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Sommerfeld quantisation:relation between energy and potential

  1. Feb 22, 2010 #1
    I found the following question in a question paper.

    If the potential experienced by a particle in one dimension is given by [tex]ax^4[/tex], how does the energy of the particle depend on the parameter a? Use Bohr-Sommerfeld quantisation formula..


    I know how to find the dependence of energy on n, the integer that occurs in the quantisation.

    But how does it depend on a?

    Since
    Hamiltonian is [tex]\ K+V[/tex] ,
    Energy is [tex]\ K+ax^4[/tex].

    I dont know what else I can say.
     
  2. jcsd
  3. Feb 22, 2010 #2
    For example, in the harmonic oscillator, the potential V is,
    [tex]V = ax^2 = \frac{1}{2}m \omega^{2}x^2[/tex]
    So, the total energy is,
    [tex]E= K+V = \frac{p_{x}^2}{2m}+\frac{1}{2}m \omega^{2}x^2(=ax^2)[/tex]

    As shown in this thread, The Bohr-Sommerfeld quantization formula is, (in a quarter period)
    [tex]\oint p_{x}dx = nh \to \sqrt{2m}\int_{0}^{\sqrt{\frac{E}{a}}}\sqrt{E-ax^2}dx = \frac{1}{4}nh[/tex]

    change the variable to [tex]\theta[/tex] as follows,
    [tex]\sqrt{\frac{a}{E}}x=sin\theta, \quad E\sqrt{\frac{2m}{a}}\int_{0}^{\frac{\pi}{2}}cos^2\theta d\theta =\frac{\sqrt{2m}E\pi}{4\sqrt{a}}=\frac{1}{4}nh[/tex]

    So if we erase the a, the the total energy E is,
    [tex]E=\frac{nh\sqrt{a}}{\sqrt{2m}\pi}=\frac{nh\omega}{2\pi}=n\hbar\omega[/tex]

    This result shows the interval of the energy E is [tex]\hbar\omega[/tex]

    But as shown in the ground state electron of the Bohr hydrogen model, the electron has the orbital angular momentum 1 (x [tex]\hbar[/tex]) instead of the spin 1/2. So we have to consider the minimum rotation (not 1-D) in the Bohr model.

    This magnetic moment which can be measured by the experiment is the same as the QM hydrogen model because the magnetic moments(angular momentum x g-factor) are 1 x 1 (in the Bor model) = 1/2 x 2 (in the QM model).
    If the angular momentum is [tex]\hbar[/tex], the electron naturally return by one rotation as follows, (also in the result of this experiment)

    [tex]L_{z}(=-i\hbar\frac{d}{d\varphi})\phi_{m}(\varphi) = m\hbar\phi_{m}(\varphi), \to \quad \phi_{m}(\varphi)=e^{im\varphi}[/tex]

    Sorry, the case of [tex]V=ax^4[/tex] is a little complicated, so I'm glad if you calculate like this, and tell me the result.
     
  4. Feb 27, 2010 #3
    My question was not about the dependence on the quantum number n obtained from the quantisation condition....

    It was about the dependence of the energy on the parameter a in the Hamiltonian..that is the potential energy being given as [tex]a x^4[/tex]...



    As for dependence on n, it can be easily accomplished..see this video...at around the thirty minute mark...

     
    Last edited by a moderator: Sep 25, 2014
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