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Chapter 5: Problem 27

Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using $$ \langle x \mid \tilde{0}\rangle=e^{-\beta|x|} $$ as a trial function with \(\beta\) to be varied.

Short Answer

Expert verified
The estimated ground-state energy is \( E = \frac{1}{2} \hbar \omega \).

Step by step solution

01

Understand the Simple Harmonic Oscillator

A one-dimensional simple harmonic oscillator has potential energy given by \( V(x) = \frac{1}{2}m\omega^2 x^2 \), where \( m \) is the mass and \( \omega \) is the angular frequency. The aim is to estimate its ground-state energy using a trial wave function and variational principle.
02

Define the Trial Wave Function

The provided trial wave function is \( \langle x | \tilde{0} \rangle = e^{-\beta |x|} \). This function depends on the parameter \( \beta \) which needs to be optimized. The goal is to compute the expectation value of the energy \( E(\beta) \) using this function and find the value of \( \beta \) that minimizes this energy.
03

Write the Energy Expectation Value

The expectation value of the energy \( E(\beta) \) in terms of the trial wave function is given by the integral \( E(\beta) = \int \psi^* (x) \hat{H} \psi(x) \, dx \), where \( \hat{H} \) is the Hamiltonian operator \( \hat{H} = \frac{-\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 \).
04

Compute the Kinetic Energy Expectation

The kinetic energy expectation value is calculated from \( \langle T \rangle = \int^{\infty}_{-\infty} \psi^*(x) \left( \frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \right) \psi(x) \, dx \). After taking the second derivative of the trial function and integrating, this results in \( \langle T \rangle = \frac{\hbar^2 \beta^2}{2m} \).
05

Compute the Potential Energy Expectation

The potential energy expectation is \( \langle V \rangle = \int^{\infty}_{-\infty} \psi^*(x) \left( \frac{1}{2} m \omega^2 x^2 \right) \psi(x) \, dx \). By substituting the trial wave function and evaluating the integrals, this yields \( \langle V \rangle = \frac{m \omega^2}{4 \beta^2} \).
06

Sum Kinetic and Potential Energies for Total Energy

The total expectation value of energy is obtained by adding the kinetic and potential energies: \( E(\beta) = \langle T \rangle + \langle V \rangle = \frac{\hbar^2 \beta^2}{2m} + \frac{m \omega^2}{4 \beta^2} \). The next step is to minimize this expression with respect to \( \beta \).
07

Minimize Energy to Find Optimal \(\beta\)

To find the optimal \( \beta \), differentiate the expression for \( E(\beta) \) with respect to \( \beta \) and set the derivative equal to zero. Solving this yields \( \beta = \sqrt{\frac{m\omega}{2\hbar}} \).
08

Substitute Optimal \(\beta\) into Energy Expression

Substituting \( \beta = \sqrt{\frac{m\omega}{2\hbar}} \) back into the energy expression, calculate \( E(\beta) \) to get \( E = \frac{1}{2} \hbar \omega \), which is the estimated ground-state energy of the harmonic oscillator.

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Key Concepts

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

Simple Harmonic Oscillator
The concept of a simple harmonic oscillator is foundational in physics. This system consists of a particle or object that experiences a restoring force proportional to its displacement from equilibrium. For a one-dimensional simple harmonic oscillator, the potential energy is described by the formula \( V(x) = \frac{1}{2}m\omega^2 x^2 \). Here, \( m \) represents the mass of the particle, and \( \omega \) is the angular frequency. This potential energy function illustrates how the system oscillates back and forth through equilibrium, much like a mass attached to a spring or a pendulum swinging smoothly.

If you've ever watched a swing, you've seen something similar to a simple harmonic oscillator in action. The swing moves faster as it approaches the lowest point and slows down as it reaches the peak before stopping and reversing direction. This elegant and predictable motion is the hallmark of simple harmonic oscillators. Understanding this helps us tackle more complex quantum problems, such as estimating ground-state energy using variational principles.
Ground-State Energy Estimation
Estimating the ground-state energy of a system like a simple harmonic oscillator is crucial in quantum mechanics. The ground state is the lowest energy state of a quantum system, and it's essential to predict this value as accurately as possible. One powerful method to estimate this is the variational principle.

The variational principle states that for any trial wave function chosen, the expectation value of the Hamiltonian, \( E(\beta) = \int \psi^*(x) \hat{H} \psi(x) \, dx \), will always be greater than or equal to the true ground-state energy of the system. Here, \( \psi(x) \) is our trial wave function, and \( \hat{H} \) is the Hamiltonian operator representing the total energy of the system including both kinetic and potential energy.

Estimating ground-state energy involves selecting an appropriate trial wave function and adjusting its parameters to minimize the energy. This minimization helps in finding the best possible approximation for \( \beta \) that brings \( E(\beta) \) as close as possible to the actual ground-state energy. In our case, by finding \( \beta = \sqrt{\frac{m\omega}{2\hbar}} \), we achieve an estimated ground-state energy of \( \frac{1}{2} \hbar \omega \), demonstrating the efficiency of the variational method.
Trial Wave Function
A trial wave function is a crucial concept in variational calculations. It is an educated guess about the form of the system's wave function that allows us to compute an upper bound on the ground-state energy. For the simple harmonic oscillator exercise, the trial wave function provided is \( \langle x | \tilde{0} \rangle = e^{-\beta |x|} \).

This wave function is chosen based on symmetry and expected behavior of the oscillator. It depends on a parameter \( \beta \), which is used to adjust the function to best replicate the conditions of the real system. The goal is to optimize \( \beta \) to minimize the energy, a typical strategy in the variational method.

