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In Mathematica's documentation, the Spherical Harmonics are said to be defined as follows, for $l \geq 0$:

enter image description here

Furthermore, we know that $\cos(x)=\cos(-x)$, hence one can be led to believe that $Y_l^m(-\theta,\phi)=Y_l^m(\theta,\phi)$.

A quick check with mathematica shows us that might not be the case

Table[Table[{SphericalHarmonicY[l, m, -(\[Pi]/2), 0], 
   SphericalHarmonicY[l, m, \[Pi]/2, 0]}, {m, -l, l}], {l, 0, 2}]

enter image description here

as the $l=1=m$ values differ.

Am I doing something wrong? Does the value of $\theta$ need to be in the canonical range $[0,\pi]$. If so, how do I relate $-\frac{\pi}{2}$ to something in that range. The formulae I know allow me to relate it to $\frac{3\pi}{2}$, which is still outside the range.

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  • $\begingroup$ Indeed, something weird is going on: Simplify[Table[SphericalHarmonicY[l, m, th, ph] == Sqrt[(2 l + 1)/(4 Pi)] (Sqrt[Gamma[l - m + 1]]/Sqrt[Gamma[l + m + 1]]) Exp[I ph m] LegendreP[l, m, 2, Cos[th]], {l, 0, 2}, {m, -l, l}], 0 <= th <= Pi && 0 <= ph <= 2 Pi] $\endgroup$ – J. M.'s torpor♦ Jan 6 at 18:51
  • $\begingroup$ @J.M. That remaining equation is an identity under the conditions you specified though, isn't it? At least, the lhs does evaluate to $0$ for all positive values of th. That's more a failure of plain Simplify than of the definitions. Indeed, if you change to FullSimplify, and add ComplexExpand as a TransformationFunction (which should be ok, since variables are real), then all results are True. Or am I missing something else? $\endgroup$ – MarcoB Jan 6 at 20:07
  • $\begingroup$ @J.M. What is this fourth argument in LegendreP ? The documentation allows only 3. And why the extra phase factor? Are the Gamma functions doing something under the hood, as for integer arguments, they are equvalent to factorials. $\endgroup$ – ThunderBiggi Jan 6 at 21:28
  • $\begingroup$ @Thunder, the extra argument in LegendreP[] modifies the branch cut convention of the Legendre function used. Please have a look at the docs for more details, but in brief, the discrepancy apparently happens since we are using "type 2" Legendre functions. $\endgroup$ – J. M.'s torpor♦ Jan 8 at 4:43
  • $\begingroup$ @Marco, ah you are right! Things work out when I change one of the conditions to 0 < th <= Pi. $\endgroup$ – J. M.'s torpor♦ Jan 8 at 4:48
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The definition with Cos[phi]is a bit misleading. Consider e.g.

SphericalHarmonicY [1,1,phi,theta] == ... LegendreP[1,1,Cos[phi]] ..

Now the associated Legendre Polynomial LegendreP[1,1,x]is defined by:

LegendreP[1, 1, x] == -Sqrt[1 - x^2]

and

LegendreP[1,1,Cos[phi]] ==  -Sqrt[1 - Cos[phi]^2] ==  -Sqrt[Sin[phi]^2] == -Sin[phi]

Therefore, we get for the full blown function:

SphericalHarmonicY [1,1,[phi],theta] == [![enter image description here][1]][1]

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  • $\begingroup$ Ok, you are making a choice of sign for the square root, which is ok if you are on $0\leq \theta\leq \pi$. And I guess the definition in terms of Associated Legendre Polynomials is probably only defined on the standard range of coordinates on $S^2$. How do I then transform my $-\frac{\pi}{2}$ Spherical harmonic to one in that range? $\endgroup$ – ThunderBiggi Jan 7 at 12:14
  • $\begingroup$ What do you think about taking the absolute value of -Pi/2 and add Pi tp ϕ? $\endgroup$ – Daniel Huber Jan 7 at 12:48
  • $\begingroup$ I guess that should work. Does it work for any values of $\theta$ and $\phi$? $\endgroup$ – ThunderBiggi Jan 7 at 13:14
  • $\begingroup$ You would have to ask Wolfram for this. $\endgroup$ – Daniel Huber Jan 7 at 13:19
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    $\begingroup$ @Thunder, as you might have already noticed, if you use the explicit definition of the spherical harmonics that uses "type-2" Legendre functions, for l = m = 1, there is a dangling factor of $\frac{\cos \theta-1}{\sqrt{1-\cos\theta}}$. The result you were expecting would be valid if the "type 1" Legendre function, LegendreP[l, m, 1, Cos[th]], had been used instead. $\endgroup$ – J. M.'s torpor♦ Jan 8 at 4:54

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