Normalization ((Subscript[N, ab])) and θ dependence ((Subscript[F, ab]))
for the qq ->  γγ  and gg -> γγ processes

Packages

FeynArts

AppendTo[$Path,  "/home2/balazs/HighEnergyPhysics"];
AppendTo[$Path,  "/home2/balazs/HighEnergyPhysics/FeynArts12"];
      
<< FeynArts12`FeynArts`

FeynArts`Models  =  
        { SM,  QED,  QCD,  QE,  QC,  WEYL,  YOT,  BGF,  SMbgf, SuSy};
      
$ScreenSymbolFont  =  {"Symbol", 10};
$SmallScreenSymbolFont  =  {"Symbol", 8};

AppendTo[$Path,  "/home2/balazs/Library/Mathematica/Packages"];

<< CsB`CreateAmplitude1`

FeynCalc

AppendTo[$Path,  "/home2/balazs/HighEnergyPhysics"];
<< fc.m

Please click on the Cell menu, then go to the Default Output Format Type item and switch to TraditionalForm.

AppendTo[$Path,  "/home2/balazs/Library/Mathematica/Packages"];

<< CsB`GetAmplitude`
<< CsB`SumPolarization`
<< CsB`FC`AmplitudeSquare`

\(TraditionalForm\`q  q\&_\) -> \(TraditionalForm\`\(W\^+\)\) -> \(e\^+\)(Subscript[ν, e])

The three amplitude of \(TraditionalForm\`u  d\&_\) -> \(TraditionalForm\`\(W\^+\)\) -> \(e\^+\)(Subscript[ν, e]) is:

CreateAmplitude1[{SM, QCD}, {u, D}, {E, ne}, 0]

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(d\&_\)(p\_2)\)\) -> \(TraditionalForm\`\(W\^+\)\) -> \(e\^+\)((Subscript[k, 1]))(Subscript[ν, e])((Subscript[k, 2])).
We use 't Hooft-Feynman gauge.

(Subscript[M, \(u  d\&_\)->\(e\^+\)(Subscript[ν, e])])=("uD->Ene" // GetAmplitude) /. (Subscript[m, x])_ :> 0 /; x =!= W

The Mandelstam variables are defined such that the fermion comes from the proton:

SetMandelstam[s, t, u, (Subscript[p, 1]), (Subscript[p, 2]), -(Subscript[k, 1]), -(Subscript[k, 2]), 0, 0, 0, 0];

The amplitude squared is:

(Subscript[(M^2), \(u  d\&_\)->\(e\^+\)(Subscript[ν, e])])=((Subscript[M, \(u  d\&_\)->\(e\^+\)(Subscript[ν, e])])//AmplitudeSquare)[[1, 1]]

In the W CM frame:

\(\(Momenta = 
        {
p\_1 -> \(\@s\/2\) {I, 0, 0, 1}, 
          p\_2 -> \(\@s\/2\) {I, 0, 0, \(-1\)}, 

          k\_1 -> \(\@s\/2\) {I, 0, Sin[θ], Cos[θ]}, 

          k\_2 -> \(\@s\/2\) {I, 0, \(-Sin[θ]\), \(-Cos[θ]\)}};
      \)\)
\(MandelstamVariables = 
      {
s -> \(-\((p\_1 + p\_2)\) . \((p\_1 + p\_2)\)\) /. Momenta; 

        t -> \((\(\(\(-\((p\_1 - k\_1)\) . \((p\_1 - k\_1)\)\) /. Momenta\) /. 
                  Sin[θ]\^2 -> 1 - Cos[θ]\^2 // Expand\) // 
              Factor)\), 

        u -> \((\(\(\(-\((p\_1 - k\_2)\) . \((p\_1 - k\_2)\)\) /. Momenta\) /. 
                  Sin[θ]\^2 -> 1 - Cos[θ]\^2 // Expand\) // 
              Factor)\)}\)

(Subscript[(M^2), \(u  d\&_\)->\(e\^+\)(Subscript[ν, e])])/.MandelstamVariables

\(Amp2[uD -> \(e\^+\) ν\_e]\)=(((e^4) (s^2) (((1 - cos(θ)))^2) (Subscript[V, ud]^2))/(4 (((s - (Subscript[m, W]^2)))^2) (Subscript[s, w]^4)));

This agrees with my hand calculations (notes: Total Cross Section) and C.-P.'s notes (Zeroth Order) and with CompHEP.

The averaging factors: initial state color ((1/3)) and spin ((1/2) (1/2)).

