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Old quantum mechanics

  1. Dec 23, 2009 #1
    "Old" quantum mechanics

    There is famous Bohr Sommerfeld quantization rule which says that [tex]\oint p dx = n h [/tex]. Is it possible to apply this rule to three dimensonal system e.g. hydrogen atom.
    The integral will probably look as follow:
    \int \sqrt{2E - \frac{l^2}{r^2}+\frac{\alpha c \hbar}{r}} = n h
    But I don't know which the contour I should chose.
  2. jcsd
  3. Dec 23, 2009 #2
    Re: "Old" quantum mechanics

    In the hydrogen of the Bohr-Sommerfeld model, the orbital becomes elliptical or circular.
    So simply I use the two-dimensonal system.
    The orbital length becomes a integer times the de Broglie's wavelength ([tex]\lambda =h/mv =h/p [/tex]).

    For example, when the orbital is elliptical, we divide the electron's movement into the two directions at short time intervals.

    The [tex]\perp[/tex] is the direction of the angular momentum(tangential), and the [tex]r[/tex] is the radial direction.
    The two directions are rectangular at each point.

    [tex] \oint p_{\perp} dq_{\perp} = n_{1}h, \qquad \oint p_{r} dq_{r} = n_{2}h[/tex]

    At each point, the de Broglie's wavelengths are [tex]\lambda_{\perp}= h/p_{\perp}, \lambda_{r}= h/p_{r} [/tex]
    The wavelengths are changing at each point. So, the above equations are equal to,

    [tex] \oint dq_{\perp}/\lambda_{\perp} = n_{1}, \qquad \oint dq_{r}/\lambda_{r}= n_{2} [/tex]

    This means that in the Bohr-Sommerfeld model, the sum of the number of the de Broglie's waves contained in each short segment becomes a interger.

    In the case of the three-dimensional system or more complex system, we had better use the computer or something, I think.
    Last edited: Dec 23, 2009
  4. Dec 25, 2009 #3
    Re: "Old" quantum mechanics

    Sorry. I should have added one more thing in the above statement(#2).

    I divided the electron's motion into the two directions which are rectangular at each point on the elliptical orbital.

    But actually the electron is moving in one direction.
    And its momentum [tex]p[/tex] satisfies [tex]p^2 = p_{\perp}^2 + p_{r}^2[/tex] at each point.

    So in a short time([tex]dt[/tex]), the electron moves [tex]dq=\frac{p}{m} dt [/tex].
    Of course this satisfies,

    [tex] dq^2 = dq_{\perp}^2 + dq_{r}^2 [/tex]

    The number of de Broglie's waves ([tex] \lambda [/tex] in length) contained in the short segment([tex]dq[/tex]) is,

    [tex] dq/\lambda=dq/(\frac{h}{p})=p dq /h = p^2dt /hm [/tex]

    As I said above, due to [tex]p^2 = p_{\perp}^2 + p_{r}^2[/tex], we arrive at the following relation,

    [tex] dq /\lambda = dq_{\perp}/\lambda_{\perp} + dq_{r}/\lambda_{r} [/tex]

    As a result, the number of de Broglie's waves contained in one-round of the orbital satisfies,

    [tex] n = n_{1} + n_{2} [/tex]

    For example, in the energy level (n=2) of the hydrogen(the Bohr model), there are two patterns as follows,
    n1=1, n2=1 ----- elliptical (angular momentum=1)
    n1=2, n2=0 ----- circular (angular momentum =2)
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