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A free particle moving in one dimension has wave function $$\Psi(x, t) = A[e^{i(kx-vt)} - e^{i(2kx-4vt)}]$$ 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 \(\mid \Psi(x, t) \mid ^2\) is a maximum? (b) Repeat part (a) for time \(t = 2\pi/\omega\). (c) Calculate \(v_{av}\) as the distance the maxima have moved divided by the elapsed time. Compare your result to the expression \(v_{av} = (\omega_2 - \omega_1)/(k_2 - k_1)\) from Example 40.1.

Short Answer

Expert verified
(a) \(x = \frac{\pi}{k}\), \(x = \frac{3\pi}{k}\); (b) Same \(x\) values; (c) \(v_{av} = \frac{\omega}{k}\).

Step by step solution

01

Simplify the Wave Function at t = 0

Set time \( t = 0 \) in the wave function: \[\Psi(x, 0) = A \left[ e^{ikx} - e^{i2kx} \right] = A (e^{ikx} - e^{i2kx}).\]This expression is a linear combination of complex exponentials, which can be simplified using Euler’s formula \( e^{ix} = \cos(x) + i\sin(x) \). The probability density function is then expressed as:\[|\Psi(x,0)|^2 = |A|^2 |e^{ikx} - e^{i2kx}|^2.\]
02

Find the Probability Density Function at t = 0

To obtain the probability density function, compute the modulus squared:\[|\Psi(x, 0)|^2 = |A|^2 (e^{ikx} - e^{i2kx})(e^{-ikx} - e^{-i2kx}).\]Expanding, this gives:\[|\Psi(x, 0)|^2 = |A|^2 (2 - e^{ikx} - e^{-ikx}) = 2|A|^2(1 - \cos(kx)).\]
03

Determine x for Maximum Probability at t = 0

The probability density function \(|\Psi|^2\) is maximized when \(\cos(kx) = -1\). This occurs at:\[kx = (2n + 1)\pi,\]where \(n\) is an integer. Solving for \(x\):\[x = \frac{(2n+1)\pi}{k}.\]The two smallest positive values for \(x\) are \(x = \frac{\pi}{k}\) and \(x = \frac{3\pi}{k}\).
04

Simplify the Wave Function at t = 2\pi/\omega

For \(t = \frac{2\pi}{\omega}\), substitute into the wave function:\[\Psi(x, \frac{2\pi}{\omega}) = A \left[ e^{i(kx- \frac{2\pi}{\omega}v)} - e^{i(2kx - 4\frac{2\pi}{\omega}v)} \right].\]Since \(v = \omega/k\), this simplifies to:\[\Psi(x, \frac{2\pi}{\omega}) = A \left[ e^{ikx - i2\pi} - e^{i2kx - i8\pi} \right].\] Since exponentials of multiples of \(-2\pi i\) are equal to 1, this yields:\[\Psi(x, \frac{2\pi}{\omega}) = A(e^{ikx} - e^{i2kx}).\]
05

Find the Probability Density Function at t = 2\pi/\omega

Evaluate the modulus squared as at \(t = 0\):\[|\Psi(x, \frac{2\pi}{\omega})|^2 = 2|A|^2(1 - \cos(kx)).\]This is identical to the expression at \(t = 0\), so the maxima occur at the same values of \(x\).
06

Calculate Average Velocity of Maximum Positions

The maxima move from \(x_1 = \frac{\pi}{k}\) to \(x_2 = \frac{3\pi}{k}\) over time \(t = \frac{2\pi}{\omega}\). Calculate the distance moved:\[\Delta x = \frac{3\pi}{k} - \frac{\pi}{k} = \frac{2\pi}{k}.\]The average velocity \(v_{av}\) is given by:\[v_{av} = \frac{\Delta x}{\Delta t} = \frac{\frac{2\pi}{k}}{\frac{2\pi}{\omega}} = \frac{\omega}{k}.\]Compare this with \(v_{av} = \frac{\omega_2 - \omega_1}{k_2 - k_1}\), which for this problem also results in \(\frac{\omega}{k}\).

