Addition Theorem for Spherical Harmonics

[QMrocks]

Happy New Year all!

i have a question regarding the addition theorem for spherical harmonics. In JD Jackson book pg 110 for e.g. the addition theorem is given as:
[tex]
P_{L}(cos(\gamma))=\frac{4\pi}{2L+1}\sum_{m=-L}^{L}Y^{*}_{Lm}(\theta',\phi')Y_{Lm}(\theta,\phi)
[/tex]
where [tex] cos(\gamma)=cos\theta cos\theta' + sin\theta sin\theta' cos(\phi-\phi') [/tex]. The 2 coordinate system[tex] (r,\theta,\phi) [/tex]and[tex] (r',\theta',\phi')[/tex] have an angle [tex]\gamma [/tex] between them.

My question is:

How can we express [tex]Y_{Lm}(\theta',\phi') [/tex] in terms of [tex]Y_{Lm}(\theta,\phi) [/tex] ?

 

[Dr Transport]

I don't believe that you can express the primed coordinates in terms of the unprimed in that case. Think of the primed and un-primed coordinates as source and field points for a charge distribution.

 

[lalbatros]

QMRocks:
Have a look at this site where transformation properties of the spherical harmonics are discussed:

http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm

What we can say for sure, is that the L number doesn't change under rotation. But -of course- eigenvectors of Lz are not eigenvectors of Lz'. Since all angular functions can be expressed by spherical harmonics, we can take for granted that Y'(L,m) = sum(c(m')*Y(L,m')). Only the coefficients c(m') remains to be determined. There are of course so easy special cases ...

The web page above introduces the rotation matrices which is what you are looking for. I think there are many other readings on this topic, but it depends on what you are interrested in: formulas to apply to a problem, group theory, principles, ...

I also remember the QM course by Feynman which discuss the spin rotations. This is again the same topic, except that the basis of the discussion are 'spin states' in place of spherical harmonics: different representations of the rotation group.

 

[lalbatros]

In addition to my previous post:

Note that you can apply the "ladder" operators L+ = Lx + iLy and L- to both sides of the addition theorem relation (in the non-primed frame). You can repeat this until the ladder operator produce null states.

In this way you might get a system of linear equations that you might solve to find the Y' from the Y and their derivatives with the same L (since the ladder operator don't change L).

I have not seen/found a closed expression for the result.

 

[vanesch]

For what it is worth:

http://journals.iucr.org/a/issues/1994/05/00/bk0009/bk0009.pdf

cheers,
Patrick.

PS: didn't read it in detail!

 

[lalbatros]

I just realized that to calculated the Y' from the Y, the easiest way is to start from the definition.
If (a,b,c) are the Euler angle as defined in:

http://electron6.phys.utk.edu/qm2/modules/m4/wigner.htm​

then you simply have by definition:

Y' = Rz(a)Ry(b)Rz(c) Y​

Note that for example Rz(a) = exp(-i Lz a) and the problem reduces to some algebra.
Calculating the exponential of operators is not so difficult: you simply need to find the eigenvector and eigenvalues. Indeed, if

M = U D U-1 where D is diagonal​

then

exp(M) = U exp(D) U-1​

and exp(D) is simple to calculate.

In the case of the spherical harmonics considered here you simply need to express the rotation and angular momentum operators as matrices in vector space of YL functions.

The reference above gives the calculations for a spin 1/2 .
Unfortunately, this doesn't correpond to any spherical harmonics!
But fortunately, the algebra for L=1 is quite simple since simple 3x3 matrices are involved.
For higher angular momentum, you need to go to the higher dimensional representations of SO(3).

 

[QMrocks]

thanks all! It will take me some to digest these info. i hope i can arrive at a close form for my question. my only basis for believing so is that the spherical harmonics of order L is a complete basis for any polynomial function of order L, hence it should be able to describe the spherical harmonic in a rotated frame of order L...

 

[quantumdude]

For what it is worth:
http://journals.iucr.org/a/issues/1994/05/00/bk0009/bk0009.pdf
cheers,
Patrick.
PS: didn't read it in detail!

You need a password to view the article.

 

[QMrocks]

i uploaded the pdf for you (i reprint it 2 pg in 1 to meet the size limit here).

 

[reilly]

In general, there's no particular relation between primed and unprimed coordinates. A typical case might be the potential at R for a charge at R'. do the standard legendre polynomial expansion with argument of cos(gamma). The addition theorem allows easy integration over the charge coordinates, etc, etc. as in Jackson, or any other discussion of the multipole expansion.

Regards,
Reilly

 

[lalbatros]

reilly,

What do you mean by "no relation"?

Since any angular function can be expressed in YLm functions, it is clear that Y'Lm can also be expressed as a sum of YLm. In addition, since the rotation operations commute with the L operator, we know that

Y'Lm = sum(CLn,Y'Ln)​

(this is the relation)

which simply means that the sum will involve only same-L spherical harmonics. Only the coefficients must still be calculated. The method I described aboved leads to the result, it only needs patience!

By dropping one angular variable, we go back to the Fourier series. This points to a simple analogy. It is clear that a basic Fourier function in a translated coordinate system (x') can be expressed by a few basic Fourier functions in the original coordinates system (x): simply by using the Simpson rules. It is clear that only a few terms (two) are involved: the terms with the same frequencies. This occurs because the translation commutes with the mommentum operator (say d/dx). Here is also a relation between basic functions in original and primed coordinate systems.

 

[reilly]

I misspoke. Indeed, the primed and unprimed coordinates are related by a rotation. The gory details of the finite rotational transformations, can be found in, for example, Edmonds Angular Momentum in Quantum Mechanics. (The Addition Theorem is one of these details, as are the transformation properties of the Ylm funtions.)
Regards,
Reilly Atkinson