Differential equation similar to Legendre

  • #1

Main Question or Discussion Point

I am trying to solve the following differential equation:

[tex]
(\frac{L^2}{6k^2}+\frac{w\sqrt{3}}{2}\sin^2\theta\cos 2\phi)\psi=E\psi
[/tex]

where [tex]L^2 [/tex]is the angular momentum given by:
[tex] \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}(sin\theta\frac{partial}{\partial\theta})-\frac{1]{sin^2\theta}\frac{\partial^2}{\partial\phi^2}
[/tex]. [tex] \theta [/tex] goes from 0 to [tex] \pi [/tex] while [tex] \phi [/tex] goes from 0 to 2 [tex] \pi [/tex]. [tex] k [/tex] and [tex] w[/tex] are constants and E is the energy of the system.. This differential equation seems non separable. Any ideas how to solve it...I also realised that the term [tex] sin^2\theta\cos 2\phi [/tex] is a combination of [tex] (Y_{2,-2]+ Y_{2,2}) [/tex]. But then how to continue?

Thanks
 

Answers and Replies

  • #2
Sorry L^2 is:
[tex]
\frac{1}{\sin\theta}\frac{\partial}{\partial\theta} (\sin\theta\frac{partial}{\partial\theta})-\frac{1]{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}

[/tex]
 
  • #3
173
0
You mean

[tex]L^2 = \frac{1}{\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial}{\partial\theta}\right)-\frac{1}{\sin^2\theta}\frac{\partial^2}{\partial\phi^2}.[/tex]

Some ideas: apply the operator to [itex]Y_L^M[/itex] with arbitary values of [itex]L,\,M[/itex] and see if you can determine [itex]L,\,M[/itex] using the properties of the polynomials, or maybe using a linear combination of Legendre polynomials, or appliyng the operator to [itex]f(\theta)Y_L^M,\,g(\phi)Y_L^M,\,h(\theta,\phi)Y_L^M[/itex] and use the properties of the polynomials to simplify the equation and determine [itex]f,\,g,\,h[/itex].

Just ideas.
 
Last edited:

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