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epsilonjon
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Homework Statement
http://img109.imageshack.us/img109/1065/87070684.png
Homework Equations
1) [tex]L_{\pm}=\pm\hbar e^{\pm i \phi}(\frac{\partial}{\partial\theta}\pm i cot\theta \frac{\partial}{\partial\phi})[/tex]
2) [tex]L_{\pm}Y^m_l = \hbar\sqrt{(l \mp m)(l \pm m+1)}Y^{m \pm 1}_{l}[/tex]
3)Answer given in book: [tex]Y^m_l(\theta, \phi) = (-1)^m \sqrt{\frac{(2l+1)(l- |m|)!}{4\pi(l+ |m|)!}} e^{im\phi}P^m_l (cos\theta)[/tex]
The Attempt at a Solution
Eq 1 above gives
[tex]L_{+}Y^{m}_{l} = \hbar e^{i \phi}(B^m_l e^{im \phi} \frac{\partial P^m_l (cos\theta)}{\partial \theta} + icot\theta B^m_l P^m_l (cos\theta)ime^{im \phi})[/tex]
[tex]=\frac{-\hbar B^m_l e^{i(m+1)\phi}}{sin\theta}[sin^2 \theta \frac{d P^m_l (cos\theta)}{d cos\theta} + mcos\theta P^m_l (cos\theta)][/tex]
Using the formula for the derivatives of associated Legendre functions (given in the question), this becomes
[tex]L_{+}Y^m_l = \frac{-\hbar B^m_l e^{i(m+1)\phi}}{sin\theta}[\sqrt{1-cos^2 \theta}P^{m+1}_l(cos\theta) - mcos\theta P^m_l(cos\theta)+mcos\theta P^m_l(cos\theta)][/tex]
[tex]= - \hbar B^m_l e^{i(m+1)\phi}P^{m+1}_l(cos\theta)[/tex]
But eq 2 says that
[tex]= \hbar \sqrt{(l-m)(l+m+1)} B^{m+1}_{l} e^{i(m+1)\phi} P^{m+1}_l (cos\theta)[/tex]
so combining these we get
[tex]B^{m+1}_l = \frac{-B^m_l}{\sqrt{(l-m)(l+m+1)}}[/tex]
I've checked that a few times and I'm pretty sure it's correct. Could someone confirm this please? I've then gone on to try to solve this recursion formula and got an answer, but unfortunately it's different from the one given in the book
I have noticed one thing though: when I plug in {m=-1, l=2} and {m=-2, l=2} into the book answer (given above) I get
[tex]B^{-1}_2 = \frac{1}{2}\sqrt{\frac{5}{6\pi}}[/tex]
[tex]B^{-2}_2 = \frac{1}{4}\sqrt{\frac{5}{6\pi}} = -\frac{1}{2}B^{-1}_{2}[/tex]
But these do not obey the recursion formula!
Thanks!
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