# Homework Help: Normalization of spherical harmonics

1. Jan 28, 2005

### Pietjuh

There is this excersise in Griffith's QM text that I can't seem to solve. It's about the calculation of the normalization factor of the spherical harmonic functions using the angular momentum step up operator.

These definitions/results are given:

$$Y_l^m = B_l^m e^{im\phi} P_l^m (\cos\theta )$$
$$L_{+} = \hbar e^{i\phi}\left(\frac{\partial}{\partial\theta} +i\cot\theta\frac{\partial}{\partial \phi}\right)$$
$$L_+ Y_l^m = A_l^m Y_l^{m+1}$$
$$A_l^m = \hbar \sqrt{l(l+1) - m(m+1)}$$

The problem is to calculate $$B_l^m$$. The approach suggested by Griffiths is to calculate $$L_+ Y_l^m$$ to get a recurrence relation for $$B_l^m$$. This is the point where I get stuck. I guess it has something to do with the deravitive of $$P_l^m$$. Anyhow, I'll give you my calculation until the point I got stuck and hope for the best that someone sees a mistake :)

$$L_+ Y_l^m = \hbar e^{i\phi}\left(\frac{\partial}{\partial\theta} +i\cot\theta\frac{\partial}{\partial \phi}\right) B_l^m e^{im\phi} P_l^m (\cos\theta )$$

$$= \hbar e^{i\phi} B_l^m \left [ e^{im\phi}\frac{\partial}{\partial\theta}\left(P_l^m (\cos\theta )\right) + i\cot\theta P_l^m (\cos\theta ) \frac{\partial}{\partial\phi} e^{im\phi} \right]$$

Now I calculated the derivative of $$P_l^m$$ using the following formula:

$$(1-x^2)\frac{dP_l^m}{dx} = \sqrt{1-x^2}P_l^{m+1} - mxP_l^m$$

So using the chain rule $$\frac{d P_l^m(\cos\theta)}{d\theta} = \frac{d P_l^m}{dx}\frac{dx}{d\theta}$$ with $$x=\cos\theta$$ I got:

$$\frac{d}{d\theta} P_l^m = - \left[\frac{1}{\sqrt{1-\cos^2\theta}}P_l^{m+1} - \frac{m\cos\theta}{\sin^2\theta}P_l^m\right]\sin\theta = m\frac{\cos\theta}{\sin\theta}P_l^m - P_l^{m+1}$$

Plugging all this in I got the following result:

$$\hbar B_l^m \left[ (m-1)P_l^m\cot\theta - P_l^{m+1}\right] = A_l^m B_l^{m+1}P_l^{m+1}$$

but when I try to solve this for $$B_l^m$$ I get an exploding expression :(

2. Jan 28, 2005

### dextercioby

What do you mean??It should look pretty bad indeed...Do you have the expression in Griffiths to compare your result to...??

Daniel.

3. Jan 28, 2005

### Pietjuh

I' m sorry my english isn't that good!
From the last equation I am trying to iterate $$B_l^m$$ so I will get $$B_l^m = A B_l^0$$

Griffiths says it should be $$B_l^m = (-1)^m\sqrt{\frac{(2l+1)}{4\pi}\frac{(l-|m|)!}{(l+|m|)!}$$

4. Jan 28, 2005

### dextercioby

Check pages 678 pp 690 from Cohen-Tannoudji.

Daniel.

5. Jan 28, 2005

### Pietjuh

I don't have that book. The only QM text I have is the one from Griffiths.

6. Jan 28, 2005

### dextercioby

Well,i'm not gonna type 100 formulas for you,i'll give you some hints:
#1
$$\hat{L}_{-} Y_{l,-l}(\vartheta,\varphi) =0$$

Okay??

Solve this equation and find
$$Y_{l,-l} (\vartheta,\varphi)$$

and tell me what u get.Normelize the solution and give the constant of integration as well.

Then i'll guide through step #2.

Daniel.

7. Jan 28, 2005

### Pietjuh

These are a few results I'm getting but again it seems I'm stuck on the same sort of problem again :(

$$L_{-} Y_l^{-l} = -\hbar e^{-i\phi}\left(\frac{\partial Y_l^{-l}}{\partial\theta} - \cot\theta \frac{\partial Y_l^{-l}}{\partial \phi}\right)=0$$

So $$\frac{\partial Y_l^{-l}}{\partial\theta} = \cot\theta \frac{\partial Y_l^{-l}}{\partial \phi}$$

$$\frac{\partial Y_l^{-l}}{\partial \phi} = -il B_l^{-l} e^{-il\phi} P_l^{-l}$$

$$\frac{\partial Y_l^{-l}}{\partial \theta} = B_l^{-l} e^{-il\phi}\left(P_l^{-l+1} + l\frac{\cos\theta}{\sin^2\theta}P_l^{-l}\right)$$

so from this I get

$$P_l^{-l+1} + l\frac{\cos\theta}{\sin^2\theta}P_l^{-l} = -il\cot\theta P_l^{-l}$$

But I can't see how I should continue....
Forgive me for my seemingly stupidity

8. Jan 28, 2005

### dextercioby

So
$$\hat{L}_{-} =\hbar e^{-i\varphi}(-\frac{\partial}{\partial \vartheta}+i\cot\vartheta \frac{\partial}{\partial \varphi})$$ (1)

Okay??
$$Y_{l,-l}(\vartheta,\varphi)=P_{l,-l}(\vartheta)e^{-il\varphi}$$(2)

Apply the operator (1) on the function (2) and give me the differential equation for
$$P_{l,-l}(\vartheta)$$

Daniel.

