Re: is functionalD is automatically in D-dimension

Name: Rolf Mertig (email_not_shown)
Date: 10/24/06-02:41:39 AM Z


Well, without explicit SUN-indices yes. It does not really matter if
it is 4- or D-dimensional, right? Since never two lorentz indices are contracted and I think there is no Feynman rule where the dimension
shows up explictly (or? I am out of sync of course).

So, if you switch off the default loading of Phi ( I do this since
Frederik introduced some MakeBoxes rules which will not display
something like QuantumField[h, {li1, li2}] correctly ), and
forget about SUN-indices for the moment, then the following just works.

If you have specific questions about the whole process with
explicit SUN-indices, please ask Frederik (though he might just have
no time to answer).

Rolf

$LoadPhi = False;
<< HighEnergyPhysics`fc`;
QuantumField /:
  MakeBoxes[QuantumField[f_, lori : lo_[_, ___] .., suni : sun_[_] ...],
    TraditionalForm] :=
  SubsuperscriptBox[Tbox[f], Tbox[lori], Tbox[suni]] /;
    MatchQ[lo, LorentzIndex | Momentum] && sun === SUNIndex;
v3gt = QuantumField[h, {li1, li2}].QuantumField[PartialD[li1],
      h, {li3, li4}].QuantumField[PartialD[li2], h, {li3, li4}];
v3gtD = ChangeDimension[v3gt, D];
g3Field = {QuantumField[h, {\[Mu], \[Nu]}][k1],
    QuantumField[h, {\[Rho], \[Sigma]}][k2],
    QuantumField[h, {\[Lambda], \[Tau]}][k3]};
g3FieldD = ChangeDimension[g3Field, D];
t3gExpD = ExpandPartialD[v3gtD];
ft = FunctionalD[t3gExpD, g3FieldD]
ft2 = ft // Calc
ft2 // FCE // InputForm



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