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Consider a particle of mass \(m\) subject to a one-dimensional potential of the following form: \\[ V=\left\\{\begin{array}{ll} \frac{1}{2} k x^{2} & \text { for } x>0 \\ \infty & \text { for } x<0 \end{array}\right. \\] (a) What is the ground-state energy? (b) What is the expectation value \(\left\langle x^{2}\right\rangle\) for the ground state?

Short Answer

Expert verified
Ground-state energy: \( E_0 = \frac{3}{4} \hbar \omega \). Expectation value: \( \langle x^2 \rangle = \frac{\hbar}{4m\omega} \).

Step by step solution

01

Identify the Potential Energy Problem

The potential energy function given is a harmonic oscillator for \(x > 0\) with \( V(x) = \frac{1}{2}kx^2 \) and an infinite potential wall at \( x < 0 \). This is a half-harmonic oscillator problem.
02

Determine the Ground-State Energy for Half-Harmonic Oscillator

For a full harmonic oscillator, the ground state energy is given by \( E_0 = \frac{1}{2} \hbar \omega \), where \( \omega = \sqrt{\frac{k}{m}} \). However, due to the infinite barrier at \(x < 0\), the wavefunction must be zero at \(x = 0\), doubling the energy of the ground state: \( E_0 = \frac{3}{4} \hbar \omega \).
03

Calculate the Expectation Value \( \langle x^2 \rangle \)

For a full harmonic oscillator in the ground state, \( \langle x^2 \rangle = \frac{\hbar}{2m\omega} \). For the half-harmonic oscillator, the normalization changes, resulting in \( \langle x^2 \rangle = \frac{\hbar}{4m\omega} \).
04

Re-examine Assumptions and Assertions

Revalidate assumptions involved, such as symmetry breaking due to the potential wall, and confirm calculations align with physical expectations, such as wavefunction behavior near \(x = 0\).

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Key Concepts

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

half-harmonic oscillator
In quantum mechanics, the concept of a half-harmonic oscillator is a fascinating topic that delves into the behavior of particles in a constrained potential. Typically, a harmonic oscillator refers to a system like a spring, where the potential energy is represented as \( V(x) = \frac{1}{2} k x^2 \). This is valid across all space and reflects the symmetric nature of such systems. However, adding a twist, if we have an infinite potential barrier at \( x < 0 \), the system becomes a half-harmonic oscillator.

In this scenario, the particle is only allowed to move in the positive direction because the left side of the potential \( x<0 \) acts as a wall that the particle cannot penetrate. This constraint modifies how we describe the particle’s wavefunction. For \( x>0 \), the potential is still parabolic like the standard harmonic oscillator. Yet, at \( x = 0 \), the wavefunction must be zero, a condition imposed by the infinite potential barrier. This leads to distinct differences in the particle's energy levels and wavefunctions compared to the traditional harmonic oscillator.

Understanding the half-harmonic oscillator helps illustrate how boundaries or potential barriers alter quantum systems. It demonstrates the importance of potential constraints in defining the scope of particle movement and influences what the ground and excited states look like.
ground-state energy
The ground-state energy of a quantum system is its lowest possible energy state. This is a key concept in quantum mechanics, as it represents the stable state to which a system naturally tends. For a standard harmonic oscillator, the ground-state energy is given by the equation \( E_0 = \frac{1}{2} \hbar \omega \), where \( \hbar \) is the reduced Planck's constant and \( \omega \) is the angular frequency of the oscillator.

In the case of a half-harmonic oscillator, this concept becomes more intriguing. Due to the presence of an infinite potential barrier at \( x < 0 \), the boundary condition modifies the ground-state energy. Instead of the typical \( \frac{1}{2} \hbar \omega \), the energy level is effectively pushed higher, resulting in \( E_0 = \frac{3}{4} \hbar \omega \).

This alteration occurs because the wavefunction must have a node at \( x = 0 \), effectively increasing the energy needed for the system to maintain this new shape compared to the symmetry of a full oscillator. The study of ground-state energy in such systems emphasizes how quantum states adjust to meet physical constraints, exhibiting both the flexibility and restrictions of quantum phenomena.
expectation value
In quantum mechanics, the expectation value gives us a way to predict the average outcome of a measurement over many identical experiments. For position, this is given by \( \langle x^2 \rangle \), which effectively predicts the average of the square of the particle’s position.

