Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 41

Show that \(\Psi(x, t)=A \sin (k x-\omega t) \quad\) and \(\Psi(x, t)=A \cos (k x-\omega t) \quad\) do not obey Schrödinger's time-dependent equation.

Short Answer

Expert verified
By substituting both the given wavefunctions into the time-dependent Schrödinger equation and showing that the resulting equations are not satisfied for any potential, we are able to demonstrate that these wavefunctions do not obey the time-dependent Schrödinger equation.

Step by step solution

01

Consider the first wavefunction and its derivatives

The first wavefunction is \(\Psi_1(x, t)=A \sin (k x-\omega t)\). Compute its first and second time and position derivatives. For time derivative: \(\frac{\partial \Psi_1}{\partial t} = -A\omega \cos(kx - \omega t)\). For position derivative: \(\frac{\partial \Psi_1}{\partial x} = Ak\cos(kx - \omega t)\), and \(\frac{\partial^2 \Psi_1}{\partial x^2} = -Ak^2\sin(kx - \omega t)\).
02

Substitute into Schrödinger's equation

Substitute these derivatives into the time-dependent Schrödinger equation. It gives \(i\hbar A\omega \cos(kx - \omega t) = -\frac{\hbar^2}{2m}Ak^2\sin(kx - \omega t) + VA\sin(kx - \omega t)\). This equation is not automatically satisfied for any potential V, hence the wavefunction \(\Psi_1(x, t)\) does not satisfy the time-dependent Schrödinger equation.
03

Repeat steps for the second wavefunction

Repeat the process for the second wavefunction \(\Psi_2(x, t)=A \cos(k x-\omega t)\). First calculate the derivatives: \(\frac{\partial \Psi_2}{\partial t} = -A\omega \sin(kx - \omega t)\), \(\frac{\partial \Psi_2}{\partial x} = -Ak\sin(kx - \omega t)\), and \(\frac{\partial^2 \Psi_2}{\partial x^2} = -Ak^2\cos(kx - \omega t)\). Substituting these results into Schrödinger's equation leads to \(i\hbar A\omega \sin(kx - \omega t) = -\frac{\hbar^2}{2m}Ak^2\cos(kx - \omega t) - VA\cos(kx - \omega t)\), which is not an identity for all potentials V. Hence, the wavefunction \(\Psi_2(x, t)\) also does not obey the time-dependent Schrödinger equation.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Measurements indicate that an atom remains in an excited state for an average time of 50.0 ns before making a transition to the ground state with the simultaneous emission of a 2.1 -eV photon. (a) Estimate the uncertainty in the frequency of the photon. (b) What fraction of the photon's average frequency is this?

If a quantum particle is in a stationary state, does it mean that it does not move?

A free proton has a wave function given by \(\Psi(x, t)=A e^{i\left(5.02 \times 10^{11} x-8.00 \times 10^{15} t\right)}\) The coefficient of \(x\) is inverse meters \(\left(\mathrm{m}^{-1}\right)\) and the coefficient on \(t\) is inverse seconds \(\left(\mathrm{s}^{-1}\right) .\) Find its momentum and energy.

Can the magnitude of a wave function \((\Psi *(x, t) \Psi(x, t))\) be a negative number? Explain.

Find the expectation value of the position squared when the particle in the box is in its third excited state and the length of the box is \(L\)
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Radiation

Read Explanation

Space Physics

Read Explanation

Particle Model of Matter

Read Explanation

Classical Mechanics

Read Explanation

Electricity and Magnetism

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.

Sign-up for free