# Normalizing the spherical harmonics

1. May 15, 2012

### epsilonjon

1. The problem statement, all variables and given/known data

http://img109.imageshack.us/img109/1065/87070684.png [Broken]

2. Relevant equations

1) $$L_{\pm}=\pm\hbar e^{\pm i \phi}(\frac{\partial}{\partial\theta}\pm i cot\theta \frac{\partial}{\partial\phi})$$
2) $$L_{\pm}Y^m_l = \hbar\sqrt{(l \mp m)(l \pm m+1)}Y^{m \pm 1}_{l}$$
3)Answer given in book: $$Y^m_l(\theta, \phi) = (-1)^m \sqrt{\frac{(2l+1)(l- |m|)!}{4\pi(l+ |m|)!}} e^{im\phi}P^m_l (cos\theta)$$

3. The attempt at a solution

Eq 1 above gives

$$L_{+}Y^{m}_{l} = \hbar e^{i \phi}(B^m_l e^{im \phi} \frac{\partial P^m_l (cos\theta)}{\partial \theta} + icot\theta B^m_l P^m_l (cos\theta)ime^{im \phi})$$
$$=\frac{-\hbar B^m_l e^{i(m+1)\phi}}{sin\theta}[sin^2 \theta \frac{d P^m_l (cos\theta)}{d cos\theta} + mcos\theta P^m_l (cos\theta)]$$

Using the formula for the derivatives of associated Legendre functions (given in the question), this becomes

$$L_{+}Y^m_l = \frac{-\hbar B^m_l e^{i(m+1)\phi}}{sin\theta}[\sqrt{1-cos^2 \theta}P^{m+1}_l(cos\theta) - mcos\theta P^m_l(cos\theta)+mcos\theta P^m_l(cos\theta)]$$
$$= - \hbar B^m_l e^{i(m+1)\phi}P^{m+1}_l(cos\theta)$$

But eq 2 says that
$$= \hbar \sqrt{(l-m)(l+m+1)} B^{m+1}_{l} e^{i(m+1)\phi} P^{m+1}_l (cos\theta)$$

so combining these we get

$$B^{m+1}_l = \frac{-B^m_l}{\sqrt{(l-m)(l+m+1)}}$$

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

$$B^{-1}_2 = \frac{1}{2}\sqrt{\frac{5}{6\pi}}$$
$$B^{-2}_2 = \frac{1}{4}\sqrt{\frac{5}{6\pi}} = -\frac{1}{2}B^{-1}_{2}$$

But these do not obey the recursion formula!
Thanks!!

Last edited by a moderator: May 6, 2017
2. May 17, 2012

### vela

Staff Emeritus
What are equations 4.120, 4.121, and 4.130?

3. May 17, 2012

### Steely Dan

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.

Last edited: May 17, 2012
4. May 18, 2012

### epsilonjon

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.
Okay, but I'm warning you it's a bit long-winded! :yuck: 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...

Last edited: May 18, 2012
5. May 18, 2012

### Steely Dan

Everything up until here is correct, as far as I can tell (you went to the trouble of posting everything, even though it's not exactly what I was asking for, so I tried to follow everything as best as I could).

There is an error here, however. What you have actually written down is the associated Legendre polynomial $P^{l}_{l}$; there is an additional constant factor out front for when the $m$ number is negative, that you are missing here.

6. May 19, 2012

### epsilonjon

Ah yeah sorry, for some reason my book omits the phase $(-1)^m$ from the Legendre function so I forgot about it. But I think all that does is change the final answer to

$$B^m_l = \frac{(-1)^{l}}{2(2l)!} \sqrt{\frac{2l+1}{\pi}} \sqrt{\frac{(l-m)!}{(l+m)!}}$$
I've still got that pesky $\frac{1}{(2l)!}$ in there. I am still confused as to why the book answer does not satisfy the recursion formula though It indicates to me that the recursion formula is wrong, but I've checked it like 10 times and I swear it's correct!

7. May 19, 2012

### Steely Dan

When I first looked at it I assumed you had calculated the coefficients incorrectly, but I just took another look at it and maybe I was wrong. I'll take a closer look later on. The negative sign in the recursion relation definitely worries me; it shouldn't be there, I think.

Last edited: May 19, 2012
8. May 19, 2012

### epsilonjon

Thanks mate, really appreciate your help.

9. May 24, 2012

### epsilonjon

Sorry to bump this but was wondering whether you (or anyone else) had managed to figure it out or could give me a hint? I've gone over it again and still get the same recursion formula. It's really bugging me now

10. May 25, 2012

### vela

Staff Emeritus
Your recursion formula is correct. It's the normalization of the associated Legendre polynomials that is confusing you, I think.

Try looking up the $Y_2^m$'s and $P_2^m$'s and explicitly working out what $B_2^m$ is equal to for the various allowed values of m. You should be able to see what's going on.