9. Jan 28, 2005

### Pietjuh

$$L_{-} Y_l^{-l} = \hbar e^{-i(l+1)\phi}\left(\cot\theta P_l^{-l} - \frac{\partial P_l^{-l} } {\partial \theta} \right) = 0$$

so $$\frac{\partial P_l^{-l} } {\partial \theta} = \cot\theta P_l^{-l}$$

10. Jan 28, 2005

### dextercioby

Almost,u missed the "l" when differentiating wrt $\varphi$.It's not PD anymore (one variable only,$\vartheta$):

$$\frac{dP_{l,-l}(\vartheta)}{d\vartheta}=l\cot\vartheta P_{l,-l}(\vartheta)$$

Solve this ODE.Then write the spherical harmonic...

Daniel.

11. Jan 28, 2005

### Pietjuh

Integrating factor is $$\frac{1}{\sin^l\theta}$$, so
$$P_l^{-l} = C \sin^l\theta$$

So $$Y_l^{-l} = C \sin^l\theta e^{-il\phi}$$

#edit# I have to go now. Much thanks for your help! :)

Last edited: Jan 28, 2005
12. Jan 28, 2005

### dextercioby

Perfect.Now normalize it and tell me what is the normalization constant...Chose the phase equal to 1.

Daniel.

13. Jan 28, 2005

### Pietjuh

$$\int d\theta d\phi C^2 \sin^{2l}\theta = 2\pi C^2 \int_0^\pi d\theta sin^{2l}\theta$$

since $$\int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta$$

the integral is equal to

$$\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}$$
if n>=2 and n is even and this is equal to

$$\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{2}{3}$$

if n>=3 and n is odd.
But because 2l is always even and >= 2 we only need the first one,
so

$$C = \frac{1}{\sqrt{2\pi}} \sqrt{\frac{2l-1}{2l}\frac{2l-3}{2l-2}\cdots \frac{1}{2} \frac{\pi}{2}}$$

14. Jan 28, 2005

### dextercioby

How did you get that result??Mine is totally different.

BTW,the site which had Abramowitz & Stegun is not working...

Daniel.

15. Jan 29, 2005

### Pietjuh

The formula $$\int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta$$ was on the backcover of my calculus book and the result I got in my previous post was just a sort of copy from an example in the chapter on partial integration.

16. Jan 29, 2005

### dextercioby

Then why do i get the feeling that my result is correct
$$C_{l}=\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{2^{l} \ l!}$$

and yours,which i cannot see how it can be put in a form similar to mine,is wrong...

Daniel.

P.S.BTW,mine coincides with the one given by Cohen-Tannoudji,page 682,#29.

17. Jan 29, 2005

### Pietjuh

I think you are absolutely correct, which makes me feel even more stupid because I can't seem to evaluate integrals properly anymore

I'm going to type it out step by step now, and hopefully you can discover (or myself) where I'm going the wrong way.

$$\int_0^{2\pi}\int_0^\pi |Y_l^m|^2 \sin\theta d\theta d\phi = 1$$

$$\int_0^{2\pi}\int_0^\pi |Y_l^m|^2 \sin\theta d\theta d\phi = \int_0^{2\pi}\int_0^\pi C^2 \sin^{2l+1}\theta d\theta d\phi = 2\pi C^2 \int_0^\pi \sin^{2l+1} d\theta = 1$$

Using the iteration formula $$\int_0^\pi d\theta sin^n \theta = \frac{n-1}{n}\int_0^\pi d\theta sin^{n-2} \theta$$ I get:

$$\int_0^\pi \sin^{2l+1} d\theta = \frac{2l}{2l+1} \frac{2l-2}{2l-1} \cdots \frac{2}{3} \int_0^\pi \sin\theta d\theta = \frac{2l}{2l+1} \frac{2l-2}{2l-1} \cdots \frac{2}{3} 2$$

So

$$C^2 = \frac{1}{4\pi} \frac{2l+1}{2l} \frac{2l-1}{2l-2}\cdots \frac{3}{2}$$

But I can't seem to rewrite this in your form

18. Jan 29, 2005

### dextercioby

Maybe i can

$$|C_{l}|^{2}=\frac{1}{4\pi} \frac{(2l+1)!!}{(2l)!!}=\frac{1}{4\pi}\frac{(2l+1)!}{[(2l)!!]^{2}}$$

Then
$$|C_{l}|=\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{(2l)!!}$$

Writing
$$(2l)!!=2^{l} \ l!$$
and chosing the phase
$$\phi=(-1)^{l}$$

,we get
$$C_{l}=(-1)^{l}\frac{1}{\sqrt{4\pi}}\frac{\sqrt{(2l+1)!}}{2^{l} \ l!}$$

which coincides with the formula #14,page 679,Cohen-Tannoudji.

Daniel.

19. Jan 29, 2005

### Pietjuh

ah ok :)
I never heard of this double factorial, so I was wondering how to write (2l+1)(2l-1)..... as a factorial :)