vela said:
What are equations 4.120, 4.121, and 4.130?
Ah sorry, I forgot to mention that! Eqs 4.121 and 4.130 are the first two 'relevant equations' I provided in my original post. Eq 4.120 is basically the same as Eq. 4.121 so I haven't bothered putting it.
Steely Dan said:
Your math to determine that recursion relation seems solid. Walk us through your reasoning from how you got from this to the solution for the coefficients.
Okay, but I'm warning you it's a bit long-winded! Here goes...
First I started with the 'bottom rung' and used the recursion formula to work out the higher rung coefficients in terms of this:
B^{-l+1}_l = \frac{-B^{-l}_l}{\sqrt{l-(-l)} \sqrt{l-l+1}} = \frac{-B^{-l}_l}{\sqrt{2l} \sqrt{1}}
B^{-l+2}_l = \frac{-B^{-l+1}}{\sqrt{l-(-l+1)} \sqrt{l-l+1+1}} = \frac{-B^{-l+l}}{\sqrt{2l-1} \sqrt{2}} = \frac{B^{-l}_l}{\sqrt{(2l)(2l-1)} \sqrt{2*1}}
B^{-l+3}_l = \frac{-B^{-l+2}_l}{\sqrt{l-(-l+2)} \sqrt{l-l+2+1}} = \frac{-B^{-l+2}_l}{\sqrt{2l-2} \sqrt{3}} = \frac{-B^{-l}_l}{\sqrt{(2l)(2l-1)(2l-2)} \sqrt{3*2*1}}
etc. Observing the pattern, we see that
B^m_l = \frac{(-1)^{m+l}}{ \sqrt{\frac{(2l)!}{[2l-(m+l)]!}} \sqrt{(m+l)!}}B^{-l}_l = \frac{(-1)^{m+l}}{\sqrt{(2l)!}}\sqrt{\frac{(l-m)!}{(l+m)!}}B^{-l}_l
So now we need to find B^{-l}_l. Using the fact that L_z = \frac{\hbar}{i}\frac{\partial}{\partial \phi} we get
L_z Y^{-l}_l = - \hbar l Y^{-l}_l = \frac{\hbar}{i}\frac{\partial Y^{-l}_l}{\partial \phi} \Rightarrow \frac{\partial Y^{-l}_l}{\partial \phi} = -il Y^{-l}_l \Rightarrow Y^{-l}_l = f(\theta)e^{-il \phi}
Using eqs 1 and 2 from my relevant equations above,
L_{-}Y^{-l}_{l} = - \hbar e^{-i \phi}(\frac{\partial}{\partial \theta} - icot \theta \frac{\partial}{\partial \phi})Y^{-l}_{l} = 0 \Rightarrow \frac{\partial Y^{-l}_l}{\partial \theta} = icot \theta \frac{\partial Y^{-l}_l}{\partial \phi}
Combining this with Y^{-l}_l = f(\theta)e^{-il \phi}, we get
\frac{df}{d \theta} = lcot \theta f(\theta) \Rightarrow \int \frac{1}{f}df = l \int \frac{cos \theta}{sin \theta}d \theta \Rightarrow Ln(f) = l Ln(sin \theta) + constant \Rightarrow f(\theta) = A(sin \theta)^l \Rightarrow Y^{-l}_l = A(sin \theta)^l e^{-il \phi}
We now need to normalize this to find A:
1 = |A|^2 \int (sin \theta)^{2l} sin \theta d \theta d \phi = 2 \pi |A|^2 \int (sin \theta)^{2l+1} d \theta = 2 \pi |A|^2 2 \frac{(2*4*6*...*(2l))}{1*3*5*...*(2l+1)}
= 4 \pi |A|^2 \frac{(2*4*6*...*(2l))^2}{1*2*3*4*...*(2l)*(2l+1)} = 4 \pi |A|^2 \frac{(2^l l!)^2}{(2l+1)!} \Rightarrow A = \frac{1}{2^{l+1}l!}\sqrt{\frac{(2l+1)!}{\pi}}
So we have Y^{-l}_l = \frac{1}{2^{l+1}l!}\sqrt{\frac{(2l+1)!}{\pi}}(sin \theta)^l e^{-il \phi}. Call this equation (#).
From the original form of Y^m_l we have Y^{-l}_l = B^{-l}_l e^{-il \phi} P^{-l}_l(cos \theta) where P^{-l}_l(cos \theta) is an associated Legendre function, defined by
P^{-l}_{l}(x) = (1-x^2)^{l/2}(\frac{d}{dx})^l[\frac{1}{2^l l!} (\frac{d}{dx})^l (x^2-1)^l] = (1-x^2)^{l/2} \frac{1}{2^l l!} (\frac{d}{dx})^{2l}(x^2-1)^l
= (1-x^2)^{l/2} \frac{1}{2^l l!} (\frac{d}{dx})^{2l}[x^{2l}+...] = (1-x^2)^{l/2} \frac{1}{2^l l!} (2l)! \Rightarrow P^{-l}_l(cos \theta) = (sin \theta)^l \frac{(2l)!}{2^l l!}
Hence Y^{-l}_l = B^{-l}_l e^{-il \phi} (sin \theta)^l \frac{(2l)!}{2^l l!}. Combining this with equation (#) above, we get
B^{-l}_l \frac{(2l)!}{2^l l!} = \frac{1}{2^{l+1}l!} \sqrt{\frac{(2l+1)!}{\pi}} \Rightarrow B^{-l}_l = \frac{1}{2(2l)!} \sqrt{\frac{(2l+1)!}{\pi}}
Plugging this equation for B^{-l}_l into our equation for B^m_l we finally end up with
B^m_l = \frac{(-1)^{m+l}}{\sqrt{(2l)!}}\sqrt{\frac{(l-m)!}{(l+m)!}}\frac{1}{2(2l)!} \sqrt{\frac{(2l+1)!}{\pi}} = \frac{(-1)^{m+l}}{2(2l)!} \sqrt{\frac{2l+1}{\pi}} \sqrt{\frac{(l-m)!}{(l+m)!}}
which is close to the answer given in the book, but not the same
However, I checked it for a few different combinations of m and l, and it does obey the recursion relation (something which the book answer doesn't seem to do!).
Thanks for your help. I'm going to lie down in a dark room...
