# Spherical Harmonics Parity

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

• 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] – J. M.'s torpor Jan 6 at 18:51
• @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? – MarcoB Jan 6 at 20:07
• @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. – ThunderBiggi Jan 6 at 21:28
• @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. – J. M.'s torpor Jan 8 at 4:43
• @Marco, ah you are right! Things work out when I change one of the conditions to 0 < th <= Pi. – J. M.'s torpor Jan 8 at 4:48

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: • 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? – ThunderBiggi Jan 7 at 12:14
• What do you think about taking the absolute value of -Pi/2 and add Pi tp ϕ? – Daniel Huber Jan 7 at 12:48
• I guess that should work. Does it work for any values of $\theta$ and $\phi$? – ThunderBiggi Jan 7 at 13:14
• You would have to ask Wolfram for this. – Daniel Huber Jan 7 at 13:19
• @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. – J. M.'s torpor Jan 8 at 4:54