{~m} .\) Calculate the three lowest values of \(E\) for which \(T=1\)

In your research on new solid-state devices, you are studying a solid-state structure that can be modeled accurately as an electron in a one-dimensional infinite potential well (box) of width \(L\). In one of your experiments, electromagnetic radiation is absorbed in transitions in which the initial state is the \(n=1\) ground state. You measure that light of frequency \(f=9.0 \times 10^{14} \mathrm{~Hz}\) is absorbed and that the next higher absorbed frequency is \(16.9 \times 10^{14} \mathrm{~Hz}\). (a) What is quantum number \(n\) for the final state in each of the transitions that leads to the absorption of photons of these frequencies? (b) What is the width \(L\) of the potential well? (c) What is the longest wavelength in air of light that can be absorbed by an electron if it is initially in the \(n=1\) state?

A fellow student proposes that a possible wave function for a free particle with mass \(m\) (one for which the potential-energy function \(U(x)\) is zero) is $$ \psi(x)=\left\\{\begin{array}{ll} e^{+\kappa x}, & x<0 \\ e^{-\kappa x}, & x \geq 0 \end{array}\right. $$ where \(\kappa\) is a positive constant. (a) Graph this proposed wave function. (b) Show that the proposed wave function satisfies the Schr枚dinger equation for \(x<0\) if the energy is \(E=-\hbar^{2} \kappa^{2} / 2 m-\) that is, if the energy of the particle is negative. (c) Show that the proposed wave function also satisfies the Schr枚dinger equation for \(x \geq 0\) with the same energy as in part (b). (d) Explain why the proposed wave function is nonetheless not an acceptable solution of the Schr枚dinger equation for a free particle. (Hint: What is the behavior of the function at \(x=0 ?\) ) It is in fact impossible for a free particle (one for which \(U(x)=0\) ) to have an energy less than zero.
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Particle Model of Matter

Read Explanation

Waves Physics

Read Explanation

Modern Physics

Read Explanation

Force

Read Explanation

Geometrical and Physical Optics

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.