CompHEP result

<< "~/CompHEP/comphep_32.18/results/uD-e1n1.m"

Taking away the initial state color ((1/3)) and spin ((1/2) (1/2)) averaging factors and changing notation:

(1/((1/3) (1/2) (1/2))) 
      (Summ /. {EE  ->  e, MZ -> ((Subscript[m, W])/(Sqrt[1 - ((Subscript[s, w])^2)])), SW -> (Subscript[s, w]), 
              SC[p1, p3] -> ((-t)/2), SC[p1, p2] -> (s/2), 
              (s12^2) -> (Subscript[V, ud]) + 1, (s13^2) -> (Subscript[V, ud]) + 1} // Simplify)

% /. t -> ((1 - Cos[θ])/2)

\(TraditionalForm\`q  q\&_\) -> \(TraditionalForm\`\(W\^-\)\) -> \(e\^-\)\(ν\&_\_e\)

The three amplitude of \(TraditionalForm\`u  d\&_\) -> \(TraditionalForm\`\(W\^-\)\) -> \(e\^-\)\(ν\&_\_e\) is:

\(CreateAmplitude1[{SM, QCD}, {u\&_, d} -> {\(e\^-\), ν\&_\_e}, 0]
      \)

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(d\&_\)(p\_2)\)\) -> \(TraditionalForm\`\(W\^+\)\) -> \(e\^+\)((Subscript[k, 1]))(Subscript[ν, e])((Subscript[k, 2])).
We use 't Hooft-Feynman gauge.

(Subscript[M, \(\(u\&_\) d\)->\(e\^-\)\(ν\&_\_e\)])=("Ud->eNe" // GetAmplitude) /. (Subscript[m, x])_ :> 0 /; x =!= W

The Mandelstam variables are defined such that the fermion comes from the proton:

SetMandelstam[s, t, u, (Subscript[p, 2]), (Subscript[p, 1]), -(Subscript[k, 1]), -(Subscript[k, 2]), 0, 0, 0, 0];

The amplitude squared is:

(Subscript[(M^2), \(\(u\&_\) d\)->\(e\^-\)\(ν\&_\_e\)])=((Subscript[M, \(\(u\&_\) d\)->\(e\^-\)\(ν\&_\_e\)])//AmplitudeSquare)[[1, 1]]

In the W CM frame:

\(\(Momenta = 
        {
p\_1 -> \(\@s\/2\) {I, 0, 0, 1}, 
          p\_2 -> \(\@s\/2\) {I, 0, 0, \(-1\)}, 

          k\_1 -> \(\@s\/2\) {I, 0, Sin[θ], Cos[θ]}, 

          k\_2 -> \(\@s\/2\) {I, 0, \(-Sin[θ]\), \(-Cos[θ]\)}};
      \)\)
\(MandelstamVariables = 
      {
s -> \(-\((p\_1 + p\_2)\) . \((p\_1 + p\_2)\)\) /. Momenta; 

        t -> \((\(\(\(-\((p\_1 - k\_1)\) . \((p\_1 - k\_1)\)\) /. Momenta\) /. 
                  Sin[θ]\^2 -> 1 - Cos[θ]\^2 // Expand\) // 
              Factor)\), 

        u -> \((\(\(\(-\((p\_1 - k\_2)\) . \((p\_1 - k\_2)\)\) /. Momenta\) /. 
                  Sin[θ]\^2 -> 1 - Cos[θ]\^2 // Expand\) // 
              Factor)\)}\)

(Subscript[(M^2), \(u  d\&_\)->\(e\^+\)(Subscript[ν, e])])/.MandelstamVariables

The averaging factors: initial state color ((1/3)) and spin ((1/2) (1/2)).

CompHEP result

<< "~/CompHEP/comphep_32.18/results/uD-e1n1.m"

Taking away the initial state color ((1/3)) and spin ((1/2) (1/2)) averaging factors and changing notation:

(1/((1/3) (1/2) (1/2))) 
      (Summ /. {EE  ->  e, MZ -> ((Subscript[m, W])/(Sqrt[1 - ((Subscript[s, w])^2)])), SW -> (Subscript[s, w]), 
              SC[p1, p3] -> ((-t)/2), SC[p1, p2] -> (s/2), 
              (s12^2) -> (Subscript[V, ud]) + 1, (s13^2) -> (Subscript[V, ud]) + 1} // Simplify)

% /. t -> ((1 - Cos[θ])/2)

\(TraditionalForm\`q  q\&_\) -> \(TraditionalForm\`\(γ\^*\)\) -> \(e\^+\)\(e\^-\)

