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Perturbations and harmonic oscillator

  1. Dec 20, 2007 #1
    1. The problem statement, all variables and given/known data
    In a diatomic molecule with atom masses m1 and m2, the atoms are bound by a potential
    [tex]V(r)=V_0\big[ \big(\frac{r_0}r\big)^{12} - 2\big(\frac{r_0}r\big)^{6}\big][/tex]
    where r is the distance between the atom centra, r0 is the equilibrium istance, and V0 is a constant depending on the atoms.

    1) Expand V(r) in series close to r0 and keep only the term that corresponds to a harmonic oscillator. Find for this case the ground state wave function and the corresponding energy.

    2) Keep another term in the series of V(r) that gives a contribution to the ground state energy calculated with first order perturbation theory. Find the new (approximate) energy for this state.


    2. Relevant equations
    Wave function for harmonic oscillator, ground state:
    [tex]\psi_0(x) = \big(\frac{m \omega}{\pi \hbar}\big)^{1/4} e^{-\frac{m \omega}{2\hbar}x^2}[/tex]

    [tex]E_0^1 = <\psi_0^0 | H' |\psi_0^0>[/tex]
    where
    [tex]E_0^1=[/tex] ground state energy with first order perturbation

    [tex]\psi_0^0=[/tex] ground state wave function in unperturbed state

    [tex]H'=[/tex] the perturbation

    3. The attempt at a solution
    Serialize V (I'll drop the ordo-terms):
    [tex]V(r)= V(r_0)+V'(r_0)(r-r_0) + \frac 12 V''(r_0) (r-r_0)^2[/tex]
    V'(r0)=0 because it's the equilibrium point
    V(r0) = constant and can be removed since this doesn't change the force
    Conclusion:
    [tex]V(r) = \frac 12 V''(r_0) (r-r_0)^2[/tex]
    [tex]V''(r_0)=\frac{72}{r_0^2}V_0[/tex]
    [tex]V(r) = \frac{36}{r_0^2}V_0 (r-r_0)^2 = \alpha (r-r_0)^2[/tex]
    So compared with the harmonic oscillator's
    [tex]V(x) = \frac 12 \mu \omega^2x^2[/tex]
    I get that
    [tex]\alpha = \frac {\mu \omega^2}2 \rightarrow \omega = \frac 6r_0 \sqrt{\frac{2V_0}{\mu}} \rightarrow E=\frac 12 \hbar \omega = 3\hbar \sqrt{\frac{2V_0}{r_0^2 \mu}}[/tex]

    So far all good...

    The wave function should become
    [tex]\psi_0(r) = \big( \frac{6\sqrt{2\mu V_0}}{\pi r_0 \hbar}\big)^{1/4} e^{-\frac{3\sqrt{2\mu V_0}}{r_0 \hbar}r^2}[/tex]
    I'm not especially certain about this, becuase of two reasons:
    1) the given wave function under 2 is in x, and has been normalized with an integral from -infinity to infinity. As I'm using r, and it's a distance between two atoms, it should be 0 to infinity the get the correct wavefunction.
    2) the given wave function is for an oscillator with r0=0. Is it correct to use r in the exponential function or should it be r-r0?

    That was for part 1. Second question:
    Serialize like before but keep the third order term...
    [tex] V'''(r_0) = -\frac{1512}{r_0^3}V_0 [/tex]
    [tex] V(r) = \frac{36}{r_0^2}V_0 (r-r_0)^2 - 252\frac{V_0}{r_0^3} (r-r_0)^3[/tex]
    Last term will be H' - the perturbation.
    [tex]E_0^1 = <\psi_0^0 | H' |\psi_0^0> = -252 \frac{V_0}{r_0^3} \big(\frac{6\sqrt{2\mu V_0}}{\pi r_0 \hbar}\big)^{1/4} \int_{-\infty}^\infty e^{- \frac{6\sqrt{2\mu V_0}}{r_0 \hbar}r^2}(r-r_0)^3 dr = 252 V_0 + \frac{63\hbar}{r_0} \sqrt{\frac{V_0}{2\mu}}[/tex]
    Once again I'm confused about the integration limits.

    And this is definitely not the correct answer. If I would divide the energy from the first question with the perturbed one, I would get a constant term of 21/4 that is way too great to make sense.
     
  2. jcsd
  3. Dec 21, 2007 #2
    nevermind... I found out where I went wrong.
     
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