Graduate Spherical Harmonics: A Primer on Barton's Relations & Addition Theorem

[amjad-sh]

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.

 

[DrDu]

$cos\chi=\vec r \cdot \vec r'={cosθcosθ'+sinθsinθ'cos(Φ-Φ')}$

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 $\phi$.

 

[DrDu]

∑m=lm=−lY∗lm(Ω′)Ylm(Ω)=(2l+1/4π)Pl(⃗r⋅⃗r′)∑m=−lm=lYlm∗(Ω′)Ylm(Ω)=(2l+1/4π)Pl(r→⋅r→′)\sum_{m=-l}^{m=l}Y_{lm}^*(\Omega')Y_{lm}(\Omega)=(2l+1/4π) P_l(\vec r \cdot \vec r')

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?

 

[amjad-sh]

Can you fill in the remaining steps?

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?

 

[DrDu]

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?

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

 

[amjad-sh]

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

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

 

[DrDu]

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

Thats not always true.

 

[amjad-sh]

Thats not always true.

when it is not?

 

[DrDu]

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

 

[amjad-sh]

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

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 $\Theta_{lm}(θ)=c_{lm}P_l^m(cosθ)$?

 

[DrDu]

I found that they are equal to 1 only when m=0 and by this $P_l^m(1)=1$

... and 0 if m not equal 0.

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 $2l+1/4π$.

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

 

[amjad-sh]

No, be carefull! If cosθ=1cos⁡θ=1\cos \theta =1, cosθ′<1cos⁡θ′<1\cos \theta'

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?

 

[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 $e'=(\sin(\arccos(r\cdot r')), 0, r\cdot r')^T$