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Naïvely this is what happens and it obviously is not helpful!

In[7]:= Conjugate[SphericalHarmonicY[1, 1, θ, ϕ]]
Out[7]= -(1/2) E^(-I Conjugate[ϕ]) Sqrt[3/(2 π)] Conjugate[Sin[θ]]

So, I tried stating initially that $\theta$ and $\phi$ are reals but still that doesn't seem to have helped any bit,

In[8]:= θ ∈ Reals; ϕ ∈ Reals;

In[9]:= SphericalHarmonicY[1, 1, θ, ϕ]
Out[9]= -(1/2) E^(I ϕ) Sqrt[3/(2 π)] Sin[θ]

In[10]:= Conjugate[SphericalHarmonicY[1, 1, θ, ϕ]]
Out[10]= -(1/2) E^(-I Conjugate[ϕ]) Sqrt[3/(2 π)] Conjugate[Sin[θ]]

Kindly tell me how to do this? (I want to calculate sums like $\sum\limits_{m=-\ell}^{\ell}\left| Y_{l,m} (\theta,\phi)\right|^2$.)

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    $\begingroup$ Try Simplify[Conjugate[SphericalHarmonicY[1, 1, θ, ϕ]], θ ∈ Reals && ϕ ∈ Reals] $\endgroup$
    – J. M.'s slightly less busy ♦
    Apr 22, 2013 at 9:57
  • $\begingroup$ Simply writing θ ∈ Reals; ϕ ∈ Reals; is not the way to indicate to Mathematica that those variables are real. Try looking up $Assumptions and Assumptions. $\endgroup$
    – Sjoerd C. de Vries
    Apr 22, 2013 at 15:46

4 Answers 4

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Almost always in such situations, ComplexExpand is your friend:

Conjugate[SphericalHarmonicY[1, 1, θ, ϕ]] // ComplexExpand
(* -(Sqrt[3/(2*Pi)]*Cos[ϕ]*Sin[θ])/2 + (I/2)*Sqrt[3/(2*Pi)]*Sin[θ]*Sin[ϕ] *)
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  • $\begingroup$ Thanks! But isn't there a way to just make Mathematica understand that theta and phi are reals? I mean, I want to do many calculations like this in one file and I would like to have it declared once and for all that \theta and \phi are reals. Isn't there a way for that? $\endgroup$
    – user6818
    Apr 21, 2013 at 23:09
  • $\begingroup$ This thing of "ComplexExpand" doesn't seem to work if I say have a coefficient like of $(1/a)Y_{l,m}(\theta,\phi)$. Then ComplexExpand doesn't understand that $a$ is real and is making an unnecessary mess of the expression. $\endgroup$
    – user6818
    Apr 22, 2013 at 0:07
  • $\begingroup$ ComplexExpand does an expansion making the assumption that all variables and symbolic constants involved are real. There are not really "assertions" like \[theta] \[Element] Reals that do anything. Sometimes setting values for the global $Assumptions will do what you want, but that's only for affecting subsequent uses of certain functions such as Simplify and Integrate. $\endgroup$
    – murray
    Apr 22, 2013 at 3:18
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    $\begingroup$ I disagree with this answer, though not to the extent of downvoting. ComplexExpand does more, than just assuming all symbols to be real. As you can see, It has transformed Exp[I \[Phi]] to Cos[\[Phi]] + I Sin[\[Phi]]. I view this as undesirable behavior and would often prefer to see it drop Conjugates, with minimal alteration of the form of the expression. $\endgroup$
    – LLlAMnYP
    Apr 15, 2015 at 13:24
  • $\begingroup$ I agree with LLIAMnYP. Here is a minimal method that avoid this issue: Refine[Conjugate[SphericalHarmonicY[1, 1, θ, ϕ]], _Symbol ∈ Reals]. For an exhaustive discussion, see here. $\endgroup$
    – Jess Riedel
    Sep 27, 2017 at 18:53
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This is the spherical harmonic:

SphericalHarmonicY[1, 1, \[Theta], \[Phi]]

It returns this:

-(1/2) E^(I \[Phi]) Sqrt[3/(2 \[Pi])] Sin[\[Theta]]

And this is its complex conjugate:

SphericalHarmonicY[1, 1, \[Theta], \[Phi]] /. I -> -I

returning this:

-(1/2) E^(-I \[Phi]) Sqrt[3/(2 \[Pi])] Sin[\[Theta]]

as it can be expected.

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Rule 

{Complex[re_, im_] :> Complex[re, -im]}

seems to convert complex expressions which contain symbols which are meant to be real.

Rule

{I -> -I}

does not, even on simple example:

2 I /.{I -> -I}

2 I

the reason being that symbol I is automatically translated by Mathematica to

Complex[0, 1]

and rule above is interpreted by Mathematica as 

Complex[0, 1] -> Complex[0, -1]

However, when I apply it to a simple expression (say, 2 + 3 I), I am working with a different expression (in this case Complex[2, 3]), so the rule is not applicable.

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    $\begingroup$ Your pattern is not exhaustive. For example ArcSin[2] is a complex number that won't get picked up by your pattern. $\endgroup$
    – Greg Hurst
    Dec 9, 2014 at 22:36
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The best way about this, using again the spherical harmonics, is this:

  1. Define a symbol for the complex conjugate, e.g. Ybar

  2. Simplify the expression for the spherical harmonic:

    Y[l_, m_, θ_, ϕ_] := SphericalHarmonicY[l, m, θ, ϕ]

  3. Define Ybar

    Ybar[l_, m_, θ_, ϕ_] := SphericalHarmonicY[l, m, θ, ϕ] /. I -> -I

And that's it

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