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Chapter 37: Problem 19

Consider the energies allowed for bound states of a half-harmonic oscillator, namely, a potential that is $$ U(x)=\left\\{\begin{array}{l} \frac{1}{2} m \omega_{0}^{2} x^{2} \\ \infty \end{array}\right. \text { for }\left\\{\begin{array}{l} x>0 \\ x \leq 0 \end{array}\right. $$ Using simple arguments based on the characteristics of good wave functions, what are the energies allowed for bound states in this potential?

Short Answer

Expert verified
Answer: The allowed energies for bound states of the half-harmonic oscillator are given by \(E_{2n+1} = \hbar\omega_0(2n+1 + \frac{1}{2})\), where n is a non-negative integer. These energies correspond to the odd solutions of the full harmonic oscillator.

Step by step solution

01

Set up the Schrödinger equation for the given potential function

For the given potential function, we will set up the time-independent Schrödinger equation: \(-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + U(x) \psi(x) = E \psi(x)\) Since \(U(x)\) is a piecewise function, we have to treat the wave function \(\psi(x)\)in two intervals: 1. For x > 0: \(-\frac{\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} + \frac{1}{2} m \omega_0^2 x^2 \psi(x) = E \psi(x)\) 2. For x ≤ 0: There is no wave function because \(U(x) = \infty\).
02

Analyze the wave function for x > 0

For x > 0, we will have a harmonic oscillator-like wave function with allowed energies. We know that the wave function should satisfy good wave function properties mentioned earlier. The even and odd solutions of the harmonic oscillator can provide a guide for what the bound state wave functions may look like: Even solutions: \(\psi_{2n}(x) = A_{2n} H_{2n}(x)e^{-\frac{1}{2} x^2}\) Odd solutions: \(\psi_{2n+1}(x) = A_{2n+1} H_{2n+1}(x)e^{-\frac{1}{2} x^2}\) where \(A_n\) is a normalization constant and \(H_n(x)\) is the Hermite polynomial of order n.
03

Evaluate the continuity of the wave function at the boundary x = 0

We will now examine the wave function at the boundary of x=0. For the wave function to be continuous, it must satisfy the following condition: \(\psi(0) = 0\) This gives us a hint: odd solutions of the harmonic oscillator wave functions have this property. The odd solutions are of the form: \(\psi_{2n+1}(x) = A_{2n+1} H_{2n+1}(x)e^{-\frac{1}{2} x^2}\)
04

Determine the allowed energies of bound states

From the discussion above, we can conclude that only odd wave functions of the harmonic oscillator are suitable to be the bound state wave functions for this half-harmonic oscillator. Therefore, the allowed energies for bound states would be given by: \(E_{2n+1} = \hbar\omega_0(2n+1 + \frac{1}{2})\) where n is a non-negative integer. These energies correspond to the odd solutions of the full harmonic oscillator. The key difference is that for the full harmonic oscillator, both even and odd solutions are acceptable bound states, while for the half-harmonic oscillator, only the odd solutions are bound states due to the constraint imposed by the infinite potential barrier.

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Key Concepts

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

Schrödinger Equation
The Schrödinger equation is a fundamental part of quantum mechanics, establishing how quantum systems evolve over time. It comes in two main forms: the time-dependent and time-independent equations. For problems involving energy levels and stationary states, like with the half-harmonic oscillator, we use the time-independent Schrödinger equation.

In the half-harmonic oscillator problem, the potential energy function, denoted as U(x), splits the space into two regions: where the particle can exist (x > 0), and where it cannot due to infinite potential energy (x ≤ 0). The equation for the allowed region (x > 0) closely resembles that of a standard harmonic oscillator, but with the caveat of a unidirectional constraint.
Wave Function Properties
Wave functions in quantum mechanics describe the probability amplitude of particle positions. They must adhere to several key properties: normalizability, continuity, and, in the case of bound states, they must vanish at infinity.

In the case of the half-harmonic oscillator, we are particularly interested in the continuity at the boundary of the potential, x = 0. This boundary condition forces us to accept only the odd eigenfunctions of the harmonic oscillator, as these inherently satisfy the condition of vanishing at the origin, while the even eigenfunctions do not. Furthermore, these eigenfunctions must be normalized to ensure that the total probability of finding the particle in the allowed region is one.
Harmonic Oscillator Energies
The energy levels of a quantum harmonic oscillator are quantized, meaning that the particle can only occupy certain allowed energy states. These states are defined by En = ℏω0(n + 1/2), where ℏ is the reduced Planck constant, ω0 is the angular frequency of the oscillator, and n is a non-negative integer representing the quantum number of the state.

In a half-harmonic oscillator, we have the additional constraint from the boundary condition at x = 0, which restricts the acceptable states to those with odd quantum numbers. As a result, the allowed energies take the form E2n+1 = ℏω0(2n+1 + 1/2). This reflects a spectrum where only every other energy level of the full harmonic oscillator is permitted.

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Most popular questions from this chapter

Is the superposition of two wave functions, which are solutions to the Schrödinger equation for the same potential energy, also a solution to the Schrödinger equation? a) no b) yes c) depends on potential energy d) only if \(\frac{d^{2} \psi(x)}{d x^{2}}=0\)

True or False: The larger the amplitude of a Schrödinger wave function, the larger its kinetic energy. Explain your answer.

A beam of electrons moving in the positive \(x\) -direction encounters a potential barrier that is \(2.51 \mathrm{eV}\) high and \(1.00 \mathrm{nm}\) wide. Each electron has a kinetic energy of \(2.50 \mathrm{eV},\) and the electrons arrive at the barrier at a rate of 1000 electrons/s (1000. electrons every second). What is the rate \(\mathrm{I}_{\mathrm{T}}\) in electrons/s at which electrons pass through the barrier, on average? What is the rate \(\mathrm{I}_{\mathrm{R}}\) in electrons/s at which electrons reflect back from the barrier, on average? Determine and compare the wavelengths of the electrons before and after they pass through the barrier.

An electron is confined between \(x=0\) and \(x=L\). The wave function of the electron is \(\psi(x)=A \sin (2 \pi x / L)\). The wave function is zero for the regions \(x<0\) and \(x>L\) a) Determine the normalization constant \(A\). b) What is the probability of finding the electron in the region \(0 \leq x \leq L / 3 ?\)

Consider an attractive square-well potential, \(U(x)=0\) for \(x<-\alpha, U(x)=-U_{0}\) for \(-\alpha \leq x \leq \alpha\) where \(U_{0}\) is a positive constant, and \(U(x)=0\) for \(x>\alpha .\) For \(E>0,\) the solution of the Schrödinger equation in the 3 regions will be the following: For \(x<-\alpha, \psi(x)=e^{i \kappa x}+R e^{-i \kappa x}\) where \(\kappa^{2}=2 m E / \hbar^{2}\) and \(R\) is the amplitude of a reflected wave. For \(-\alpha \leq x \leq \alpha, \psi(x)=A e^{i \kappa^{\prime} x}+B e^{-i \kappa^{\prime} x}\) and \(\left(\kappa^{\prime}\right)^{2}=2 m\left(E+U_{0}\right) / \hbar^{2}\). For \(x>\alpha, \psi(x)=T e^{i \kappa x}\) where \(T\) is the amplitude of the transmitted wave. Match \(\psi(x)\) and \(d \psi(x) / d x\) at \(-\alpha\) and \(\alpha\) and find an expression for \(R\). What is the condition for which \(R=0\) (that is, there is no reflected wave)?
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