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A Spherical harmonics

  1. Oct 11, 2017 #1
    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. jcsd
  3. Oct 11, 2017 #2

    DrDu

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    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##.
     
  4. Oct 11, 2017 #3

    DrDu

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    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?
     
  5. Oct 11, 2017 #4
    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?
     
  6. Oct 11, 2017 #5

    DrDu

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    Yes. What is the value of ##P_l^m(1)##?
     
    Last edited: Oct 12, 2017
  7. Oct 13, 2017 #6
    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##
     
  8. Oct 13, 2017 #7

    DrDu

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    Thats not always true.
     
  9. Oct 13, 2017 #8
    when it is not?
     
  10. Oct 13, 2017 #9

    DrDu

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    Start to check for the lowest values of l an m.
     
  11. Oct 13, 2017 #10
    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
  12. Oct 13, 2017 #11

    DrDu

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    ... and 0 if m not equal 0.

    No, be carefull! If ##\cos \theta =1##, ##\cos \theta' <1## in general.
     
  13. Oct 16, 2017 #12
    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?
     
  14. Oct 16, 2017 #13

    DrDu

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    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##
     
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