Choose the spherical harmonic for which 11 and m normalized
Solution
Answer:
All we have to show is that the integral of the square value of the harmonic function for given m,l is equal to one.
In other words:
\\int_{0}^{2\\pi} \\int_{0}^{\\pi} (\\left |Y_{1}^{-1}\ ight |)^2 sin(\\theta) d(\\theta) d(\\phi) =1
But Y_{1}^{-1}(\\phi,\\theta) = \\frac{1}{2}\\sqrt{\\frac{3}{2\\pi}}e^{-i\\phi}sin(\\theta)
So we input the above integral into Wolfram Alpha or Mathematica and we get what we want, that it\'s equal to 1.
The code needed for Mathematica/WolframAlpha is the following:
Integrate[(Abs[SphericalHarmonicY[1,-1,theta,phi]])^2*Sin[theta],{phi,0,2*Pi},{theta,0,Pi}]
Notice that you\'ll have to get the absolute value of the harmonic and then square it to get the correct value of 1 for the integral.
So, if we insert the above code to WolframAlpha or Mathematica we get what we wanted, 1, and thus the spherical armonic for this l and m is normal.
