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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.

Short Answer

Expert verified
Increasing \( \alpha \) decreases position uncertainty and increases momentum uncertainty.

Step by step solution

01

Understand the Wave Function

The wave function given is \( \psi(x) = Ae^{-\alpha x^2} \). This is a Gaussian wave function, representing the position probability distribution of a quantum particle.
02

Observation about the Width of the Wave Packet

The parameter \( \alpha \) affects the width of the Gaussian wave packet. As \( \alpha \) increases, the width of the wave packet becomes narrower in real space.
03

Relate Width to Uncertainty in Position

Using the relation \( \Delta x \approx \frac{1}{\sqrt{\alpha}} \), we find that the uncertainty in position \( \Delta x \) is inversely proportional to \( \sqrt{\alpha} \). Thus, increasing \( \alpha \) decreases \( \Delta x \), meaning the uncertainty in position decreases.
04

Apply the Heisenberg Uncertainty Principle

The Heisenberg Uncertainty Principle states \( \Delta x \Delta p \geq \frac{\hbar}{2} \). If \( \Delta x \) decreases, \( \Delta p \), the uncertainty in momentum, must increase to satisfy this inequality because of the inverse relationship between position and momentum uncertainties.
05

Determine Effect on Particle's Uncertainty in Momentum

Thus, increasing \( \alpha \) results in an increase in \( \Delta p \), the uncertainty in momentum, due to the reciprocal relationship derived from the Heisenberg Uncertainty Principle.

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Key Concepts

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

Wave Function
The wave function \(\psi(x) = Ae^{-\alpha x^2}\) is at the heart of quantum mechanics and describes the state of a quantum particle. This particular wave function is Gaussian, which means it forms a bell-shaped curve when graphed. The wave function itself doesn't give us direct information like the classical wave would; instead, it provides the probability distribution of finding a particle at a certain position. The presence of \(\alpha\) influences how spread out this wave function is. When you see a more spread-out wave function, the particle's location is less certain.
  • Amplitude (A): Determines the wave's height but not the importance of spread.
  • Gaussian Shape: The hallmark of this specific wave function, which implies rapid decrease as you move from the center.
Understanding the wave function is key because it quantitatively embodies how strange and wonderful quantum particles behave.
Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle is a fundamental concept in quantum mechanics. It tells us that we can't exactly know both the position and momentum of a particle at the same time. Mathematically, it's represented as \(\Delta x \Delta p \geq \frac{\hbar}{2}\). This means if we become more certain about where a particle is (position), we become less certain about how it's moving (momentum), and vice versa. Sometimes this principle can be confusing:
  • Position Uncertainty (\Delta x): How unsure we are about the particle's location.
  • Momentum Uncertainty (\Delta p): How unsure we are about the particle's momentum.
In the exercise, as \(\alpha\) increases, the position uncertainty decreases, forcing the momentum uncertainty to increase due to this principle.
Momentum Uncertainty
Momentum uncertainty, represented as \(\Delta p\), is a measure of how much we don't know about a particle's momentum. In quantum mechanics, due to the Heisenberg Uncertainty Principle, knowing a particle's position more precisely means knowing less about its momentum. When the \(\alpha\) in the wave function increases, the wave packet gets narrower, reducing position uncertainty. This inverse relationship means:
  • As \(\alpha\) gets larger, the particle's momentum uncertainty, \(\Delta p\), increases to maintain the principle balance.
  • This explains why it's challenging to measure exact momentum if a position is accurately known.
Conceptually, if the particle's position becomes tightly located, its "possible" velocities spread widely.
Position Uncertainty
Position uncertainty, denoted as \(\Delta x\), indicates the range in which a particle's position could be found. With the Gaussian wave function, \(\Delta x\) is related to \(\alpha\) as \(\Delta x \approx \frac{1}{\sqrt{\alpha}}\). This shows:
  • Position uncertainty decreases as \(\alpha\) increases: the wave packet tightens.
  • More \(\alpha\) makes the particle's possible location more precise.
In our given exercise, when we increase \(\alpha\), the position uncertainty shrinks, concentrating the area where we're likely to find the particle. However, this shift results in increased momentum uncertainty due to the Heisenberg Uncertainty Principle, emphasizing the trade-off nature central to quantum mechanics.

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Most popular questions from this chapter

Consider the wave packet defined by $$\psi(x) = \int ^\infty_0 B(k)cos kx dk$$ Let \(B(k) = e^{-a^2k^2}\). (a) The function \(B(k)\) has its maximum value at \(k\) = 0. Let \(k_h\) be the value of \(k\) at which \(B(k)\) has fallen to half its maximum value, and define the width of \(B(k)\) as \(w_k = k_h\) . In terms of \(\alpha\), what is \(w_k\) ? (b) Use integral tables to evaluate the integral that gives \(\psi(x)\). For what value of \(x\) is \(\psi(x)\) maximum? (c) Define the width of \(\psi(x)\) as \(w_x = x_h\) , where \(x_h\) is the positive value of \(x\) at which \(\psi(x)\) has fallen to half its maximum value. Calculate \(w_x\) in terms of \(\alpha\). (d) The momentum \(p\) is equal to \(hk/2\pi\), so the width of \(B\) in momentum is \(w_p = hw_k /2\pi\). Calculate the product \(w_p w_x\) and compare to the Heisenberg uncertainty principle.

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\(^{14}\) Hz is absorbed and that the next higher absorbed frequency is 16.9 \(\times\) 10\(^{14}\) 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?

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 eV and a width of 1.8 \(\times\) 10\(^{-10}\) m. Calculate the three lowest values of \(E\) for which \(T\) = 1.

An electron is in the ground state of a square well of width \(L = 4.00 \times 10^{-10}\) m. The depth of the well is six times the ground-state energy of an electron in an infinite well of the same width. What is the kinetic energy of this electron after it has absorbed a photon of wavelength 72 nm and moved away from the well?

A harmonic oscillator consists of a 0.020-kg mass on a spring. The oscillation frequency is 1.50 Hz, and the mass has a speed of 0.480 m/s as it passes the equilibrium position. (a) What is the value of the quantum number n for its energy level? (b) What is the difference in energy between the levels \(E_n\) and \(E_{n+1}\)? Is this difference detectable?

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