# Homework Help: Expansion in spherical harmonics

1. Nov 24, 2007

### ehrenfest

1. The problem statement, all variables and given/known data
I want to expand 1+sin(phi)sin(theta) in the spherical harmonics. I am not sure if this will be an infinite series or not? If it were infinite that would seem rather difficult because the spherical harmonics get really complicated when l > 3. Also, all of the sine terms in the spherical harmonics are imaginary, so how is this possible?

2. Relevant equations

3. The attempt at a solution

2. Nov 25, 2007

### malawi_glenn

what have you tried so far?

a hint is tho rewrite $$e^{i \phi}$$ as $$\cos \phi + i \sin \phi$$

And then you can get rid of the $$i$$ by doing a linear combination of two spherical harmonics with same $$l$$ but different $$m$$.

The series will be finite, it will contain 3 different $$Y_{l,m}$$

Now go ahead and try

3. Nov 25, 2007

### ehrenfest

OK. We can get take care of the constant term with Y0,0. For the other term I tried Y2,2 and Y2,1 but that does not get rid of the i. Are you saying I can find two spherical harmonics with exactly the same imaginary term up to a constant? Don't each of the spherical harmonics with the same l have different imaginary parts?
Do I need to use negative m values?

4. Nov 25, 2007

### malawi_glenn

why try l = 2 first?.. why not l = 1?..

5. Nov 25, 2007

### ehrenfest

Because Y1,0 does not have an imaginary part, it could not possibly cancel with the imaginary part of Y1,1.

6. Nov 26, 2007

### malawi_glenn

Take a look again at the l = 1 spherical harmonics. For each l, you have 2l + 1 spherical harmonics.

Spherical harmonics with l = 2 have sin(theta)^2 ; so can IMPOSSIBLE do anything with those. Spherical harmonics with l = 1 have sin(theta); so I think it is quite obvious..

Last edited: Nov 26, 2007
7. Nov 26, 2007

### ehrenfest

How stupid of me! I see now. Thanks.

8. Nov 28, 2007

### malawi_glenn

The answer will be:

$$1 + \sin \theta \sin \phi = \sqrt{4\pi} + \sqrt{\frac{2\pi}{3}}i \cdot (Y_{1,+1} + Y_{1,-1})$$

If i did it correct, you got the same?

[edited]

Last edited: Nov 28, 2007
9. Nov 28, 2007

### ehrenfest

I got
$$1 + \sin \theta \sin \phi = \sqrt{4\pi} + \sqrt{\frac{2\pi}{3}}i \cdot (Y_{1,+1} + Y_{1,-1})$$

10. Nov 28, 2007

### malawi_glenn

yes that is correct, because I had i in the denominator on my papers, but wrote wrong here in TeX. Good job!

11. Dec 5, 2007

### Gerenuk

I could see that one, but there's something silly I'm missing :(

How would I expand
sin(theta)?

To me it seems only Y^0_1 has the right order, but its a cos(theta) that I cannot convert?!

12. Dec 5, 2007

### malawi_glenn

silly? now are you asking just for curiosity or is it homework-related?

13. Dec 5, 2007

### Gerenuk

Well, its homework. But given by myself.
I was wondering about this splitting of cubic symmetry along the 3 fold diagonal axis. It splits into 2+2+1 energylevels, but the basis system is is along the diagonal and not the z axis anymore so I was trying to project
(3cos^2(theta+45°)-1)*exp(-i*(psi+45°))
back into the usual system.
But now I'm stuck with it. Maybe it's been a hard day today. Just can't see what I'm doing wrong :(

14. Dec 5, 2007

### malawi_glenn

I dont know either. Try posting the whole problem in its entire lenght in a new thread. Showing some of the relations you know of and your attempt to solve it. Then someone might be able to help you =)

Btw, I just think you can write $$\sin ^2\theta = 1 - \cos ^2\theta$$ and take it from there.

15. Dec 5, 2007

### Gerenuk

OK. Maybe its something more fundamental.
I can switch between sin^2(a) and cos^2(a), but not to sin(a)cos(a)
But I think I know all trigonometric relations. Yesterday I derived for fun:
sin(a+b+c+d)/cos(a)/cos/(b)/cos(c)/cos(d)=tan(a)+tan(b)+tan(c)+tan(d)+tan(a)tan(b)tan(c)tan(d)(1/tan(a)+1/tan(b)+1/tan(c)+1/tan(d))
Its pretty useless but nice.

16. Dec 5, 2007

### malawi_glenn

I was rather thinking of

$$\sin ^2\theta = 1 - \cos ^2\theta$$

$$\sin \theta = \sqrt{4\pi Y_{0,0}^2-\frac{4\pi }{3}Y_{1,0}^2}$$

17. Dec 5, 2007

### Gerenuk

Yes, OK. But thats not linear in the spherical harmonics.
Mr Wiki Pedia says
http://en.wikipedia.org/wiki/Spherical_harmonics
one can expand linearly. There is also an equation for the coefficients. I'm not in the state of doing integrals now, but I suspect I get zero for all of them.

18. Dec 5, 2007

### malawi_glenn

no ofcourse its not a linear expansion.

You should in principle be able to expand ANY angular function as a combination of spherical harms-