Reducing angular Schrodinger equation to eigenvalue problem

[perishingtardi]

Homework Statement

The angular part of the Schrodinger equation for a positron in the field of an electric dipole moment {\bf d}=d{\bf \hat{k}} is, in spherical polar coordinates (r,\vartheta,\varphi),
\frac{1}{\sin\vartheta}\frac{\partial}{\partial\vartheta} \left( \sin\vartheta\frac{\partial Y}{\partial\vartheta} \right) + \frac{1}{\sin^2 \vartheta}\frac{\partial^2 Y}{\partial\varphi^2} - 2dY\cos\vartheta + \lambda Y=0.
By considering the ansatz Y=Y_m (\vartheta,\varphi) = \sum_{l'=|m|}^{\infty} C_{l'} Y_{l'm}(\vartheta,\varphi),
where Y_{l'm} are spherical harmonics, show that the problem of finding \lambda reduces to a matrix eigenvalue problem of the form \sum_{l'=|m|}^{\infty} A_{ll'}C_{l'} = \lambda C_{l}.

Homework Equations

\frac{1}{\sin\vartheta}\frac{\partial}{\partial\vartheta} \left( \sin\vartheta\frac{\partial Y_{l'm}}{\partial\vartheta} \right) + \frac{1}{\sin^2 \vartheta}\frac{\partial^2 Y_{l'm}}{\partial\varphi^2} = -l'(l'+1)Y_{l'm}
\int_0^{2\pi}\int_0^\pi Y_{lm}^* Y_{l'm} \sin\vartheta\,d\vartheta\,d\varphi = \delta_{ll'}
Y_{lm}^* = (-1)^m Y_{l,-m}
\int_0^{2\pi}\int_0^\pi Y_{l_1 m_1}Y_{l_2 m_2}Y_{l_3 m_3} \sin\vartheta \, d\vartheta \, d\varphi = \sqrt{\frac{(2l_1 +1)(2l_2 + 1)(2l_3 + 1)}{4\pi}}\begin{pmatrix}l_1&l_2&l_3\\ 0&0&0\end{pmatrix}\begin{pmatrix}l_1&l_2&l_3\\ m_1&m_2&m_3\end{pmatrix} where the arrays in parentheses are Wigner 3jm symbols.
\cos\vartheta = \sqrt{\frac{4\pi}{3}}Y_{10}

The Attempt at a Solution

By inserting the ansatz into the angular Schrodinger equation and using the first of the 'relevant equations' I got
\sum_{l'} C_{l'} \{ -l'(l'+1)Y_{l'm} - 2d\cos\vartheta\,Y_{l'm} + \lambda Y_{l'm} \}=0.
Then by multiplying through by Y_{lm}^* and integrating over the unit sphere and using the other equations given, I got (after some manipulation)
2d(-1)^m \sum_{l'=|m|}^{\infty} \sqrt{(2l+1)(2l'+1)} \begin{pmatrix}1&l&l'\\0&0&0\end{pmatrix}\begin{pmatrix}1&l&l'\\0&-m&m\end{pmatrix}C_{l'} = [\lambda - l(l+1)]C_l.
This is almost in the form required, I think, except that there is an extra term of -l(l+1) on the RHS, which I don't know how to get rid of. Any help would be very much appreciated.

 

[DrClaude]

This is almost in the form required, I think, except that there is an extra term of -l(l+1) on the RHS, which I don't know how to get rid of. Any help would be very much appreciated.

Just move it to the LHS in the form of a diagonal term (you can use $\delta_{l,l'}$ if you want to put it in the sum).