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Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using \\[ \langle x | \tilde{0}\rangle=e^{-\beta|x|} \\] as a trial function with \(\beta\) to be varied. You may use \\[ \int_{0}^{\infty} e^{-\alpha x} x^{n} d x=\frac{n !}{\alpha^{n+1}} \\]

Short Answer

Expert verified
Minimize energy: \( \beta = m \omega / \hbar \). Estimated energy: \( \hbar \omega / 2 \).

Step by step solution

01

Understanding the Problem

We need to estimate the ground-state energy of a one-dimensional simple harmonic oscillator using a variational method. Our trial wave function is given by \( \langle x | \tilde{0}\rangle = e^{-\beta |x|} \). We will calculate the expectation value of energy and minimize it with respect to \( \beta \).
02

Define Expectations

The Hamiltonian for a one-dimensional harmonic oscillator is \( H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 \). The expectation value of energy is \( \langle H \rangle = \langle \psi | H | \psi \rangle \). We'll find \( \langle T \rangle \) for the kinetic energy and \( \langle V \rangle \) for the potential energy.
03

Calculate Kinetic Energy Expectation

The expectation value of kinetic energy is given by \( \langle T \rangle = \int_{-\infty}^{\infty} \psi^*(x) \left( -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \right) \psi(x) \, dx \). Compute the second derivative of \( \psi(x) = e^{-\beta |x|} \), and then apply the kinetic energy operator.
04

Calculate Potential Energy Expectation

The expectation value of potential energy is \( \langle V \rangle = \int_{-\infty}^{\infty} \psi^*(x) \left( \frac{1}{2}m\omega^2 x^2 \right) \psi(x) \, dx \). Substitute the trial wave function and simplify the expression.
05

Simplify Integrals Using Given Formula

Break down the integrals for \( \langle T \rangle \) and \( \langle V \rangle \) into simpler terms using the formula \( \int_{0}^{\infty} e^{-\alpha x} x^{n} \, dx = \frac{n!}{\alpha^{n+1}} \). This will help us compute these integrals analytically.
06

Minimization of Total Energy

Combine results from the kinetic and potential energy expectations to get \( \langle H \rangle = \langle T \rangle + \langle V \rangle \). Then differentiate this total energy with respect to \( \beta \) and set the derivative equal to zero to find the optimal \( \beta \).
07

Solve for \( \beta \) and Energy

Solve the derivative equation obtained from minimization to find the optimal value of \( \beta \). Substitute this \( \beta \) back into \( \langle H \rangle \) to find the estimated ground-state energy.

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Key Concepts

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

Ground-State Energy
Ground-state energy refers to the lowest possible energy that a quantum mechanical system can have. When you're looking at a system like the simple harmonic oscillator, determining this energy is crucial because it presents the most stable state of the system.

This energy is found using the variational method, an optimization technique that allows us to make educated guesses about the system's energy. You select a trial wave function, adjust parameters within it, and calculate the expectation value of the energy. By minimizing this calculated energy with respect to the parameters in your trial function, you can estimate the ground-state energy.

In the context of this exercise, finding the ground-state energy requires us to engage in this minimization, revolving around a suitable trial function.
Simple Harmonic Oscillator
A simple harmonic oscillator is a fundamental concept in physics, referring to a system that moves back and forth around an equilibrium point. Imagine a mass on a spring; it wants to return to its central position when displaced, causing it to oscillate.

The behavior of this system can be described with a Hamiltonian, representing the total energy of the system. The Hamiltonian for a simple harmonic oscillator is:
  • \( H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2 x^2 \)

Here, there are two primary components:
- **Kinetic Energy (\( \frac{p^2}{2m} \))**: This part accounts for the energy due to motion.
- **Potential Energy (\( \frac{1}{2}m\omega^2 x^2 \))**: This part arises from the system's position, analogous to energy stored in a stretched or compressed spring.

In quantum mechanics, studying the properties of a harmonic oscillator helps to understand more complex quantum behaviors encountered in real-world systems.
Expectation Value
The expectation value in quantum mechanics provides an average value for a measurable parameter, like position or energy, that you would expect to find if you performed many measurements on a quantum system.

To estimate the ground-state energy of the harmonic oscillator, you calculate the expectation values of kinetic and potential energies using the trial wave function:
  • **Kinetic Energy Expectation (\( \langle T \rangle \))**: This involves integrating the kinetic energy operator with the trial wave function.
  • **Potential Energy Expectation (\( \langle V \rangle \))**: For this, integrate the potential energy operator using the trial function.

