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Mean energy of a 3D (quantum) harmonic oscillator

  1. Mar 10, 2010 #1
    This is not really homework, just a project I'm toying with in my sparetime. I'm doing some Path Integral Monte Carlo simulations, for now just for the 1D quantum harmonic oscillator. Anyways, currently I compare my results to the analytic mean energy of a 1D quantum harmonic oscillator, given by the Hamiltonion

    [tex]\hat H = \frac{{{\hbar ^2}}}{{2m}}\frac{{{\partial ^2}}}{{\partial {x^2}}} + \frac{1}{2}m{\omega ^2}{x^2}[/tex]

    with the trace given as

    [tex]\int\limits_{ - \infty }^\infty {\left\langle x \right|\exp \left( { - \beta \hat H} \right)\left| x \right\rangle } = \frac{{\exp \left( { - 0.5\beta \hbar \omega } \right)}}{{1 - \exp \left( { - 0.5\beta \hbar \omega } \right)}}[/tex].

    Beta is representing inverse temperature. From the mean energy in terms of the trace we get

    [tex]\left\langle H \right\rangle = - \frac{\partial }{{\partial \beta }}\ln \sum\limits_x {\left\langle x \right|\exp \left( { - \beta \hat H} \right)\left| x \right\rangle } [/tex],

    which finally amounts to

    [tex]\left\langle H \right\rangle = \frac{1}{2}\hbar \omega \coth \left( {\frac{1}{2}\beta \hbar \omega } \right)[/tex].

    My next step is to expand my program to 3 dimensions. I know that the energy of the classical harmonic oscillator in 3D is simply the result from 1D times the degrees of freedom (3 translational, 2 rotational and 1 vibrational), but does this also hold for the Quantum version of the harmonic oscillator?

    I.e. is

    [tex]\left\langle {{H_{3D}}} \right\rangle = 3\hbar \omega \coth \left( {\frac{1}{2}\beta \hbar \omega } \right)[/tex]

    true?

    Thanks in advance!
     
  2. jcsd
  3. Mar 10, 2010 #2
    Actually it should only have 3 degrees of freedom, I was thinking of a diatomic molecule. A single particle described as a 3D QHO got 3 degrees of freedom.
     
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