The three amplitude of \(TraditionalForm\`u  u\&_\)  -> \(TraditionalForm\`\(γ\^*\)\) -> \(e\^+\)\(e\^-\) is (the  \(TraditionalForm\`u  u\&_\)  -> Z -> \(e\^+\)\(e\^-\) process was deleted by hand):

CreateAmplitude1[{SM, QCD}, {u, U}, {e, E}, 0]

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(d\&_\)(p\_2)\)\) -> \(TraditionalForm\`\(W\^+\)\) -> \(e\^+\)((Subscript[k, 1]))(Subscript[ν, e])((Subscript[k, 2])).
We use 't Hooft-Feynman gauge. We made tha quark charge explicit.

FAL = "uU->eE" // GetAmplitude;
Amp = 
      (((Subscript[Q, f])/((2/3))) FAL[[1, 2]] // Contract) /. 
          Conjugate[QMM11] -> (Subscript[V, ud]) /. {MU  ->  0, ME -> 0, MLA -> 0}

The Mandelstam variables are defined in the standard way:

SetMandelstam[s, t, u, p1, p2, -k1, -k2, 0, 0, 0, 0];

The amplitude squared is:

\(Amp2[0]\  = 
      \(\(\(\((\(ComplexConjugate[Amp] Amp // 
	FermionSpinSum\) // 
	
                    DiracSimplify)\) /. 
	DiracTrace -> Tr // 
	
              Simplify\) // 
	Contract\) // PropagatorDenominatorExplicit\) 
        // Simplify\)

In the W CM frame:

\(p\_1 = \(\@s\/2\) {I, 0, 0, 1}; p\_2 = \(\@s\/2\) {I, 0, 0, \(-1\)};

    k\_1 = \(\@s\/2\) {I, 0, Sin[θ], Cos[θ]}; 
    k\_2 = \(\@s\/2\) {I, 0, \(-Sin[θ]\), \(-Cos[θ]\)};

    s -> \(-\((p\_1 + p\_2)\) . \((p\_1 + p\_2)\)\)\)
t -> -((Subscript[p, 1]) - (Subscript[k, 1])) . ((Subscript[p, 1]) - (Subscript[k, 1])) // Simplify
u -> -((Subscript[p, 1]) - (Subscript[k, 2])) . ((Subscript[p, 1]) - (Subscript[k, 2])) // Simplify

Mandelstam = 
        {t -> s (( 1 - Cos[θ])/2), u -> s (( 1 + Cos[θ])/2)};

Amp2[0] /. Mandelstam // Expand // Factor

\(Amp2[qQ -> \(e\^+\) \(e\^-\)]\)=4 (e^4) ((cos^2)(θ) + 1) (Subscript[Q, f]^2);

This agrees with CompHEP.

The averaging factors: initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)).
Summation over colors has not been done yet.

CompHEP result

<< "~/CompHEP/comphep_32.18/results/uU-eE.m"

Taking away the initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) averaging factors, making the quark charge and the color sum explicit  and changing notation:

(1/(((((1/3)))^2) ((((1/2)))^2))) (((Subscript[Q, f])^2)/((((2/3)))^2)) 
        ((Subscript[N, C])/3) 
        (Summ /. {EE  ->  e, SC[p1, p3] -> ((-t)/2), SC[p1, p2] -> (s/2)} 
            // Simplify) // TrickMandelstam[#, {s, t, u, 0}] &

\(TraditionalForm\`q  q\&_\) -> γ γ

FeynCalc

The three amplitude of \(TraditionalForm\`u  u\&_\) -> γ γ is:

CreateAmplitude1[{SM, QCD}, {u, U}, {A, A}, 0]

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(u\&_\)(p\_2)\)\) -> γ((Subscript[k, 1]))γ((Subscript[k, 2])).
We use 't Hooft-Feynman gauge. The u quark charge is included by FA and taken made explicit.

