# Worksheet 8 Solutions

### Solution

Name: ______________________________

Section: _____________________________

Student ID#:__________________________

Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

The variation method approximates the lowest energy eigenvalue, $$E_\phi$$, and eigenfunction, $$\phi$$, for a quantum mechanical system by guessing a function that is well-behaved over the limits of the system and minimizing the energy. For the trial wavefunction  $$| \phi(x)\rangle =\dfrac{1}{1+\beta x^2}$$, the energy is

$E_\phi=\dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi |\phi\rangle}$

where $$|\phi \rangle$$ is the wavefunction that we guess and $$\hat H$$ is the Hamiltonian for the system. The variational theory argues that when the energy is minimized, then $$E_\phi\geq E_{actual}$$. Some equations you may find useful for a harmonic oscillator:

$\hat H_{HO}=-\dfrac{\hbar^2}{2\mu}\dfrac{d^2}{dx^2}+\dfrac{kx^2}{2}$

$E_{n,HO}=h\nu \left(n+\dfrac{1}{2} \right)=\hbar \omega \left(n+\dfrac{1}{2}\right)$

$\underset{\text{lowest energy eigenstate}}{| \psi(x) \rangle}=\left(\dfrac{a}{\pi}\right)^{1/4}e^{-\dfrac{ax^2}{2}} ﻿﻿$

### Q1

From the previous groupwork, we got

$E_\phi=\dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi |\phi\rangle}=\dfrac{\hbar^2\beta}{4\mu}+\dfrac{k}{2\beta}$

To get the best value for the energy, we must minimize $$E_\phi(\beta)$$ with respect to $$\beta$$.

What is, $$\dfrac{dE_\phi}{d\beta}=0$$.

$\dfrac{dE_\phi}{d\beta}= \dfrac{\hbar^2}{4\mu}-\dfrac{k}{2\beta^2}$

What value for $$\beta$$ fulfills the minimized $$E_\phi$$?

$\beta_{min} = \sqrt{\dfrac{4\mu k}{2\hbar^2}} =\sqrt{\dfrac{2\mu k}{\hbar^2}}$

Using the value for $$\beta$$ that fulfills the minimization, what is $$E_\phi(\beta)$$?

$E_\phi ^{min}=\dfrac{\hbar^2\beta_{min}}{4\mu}+\dfrac{k}{2\beta_{min}}=\dfrac{\hbar^2\sqrt{\dfrac{2\mu k}{\hbar^2}}}{4\mu}+\dfrac{k}{2\sqrt{\dfrac{2\mu k}{\hbar^2}}} = \dfrac{\hbar\sqrt{2\mu k}}{4\mu}+\dfrac{k \hbar }{2\sqrt{2\mu k}} = \dfrac{\hbar\sqrt{2}}{4} \sqrt {\dfrac{k}{\mu}}+ \dfrac{\hbar}{2\sqrt{2}} \sqrt {\dfrac{k}{\mu}} = \dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}} ( \dfrac{\sqrt{2}}{2} + \dfrac{1}{\sqrt{2}}) = \dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}} ( \dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2})= \dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}} \sqrt{2}$

The variation method approximates the ground state energy for the system. What is the expression for the exact energy of the harmonic oscillator?

$\hbar \sqrt {\dfrac{k}{\mu}} (\frac{1}{2} +v)$

where v = 0,1,2,... is a quantum number for a vibrational state.

### Q2

What is the quantum number for the ground state of the harmonic oscillator?

$v=0$

What is the exact energy for the ground state of the harmonic oscillator?

$\frac{1}{2} \hbar \sqrt {\dfrac{k}{\mu}}$

How well does the guess of $$| \phi(x) \rangle =\dfrac{1}{1+\beta x^2}$$ approximate the lowest energy harmonic oscillator eigenstate?

$\frac {E_{trial} - E_{true}}{E_{true}} = \frac {\dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}} \sqrt{2} - \dfrac{1}{2} \hbar \sqrt {\dfrac{k}{\mu}}}{\frac{1}{2} \hbar \sqrt {\dfrac{k}{\mu}}} = \frac {\dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}} (\sqrt{2} - 1) }{\dfrac{\hbar}{2} \sqrt {\dfrac{k}{\mu}}} = 0.41$

The estimate with 41% error is not very good.  The error of 5-10% would be considered acceptable.

Given the knowledge you have of the true harmonic oscillator wavefunction, how well would the new trial wavefunction, $$|\phi(x) \rangle =e^{-\beta x^2}$$ approximate the solution for the lowest energy state of the harmonic oscillator?

Very well, since the functional form is the same for trial wavefunction and true wavefunction of harmonic oscillator.