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A particle is described by a wave function \(\psi(x)=A e^{-\alpha x^{2}}\) 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 the uncertainty in position and increases the uncertainty in momentum.

Step by step solution

01

Identify the Wave Function

The wave function of the particle is given by \( \psi(x) = A e^{-\alpha x^2} \). This describes a Gaussian function, which is centered around zero and its spread is determined by the parameter \( \alpha \).
02

Effect on Position Uncertainty

The uncertainty in position, \( \Delta x \), for a Gaussian function is inversely related to the square root of the parameter \( \alpha \). Mathematically, \( \Delta x \propto \frac{1}{\sqrt{\alpha}} \). Hence, as \( \alpha \) increases, \( \Delta x \) decreases, meaning the particle is more localized.
03

Effect on Momentum Uncertainty

The uncertainty in momentum \( \Delta p \) is related to the uncertainty in position due to the Heisenberg Uncertainty Principle, \( \Delta x \Delta p \geq \frac{\hbar}{2} \). With \( \Delta x \propto \frac{1}{\sqrt{\alpha}} \), \( \Delta p \propto \sqrt{\alpha} \). Therefore, increasing \( \alpha \) leads to an increase in \( \Delta p \).

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Key Concepts

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

Uncertainty Principle
The uncertainty principle is a fundamental concept in quantum mechanics. It states that certain pairs of physical properties, like position and momentum, cannot be simultaneously measured with arbitrary precision. For any particle, if you try to measure its position very accurately, you will find that you cannot measure its momentum to the same degree of accuracy, and vice versa.

In mathematical terms, the product of the uncertainties in position (\(\Delta x\)) and momentum (\(\Delta p\)) is always greater than or equal to \(\frac{\hbar}{2}\), where \(\hbar\) is the reduced Planck's constant:
  • \(\Delta x \Delta p \geq \frac{\hbar}{2}\)
This implies a fundamental limit on the precision of such measurements.

The principle highlights the intrinsic limitations of our ability to know all the properties of a particle at once, underscoring the nature of quantum uncertainty.
Wave Function
In quantum mechanics, a wave function is a mathematical description of the quantum state of a particle or a system of particles. It encapsulates all the possible information about a system. The wave function is typically denoted as \(\psi(x)\), where \(x\) represents the position.

The wave function provides the probability amplitude of finding a particle at a particular position. By squaring the wave function, \(|\psi(x)|^2\), one obtains the probability density. This is crucial because it tells us where a particle is likely to be found upon measurement.

A wave function must satisfy certain conditions to be physically meaningful. It should be normalizable, meaning the total probability of finding the particle somewhere in space must be 1:
  • \(\int\limits_{-\infty}^{\infty} |\psi(x)|^2 \, dx = 1\)
Thus, the wave function is central to understanding and predicting a particle’s behavior.
Gaussian Function
A Gaussian function is a specific type of mathematical function that appears frequently in statistics and quantum mechanics. It has the general form \(A e^{-\alpha x^2}\), where \(A\) and \(\alpha\) are constants.

In quantum mechanics, a Gaussian function often represents a wave function with a peak centered at a certain position with a spread determined by \(\alpha\). The spread of the Gaussian function is related to the uncertainty in position. Specifically:
  • As \(\alpha\) increases, the function becomes narrower, reducing the position uncertainty.
  • Conversely, a smaller \(\alpha\) results in a wider spread, indicating greater uncertainty in position.
Gaussian functions are very useful because they simplify many calculations and are easy to manipulate mathematically due to their symmetrical properties.
Heisenberg Uncertainty Principle
The Heisenberg Uncertainty Principle is a cornerstone of quantum mechanics formulated by Werner Heisenberg. It provides a quantitative relationship between uncertainties in position and momentum.

This principle does not arise due to technical imperfections in measurement but is instead an inherent property of quantum systems. It can be represented as:
  • \(\Delta x \Delta p \geq \frac{\hbar}{2}\)
According to this, if a particle's position is very precisely measured (small \(\Delta x\)), the uncertainty in its momentum (\(\Delta p\)) becomes larger.

In the context of the question, increasing \(\alpha\) makes \(\Delta x\) smaller, which implies, by the uncertainty principle, that \(\Delta p\) must increase. This highlights the trade-off between knowing a particle's exact position and its momentum, a fundamental characteristic of the quantum realm.

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Most popular questions from this chapter

A beam of 40 -V electrons traveling in the \(+x\) direction passes through a slit that is parallel to the \(y\) -axis and 5.0\(\mu \mathrm{m}\) wide. The diffraction pattern is recorded on a screen 2.5 \(\mathrm{m}\) from the slit. (a) What is the de Broglie wavelength of the electrons? (b) How much time does it take the electrons to travel from the slit to the screen? (c) Use the width of the central diffraction pattern to calculate the uncertainty in the \(y\) -component of momentum of an electron just after it has passed through the slit. (d) Use the result of part (c) and the Heisenberg uncertainty principle (Eq, 39.11 for \(y\) ) to estimate the minimum uncertainty in the \(y\) -coordinate of an electron just after it has passed through the slit. Compare your result to the width of the slit.

A particle moving in one dimension (the \(x\) -axis) is described by the wave function $$ \psi(x)=\left\\{\begin{array}{ll} A e^{-b x}, & \text { for } x \geq 0 \\ A e^{k x}, & \text { for } x<0 \end{array}\right. $$ 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 cach 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?), (iit) betwcen \(x=0.500 \mathrm{mand} x=1.00 \mathrm{~m}\)

(a) What is the de Broglie wavelength of an electron accelerated from rest through a potential increase of 125 \(\mathrm{V} ?\) (b) What is the de Broglie wavelength of an alpha particle \((q=+2 e,\) \(m=6.64 \times 10^{-27} \mathrm{kg} )\) accelerated from rest through a potential drop of 125 \(\mathrm{V} ?\)

You want to study a biological spocimen by mcans of a wavelength of \(10.0 \mathrm{nm},\) and you have a choice of using electromagnetic waves or an electron microscope. (a) Calculate the ratio of the energy of a 10.0 -nm- wavelength photon to the kinetic energy of a 10.0 -nm-wavelength electron. (b) In view of your answer to part (a), which would be less damaging to the specimen you are studying; photons or electrons?

An atom in a metastable state has a lifetime of 5.2 \(\mathrm{ms}\) . What is the uncertainty in energy of the metastable state?

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