The selection of a trial wave function is both an art and a science. It requires intuition about the system's physical behavior, mathematical skills to efficiently calculate expectation values, and optimization techniques to find the parameter's best value. This careful selection can greatly influence the accuracy of the energy estimation, highlighting the importance of a well-chosen trial function in this quantum exercise.

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Most popular questions from this chapter

This problem highlights anomalies in the "exponential" decay of a state. It is inspired by Winter, Phys. Rev., 123 (1961) 1503 , but modern computer applications make it straightforward to directly evaluate the integrals numerically. Consider a particle of mass \(m\) that is initially inside a "well" bounded by an infinite wall to the left and a \(\delta\)-function potential on the right: The infinite wall is located at \(x=-a\), and the potential at \(x=0\) is \(U \delta(x)\) where \(U\) is a positive constant. The figure also shows a plausible "ground state" initial wave function \(\Psi(x, t=0)=(2 / a)^{1 / 2} \sin (n \pi x / a)\) with \(n=1\). a. Show that the wave function at all times can be written as $$ \Psi(x, t)=2 n\left(\frac{2}{a}\right)^{1 / 2} \int_{0}^{\infty} d q \frac{e^{-i T q^{2}} q \sin q[q \sin (l+1) q+f]}{\left(q^{2}-n^{2} \pi^{2}\right)\left(q^{2}+G q \sin 2 q+G^{2} \sin ^{2} q\right)} $$ where \(q \equiv\left[a(2 m E)^{1 / 2}\right] / \hbar\) for a particle with energy \(E, T \equiv \hbar t / 2 m a^{2}, l \equiv x / a\), and \(G \equiv 2 m a U / \hbar^{2}\) are all dimensionless quantities, and \(f=0\) for \(-a \leq x \leq 0\) and \(f=G \sin q \sin l q\) for \(x \geq 0\). This is most easily done by expanding the wave function in energy eigenstates \(|E\rangle\), as $$ \Psi(x, t)=\int_{0}^{\infty} d E \phi_{E}(x) e^{-i E t / \hbar} $$ where \(\phi_{E}(x)\) is an energy eigenfunction and \(\left\langle E \mid E^{\prime}\right\rangle=\delta\left(E-E^{\prime}\right)\). b. Write a computer program to (numerically) integrate the probability of finding the particle inside the well. Carry out the integration for a series of values of \(T\) between zero and 12 , using the same parameters as Winter, namely \(n=1\) and \(G=6\). Plotting these probabilities as a function of \(T\) should resemble Figure 2 of Winter's paper. Fit the points for \(2 \leq T \leq 8\) to an exponential, and compare the decay time to Winter's value of \(0.644\). c. Examine the behavior for \(T \geq 8\), and compare to the behavior Winter found for the current at \(x=0\). This suggests an experimental measurement. See Norman et al., Phys. Rev. Lett., \(\mathbf{6 0}\) (1988) \(2246 .\)

Suppose the electron had a very small intrinsic electric dipole moment analogous to the spin magnetic moment (that is, \(\mu_{e l}\) proportional to \(\sigma\) ). Treating the hypothetical \(-\mu_{e l} \cdot \mathbf{E}\) interaction as a small perturbation, discuss qualitatively how the energy levels of the Na atom \((Z=11)\) would be altered in the absence of any external electromagnetic field. Are the level shifts first order or second order? State explicitly which states get mixed with each other. Obtain an expression for the energy shift of the lowest level that is affected by the perturbation. Assume throughout that only the valence electron is subjected to the hypothetical interaction.

Consider a two-level system with \(E_{1}0\) by exactly solving the coupled differential equation $$ i \hbar \dot{c}_{k}=\sum_{n=1}^{2} V_{k n}(t) e^{i \omega_{k n} t} c_{n} \quad(k=1,2) $$ b. Do the same problem using time-dependent perturbation theory to lowest nonvanishing order. Compare the two approaches for small values of \(\gamma\). Treat the following two cases separately: (i) \(\omega\) very different from \(\omega_{21}\) and (ii) \(\omega\) close to \(\omega_{21}\). Answer for (a): (Rabi's formula) $$ \begin{aligned} &\left|c_{2}(t)\right|^{2}=\frac{\gamma^{2} / \hbar^{2}}{\gamma^{2} / \hbar^{2}+\left(\omega-\omega_{21}\right)^{2} / 4} \sin ^{2}\left\\{\left[\frac{\gamma^{2}}{\hbar^{2}}+\frac{\left(\omega-\omega_{21}\right)^{2}}{4}\right]^{1 / 2} t\right\\} \\ &\left|c_{1}(t)\right|^{2}=1-\left|c_{2}(t)\right|^{2} \end{aligned} $$

A simple harmonic oscillator (in one dimension) is subjected to a perturbation $$ H_{1}=b x $$ where \(b\) is a real constant. a. Calculate the energy shift of the ground state to lowest nonvanishing order. b. Solve this problem exactly and compare with your result obtained in (a).

Compute the Stark effect for the \(2 s_{1 / 2}\) and \(2 p_{1 / 2}\) levels of hydrogen for a field \(\mathscr{E}\) sufficiently weak so that \(e \mathscr{E} a_{0}\) is small compared to the fine structure, but take the Lamb \(\operatorname{shift} \delta(\delta=1057 \mathrm{MHz})\) into account (that is, ignore \(2 p_{3 / 2}\) in this calculation). Show that for \(e \mathscr{E} a_{0} \ll \delta\), the energy shifts are quadratic in \(\mathscr{E}\), whereas for \(e \mathscr{E} a_{0} \gg \delta\) they are linear in \(\mathscr{E}\). Briefly discuss the consequences (if any) of time reversal for this problem. This problem is from Gottfried (1966), Problem 7-3.
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