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Key Concepts

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

Wave Function
Wave functions are fundamental in quantum mechanics, offering a mathematical depiction of a quantum system. They encapsulate the system's state and information about a particle's position and momentum. For a free particle moving in one dimension, the wave function is often expressed as a combination of complex exponentials. In our problem, the wave function \[\Psi(x, t) = A[e^{i(kx-vt)} - e^{i(2kx-4vt)}]\] is used.Understanding this involves a few key points:- **Complex Exponentials**: These are parts of the wave function that describe oscillatory behavior, ensuring we can capture properties such as frequency and wavelength.- **Linear Combinations**: The function is a blend or superposition of two exponential terms, which is critical in quantum mechanics because it reflects the principle of superposition, allowing particles to exist in multiple states simultaneously.Ultimately, the wave function allows us to compute probabilities of finding particles in particular states, crucial for solving quantum mechanics problems.
Probability Density
Probability density functions are derived from wave functions and give the likelihood of locating a particle at a specific position. For a wave function \( \Psi(x, t) \), the probability density is given by the modulus squared, \( |\Psi(x, t)|^2 \).In the exercise, at time \( t = 0 \) and \( t = \frac{2\pi}{\omega} \), the resulting probability density is:\[ |\Psi(x, t)|^2 = 2|A|^2(1 - \cos(kx)). \]Here are a few important aspects:- **Maximizing Probability**: The density is maximized when \( \cos(kx) = -1 \), leading to specific quantized locations \( x \) where the probability is highest.- **Physical Meaning**: These maxima correspond to where a particle is most likely to be found. Calculating these points is essential for predicting particle behavior.This reflects the probabilistic nature of quantum mechanics, unlike classical physics, where predictions are deterministic.
Average Velocity
The concept of average velocity in quantum mechanics for probability maxima provides insight into how wave peaks move over time. In this problem, we determine the average velocity \( v_{av} \) using the relationship:\[ v_{av} = \frac{\Delta x}{\Delta t}. \]A few key points about this are:- **Maxima Movement**: By noting the shift from \( x_1 = \frac{\pi}{k} \) to \( x_2 = \frac{3\pi}{k} \) over a time period \( \Delta t = \frac{2\pi}{\omega} \), we calculate how fast the maxima move.- **Time and Spacetime Averaging**: These calculations average position changes over specified time intervals, echoing concepts of velocity known from classical physics but adapted to quantum wave dynamics.The calculated average velocity aligns with hypothetical velocity expressions from theoretical examples, reinforcing the mathematical consistency of quantum mechanics.
Complex Exponentials
Complex exponentials are fundamental in quantum physics, often emerging from wave function solutions. These functions, expressed in Euler's form \( e^{ix} = \cos(x) + i\sin(x) \), define oscillating phenomena key to understanding wave mechanics.When applied, complex exponentials:- **Enable Representation**: They allow wave functions to efficiently represent periodic behaviors in quantum systems.
  • **Concisely encode information**: Amplitude and phase are represented compactly.
  • **Facilitate Calculations**: Operations like differentiation and integration are simplified.
- **Interference and Superposition**: They underpin these key quantum mechanics concepts, essential for describing how waves interact through constructive and destructive interference.Thus, they play a critical role in both mathematical formulations and in elucidating physical phenomena involving waves and particles in quantum mechanics.

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Most popular questions from this chapter

(a) The wave nature of particles results in the quantum-mechanical situation that a particle confined in a box can assume only wavelengths that result in standing waves in the box, with nodes at the box walls. Use this to show that an electron confined in a one-dimensional box of length \(L\) will have energy levels given by $$E_n = {n^2h^2 \over8mL^2}$$ (\(Hint\): Recall that the relationship between the de Broglie wavelength and the speed of a nonrelativistic particle is \(mv = h/\lambda\). The energy of the particle is \({1\over2} mv^2\).) (b) If a hydrogen atom is modeled as a one-dimensional box with length equal to the Bohr radius, what is the energy (in electron volts) of the lowest energy level of the electron?

The \(penetration\) \(distance\) \(\eta\) in a finite potential well is the distance at which the wave function has decreased to 1/\(e\) of the wave function at the classical turning point: $$\psi(x = L + \eta) = {1\over e} \psi(L)$$ The penetration distance can be shown to be $$\eta = {\hslash \over \sqrt{2m(U_0 - E)}}$$ The probability of finding the particle beyond the penetration distance is nearly zero. (a) Find \(\eta\) for an electron having a kinetic energy of 13 eV in a potential well with \(U_0\) = 20 eV. (b) Find \(\eta\) for a 20.0-MeV proton trapped in a 30.0-MeV-deep potential well.

A particle is described by a wave function \(\psi(x) = Ae^{-\alpha x^2}\), where A and \(\alpha\) are real, positive constants. If the value of \(\alpha\) is increased, what effect does this have on (a) the particle's uncertainty in position and (b) the particle's uncertainty in momentum? Explain your answers.

A particle of mass \(m\) in a one-dimensional box has the following wave function in the region \(x\) = 0 to \(x = L\) : $$\Psi(x, t) = {1\over \sqrt2} \psi_1(x)e^{-iE_1t/\hslash} + {1\over \sqrt 2} \psi_3(x)e^{-iE_3t/\hslash}$$ 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.

An electron in a one-dimensional box has ground-state energy 2.00 eV. What is the wavelength of the photon absorbed when the electron makes a transition to the second excited state?

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