\(\(FAL = "uU->AA" // GetAmplitude;\)

    Amp[1] = 
      \(Q\_f\^2\/\((2\/3)\)\^2\) FAL[\([1, 2]\)] /. 
        {MU\  -> \ 0, MD -> 0, ME -> 0}\)
Amp[2] = 
      (((Subscript[Q, f])^2)/((((2/3)))^2)) FAL[[2, 2]] /. 
        {MU  ->  0, MD -> 0, ME -> 0}

\(\(\(

    \)\(SetMandelstam[s, t, u, p1, p2, \(-k1\), \(-k2\), 0, 0, 0, 0];\)\)\)

\(\(Do[
        Do[
	Amp2[i, j] = 
            \(\((\(\(\(\(\(ComplexConjugate[Amp[i]] Amp[j] // Simplify\) 
                              // 
	SumPolarization[#, k1] &\) // 
                            SumPolarization[#, k2] &\) // 
	Contract\) 
                        // 
	FermionSpinSum\) // 
	DiracSimplify)\) 
                  /. 
\ DiracTrace -> Tr // Simplify\) // 
	
              TrickMandelstam[#, {s, t, u, 0}] &
, {i, 2}], {j, 2}];\)\)
\(Amp2[3] = 
      \(∑\+\(i = 1\)\%2\(∑\+\(j = 1\)\%2 Amp2[i, j]\) 
          // 
		PropagatorDenominatorExplicit\) // Simplify\)

This must still be summed over (Subscript[N, C]).

In the γγ CM frame:

\(p\_1 = \(\@s\/2\) {I, 0, 0, 1}; p\_2 = \(\@s\/2\) {I, 0, 0, \(-1\)};

    k\_1 = \(\@s\/2\) {I, 0, Sin[θ], Cos[θ]}; 
    k\_2 = \(\@s\/2\) {I, 0, \(-Sin[θ]\), \(-Cos[θ]\)};

    s -> \ \(-\((p\_1 + p\_2)\) . \((p\_1 + p\_2)\)\)\)
t -> -((Subscript[p, 1]) - (Subscript[k, 1])) . ((Subscript[p, 1]) - (Subscript[k, 1])) // Simplify
u -> -((Subscript[p, 1]) - (Subscript[k, 2])) . ((Subscript[p, 1]) - (Subscript[k, 2])) // Simplify

Mandelstam = 
        {t -> s (( 1 - Cos[θ])/2), u -> s (( 1 + Cos[θ])/2)};

Amp2[3] /. Mandelstam // Simplify[#, Trig -> False] &

Amp2[qQ -> γγ]=-((16 (e^4) ((cos^2)(θ) + 1) 
                  (Subscript[Q, f]^4))/((cos^2)(θ) - 1));

This agrees with both Ohnemus and Owens and CompHEP.

The averaging factors: initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) and the identical final state particles factor ((1/2)).

CompHEP

<< "~/CompHEP/comphep_32.18/results/sum1.m"

To obtain the uanveraged amplitude square we take away the initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) averaging factors and the identical final state particles factor ((1/2)), make explicit the quark charges ((Subscript[Q, f]) = (2/3)) and the sum over colors ((Subscript[N, C]) = 3), and change notation:

(1/(((((1/3)))^2) ((((1/2)))^2) (1/2))) 
            (((Subscript[Q, f])^4)/((((2/3)))^4)) ((Subscript[N, C])/3) Summ /. 
          {EE  ->  e, SW -> (Subscript[s, w]), SC[p1, p3] -> ((-t)/2), 
            SC[p1, p2] -> (s/2)} // Factor // 
      TrickMandelstam[#, {s, t, u, 0}] &

This agrees with both Ohnemus and Owens and FC.

Papageno

In Papageno the following is coded:

      CONST=AEM**2*16.*PISQ
      CONST1=QU4
      CONST2=QD4
      ....
      A4=CONST*2.*(U2+T2)/U/T/3.
      A5=CONST*AL**2*11.**2/9**2.*2./4./8./8.*XMAT_PHOT(S,T,U)
      WT=((SF(23)+SF(24))*CONST1+(SF(27)+SF(28))*CONST2)*A4/2.
     1 +SF(9)*A5/2./16./PISQ

That is

\(\(\(\ 
    \)\(WT\_\(q  q\&_ -> γγ\) = 
        Q\_q\^4\ α\_EM\^2\ 16\ π\^2\ 2 
            \(\( u\^2 + t\^2\)\/\(u\ t\)\) \(1\/3\) 1\/2 /. 
          α\_EM -> e\^2\/\(4  π\);\)\)\)

After taking out the initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) averaging factors and the identical final state particles factor ((1/2)), and making the color sum explicit, for the amplitude square (before averaging) we obtain:

\(\(\(\(1\/D\_F\^2\) \(1\/\((N - 2)\)\^2\) 
            1\/N\_identical\)\/\(\(\((1\/3)\)\^2\) \(\((1\/2)\)\^2\) 1\/2\)\) 
      \(Abs[M\_\(q  q\&_ -> γγ\)[s, t, u]]\^2\/\(\(8\ e\^4\ \((t\^2 + u\^2)\)\ Q\_q\%4\)\/\(t\ u\)\)\) \(N\_C\/3\) 
      WT\_\(q  q\&_ -> γγ\)\)

which agrees with both FC and CompHEP.

