# Worksheet 8 Solutions

- Page ID
- 95490

## 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.