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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=2π/ω

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.

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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.

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Most popular questions from this chapter

A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x=0\) to \(x=L:\) $$\Psi(x, t)=\frac{1}{\sqrt{2}} \psi_{1}(x) e^{-i E_{1} t / \hbar}+\frac{1}{\sqrt{2}} \psi_{3}(x) e^{-i E_{3} t / \hbar}$$ Here \(\psi_{1}(x)\) and \(\psi_{3}(x)\) are the normalized stationary-state wave functions for the \(n=1\) and \(n=3\) levels, and \(E_{1}\) and \(E_{3}\) are the energies of these levels. The wave function is zero for \(x<0\) and for \(x>L\) (a) Find the value of the probability distribution function at \(x=L / 2\) as a function of time. (b) Find the angular frequency at which the probability distribution function oscillates.

(a) Show by direct substitution in the Schrödinger equation for the one- dimensional harmonic oscillator that the wave function \(\psi_{1}(x)=A_{1} x e^{-\alpha^{2} x^{2} / 2},\) where \(\alpha^{2}=m \omega / \hbar,\) is a solution with energy corresponding to \(n=1\) in Eq. \((40.46) .\) (b) Find the normalization constant \(A_{1}\). (c) Show that the probability density has a minimum at \(x=0\) and maxima at \(x=\pm 1 / \alpha,\) corresponding to the classical turning points for the ground state \(n=0\).

A particle moving in one dimension (the \(x\) -axis) is described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} A e^{-b x}, & \text { for } x \geq 0 \\ A e^{b x}, & \text { for } x<0 \end{array}\right. $$ where \(b=2.00 \mathrm{~m}^{-1}, A>0,\) and the \(+x\) -axis points toward the right. (a) Determine \(A\) so that the wave function is normalized. (b) Sketch the graph of the wave function. (c) Find the probability of finding this particle in each of the following regions: (i) within \(50.0 \mathrm{~cm}\) of the origin, (ii) on the left side of the origin (can you first guess the answer by looking at the graph of the wave function?), (iii) between \(x=0.500 \mathrm{~m}\) and \(x=1.00 \mathrm{~m}\).

(a) An electron with initial kinetic energy \(32 \mathrm{eV}\) encounters a square barrier with height \(41 \mathrm{eV}\) and width \(0.25 \mathrm{nm}\). What is the probability that the electron will tunnel through the barrier? (b) A proton with the same kinetic energy encounters the same barrier. What is the probability that the proton will tunnel through the barrier?

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box of width \(0.360 \mathrm{nm}\). (b) The electron makes a transition from the \(n=1\) to \(n=4\) level by absorbing a photon. Calculate the wavelength of this photon.

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