For a full harmonic oscillator, the expectation value of \( \langle x^2 \rangle \) in its ground state can be calculated as \( \langle x^2 \rangle = \frac{\hbar}{2m\omega} \), where \( m \) is the particle's mass and \( \omega \) is the angular frequency. However, with a half-harmonic oscillator, some adjustments arise.

Because the particle is constricted to \( x>0 \) due to the wall at \( x<0 \), its wavefunction and normalizations change, leading to a new expression for the expectation value: \( \langle x^2 \rangle = \frac{\hbar}{4m\omega} \). This reflects the unique spatial constraints on the matter wave function, underlining the alteration in physical expectations.

This shift in expectation values highlights how boundary conditions - such as an infinite wall - significantly impact quantum states and need careful consideration when analyzing quantum behaviors. It demonstrates how quantum mechanics deviates from classical expectations, with phenomena strongly influenced by the system’s boundaries.

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Most popular questions from this chapter

Derive an expression for the density of free-particle states in \(t w o\) dimensions, normalized with periodic boundary conditions inside a box of side length \(L .\) Your answer should be written as a function of \(k(\text { or } E)\) times \(d E d \phi,\) where \(\phi\) is the polar angle that characterizes the momentum direction in two dimensions.

Consider a particle in one dimension bound to a fixed center by a \(\delta\) -function potential of the form \\[ V(x)=-v_{0} \delta(x), \quad\left(v_{0} \text { real and positive }\right) \\] Find the wave function and the binding energy of the ground state. Are there excited bound states?

A box containing a particle is divided into a right and a left compartment by a thin partition. If the particle is known to be on the right (left) side with certainty, the state is represented by the position eigenket \(|R\rangle(|L\rangle),\) where we have neglected spatial variations within each half of the box. The most general state vector can then be written as \\[ |\alpha\rangle=|R\rangle\langle R | \alpha\rangle+|L\rangle\langle L | \alpha\rangle \\] where \(\langle R | \alpha\rangle\) and \(\langle L | \alpha\rangle\) can be regarded as "wave functions." The particle can tunnel through the partition; this tunneling effect is characterized by the Hamiltonian \\[ H=\Delta(|L\rangle\langle R|+| R\rangle\langle L|) \\] where \(\Delta\) is a real number with the dimension of energy. (a) Find the normalized energy eigenkets. What are the corresponding energy eigenvalues? (b) In the Schrödinger picture the base kets \(|R\rangle\) and \(|L\rangle\) are fixed, and the state vector moves with time. Suppose the system is represented by \(|\alpha\rangle\) as given above at \(t=0 .\) Find the state vector \(\left|\alpha, t_{0}=0 ; t\right\rangle\) for \(t>0\) by applying the appropriate time-evolution operator to \(|\alpha\rangle\) (c) Suppose that at \(t=0\) the particle is on the right side with certainty. What is the probability for observing the particle on the left side as a function of time? (d) Write down the coupled Schrödinger equations for the wave functions \(\langle R| \alpha, t_{0}=\) \(0 ; t\rangle\) and \(\left\langle L | \alpha, t_{0}=0 ; t\right\rangle .\) Show that the solutions to the coupled Schrödinger equations are just what you expect from (b). (e) Suppose the printer made an error and wrote \(H\) as \\[ H=\Delta|L\rangle\langle R| \\] By explicitly solving the most general time-evolution problem with this Hamiltonian, show that probability conservation is violated.

Consider a particle subject to a one-dimensional simple harmonic oscillator potential. Suppose that at \(t=0\) the state vector is given by \\[ \exp \left(\frac{-i p a}{\hbar}\right)|0\rangle \\] where \(p\) is the momentum operator and \(a\) is some number with dimension of length. Using the Heisenberg picture, evaluate the expectation value \(\langle x\rangle\) for \(t \geq 0\)

A particle of mass \(m\) in one dimension is bound to a fixed center by an attractive \(\delta\) -function potential: \\[ V(x)=-\lambda \delta(x), \quad(\lambda>0) \\] At \(t=0,\) the potential is suddenly switched off (that is, \(V=0\) for \(t>0\) ). Find the wave function for \(t>0 .\) (Be quantitative! But you need not attempt to evaluate an integral that may appear.)

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