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Consider the following matrix:

Pmatrix = 
  1/Sqrt[2] {{π0/Sqrt[2] + η/Sqrt[3] + ηpr/Sqrt[
      6], πplus, 
     Kplus}, {πminus, -π0/2 + η/Sqrt[3] + ηpr/Sqrt[
      6], K0}, {Kminus, 
     K0bar, -η/Sqrt[3] + 2 ηpr/Sqrt[6]}};
mmatrix = DiagonalMatrix[{mu, md, ms}];
ΣmatrixExpanded2[fπ_] = 
  IdentityMatrix[3] + 2*I*Pmatrix/fπ + 
     1/2 ((2*I)/fπ)^2*Pmatrix . Pmatrix // Expand // Simplify;

And the following list of rules:

rule := {Conjugate[πplus] :> πminus, 
  Conjugate[πminus] -> πplus, Conjugate[Kminus] -> Kplus, 
  Conjugate[Kplus] -> Kminus, Conjugate[π0] -> π0, 
  Conjugate[η] -> η, Conjugate[ηpr] -> ηpr, 
  Conjugate[K0] -> K0bar, Conjugate[K0bar] -> K0, 
  Conjugate[fπ] -> fπ}

I would like to calculate the trace Trace[mmatrix.ConjugateTranspose[ΣmatrixExpanded2[fπ]]]. This is how I do it:

(ConjugateTranspose[ΣmatrixExpanded2[fπ]] // 
   Expand) /. Evaluate[rule]

However, some of the matrix elements are not expanded and have the (...)* form:

enter image description here

Could you please tell me how to compute this product properly and apply rule?

Edit

Okay, so it works with TagDelayed, but not with the rule.

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1 Answer 1

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This appears to be the same as your immediate previous question, but with more detail. A potential approach is the same, through TagSetDelayed:

πplus /: Conjugate[πplus] := πminus
πminus /: Conjugate[πminus] := πplus
Kminus /: Conjugate[Kminus] := Kplus
Kplus /: Conjugate[Kplus] := Kminus
π0 /: Conjugate[π0] := π0
η /: Conjugate[η] := η
ηpr /: Conjugate[ηpr] := ηpr
K0 /: Conjugate[K0] := K0bar
K0bar /: Conjugate[K0bar] := K0
fπ /: Conjugate[fπ] := fπ

Your trace then runs, but some things are not expanded, mostly because they are inside ConjugateTranspose. The result is pretty ugly, but there are no more Conjugates:

Trace[mmatrix . ConjugateTranspose[ΣmatrixExpanded2[fπ]]]
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  • 1
    $\begingroup$ Great answer, didn't know this was possible! $\endgroup$
    – Hans Olo
    Jul 27, 2023 at 15:21

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