• Support PF! Buy your school textbooks, materials and every day products Here!

Reducing angular Schrodinger equation to eigenvalue problem

  • #1

Homework Statement


The angular part of the Schrodinger equation for a positron in the field of an electric dipole moment [itex]{\bf d}=d{\bf \hat{k}}[/itex] is, in spherical polar coordinates [itex](r,\vartheta,\varphi)[/itex],
[tex]\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.[/tex]
By considering the ansatz [tex]Y=Y_m (\vartheta,\varphi) = \sum_{l'=|m|}^{\infty} C_{l'} Y_{l'm}(\vartheta,\varphi),[/tex]
where [itex]Y_{l'm}[/itex] are spherical harmonics, show that the problem of finding [itex]\lambda[/itex] reduces to a matrix eigenvalue problem of the form [tex]\sum_{l'=|m|}^{\infty} A_{ll'}C_{l'} = \lambda C_{l}.[/tex]

Homework Equations


[tex]\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}[/tex]
[tex]\int_0^{2\pi}\int_0^\pi Y_{lm}^* Y_{l'm} \sin\vartheta\,d\vartheta\,d\varphi = \delta_{ll'}[/tex]
[tex]Y_{lm}^* = (-1)^m Y_{l,-m}[/tex]
[tex]\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} [/tex] where the arrays in parentheses are Wigner 3jm symbols.
[tex]\cos\vartheta = \sqrt{\frac{4\pi}{3}}Y_{10}[/tex]

The Attempt at a Solution



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

Answers and Replies

  • #2
DrClaude
Mentor
7,269
3,424
This is almost in the form required, I think, except that there is an extra term of [itex]-l(l+1)[/itex] 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).
 
  • Like
Likes 1 person

Related Threads on Reducing angular Schrodinger equation to eigenvalue problem

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
23
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
3K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
992
Replies
8
Views
2K
Replies
8
Views
5K
Top