Berger et.al. NPB239 (1984) 52

On page 56 they have

\(\(dσ\_\(q  q\&_ -> γγ\) = 
        π\ α\_EM\^2\ \(1\/3\) 
            2\/s\^2\ \(\(u\^2 + t\^2\)\/\(u\ t\)\) dt /. 
          α\_EM -> e\^2\/\(4  π\);\)\)

After taking out the initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) averaging factors and the identical final state particles factor ((1/2)), and making the color sum explicit, for the amplitude square (before averaging) we obtain:

\(\(1\/\(\(\((1\/3)\)\^2\) \(\((1\/2)\)\^2\) 1\/2\)\) \(N\_C\/3\) 
      dσ\_\(q  q\&_ -> γγ\)\/dt\)

There's a ((π (s^2))/8)factor which I don't understand.

g g -> γ γ

FeynCalc

The box contribution to the O((α^2) ((Subscript[α, S])^2)) amplitude of g g -> γ γ is:

CreateAmplitude1[{SM, QCD}, {G, G}, {A, A}, 1,  Boxes  ->  True]

The triangular contribution to the O((α^2) ((Subscript[α, S])^2)) amplitude of g g -> γ γ is:

CreateAmplitude1[{SM, QCD}, {G, G}, {A, A}, 1,  Triangles  ->  True]

For all of the above diagrams:
- each diagram comes with a factor of 2, since the quark can run either in one or in the other direction in the loop,
- all six flavors can run around the loop.

Papageno

We want to find out the normalization of the g g -> γ γ process with respect to the q \(TraditionalForm\`q\&_\) -> γ γ process from Papageno.
In Papageno the following is coded:

      CONST=AEM**2*16.*PISQ
      CONST1=QU4
      CONST2=QD4
      ....
      A4=CONST*2.*(U2+T2)/U/T/3.
      A5=CONST*AL**2*11.**2/9**2.*2./4./8./8.*XMAT_PHOT(S,T,U)
      WT=((SF(23)+SF(24))*CONST1+(SF(27)+SF(28))*CONST2)*A4/2.
     1 +SF(9)*A5/2./16./PISQ
     
That is

(Subscript[WT, gg -> γγ]) = 
        ((Subscript[α, EM])^2) 16  (π^2) ((Subscript[α, S])^2) 
          ((((11/9)))^2) 2  (1/4) (1/8) (1/8) 
          Abs[(Subscript[M, gg -> γγ])[s, t, u]]^2 (1/2) 
          (1/16) (1/(π^2));

After taking out the initial state color ((1/8) (1/8)) and spin ((1/2) (1/2)) averaging factors and the identical final state particles factor ((1/2)), the factor ((((Subscript[(∑_(i = 1)^n), f]) ((Subscript[Q, q])^2)))^2) (without the top quark)  and making the factor from the color sum/traces ((Subscript[F, color])= 8 (1/2) (1/2)) explicit we obtain:

(((1/((Subscript[D, A])^2)) (1/(((N - 2))^2)) 
            (1/(Subscript[N, identical])))/(((((1/8)))^2) ((((1/2)))^2) 
            (1/2)))((((((Subscript[(∑_(i = 1)^n), f]) ((Subscript[Q, q[i]])^2)))^2))/(((((11/9)))^2)))(((Subscript[F, color]))/(8 (1/2) (1/2))) (Subscript[WT, gg -> γγ])

The factor (Subscript[(∑_(i = 1)^n), f]) ((Subscript[Q, q])^2) (without and with the top quark) is

\({\ 2 \((2\/3)\)\^2\  + \ 3 \((\(-\(1\/3\)\))\)\^2
, 
      3 \((2\/3)\)\^2\  + \ 3 \((\(-\(1\/3\)\))\)\^2}\)

ResBos

In RES.FOR:
- For H production we have: 	\(TraditionalForm
      \`HCOUPL*\((\(ALPI(C2*S_Q)\)*Q_V)\) ** 2 = 
        \(\(\(\@2\) \(G\_μ\) π\)\/576\) 
          \((\(\(α\_S\)(\(C\_2\) Q)\)\/π)\)\^2\)

