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I have a question: Is't possible Mathematica doesn't plot the function when the values of l1 and l2 are the same? What should I add to the cycle?

In[1]:= Psi1[l1_,m1_,l2_,m2_]:=Sqrt[(2*l1+1)/(4*Pi)]*LegendreP[l1,m1,Cos[theta1]]*Sqrt[(2*l2+1)/(4*Pi)]*LegendreP[l2,m2,Cos[theta2]]-Sqrt[(2*l1+1)/(4*Pi)]*LegendreP[l1,m1,Cos[theta2]]*Sqrt[(2*l2+1)/(4*Pi)]*LegendreP[l2,m2,Cos[theta1]]

In[2]:= Den1[l1_,m1_,l2_,m2_]:=1/2*Psi1[l1,m1,l2,m2]^2

In[3]:= Dens1[l1_,m1_,l2_,m2_]:=Plot3D[Integrate[Den1[l1,m1,l2,m2],{phi1,0,2*Pi},{phi2,0,2*Pi}]*Sin[theta1]*Sin[theta2],{theta1,0,Pi},{theta2,0,Pi}]

In[4]:= Do[Print[{l1,l2,Dens1[l1,m1,l2,m2]}],{l1,0,5},{l2,0,5},{m1,0,0},{m2,0,0}]
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Your function vanishes when l1=l2 and m1=m2 as

Psi1[l, m, l, m] $\mapsto$ 0.

If you want non-vanishing same-$l$ situations, you should distinguish $m1$ and $m2$.

Let's add all labeling to the plot just so it's clear:

Dens1[l1_, m1_, l2_, m2_] := 
 Plot3D[Integrate[
    Den1[l1, m1, l2, m2], {phi1, 0, 2*Pi}, {phi2, 0, 2*Pi}]*
   Sin[theta1]*Sin[theta2], {theta1, 0, Pi}, {theta2, 0, Pi}, 
  PlotLabel -> 
   Grid[{{Subscript[l, 1] -> l1, 
      Subscript[m, 1] -> m1}, {Subscript[l, 2] -> l2, 
      Subscript[m, 2] -> m2}}]]

Now physically $m$ can only range from -$l$ to $l$. If you only want to print when it isn't uniformly zero you can add an if statement:

Do[Do[If[l1 =!= l2 || m1 =!= m2, 
   Print[{l1, l2, Dens1[l1, m1, l2, m2], m1, m2}]], {m1, -l1, 
   l1}, {m2, -l2, l2}], {l1, 0, 5}, {l2, 0, 5}]

yielding e.g.

l1l1m-1m0 plot

The logic in the if statement is probably clear, but in english, only evaluate if the $l$s are different, or if the $l$s are the same as long as the $m$s are different.

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  • $\begingroup$ That's right. For this reason, I'm looking some Mathematica tool that excludes from the calculation all cases when l1=l2. $\endgroup$ – Arturo García Flores Aug 19 '17 at 3:32
  • $\begingroup$ Are you looking for the right place to put an If statement? i.e. Only do the evaluation if ls are different or if m's are different? I've updated my answer. $\endgroup$ – John Joseph M. Carrasco Aug 19 '17 at 6:04
  • $\begingroup$ Yes, man, that's what I was looking for. Since the function is antisymmetric, it makes no sense to plot the density when l1=l2, or to run another type of computation involving these cases. For a symmetric function, I'm interested in cases when l1 and l2 are the same (and different, of course) but there I have no problem. On the other hand, at this moment I'm interested in cases when m1=m2=0. Thank you very much! $\endgroup$ – Arturo García Flores Aug 19 '17 at 7:45

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