/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 Compute \(|\Psi|^{2}\) for \(\Ps... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 39: Problem 30

Compute \(|\Psi|^{2}\) for \(\Psi=\psi\) sin \(\omega t,\) where \(\psi\) is time independent and \(\omega\) is a real constant. Is this a wave function for a stationary state? Why or why not?

Short Answer

Expert verified
No, it is not a stationary state because \(|\Psi|^2 = \psi^2 \, \sin^2(\omega t)\) is time-dependent.

Step by step solution

01

Understand \\\\(|\\\\\\\\Psi|^{2}\\\\\\\\) Calculation

The exercise requires you to calculate the squared modulus of the wave function \(\Psi\), which is represented as \(\Psi = \psi \sin(\omega t)\). \(\psi\) is time-independent and \(\omega\) is a real constant. The squared modulus of any wave function \(f\) is given by \(|f|^2 = f^* \cdot f\), where \(f^*\) is the complex conjugate of \(f\).
02

Find the Complex Conjugate of \\\\(|\\\\\\\\Psi|\)

For this function, \(\Psi^* = \psi^* \sin(\omega t)\). Since \(\psi\) is time-independent, we assume it is real or its square will suffice the calculation without the complex component. Thus, \(\Psi^* = \psi \sin(\omega t)\).
03

Calculate Squared Modulus \\\\(|\\\\\\\\Psi|^{2}\\\\\\\\)

The squared modulus \(|\Psi|^2\) is \(\Psi^* \cdot \Psi = (\psi \sin(\omega t))^* \cdot (\psi \sin(\omega t)) = \psi^2 \sin^2(\omega t)\).
04

Analyze Stationary State Characteristics

A wave function represents a stationary state if the probability density \(|\Psi|^2\) is time-independent. Here, \(|\Psi|^2 = \psi^2 \sin^2(\omega t)\), which is clearly time-dependent. Thus, \(\Psi\) is not a stationary state.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Complex Conjugate
In the realm of quantum mechanics, the complex conjugate plays an essential role in analyzing wave functions. Given any complex number, its conjugate is formed by changing the sign of the imaginary part. For a wave function \( \Psi = \psi \sin(\omega t) \) in the given problem, we first find the complex conjugate \( \Psi^* \).
  • If \psi is purely real and time-independent, \Psi and its conjugate are identical.
  • This simplifies calculations as dealing with complex conjugates essentially involves considering mirror images of complex parts.
Ultimately, the role of the complex conjugate is to determine the squared modulus by forming a pair: the original function and its complex conjugate.
Squared Modulus
The squared modulus \(|\Psi|^2\) reflects the core concept of normalizing the wave function in quantum mechanics. It helps us understand how the wave function behaves concerning probability. Given our function, \(|\Psi|^2\) is calculated as \(\Psi^* \cdot \Psi\).
  • If \Psi is \psi \sin(\omega t), then its squared modulus becomes: \(\psi^2 \sin^2(\omega t)\).
  • Since \psi is assumed to be real and time-independent here, any complex part drops out, simplifying the squared modulus.
This calculated value often informs us about physical properties in quantum mechanics, like uniformity and distribution of states.
Stationary State
A stationary state in quantum mechanics is a special kind of quantum state whose probability density does not change over time. In terms of our problem, this involves assessing whether \(|\Psi|^2\), the squared modulus, has any time dependence.Stationary states have
  • Time-independent probability density, which remains constant as time progresses.
  • No observable change or transformation, highlighting stability in energy levels.
When \(|\Psi|^2 = \psi^2 \sin^2(\omega t)\),alongside the oscillating sine function, it implies time dependence. Hence, the wave function in question does not represent a stationary state as it evolves with time.
Probability Density
The concept of probability density is integral to understanding quantum mechanics. It tells us the likelihood of finding a particle at a specific location at a specific time. It is derived from the squared modulus \(|\Psi|^2\).
  • Probability density explains and predicts where a particle is likely to be detected.
  • For \(|\Psi|^2 = \psi^2 \sin^2(\omega t)\),it changes over time, indicating a non-stationary behavior of this wave function.
This time variability implies the probability of finding a particle is not constant but oscillates, reflecting the dynamics of such a system in quantum mechanics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the de Broglie wavelength for an electron with speed (a) \(v=0.480 c\) and $(b) v=0.960 c ?(\text { Hint } \text { Use the correct rela- }tivistic expression for linear momentum if necessary.)

Particle \(A\) is described by the wave function \(\psi(x, y, z)\) . Particle \(B\) is described by the wave function \(\psi(x, y, z) e^{i \phi}\) , where \(\phi\) is a real constant. How does the probability of finding particle \(A\) within a volume \(d V\) around a certain point in space compare with the probability of finding particle \(B\) within this same volume?

(a) A particle with mass \(m\) has kinetic energy equal to three times its rest encrgy. What is the de Broglic wavelength of this particle? (Hint: You must use the relativistic expressions for momentum and kinetic energy: \(E^{2}=(p c)^{2}+\left(m c^{2}\right)^{2}\) and \(K=E-\) \(m c^{2} \cdot(b)\) Determine the numerical value of the kinetic energy (in Mev) and the wavelength (in meters) if the particle in part (a) is (i) an electron and (i) a proton.

Proton Energy in a Nucleus. The radii of atomic nuclei are of the order of \(5.0 \times 10^{-15} \mathrm{m}\) . (a) Estimate the minimum uncertainty in the momentum of a proton if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Fq. \((37.39),\) to obtain an estimate of the kinetic energy of a proton confined within a nucleus, (c) For a proton to remain bound within a nucleus, what must the magnitude of the (negative) potential energy for a proton be within the nucleus? Give your answer in \(\mathrm{oV}\) and in MoV. Compare to the potential energy for an electron in a hydrogen atom, which has a magnitude of a few tens of eV. (This shows why the interaction that binds the nucleus together is called the "strong nuclear force.")

(a) Approximately how fast should an electron move so it has a wavelength that makes it useful to measure the distance between adjacent atoms in typical crystals (about 0.10 \(\mathrm{nm} ) ?\) (b) What is the kinetic energy of the electron in part (a)?(c) What would be the energy of a photon of the same wavelength as the electron in part \((b) ?(d)\) Which would make a more effective probe of small-scale structures, electrons or photons? Why?
See all solutions

Recommended explanations on Physics Textbooks

Mechanics and Materials

Read Explanation

Physics of Motion

Read Explanation

Measurements

Read Explanation

Solid State Physics

Read Explanation

Electromagnetism

Read Explanation

Classical Mechanics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.