(Subscript[((σ^H)), 0]) = 
        (((Sqrt[2]) (Subscript[G, μ]) π)/576) 
          ((((Subscript[α, S])[(Subscript[C, 2]) Q]/π))^2);

- For γγ production we have:	\(TraditionalForm
      \`2*\(PI/8.0\)/8.0*PI ** 2*\(ALPI(C2*S_Q)\) ** 2 = 
        \(\(2  π\)\/64\) \(π\^2\) 
          \(\(α\_S\^2\)(\(C\_2\) Q)\)\)

(Subscript[((σ^γ)γ), 0]) = 
        ((2  π)/64) (π^2) ((Subscript[α, S])^2)[(Subscript[C, 2]) Q];

In ResBos we have:
C Identical particle final state and spin average factors
     &      1.d0/2.d0/2.d0/2.d0*
C Effective matrix element
     &      2.d0*(gWeak*sWeak)**4/(16.0*Pi**2)*
CsB t and b quark masses are non-zero in loop:
     &      GGAA(sH,tH,uH)*(SigS+SigY(0))

\(TraditionalForm\`q  q\&_\) -> Z Z

FeynCalc

The three amplitude of \(TraditionalForm\`u  u\&_\) -> Z Z is:

CreateAmplitude1[{SM, QCD}, {u, U}, {Z, Z}, 0]

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(u\&_\)(p\_2)\)\) -> γ((Subscript[k, 1]))γ((Subscript[k, 2])).
We use 't Hooft-Feynman gauge. The u quark charge is included by FA and taken made explicit.

\(\(FAL = "uU->ZZ" // GetAmplitude;\)

    \(Repl = 
        {MU\  -> \ 0, 
          DiracGamma[7]\/2 - \(2\ s\_w\%2\)\/3 -> 
            \(T\_\(3, q\)\) DiracGamma[7] - Q\_q\ s\_w\%2};\)

    Amp[1] = FAL[\([1, 2]\)] /. Repl\)
Amp[2] = FAL[[2, 2]] /. Repl

SetMandelstam[s, t, u, p1, p2, -k1, -k2, 0, 0, (Subscript[m, Z]), (Subscript[m, Z])];

\(\(Do[
        Do[
	Amp2[i, j] = 
            \(\((\(\(\(\(\(ComplexConjugate[Amp[i]] Amp[j] // Simplify\) 
                              // 
	SumPolarization[#, k1] &\) // 
                            SumPolarization[#, k2] &\) // 
	Contract\) 
                        // 
	FermionSpinSum\) // 
	DiracSimplify)\) 
                  /. 
\ DiracTrace -> Tr // Simplify\) // 
	
              TrickMandelstam[#, {s, t, u, 2\ m\_Z\^2}] &
, {i, 2}], {j, 2}];
      \)\)
\(Amp2[3] = 
      \(∑\+\(i = 1\)\%2\(∑\+\(j = 1\)\%2 Amp2[i, j]\) 
          // 
		PropagatorDenominatorExplicit\) // Simplify\)

----

This must still be summed over (Subscript[N, C]).

In the γγ CM frame:

\(p\_1 = \(\@s\/2\) {I, 0, 0, 1}; p\_2 = \(\@s\/2\) {I, 0, 0, \(-1\)};

    k\_1 = \(\@s\/2\) {I, 0, Sin[θ], Cos[θ]}; 
    k\_2 = \(\@s\/2\) {I, 0, \(-Sin[θ]\), \(-Cos[θ]\)};

    s -> \ \(-\((p\_1 + p\_2)\) . \((p\_1 + p\_2)\)\)\)
t -> -((Subscript[p, 1]) - (Subscript[k, 1])) . ((Subscript[p, 1]) - (Subscript[k, 1])) // Simplify
u -> -((Subscript[p, 1]) - (Subscript[k, 2])) . ((Subscript[p, 1]) - (Subscript[k, 2])) // Simplify

Mandelstam = 
        {t -> s (( 1 - Cos[θ])/2), u -> s (( 1 + Cos[θ])/2)};

Amp2[3] /. Mandelstam // Simplify[#, Trig -> False] &

Amp2[qQ -> γγ]=-((16 (e^4) ((cos^2)(θ) + 1) 
                  (Subscript[Q, f]^4))/((cos^2)(θ) - 1));

This agrees with both Ohnemus and Owens and CompHEP.

The averaging factors: initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) and the identical final state particles factor ((1/2)).

