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The spherical harmonic function $Y_l^m(\theta,\phi)$ is defined to be an eigenfunction of the angular part of the Laplace operator with eigenvalue $-l(l+1)$. In other words, it solves the PDE: $$\Delta_{\theta \phi}f := \frac{\partial^2 f}{\partial \theta^2} + \frac{1}{\tan \theta} \frac{\partial f}{\partial \theta} + \frac{1}{\sin^2 \theta} \frac{\partial f}{\partial \phi} = -l(l+1) f.$$ I tried to check this using the Mathematica implementation of the spherical harmonics SphericalHarmonicY. If I plug this into the left-hand side of the above equation,

D[SphericalHarmonicY[l, m, \[Theta], \[Phi]] , {\[Theta], 2}] + 
 1/Tan[\[Theta]] D[SphericalHarmonicY[l, m, \[Theta], \[Phi]] , \[Theta]]  + 1/Sin[\[Theta]]^2 D[ 
   SphericalHarmonicY[l, m, \[Theta], \[Phi]] , {\[Phi], 2}]

I get: $$-m(m+1)Y_l^m + \cdots.$$ In the dots, there are complicated expressions involving gamma functions and spherical harmonics. Now, it turns out that this actually does equal $-l(l+1) Y_l^m$ simply by trying out different values of $l$ and $m$, keeping $\theta$ and $\phi$ arbitrary. However, this is not at first obvious and I was wondering if there was a way to make Mathematica recognize this simplification?

I tried do use Refine and Element to tell Mathematica that $l$ and $m$ and integers, but this did not help.

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  • 1
    $\begingroup$ Please provide Mathematica code that you've used. $\endgroup$
    – Domen
    Commented Oct 1, 2021 at 15:47

2 Answers 2

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Too long for a comment. Indeed,

FullSimplify[D[SphericalHarmonicY[l, m, \[Theta], \[Phi]], {\[Theta], 2}] + 
1/Tan[\[Theta]] D[SphericalHarmonicY[l, m, \[Theta], \[Phi]], \[Theta]] + 
1/Sin[\[Theta]]^2 D[SphericalHarmonicY[l, m, \[Theta], \[Phi]], {\[Phi], 2}], 
Assumptions -> {l, m} \[Element] Integers]

-m (1 + m) SphericalHarmonicY[l, m, \[Theta], \[Phi]] + (E^(-2 I \[Phi]) Sqrt[ Gamma[1 + l - m]] ((1/Sqrt[Gamma[l - m]]) 2 E^(I \[Phi]) (1 + m) Cot[\[Theta]] Sqrt[Gamma[2 + l + m]] SphericalHarmonicY[l, 1 + m, \[Theta], \[Phi]] + ( Sqrt[Gamma[3 + l + m]] SphericalHarmonicY[l, 2 + m, \[Theta], \[Phi]])/Sqrt[ Gamma[-1 + l - m]]))/(Sqrt[Gamma[1 + l + m]])

, but

N[(E^(-2 I \[Phi]) Sqrt[
  Gamma[1 + l - 
    m]] ((1/Sqrt[Gamma[l - m]]) 2 E^(I \[Phi]) (1 + 
      m) Cot[\[Theta]] Sqrt[Gamma[2 + l + m]] SphericalHarmonicY[
     l, 1 + m, \[Theta], \[Phi]] + (Sqrt[
       Gamma[3 + l + m]] SphericalHarmonicY[l, 
       2 + m, \[Theta], \[Phi]])/Sqrt[Gamma[-1 + l - m]]))/(Sqrt[
 Gamma[1 + l + m]]) /. {l -> 3, m -> 2, \[Phi] -> Pi/4, \[Theta] -> Pi/3}]

0. - 2.29947 I

Maybe, a bug in FullSimplify.

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  • $\begingroup$ In version 13 the above results in 0. - 4.59893 I. $\endgroup$
    – user64494
    Commented Mar 9, 2022 at 15:10
  • $\begingroup$ In 13.3.0 the result is once again - 2.29947 I. $\endgroup$ Commented Jun 30, 2023 at 2:52
  • $\begingroup$ WRI seems to use $_2F_1$ plus GPT if ever linear second order ODE methods apply. No such formula is working as expected. Use the original operator #### $(1/ \sin \theta \partial_\theta (\sin \theta \partial_\thetal #) + 1/(\sin \theta)^2 \partial_{\phi \phi} # )/ #\&)[ e^{i m \phi} \ LegendreP[l,m , \cos \theta)]+ l(l+1) $ #### $\endgroup$
    – Roland F
    Commented Jun 30, 2023 at 5:14
  • $\begingroup$ @RolandF Your TEX failed. $\endgroup$ Commented Jul 1, 2023 at 7:08
  • $\begingroup$ Assuming[ 0 < [Theta] < [Pi] && 0 < [Phi] < 2 [Pi] && {l, m} [Element] Integers && l > 0, (1/Sin[[Theta]] D[ Sin[[Theta]] D[#, [Theta]], [Theta]] + 1/Sin[[Theta]]^2 D[#, [Phi], [Phi]] + l (l + 1) # &)[ Exp[I m [Phi]] LegendreP[l, m, Cos[[Theta]]]] // FullSimplify // FunctionExpand] // Simplify $\endgroup$
    – Roland F
    Commented Jul 1, 2023 at 8:36
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To simplify the resulting expression, for general arguments, essentially one needs to apply the following identity:

https://functions.wolfram.com/HypergeometricFunctions/LegendreP2General/17/01/01/0003/

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  • $\begingroup$ Could you elaborate your direction, giving us details and a Mathematica code? TIA. $\endgroup$
    – user64494
    Commented Oct 10, 2021 at 14:14

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