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Consider a particle in a box with rigid walls at \(x=0\) and \(x=L .\) Let the particle be in the ground level. Calculate the probability \(|\psi|^{2} d x\) that the particle will be found in the interval \(x\) to \(x+d x\) for \((\) a \() x=L / 4 ;(\) b) \(x=L / 2 ;(\) c) \(x=3 L / 4\)

Short Answer

Expert verified
(a) \(\frac{1}{L} dx\); (b) \(\frac{2}{L} dx\); (c) \(\frac{1}{L} dx\)

Step by step solution

01

Understand the Wave Function

In a one-dimensional particle in a box problem, the wave function for the particle in the ground state (n=1) is given by: \[ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) \]This formula gives the wave function of a particle in a box from \(x = 0\) to \(x = L\). We can use this to calculate the probability density.
02

Probability Density Function

The probability density function is given by \[ |\psi(x)|^2 = \left( \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) \right)^2 = \frac{2}{L} \sin^2\left(\frac{\pi x}{L}\right) \]This expression describes the probability density of finding the particle at a position \(x\) within the interval \([0, L]\).
03

Evaluate at Specific Points

Substitute the provided values of \(x\) into the probability density function to find the probability at each point:**(a) For \(x = \frac{L}{4}\):**\[ |\psi\left(\frac{L}{4}\right)|^2 = \frac{2}{L} \sin^2\left(\frac{\pi}{4}\right) = \frac{2}{L} \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{2}{L} \times \frac{1}{2} = \frac{1}{L} \]**(b) For \(x = \frac{L}{2}\):**\[ |\psi\left(\frac{L}{2}\right)|^2 = \frac{2}{L} \sin^2\left(\frac{\pi}{2}\right) = \frac{2}{L} \times 1 = \frac{2}{L} \]**(c) For \(x = \frac{3L}{4}\):**\[ |\psi\left(\frac{3L}{4}\right)|^2 = \frac{2}{L} \sin^2\left(\frac{3\pi}{4}\right) = \frac{2}{L} \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{L} \]
04

Calculate Probabilities over Small Intervals

The probability of finding the particle between \(x\) and \(x + dx\) is given by the probability density function times the small interval \(dx\):**(a) For \(x = \frac{L}{4}\):**\[ |\psi\left(\frac{L}{4}\right)|^2 dx = \frac{1}{L} dx \]**(b) For \(x = \frac{L}{2}\):**\[ |\psi\left(\frac{L}{2}\right)|^2 dx = \frac{2}{L} dx \]**(c) For \(x = \frac{3L}{4}\):**\[ |\psi\left(\frac{3L}{4}\right)|^2 dx = \frac{1}{L} dx \]

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Key Concepts

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

Probability Density
In the context of a particle in a box, probability density is a crucial concept for understanding where a particle might be located within a confined space. Essentially, the probability density function gives us a way to describe the likelihood of finding the particle at any specific position within the box. It does not give an exact location but rather a probability distribution over the range.

For a non-quantum observer, the particle might seem to bounce around randomly within the box. However, in quantum mechanics, this randomness is beautifully described by the wave function and its square, which is the probability density.
  • The probability density \(|\psi(x)|^2\) tells us how probable it is to find the particle near a point \(x\) within the box.
  • It's important to note that the probability density, multiplied by a small interval \(dx\), gives the probability of finding the particle in that interval.
Thus, by examining the probability density, one can gain insights into the behavior and expected locations of the particle, central to experiments and calculations in quantum mechanics.
Wave Function
The wave function, often denoted as \(\psi(x)\), is at the very heart of quantum mechanics, particularly in the particle in a box scenario. It captures the quantum state of a particle and provides comprehensive information about the system.

For a particle in a box with rigid walls:
  • The wave function has a specific mathematical form, especially for ground states (the lowest energy states): \[ \psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right) \]
  • This function is sinusoidal, indicating a pattern or standing wave formed by the particle trapped in the box. It is essential as it dictates how the particle can be found at various positions within the box.
The square of the wave function, \(|\psi(x)|^2\), directly relates to the probability density, serving as a bridge between the abstract concept of superposition in quantum systems and actual observable probabilities.

Understanding the wave function is crucial for mastering quantum mechanics, as it uniquely informs us of the limitations and freedoms of particles at quantum scales.
Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that describes nature at the smallest scales, such as particles and atoms. It operates in a distinctly different way compared to classical physics, with principles that seem counterintuitive but are incredibly precise in predicting experimental outcomes.

In the particle in a box model, quantum mechanics' unique principles manifest through wave functions and probability densities.
  • Here, particles exhibit wave-particle duality, meaning they have characteristics of both waves and particles.
  • The distinct boundary conditions of the box generate quantized energy levels, meaning particles can only occupy specific energy states.
  • Quantum Mechanics allows us to understand phenomena like superposition, where particles exist in all possible states simultaneously until observed.
These features challenge traditional concepts of deterministic paths and precise measurements and are critical for understanding modern technology and the universe's inherent nature. By diving into systems like the particle in a box, students can explore the quantum world's intricacies, paving the way for advanced studies and innovation in physics and engineering.

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Most popular questions from this chapter

An electron with initial kinetic energy 5.0 eV encounters a barrier with height \(U_{0}\) and width 0.60 \(\mathrm{nm} .\) What is the transmission coefficient if (a) \(U_{0}=7.0 \mathrm{eV} ;\) (b) \(U_{0}=9.0 \mathrm{eV} ;\) (c) \(U_{0}=\) 13.0 \(\mathrm{eV} ?\)

(a) Find the excitation energy from the ground level to the third excited level for an electron confined to a box that has a width of 0.125 \(\mathrm{nm}\) . (b) The electron makes a transition from the \(n=1\) to \(n=4\) level by absorbing a photon. Calculate the wave-length of this photon.

Normalization of the Wave Function. Consider a particle moving in one dimension, which we shall call the \(x\) -axis. (a) What does it mean for the wave function of this particle to be normalized? (b) Is the wave function \(\psi(x)=e^{a x},\) where \(a\) is a positive real number, normalized? Could this be a valid wave function? (c) If the particle described by the wave function \(\psi(x)=A e^{b x},\) where \(A\) and \(b\) are positive real numbers, is confined to the range \(x \geq 0\) , determine \(A\) (including its units) so that the wave function is normalized.

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?

A harmonic oscillator absorbs a photon of wavelength \(8.65 \times 10^{-6} \mathrm{m}\) when it undergoes a transition from the ground state to the first excited state. What is the ground-state energy, in electron volts, of the oscillator?

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