CompHEP

<< "~/CompHEP/comphep_32.18/results/uU-ZZ.m"

We make the initial state color ((1/3) (1/3)) and spin ((1/2) (1/2)) averaging factors and the identical final state particles factor ((1/2)), and the sum over colors ((Subscript[N, C]) = 3) explicit ((the quark charges ((Subscript[Q, f]) = (2/3)))), and change notation:

\(\((\(\(\(\(\((1\/N\_C)\)\^2\) \(\((1\/N\_qSpin)\)\^2\) 
                        1\/N\_identical\)\/\(\(\((1\/3)\)\^2\) 
                        \(\((1\/2)\)\^2\) 1\/2\)\) \(N\_C\/3\) Summ /. 
	
                {EE\  -> \ e, \((SW\^2 - 1)\) -> \(-c\_w\^2\), 
                  SW -> s\_w, 
			SC[p1, p3] -> 
                    \(-\(\(t - m\_Z\^2\)\/2\)\), SC[p1, p2] -> s\/2, 
                  MZ -> m\_Z} // Simplify\) // 
	
            TrickMandelstam[#, {s, \ t, \ u, \ 2\ m\_Z\^2}] &)\) /. 
        512\ s\_w\%8 - 768\ s\_w\%6 + 864\ s\_w\%4 - 432\ s\_w\%2 + 81 -> 
	
          16*81 \((\((T\_\(3, q\) - \(Q\_q\) s\_w\^2)\)\^4 + 
                \((\(-Q\_q\) s\_w\^2)\)\^4)\) // Simplify\)

-(-8 u t) (Subscript[m, Z]^4) - 4 t u (t + u) (Subscript[m, Z]^2) // 
      TrickMandelstam[#, { t,  u, s, 2 ((Subscript[m, Z])^2)}] &

This agrees with Ohnemus and Owens.

Notes:

16*81 ((((1/2) - (2/3) ((Subscript[s, W])^2)))^4) + 
        16*81 (((-(2/3) ((Subscript[s, W])^2)))^4) // Expand

Papageno

Berger et.al. NPB239 (1984) 52

Ratios for \(TraditionalForm\`q  q\&_\) initial state

Compared to the \(q  q\&_\)->\(γ -> \(e\^+\) \(e\^-\)\) process, in the amplitude square of \(q  q\&_\)-> γ γ there's an extra factor:

\(Amp2[qQ -> γγ]/Amp2[qQ -> \(e\^+\) \(e\^-\)]\)

This means: to obtain the resummation formula for \(q  q\&_\)-> γ γ from the resummation formula for \(q  q\&_\)->\(\(e\^+\) \(e\^-\)\) we have to inclue an extra numerical factor, include extra quark charges (Subscript[Q, f]^2), and include an extra factor of 1/(1 - (cos^2)(θ)).
An extra numerical factor ((1/2)) comes from the identical particle final state, which means that the overall extra factor (for \(q  q\&_\)-> γ γ compared to \(q  q\&_\)->\(γ -> \(e\^+\) \(e\^-\)\)) is 2.
(Note that the initial state is the same, which means that the color ((1/(3^2))) and spin ((1/(2^2))) averaging factors are the same for both process.)

In RES.FOR:
- the quark charges are adjusted.
In ResBos:
- for γγ: 	\(TraditionalForm
      \`\(\(\(1\/2\) \(1\/2\) \(1\/2\) 32 \((t\/u + u\/t)\)\)\(=\)\)\) (1/2) (1/2) (1/2) 32*2\(TraditionalForm
      \`\(1 + \(cos\^2\) θ\)\/\(1 - cosθ\)\)=8((1 + (cos^2) θ)/(1 - cosθ))is programmed,
- for \(TraditionalForm\`\(γ\^*\)\) -> ℓℓ': 	4 (1 + (cos^2) θ) 						is programmed.
This means the constant factor of 2 and the extra 1/(1 - (cos^2)(θ)) is taken into account.

Notes

Compared to the symmetric part of the \(q  q\&_\)->\(\(e\^+\) ν\_e\) process in the amplitude square of \(q  q\&_\)-> γ γ there's an extra factor:

\(Amp2[qQ -> γγ]/
      \((Amp2[uD -> \(e\^+\) ν\_e] /. 
          \((1 - Cos[θ])\)\^2 -> 1 + Cos[θ]\^2)\)\)

This means: to obtain the resummation formula for \(q  q\&_\)-> γ γ from the resummation formula for \(q  q\&_\)->\(\(e\^+\) ν\_e\) we have to set (Subscript[V, ij]) -> \(TraditionalForm\`δ\_\(i  i\&_\)\), inclue an extra numerical factor, get rid of the W propagator, include the quark charges, replace  (Subscript[g, weak]) with (Subscript[e, EM]), and include an extra factor of 1/(1 - (cos^2)(θ)).