Once you have both, the total energy expectation value, \( \langle H \rangle \), combines these values:
  • \( \langle H \rangle = \langle T \rangle + \langle V \rangle \)

By minimizing \( \langle H \rangle \) relative to the parameter in your trial wave function, you estimate the low-energy behavior of your system. The task of minimizing this expectation value highlights the versatility and importance of this concept within the variational method.
Trial Wave Function
A trial wave function is your guess about what the true wave function of the system looks like. In variational methods, you manipulate this function to get an insight into the system's properties.

The exercise utilizes the exponential trial wave function:
  • \( \langle x | \tilde{0}\rangle = e^{-\beta|x|} \)

This wave function contains the parameter \( \beta \), which can be adjusted to fit the model accurately. It's important because it illustrates general characteristics of wave behavior while being simple enough to allow calculation.

By cleverly choosing a trial wave function, you can simplify the process of calculating the expectation values. Make sure the function is normalized and compatible with boundary conditions of the physical system you're examining.

The flexibility in choosing a trial wave function emphasizes your control over approximating and understanding the system's behavior under study.

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Most popular questions from this chapter

Consider an atom made up of an electron and a singly charged \((Z=1)\) triton \(\left(^{3} \mathrm{H}\right)\) Initially the system is in its ground state \((n=1, l=0) .\) Suppose the system undergoes beta decay, in which the nuclear charge suddenly increases by one unit (realistically by emitting an electron and an antineutrino). This means that the tritium nucleus (called a triton) turns into a helium \((Z=2)\) nucleus of mass \(3\left(^{3} \mathrm{He}\right)\) (a) Obtain the probability for the system to be found in the ground state of the resulting helium ion. The hydrogenic wave function is given by \\[ \psi_{n=1, l=0}(\mathbf{x})=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} e^{-Z r / a_{0}} \\] (b) The available energy in tritium beta decay is about \(18 \mathrm{keV}\), and the size of the \(^{3}\) He atom is about 1 A. Check that the time scale \(T\) for the transformation satisfies the criterion of validity for the sudden approximation.

The unperturbed Hamiltonian of a two-state system is represented by \\[ H_{0}=\left(\begin{array}{cc} E_{1}^{0} & 0 \\ 0 & E_{2}^{0} \end{array}\right) \\] There is, in addition, a time-dependent perturbation \\[ V(t)=\left(\begin{array}{cc} 0 & \lambda \cos \omega t \\ \lambda \cos \omega t & 0 \end{array}\right) \quad(\lambda \text { real }) \\] (a) At \(t=0\) the system is known to be in the first state, represented by \\[ \left(\begin{array}{l} 1 \\ 0 \end{array}\right) \\] Using time-dependent perturbation theory and assuming that \(E_{1}^{0}-E_{2}^{0}\) is not close to \(\pm \hbar \omega,\) derive an expression for the probability that the system is found in the second state represented by \\[ \left(\begin{array}{l} 0 \\ 1 \end{array}\right) \\] as a function of \(t(t>0)\) (b) Why is this procedure not valid when \(E_{1}^{0}-E_{2}^{0}\) is close to \(\pm \hbar \omega ?\)

Consider an isotropic harmonic oscillator in \(t w o\) dimensions. The Hamiltonian is given by \\[ H_{0}=\frac{p_{x}^{2}}{2 m}+\frac{p_{y}^{2}}{2 m}+\frac{m \omega^{2}}{2}\left(x^{2}+y^{2}\right) \\] (a) What are the energies of the three lowest-lying states? Is there any degeneracy? (b) We now apply a perturbation \\[ V=\delta m \omega^{2} x y \\] where \(\delta\) is a dimensionless real number much smaller than unity. Find the zeroth-order energy eigenket and the corresponding energy to first order [that is, the unperturbed energy obtained in (a) plus the first-order energy shift] for each of the three lowest-lying states. (c) Solve the \(H_{0}+V\) problem exactly. Compare with the perturbation results obtained in (b). [You may use \(\left.\left\langle n^{\prime}|x| n\right\rangle=\sqrt{\hbar / 2 m \omega}\left(\sqrt{n+1} \delta_{n^{\prime}, n+1}+\sqrt{n} \delta_{n^{\prime}, n-1}\right) .\right]\)

A \(p\) -orbital electron characterized by \(|n, l=1, m=\pm 1,0\rangle\) (ignore spin) is subjected to a potential \\[ V=\lambda\left(x^{2}-y^{2}\right) \quad(\lambda=\text { constant }) \\] (a) Obtain the "correct" zeroth-order energy eigenstates that diagonalize the perturbation. You need not evaluate the energy shifts in detail, but show that the original threefold degeneracy is now completely removed. (b) Because \(V\) is invariant under time reversal and because there is no longer any degeneracy, we expect each of the energy eigenstates obtained in (a) to go into itself (up to a phase factor or sign) under time reversal. Check this point explicitly.

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} \\]

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