Worksheet 7B Solutions

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Solutions

Q1

We are given the solution to the variational theory for the harmonic oscillator with trial function:

\[| \phi(x)\rangle =\dfrac{1}{1+\beta x^2}\label{1}\]

\[E_\phi=\dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi |\phi\rangle}\label{2}\]

\[E_\phi=\dfrac{\langle\phi|\hat H|\phi\rangle}{\langle\phi |\phi\rangle}=\dfrac{\hbar^2\beta}{4\mu}+\dfrac{k}{2\beta}\label{3}\]

To solve for the energy we need to minimize the parameter \(\beta\)

\[\dfrac{dE_\phi}{d\beta}=0\label{4}\]

\[\dfrac{dE_\phi}{d\beta}= \dfrac{\hbar^2}{4\mu} - \dfrac{k}{2\beta^2} = 0 \label{5}\]

Solve for \(\beta\)

\[\beta^2 = \dfrac{4k\mu}{2\hbar^2} = \dfrac{2k\mu}{\hbar^2}\label{6}\]

\[\beta = \sqrt{ \dfrac{2k\mu}{\hbar^2}} \label{7}\]

Put beta back in to the formula to get the energy

\[E_\phi = \dfrac{\hbar^2 \sqrt{ \dfrac{2k\mu}{\hbar^2}}}{4\mu}+\dfrac{k}{2 \sqrt{ \dfrac{2k\mu}{\hbar^2}}} \label{8}\]

Simplify

\[E_\phi = \dfrac{\hbar}{2}\sqrt{\dfrac{k}{\mu}}(\dfrac{\sqrt{2}}{2} + \dfrac{1}{\sqrt{2}}) \label{9}\]

\[E_\phi = \dfrac{\hbar}{\sqrt{2}}\sqrt{\dfrac{k}{\mu}} \label{10}\]

I left the square roots seperate because from the Harmonic Oscillator remember the angular frequency \(\omega = \sqrt{\dfrac{k}{\mu}}\)

\[E_\phi = \dfrac{\hbar}{\sqrt{2}}\omega \label{11}\]

Compare this to the ground state Harmonic Oscillator energy

Q2

The ground state quantum number is \(n=0\)

\[E_{n=0, H.O.} = \hbar\omega(n + \dfrac{1}{2}) = \dfrac{\hbar\omega}{2} \label{12} \]

The approximation with this trail wave function is not great.

\[E_\phi = \dfrac{\hbar}{\sqrt{2}}\omega > E_{n=0, H.O.} =\dfrac{\hbar\omega}{2} \label{13} \]

The percent difference

\[\dfrac{E_\phi - E_{n=0, H.O.} }{E_{n=0, H.O.}} = \dfrac{\dfrac{\hbar\omega}{\sqrt{2}} - \dfrac{\hbar\omega}{2}}{ \dfrac{\hbar\omega}{2}} \label{14} \]

\[\dfrac{E_\phi - E_{n=0, H.O.} }{E_{n=0, H.O.}} = 2({\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}}) = \dfrac{2}{\sqrt{2}} - 1 =\sqrt{2} -1 = 0.414 \label{15} \]

\(40\%\) that's pretty bad.

If we use the trial function: \(|\phi(x) \rangle =e^{-\beta x^2}\)

You should get the exact energy for the Harmonic Oscillator since it is the exact wavefunction.