Extra numerical factor coming in from the different averaging:

(((1/3) (1/3))( (1/2) (1/2)) (1/2))/(((1/3)) ((1/2) (1/2)))

The numerical factors combined give an overall extra factor of:

64  (1/6)

Ratios for gg initial state

Based on Papageno the ratio of the weights of the g g -> γ γ and q \(TraditionalForm\`q\&_\) -> γ γ processes is:

R=((((1/((Subscript[D, A])^2)) (1/(((N - 2))^2)) 
                  (1/(Subscript[N, identical])))/(((((1/8)))^2) ((((1/2)))^2) 
                  (1/2)))((((((Subscript[(∑_(i = 1)^n), f]) ((Subscript[Q, q[i]])^2)))^2))/(((((11/9)))^2)))(((Subscript[F, color]))/(8 (1/2) (1/2))) (Subscript[WT, gg -> γγ]))/\((\(\(\(1\/D\_F\^2\) \(1\/\((N - 2)\)\^2\) 
                  1\/N\_identical\)\/\(\(\((1\/3)\)\^2\) \(\((1\/2)\)\^2\) 
                  1\/2\)\) 
            \(Abs[M\_\(q  q\&_ -> γγ\)]\^2\/\(\(8 
                      \((t\^2 + u\^2)\)\)\/\(t\ u\)\)\) \(N\_C\/3\) 
            WT\_\(q  q\&_ -> γγ\))\)

From here we see the differences:
- the (unaveraged, un-color-summed) amplitude square (not including the couplings or 	quark charges),
- the initial color averaging factors,
- the color factors,
- the couplings.

The \(TraditionalForm\`F\_\(q  q\&_\)\) factor is:

\(\(F\_\(q  q\&_\) = 
        \(\((\((\(e\ Q\_q\)\/2)\)\^2 + \((\(e\ Q\_q\)\/2)\)\^2)\)\^2\) 
          Abs[M\_\(q  q\&_ -> γγ\)]\^2\/16;\)\)

The (Subscript[F, gg]) factor then should be:

\(R*F\_\(q  q\&_\)\)

Appendix

\(TraditionalForm\`q  q\&_\) -> Z Z

The three amplitude of \(TraditionalForm\`u  u\&_\) -> Z Z is:

CreateAmplitude1[{SM, QCD}, {u, U}, {Z, Z}, 0]

The momentum assignment is (as usual with FA): \(TraditionalForm\`\(u(p\_1)\) \(\(u\&_\)(p\_2)\)\) -> γ((Subscript[k, 1]))γ((Subscript[k, 2])).
We use 't Hooft-Feynman gauge.

\(\(FAL = "uU->ZZ" // GetAmplitude;\)

    Amp[1] = FAL[\([1, 2]\)] /. {MU\  -> \ 0, MD -> 0, ME -> 0}\)
Amp[2] = FAL[[2, 2]] /. {MU  ->  0, MD -> 0, ME -> 0}

For simplicity we neglect the Z mass

SetMandelstam[s, t, u, p1, p2, -k1, -k2, 0, 0, 0, 0]

\(\(Do[
	
        Amp2[i] = 
          \((\(\(\(\(\(ComplexConjugate[Amp[i]] Amp[i] // Simplify\) // 
	
                          SumPolarization[#, k1] &\) // 
                        SumPolarization[#, k2] &\) // 
	Contract\) // 
	
                    FermionSpinSum\) // 
	DiracSimplify)\) /. 
\ 
              DiracTrace -> Tr // Simplify
, {i, 1, 2}];\)\)
Amp2[3] = 
      Amp2[1] + Amp2[2] // PropagatorDenominatorExplicit // Simplify

This must be summed over (Subscript[N, C]).

Ohnemus and Owens (PRD43,3626(1991)) has in Eq.(4) (for f = u and after taking (Subscript[M, Z]) = 0)

\(Amp2[OO] = 
      \(\(e\^4\) \((\(g\_+\)\^4 + \(g\_-\)\^4)\) 4 \((t\/u + u\/t)\) /. 
            {\(g\_+\) -> \(-\(2\/3\)\) s\_w\/c\_w, 
              \(g\_-\) -> 
                \(1\/2\) 1\/\(\(s\_w\) c\_w\) - \(2\/3\) s\_w\/c\_w} // 
	
          Expand\) // Factor\)