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A particle is in a state described by the wavefunction \(\psi=(\cos \chi) e^{i k x}+\) \((\sin \chi) \mathrm{e}^{-\mathrm{ikx}}\), where \(\chi\) (chi) is a parameter. What is the probability that the particle will be found with a linear momentum \((a)+k \hbar\), (b) \(-k \hbar ?\) What form would the wavefunction have if it were 90 per cent certain that the particle had linear momentum \(+k \hbar\) ?

Short Answer

Expert verified
The probability for momentum \(+k\hbar\) is \(\cos^2 \chi\), for momentum \( -k\hbar\) it is \(\sin^2 \chi\). If it's 90% certain the particle has momentum \(+k\hbar\), the wavefunction is \(\psi = \sqrt{0.9}e^{ikx} + \sqrt{0.1}e^{-ikx}\).

Step by step solution

01

Understand the Wavefunction

The given wavefunction \(\psi=(\cos \chi) e^{ikx}+(\sin \chi) e^{-ikx}\) represents the state of the particle as a superposition of two wavefunctions, with the term containing \( e^{ikx}\) corresponding to the momentum \(+k\hbar\) and the term containing \( e^{-ikx}\) corresponding to the momentum \( -k\hbar\).
02

Calculate the Probability for Momentum \(+k\hbar\)

The probability of finding the particle with linear momentum \(+k\hbar\) is given by the square of the amplitude of the term associated with \(e^{ikx}\), which is \(\cos^2 \chi\). Therefore, the probability \(P(+k\hbar) = \cos^2 \chi\).
03

Calculate the Probability for Momentum \( -k\hbar\)

Similarly, the probability of finding the particle with linear momentum \( -k\hbar\) is given by the square of the amplitude of the term associated with \(e^{-ikx}\), which is \(\sin^2 \chi\). Thus, the probability \(P(-k\hbar) = \sin^2 \chi\).
04

Determine Wavefunction for 90% Certainty of \(+k\hbar\)

To be 90% certain of finding the particle with momentum \(+k\hbar\), the amplitude coefficient for the term corresponding to \(+k\hbar\) must be such that its square is 0.9. Let \(\cos^2 \chi = 0.9\), then \(\sin^2 \chi = 0.1\) as the total probability must be 1 (i.e., \(\cos^2 \chi + \sin^2 \chi = 1\)). The corresponding wavefunction would then be of the form \(\psi = \sqrt{0.9}e^{ikx} + \sqrt{0.1}e^{-ikx}\).

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Key Concepts

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

Quantum Mechanics
When diving into the world of the extremely small, classical physics takes a back seat and quantum mechanics gets behind the wheel. 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. A central part of quantum mechanics is the wave-like nature of particles. Unlike a baseball that can only be at one place at a time, particles like electrons exist in a haze of probability, with mathematical functions—called wavefunctions—describing the likelihood of finding a particle at a certain location.

Every action at the quantum scale, from a particle's position to its momentum, is described by a wavefunction, typically denoted by the symbol \( \psi \). The square of the absolute value of the wavefunction, \( |\psi|^2 \), indicates the probability density of finding the particle at a specific point in space. This is why we often talk about probabilities, rather than certainties, when it comes to quantum behavior.
Superposition Principle
One counter-intuitive aspect of quantum mechanics is the superposition principle. This principle states that a particle can exist in all possible states simultaneously until it is measured. Think of it like a spinning coin that's both heads and tails at the same time—it's only when the coin lands (or in quantum terms, you make a measurement) that it 'chooses' a state.

In the case of our wavefunction \( \psi \), the particle is in a superposition of states that correspond to having momentum \( +k \hbar \) and \( -k \hbar \). It's the act of measurement—or observation—that 'collapses' the wavefunction into one of these definite states.
  • The term \( e^{ikx} \) accounts for the particle's state having momentum \( +k \hbar \) and \( (\cos \chi) \) tells us about the amplitude of this state.
  • Similarly, \( e^{-ikx} \) represents the state with momentum \( -k \hbar \) and \( (\sin \chi) \) provides the amplitude for this state.
  • The probabilities for these momenta can then be determined by squaring these respective amplitudes, as the square of the amplitude is directly proportional to a state's probability.
Momentum in Quantum Physics
In classical mechanics, momentum is simply mass times velocity. But in quantum mechanics, momentum is a bit more abstract and is associated with the wavelength of a particle's wavefunction. De Broglie showed us that particles could be described as waves, and the momentum of a particle is related to the waveness of the particle, articulated as \( p = \hbar k \), where \( p \) is the momentum, \( k \) is the wave number, and \( \hbar \) (h-bar) is the reduced Planck constant.

This relates to our problem where the wavefunction given represents the particle in states with momentum \( +k \hbar \) and \( -k \hbar \)—or, in wave terms, with wavelengths \( \frac{2\pi}{k} \) and \( -\frac{2\pi}{k} \) (because the wave number \( k \) is the spatial frequency of the wave, inversely proportional to its wavelength).

The probability of finding the particle with a certain momentum is derived by extracting the square of the amplitude of the corresponding term in the wavefunction. As such, this wave-particle duality is at the heart of quantum mechanics, describing entities that behave like both particles and waves.

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