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Quantum spherical harmonic oscillator:eigenfunctions

  1. Mar 21, 2008 #1
    Hi to everybody of PF community!!

    I have some troubles to find eigenfunctions common to [tex] H, L_{z}, L^2 [/tex] in the problem of spherical simmetric harmonic oscillator.

    I start with the Hamiltonian [tex] H=\frac{\textbf{p}^2}{2 \mu} - \frac{1}{2}k\textbf{x}^2[/tex] that in spherical coordinates become

    [tex] H=\frac{- \hbar ^2}{2 \mu} (\frac{\partial^2}{r \partial r^2} r - \frac{L^2}{r^2 \hbar^2}) - \frac{k r^2}{2} [/tex]

    now,the eigenfunctions equation is: [tex] H \psi = E \psi [/tex]

    i know the angular part of the problem that are the spherical armonics...so it remain to solve the radial equation

    [tex] (\frac{- \hbar ^2}{2 \mu} \frac{\partial^2}{r \partial r^2} r + \frac{\hbar^2 l(l+1)}{2 \mu r^2 } - \frac{k r^2}{2}) R(r) = E R(r) [/tex]

    then i don't know if to try with the hydrogen-like method (but potential is different) or with the armonic 1D oscillator (but I don't know whato to do with the centrifugal term).........so, please give me an help!!!
     
  2. jcsd
  3. Mar 21, 2008 #2
    Some of the community might be more familiar with these things, but my understanding is that it's normal to look for series solutions. The requirement that the solution has a finite area under it usually constrains the allowed eigenvalues, and gives you some series of polynomials. I forget, but I think you should get Legendre polynomials?
     
  4. Mar 21, 2008 #3
    Make the substitution [itex] u(r) = r R(r) [/itex] and you should find that the equation reduces to

    [tex] \frac{d^2 u}{dr^2} = \displaystyle \left[ \frac{l(l+1)}{r^2} - k^2 \right] u [/tex]

    For [itex] l=0 [/itex] this will reduce to the simple harmonic oscillator. For [itex] l \neq 0 [/itex] you will find solutions in the form of spherical Bessel and Neumann functions. You can look up the ode's that correspond to those functions and they'll show you how to solve them.
     
  5. Mar 22, 2008 #4

    malawi_glenn

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    Hi! The radial solution to:

    [tex] (\frac{- \hbar ^2}{2mr^2} \frac{d}{dr}(r^2\frac{d}{dr}) + \frac{\hbar^2 l(l+1)}{2 m r^2 } + U(r))R(r) = E R(r) [/tex]


    by using [itex] u(r) = r R(r) [/itex]


    and [itex] U(r) = \frac{1}{2}m\omega ^2 r^2 [/itex]

    and imposing BC:
    [itex] u(0) = 0 [/itex]
    [itex] u(r) = 0 , r \rightarrow \infty [/itex]

    is the radial Laguerre equation:

    [tex] u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\nu r^2) [/tex]

    Where L is Laguerre polynomial and
    [tex] \nu = m\omega / 2\hbar [/tex]

    Source: Nuclear Shell Model, Kris Heyde, Springer 1994
     
    Last edited: Mar 22, 2008
  6. Mar 22, 2008 #5
    if I make the substitution : [tex]u(r)=rR(r)[/tex]

    it become:

    [tex] (\frac{- \hbar ^2}{2 \mu} \frac{\partial^2}{\partial r^2} + \frac{\hbar^2 l(l+1)}{2 \mu r^2 } - \frac{k r^2}{2}) u(r) = u R(r) [/tex]

    ...isn't it??

    so... after find the solution to be [tex] u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\ny r^2) [/tex]
    ....I have to divide by a factor r to obtain R(r)??
    like [tex] R_{kl}(r) = r^{l}e^{-\nu r^2}L^{l+1/2}_k(2\ny r^2) [/tex] ???

    and another my question is:where I could find the solution to that ode???

    thanks
     
  7. Mar 22, 2008 #6
    malawi: ops....I see now the "source" of solution you give to me
     
  8. Mar 22, 2008 #7

    malawi_glenn

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    Search for Laguerre ODE
     
  9. Mar 22, 2008 #8
    but...is my post n°5 correct?
     
  10. Mar 22, 2008 #9

    malawi_glenn

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  11. Mar 24, 2008 #10
    [tex] (\frac{- \hbar ^2}{2 \mu} \frac{\partial^2}{r \partial r^2} r + \frac{\hbar^2 l(l+1)}{2 \mu r^2 } - \frac{m \omega^2 r^2}{2}) R(r) = E R(r) [/tex]

    [tex] u(r)=rR(r) [/tex]

    [tex] \frac{\partial^2}{r \partial r^2} r = \frac{\partial}{r^2 \partial r}(r^2 \frac{\partial}{\partial r})[/tex]

    [tex] u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\nu r^2) [/tex]

    [tex] \nu = m\omega / 2\hbar [/tex]

    [tex] u_{kl}(r) = r^{l+1} e^\frac{-2m\omega r^2}{2\hbar} L^{l+1/2}_k(\frac{m\omega r^2}{\hbar}) [/tex]

    [tex] R_{kl}(r) = r^{l} e^\frac{-2m\omega r^2}{2\hbar} L^{l+1/2}_k(\frac{m\omega r^2}{\hbar}) [/tex]

    [tex] R_{10}(r) = re^\frac{-m \omega r^2}{2\hbar} [/tex]

    Continue in next post……
     
  12. Mar 24, 2008 #11
    ......continue:

    and the next is what I found with another method:

    [tex] (\frac{- \hbar ^2}{2 \mu} \frac{\partial^2}{r \partial r^2} r + \frac{\hbar^2 l(l+1)}{2 \mu r^2 } - \frac{m \omega^2 r^2}{2}) R(r) = E R(r) [/tex]

    [tex] \rho=\alpha r [/tex]

    [tex] \alpha=\sqrt{m \omega/ \hbar} [/tex]

    [tex] dr=\frac{d \rho}{\alpha}[/tex]

    [tex] \frac{2 \mu E}{\hbar^2 \alpha^2} = \frac{2 E}{\hbar \omega} = \lambda [/tex]

    [tex] \frac{d^2}{r dr^2}r = \frac{d^2}{dr^2} + \frac{2d}{rdr} [/tex]

    [tex] (\frac{d^2}{d \rho^2} + \frac{2 d}{\rho d \rho} - \frac{l(l+1)}{\rho^2} - \rho^2 + \lambda) R(\rho) = 0 [/tex]

    that is a particular ODE : [tex] (\frac{d^2}{d \rho^2} + p(\rho)\frac{d}{d \rho} +q(\rho)) R(\rho) = 0 [/tex]

    and solutions are [tex] R(\rho)=\rho^{\beta} \sum^{0}_{\infty}c_{n} \rho^n [/tex] where beta is a constant and Cn is a function both dependents from the powers series of functions p(x) and q(x). The procedure is very long to write in latex, I did it in my exercise-book and the resul is different : [tex]R_{10}= \frac{-2E}{6 \hbar \omega}\rho^2[/tex]
    If anybody wants to see the procedure I'll write it ,also if it will be an hard work....

    anyway....any suggest??
     
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