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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
By increasing the value of \(\alpha\), (a) the particle's uncertainty in position decreases as the spatial distribution of the wave function becomes narrower, and (b) the particle's uncertainty in momentum increases according to Heisenberg's uncertainty principle, since a decrease in position uncertainty is compensated by an increase in momentum uncertainty.

Step by step solution

01

Understanding the given wave function

Firstly, let's understand the given wave function \(\psi(x) = A e^{-\alpha x^2}\). The parameter \(\alpha\) determines the width of the wave function, and hence influences the spatial distributio of the wave function. If \(\alpha\) were to increase, the wave function would become sharper, i.e., it would peak more sharply at its maximum value. Note that the square of the wave function gives the probability distribution of finding the particle in space.
02

Applying Heisenberg's uncertainty Principle to determine uncertainty in position

According to Heisenberg's uncertainty principle: \(\Delta x \Delta p \geq \hbar/2\), where \(\Delta x\) is the uncertainty in position, \(\Delta p\) is the uncertainty in momentum, and \(\hbar\) is the reduced Planck's constant. If the wave function becomes narrower (which happens when \(\alpha\) is increased), the position of the particle is known more precisely, i.e., the uncertainty in the position decreases.
03

Determining the uncertainty in momentum

Now let's consider the momentum uncertainty. According to Heisenberg's uncertainty principle: if the position uncertainty decreases (which happens when \(\alpha\) is increased), then the momentum uncertainty must increase to maintain the inequality. Therefore, increasing \(\alpha\) will result in an increase in the uncertainty in the momentum of the particle.

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Key Concepts

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

Wave Function
In the fascinating world of quantum mechanics, the wave function is fundamental to understanding the behavior of particles at the microscopic level. It is essentially a mathematical description of the quantum state of a particle or system, represented by the symbol \( \psi \). The wave function, \( \psi(x) = A e^{-\alpha x^2} \), given in the exercise, describes the probability amplitude of a particle's position. Probability amplitude can be a bit perplexing, but you can think of it as a way to calculate the likelihood of finding a particle in a particular place when you look for it.

The absolute square of the wave function, \( |\psi(x)|^2 \) (not to be mistaken with a simple square), gives us the probability density, which is the chance of locating the particle at point \( x \). Now, when the parameter \( \alpha \) is cranked up, the wave function narrows, meaning our particle's position becomes more predictable, assuming \( A \) remains constant. It's as if the particle's hangout spots are getting fewer and tighter.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that provides a description of the physical properties of nature at the scale of atoms and subatomic particles. It is a complex and sometimes non-intuitive framework, in which particles exhibit wavelike properties, leading to concepts like superposition, where particles can exist in multiple states at once, and entanglement, a peculiar connection between particles that Einstein famously referred to as 'spooky action at a distance'.

This branch of physics is pivotal in understanding and predicting the behavior of systems on a small scale, as it reveals phenomena that are entirely absent from the classical world we perceive daily. For instance, the wave function we talked about previously is a quantum mechanical concept, one that classical physics cannot explain. It demonstrates that the classical idea of a particle having a defined trajectory is replaced with a probability distribution in the quantum realm.
Momentum Uncertainty
Momentum uncertainty in quantum mechanics is a direct consequence of Heisenberg's uncertainty principle, which tells us that certain pairs of properties, like position and momentum, cannot both be measured precisely at the same time. The more precisely we know where a particle is (its position), the less precisely we can know how fast and in what direction it's moving (its momentum), and vice versa.

In the case of our exercise, when the wave function's parameter \( \alpha \) increases, the particle's position uncertainty decreases. This is because a sharper wave function peak means a more confined probability of finding the particle. Due to Heisenberg's principle, this confinement in position unavoidably leads to greater uncertainty in the particle's momentum. It's a bit like the universe's trade-off; when it gives us precise knowledge of one aspect (position), it demands uncertainty in another (momentum).

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Most popular questions from this chapter

(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?

Consider a wave function given by \(\psi(x)=A \sin k x,\) where \(k=2 \pi / \lambda\) and \(A\) is a real constant. (a) For what values of \(x\) is there the highest probability of finding the particle described by this wave function? Explain. (b) For which values of \(x\) is the probability zero? Explain.

An electron is in a box of width \(3.0 \times 10^{-10} \mathrm{~m} .\) What are the de Broglie wavelength and the magnitude of the momentum of the electron if it is in (a) the \(n=1\) level; (b) the \(n=2\) level; (c) the \(n=3\) level? In each case how does the wavelength compare to the width of the box?

One advantage of the quantum dot is that, compared to many other fluorescent materials, excited states have relatively long lifetimes (10 ns). What does this mean for the spread in the energy of the photons emitted by quantum dots? (a) Quantum dots emit photons of more well-defined energies than do other fluorescent materials. (b) Quantum dots emit photons of less well-defined energies than do other fluorescent materials. (c) The spread in the energy is affected by the size of the dot, not by the lifetime. (d) There is no spread in the energy of the emitted photons, regardless of the lifetime.

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

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