Expansion in spherical harmonics

In summary: The series is finite but it does contain three different Y_{l,m} values. The constant term is taken care of by the first Y_{l,m} value. and the other term is canceled by the imaginary part of the second Y_{l,m} value.
  • #1
ehrenfest
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1

Homework Statement


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?

Homework Equations


The Attempt at a Solution

 
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  • #2
what have you tried so far?

a hint is tho rewrite [tex] e^{i \phi} [/tex] as [tex] \cos \phi + i \sin \phi [/tex]

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

The series will be finite, it will contain 3 different [tex] Y_{l,m} [/tex]

Now go ahead and try
 
  • #3
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
why try l = 2 first?.. why not l = 1?..
 
  • #5
Because Y1,0 does not have an imaginary part, it could not possibly cancel with the imaginary part of Y1,1.
 
  • #6
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..
 
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  • #7
How stupid of me! I see now. Thanks.
 
  • #8
The answer will be:

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

If i did it correct, you got the same?

[edited]
 
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  • #9
I got
[tex] 1 + \sin \theta \sin \phi = \sqrt{4\pi} + \sqrt{\frac{2\pi}{3}}i \cdot (Y_{1,+1} + Y_{1,-1}) [/tex]
 
  • #10
yes that is correct, because I had i in the denominator on my papers, but wrote wrong here in TeX. Good job!
 
  • #11
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
silly? now are you asking just for curiosity or is it homework-related?
 
  • #13
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
I don't know either. Try posting the whole problem in its entire length 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 [tex] \sin ^2\theta = 1 - \cos ^2\theta [/tex] and take it from there.
 
  • #15
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
I was rather thinking of

[tex] \sin ^2\theta = 1 - \cos ^2\theta [/tex]

[tex] \sin \theta = \sqrt{4\pi Y_{0,0}^2-\frac{4\pi }{3}Y_{1,0}^2} [/tex]
 
  • #17
Yes, OK. But that's 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
no ofcourse its not a linear expansion.

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

1. What is the concept of expansion in spherical harmonics?

Expansion in spherical harmonics is a mathematical technique used to decompose a function defined on a sphere into a series of simpler functions. This allows for a more efficient and accurate representation of the original function, as well as easier manipulation and analysis.

2. How are spherical harmonics related to the geometry of a sphere?

Spherical harmonics are closely related to the geometry of a sphere, as they are defined as the eigenfunctions of the Laplace operator on a sphere. This means that they describe the spatial variation of a function on a spherical surface.

3. What are the applications of expansion in spherical harmonics?

Expansion in spherical harmonics has various applications in fields such as physics, mathematics, and computer graphics. It is commonly used to solve problems involving spherical symmetry, such as in quantum mechanics and electromagnetism. It is also used for image and signal processing, as well as in the analysis of geophysical data.

4. How are spherical harmonics calculated and represented?

Spherical harmonics are typically calculated using a recursive formula or numerical methods. They are commonly represented as a summation of terms, each with a specific coefficient and spherical harmonic function. The coefficients can be determined through various methods such as least squares fitting or spherical harmonic analysis.

5. What is the relationship between spherical harmonics and Legendre polynomials?

Spherical harmonics are closely related to Legendre polynomials, as they can be expressed as a product of Legendre polynomials and a phase factor. The Legendre polynomials represent the angular dependence of the spherical harmonics, while the phase factor captures the azimuthal dependence. This relationship is important in understanding the properties and uses of spherical harmonics.

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