NumberOfPolarizations is an option for
DoPolarizationSums. It specifies the number of
polarizations to sum over in the expression. This is relevant only for
expressions that contain terms free of polarization vectors. This may
occur e.g. if the scalar products involving polarization vectors have
already been assigned some particular values. In this case the
corresponding terms will be multiplied by the corresponding number of
polarizations.
The default value is Automatic which means that the
function will attempt to recognize the correct value automatically by
extracting the dimension dim of the polarization vectors
and putting (dim-2) for massless and (dim-1)
for massive vector bosons. Notice that if the input expression is free
of polarization vectors, the setting Automatic will fail,
and the user must specify the correct dimension by hand.
PolarizationVector[p, mu] ComplexConjugate[PolarizationVector[p, mu]]\bar{\varepsilon }^{*\text{mu}}(p) \bar{\varepsilon }^{\text{mu}}(p)
Here the setting Automatic is sufficient.
FCClearScalarProducts[];
ScalarProduct[p, p] = 0;
PolarizationVector[p, mu] ComplexConjugate[PolarizationVector[p, mu]] + xyz
DoPolarizationSums[%, p, n]\bar{\varepsilon }^{*\text{mu}}(p) \bar{\varepsilon }^{\text{mu}}(p)+\text{xyz}
\text{DoPolarizationSums: The input expression contains terms free of polarization vectors. Those will be multiplied with the number of polarizations given by }2.
2 \;\text{xyz}-2
Here it is not
DoPolarizationSums[xyz, p, n]
\text{\$Aborted}
Setting the number of polarizations by hand fixes the issue
DoPolarizationSums[xyz, p, n, NumberOfPolarizations -> 2]\text{DoPolarizationSums: The input expression contains terms free of polarization vectors. Those will be multiplied with the number of polarizations given by }2.
2 \;\text{xyz}