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Worksheet 7B Solutions

  • Page ID
    184978
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    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.


    Worksheet 7B Solutions is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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