# A Spherical harmonics

1. Oct 11, 2017

Hello.
I was recently reading Barton's book.
I reached the part where he proved that in spherical polar coordinates
$δ(\vec r - \vec r')=1/r^2δ(r-r')δ(cosθ-cosθ')δ(φ-φ')$
$=1/r^2δ(r-r')δ(\Omega -\Omega')$
Then he said that the most fruitful presentation of $δ(\Omega-\Omega')$ stems from the closure property of the spherical harmonics $Y_{lm}(\Omega)$ which constitute a complete orthonormal set over the surface of the unit sphere.
Then he said that the spherical harmonics satisfies the remarkable addition theorem:
$\sum_{m=-l}^{m=l}Y_{lm}^*(\Omega')Y_{lm}(\Omega)=(2l+1/4π) P_l(\vec r \cdot \vec r')$

My problem is that I didn't get from where he obtained this relation.
Besides he said that the angle$\chi$ between $\vec r$ and $\vec r'$ is

$cos\chi=\vec r \cdot \vec r'={cosθcosθ'+sinθsinθ'cos(Φ-Φ')}$
Where the vector where $\vec r$ and $\vec r'$ are two vectors in spherical coordinates.

If somebody can help me obtaining this relation too.

Thanks.

2. Oct 11, 2017

### DrDu

At least this relation is easy to derive from the representation of the unit vector r in terms of the anlges $\theta$ and $\phi$,
$r=(\cos \phi \sin \theta, \sin \phi \sin \theta, \cos \theta)^T$ and the addition theorems for the cosine and sine of $\phi$.

3. Oct 11, 2017

### DrDu

I think this is easierst to prove noting that the lhs is invariant under rotations. Hence we may rotate $r \to (0,0,1)^T$ and $r' \to (\sin(\theta), 0, \cos(\theta))^T$, with $r\cdot r' =\cos(\theta)$. Can you fill in the remaining steps?

4. Oct 11, 2017

Not exactly.
I still didn't get the figure yet.
$Y_{lm}(θ,φ)=(-1)^m\sqrt{(2l+1/4π)(l-m)!/(l+m)!}P_l^m(cosθ)e^{(imφ)}$ is the joint eigen function of $\hat L^2 and \hat L_z$
should I use this to obtain it?

5. Oct 11, 2017

### DrDu

Yes. What is the value of $P_l^m(1)$?

Last edited: Oct 12, 2017
6. Oct 13, 2017

its value is zero since $P_l^m(x)=(1-x^2)^{|m|/2}\frac {d^{|m|}}{dx^{|m|}}P_l(x)$
where $p_l(x)=1/(2^l l! )\frac {d^l}{dx^l}(x^2-1)^l$

7. Oct 13, 2017

### DrDu

Thats not always true.

8. Oct 13, 2017

when it is not?

9. Oct 13, 2017

### DrDu

Start to check for the lowest values of l an m.

10. Oct 13, 2017

I found that they are equal to 1 only when m=0 and by this $P_l^m(1)=1$ If I substitute in $\sum_{m=-l}^{m=l}Y_l^m(\Omega')^*Y_l^m(\Omega)$ what I will get is $(2l+1)/4π P_l^2(1)$ which is equal to $2l+1/4π$.

you said that the l's are invariant under rotation.I think because $\hat L^2$ commutes with $\hat H$, $[\hat H,\hat L^2]=0$
you mean choosing any $\vec r$ ,$\vec r'$ suffices in proving the whole relation?
but if you chose other $\vec r$ and $\vec r'$ the relation will end up with $P_l^2(\vec r \cdot \vec r')$ and not $P_l(\vec r \cdot \vec r')$ isn't this true?
and why it is $P_l(\vec r \cdot \vec r')$ and not $P_l(cosθ)$ as the solution of the legendre differential equation is $\Theta_{lm}(θ)=c_{lm}P_l^m(cosθ)$?

Last edited: Oct 13, 2017
11. Oct 13, 2017

### DrDu

... and 0 if m not equal 0.

No, be carefull! If $\cos \theta =1$, $\cos \theta' <1$ in general.

12. Oct 16, 2017

OK. So the relation will end up like this:$(2l+1)/4πP_l^0(cosθ')$.
How can I reach $P_l(\vec r \cdot \vec r')$ now?

13. Oct 16, 2017

### DrDu

Note that the Ylm span an irreducible and unitary representation of the rotation group, i.e. $Y_l^m(R^{-1} r)=\sum_m U_{m m'} Y_l^m'(r)$, where R is a 3x3 rotation matrix and U a (2l+1)x(2l+1) unitary matrix. You can use this to show that $\sum_{m=-l}^{l} Y^{m*}_l(r') Y^m_l(r)=\sum_{m=-l}^{l} Y^{m*}_l(e') Y^m_l(e_z)$, where $e_z= (0,0,1)^T$ and $e'=(\sin(\arccos(r\cdot r')), 0, r\cdot r')^T$