View Full Version : quantum spherical harmonic oscillator:eigenfunctions
folgorant
Mar21-08, 07:28 PM
Hi to everybody of PF community!!
I have some troubles to find eigenfunctions common to H, L_{z}, L^2 in the problem of spherical simmetric harmonic oscillator.
I start with the Hamiltonian H=\frac{\textbf{p}^2}{2 \mu} - \frac{1}{2}k\textbf{x}^2 that in spherical coordinates become
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}
now,the eigenfunctions equation is: H \psi = E \psi
i know the angular part of the problem that are the spherical armonics...so it remain to solve the radial equation
(\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)
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!!!
genneth
Mar21-08, 09:21 PM
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?
Kreizhn
Mar21-08, 10:49 PM
Make the substitution u(r) = r R(r) and you should find that the equation reduces to
\frac{d^2 u}{dr^2} = \displaystyle \left[ \frac{l(l+1)}{r^2} - k^2 \right] u
For l=0 this will reduce to the simple harmonic oscillator. For l \neq 0 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.
malawi_glenn
Mar22-08, 05:09 AM
Hi! The radial solution to:
(\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)
by using u(r) = r R(r)
and U(r) = \frac{1}{2}m\omega ^2 r^2
and imposing BC:
u(0) = 0
u(r) = 0 , r \rightarrow \infty
is the radial Laguerre equation:
u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\nu r^2)
Where L is Laguerre polynomial and
\nu = m\omega / 2\hbar
Source: Nuclear Shell Model, Kris Heyde, Springer 1994
folgorant
Mar22-08, 05:27 AM
if I make the substitution : u(r)=rR(r)
it become:
(\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)
...isn't it??
so... after find the solution to be u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\ny r^2)
....I have to divide by a factor r to obtain R(r)??
like R_{kl}(r) = r^{l}e^{-\nu r^2}L^{l+1/2}_k(2\ny r^2) ???
and another my question is:where I could find the solution to that ode???
thanks
folgorant
Mar22-08, 05:28 AM
malawi: ops....I see now the "source" of solution you give to me
malawi_glenn
Mar22-08, 07:29 AM
Search for Laguerre ODE
folgorant
Mar22-08, 07:36 AM
but...is my post n°5 correct?
malawi_glenn
Mar22-08, 11:18 AM
nope:
Solution is:
u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\nu r^2)
with: \nu = m\omega / 2\hbar
http://mathworld.wolfram.com/LaguerrePolynomial.html etc
folgorant
Mar24-08, 06:24 AM
(\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)
u(r)=rR(r)
\frac{\partial^2}{r \partial r^2} r = \frac{\partial}{r^2 \partial r}(r^2 \frac{\partial}{\partial r})
u_{kl}(r) = r^{l+1}e^{-\nu r^2}L^{l+1/2}_k(2\nu r^2)
\nu = m\omega / 2\hbar
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})
R_{kl}(r) = r^{l} e^\frac{-2m\omega r^2}{2\hbar} L^{l+1/2}_k(\frac{m\omega r^2}{\hbar})
R_{10}(r) = re^\frac{-m \omega r^2}{2\hbar}
Continue in next post……
folgorant
Mar24-08, 06:28 AM
......continue:
and the next is what I found with another method:
(\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)
\rho=\alpha r
\alpha=\sqrt{m \omega/ \hbar}
dr=\frac{d \rho}{\alpha}
\frac{2 \mu E}{\hbar^2 \alpha^2} = \frac{2 E}{\hbar \omega} = \lambda
\frac{d^2}{r dr^2}r = \frac{d^2}{dr^2} + \frac{2d}{rdr}
(\frac{d^2}{d \rho^2} + \frac{2 d}{\rho d \rho} - \frac{l(l+1)}{\rho^2} - \rho^2 + \lambda) R(\rho) = 0
that is a particular ODE : (\frac{d^2}{d \rho^2} + p(\rho)\frac{d}{d \rho} +q(\rho)) R(\rho) = 0
and solutions are R(\rho)=\rho^{\beta} \sum^{0}_{\infty}c_{n} \rho^n 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 : R_{10}= \frac{-2E}{6 \hbar \omega}\rho^2
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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