PolarizationVector[p, mu] denotes a 4-dimensional
ingoing polarization vector \varepsilon^\mu(p).
For the outgoing polarization vector \varepsilon^{\ast \mu}(p) use
ComplexConjugate[PolarizationVector[p, mu]]
To obtain a D-dimensional
polarization vector, just use ChangeDimension[vec, D]
In the internal representation following conventions are used
Pair[Momentum[k], Momentum[Polarization[k, I]]]
corresponds to \varepsilon^{\mu}(k),
i.e. an ingoing polarization vector
Pair[Momentum[k], Momentum[Polarization[k, -I]]]
corresponds to \varepsilon^{\ast
\mu}(k), i.e. an outgoing polarization vector
Warning: The first argument of PolarizationVector should
always be a standalone symbol denoting a momentum (e.g. p,
k1, q2 etc.). Never use symbols multiplied by
-1 or other numbers as well as products
of symbols (e.g. -p, 2*k, x*p
etc.). Doing so will inevitably lead to wrong results.
Overview, FV, Pair, Polarization, PolarizationSum, DoPolarizationSums.
A polarization vector \varepsilon^{\mu }(k) is a special 4-vector.
PolarizationVector[k, \[Mu]]\bar{\varepsilon }^{\mu }(k)
PolarizationVector[k, \[Mu]] // StandardForm
(*Pair[LorentzIndex[\[Mu]], Momentum[Polarization[k, I]]]*)Conjugate[PolarizationVector[k, \[Mu]]]\bar{\varepsilon }^{*\mu }(k)
Conjugate[PolarizationVector[k, \[Mu]]] // StandardForm
(*Pair[LorentzIndex[\[Mu]], Momentum[Polarization[k, -I]]]*)The transversality property is not automatic and must be explicitly
activated using the option Transversality
PolarizationVector[k, \[Mu]] FV[k, \[Mu]]
Contract[%]\overline{k}^{\mu } \bar{\varepsilon }^{\mu }(k)
\overline{k}\cdot \bar{\varepsilon }(k)
PolarizationVector[k, \[Mu], Transversality -> True] FV[k, \[Mu]]
Contract[%]\overline{k}^{\mu } \bar{\varepsilon }^{\mu }(k)
0
Suppose that you are using unphysical polarization vectors for
massless gauge bosons and intend to remove the unphysical degrees of
freedom at a later stage using ghosts. In this case you must not use
Transversality->True, since your polarization vectors
are not transverse. Otherwise the result will be inconsistent.
Here everything is correct, we can use the gauge trick with unphysical polarization vectors.
FCClearScalarProducts[];
SP[k1] = 0;
SP[k2] = 0;ex1 = SP[k1, Polarization[k1, I]] SP[k2, Polarization[k1, -I]] SP[k1, Polarization[k2, I]]*
SP[k2, Polarization[k2, -I]]\left(\overline{\text{k1}}\cdot \bar{\varepsilon }(\text{k1})\right) \left(\overline{\text{k2}}\cdot \bar{\varepsilon }^*(\text{k2})\right) \left(\overline{\text{k1}}\cdot \bar{\varepsilon }(\text{k2})\right) \left(\overline{\text{k2}}\cdot \bar{\varepsilon }^*(\text{k1})\right)
ex1 // DoPolarizationSums[#, k1, 0] & // DoPolarizationSums[#, k2, 0] &(\overline{\text{k1}}\cdot \overline{\text{k2}})^2
Here we erroneously set Transversality->True and
consequently obtain a wrong result. In pure QED the full result
(physical amplitude squared) would still come out right owing to the
Ward identities, but e.g. in QCD this would not be the case.
ex2 = SP[k1, Polarization[k1, I, Transversality -> True]] SP[k2, Polarization[k1, -I,
Transversality -> True]] SP[k1, Polarization[k2, I, Transversality -> True]] SP[k2,
Polarization[k2, -I, Transversality -> True]] // FCI0
ex2 // DoPolarizationSums[#, k1, 0] & // DoPolarizationSums[#, k2, 0] &0
FCClearScalarProducts[];PolarizationVector is a shortcut for 4-dimensional polarization vectors. Although
D-dimensional polarization vectors are
fully supported by FeynCalc, as of now there is no shortcut for entering
such quantities. You can either use ChangeDimension
ChangeDimension[PolarizationVector[q, \[Mu]], D]\varepsilon ^{\mu }(q)
or enter such quantities directly using the
FeynCalcInternal-notation
Pair[Momentum[Polarization[q, I], D], LorentzIndex[\[Mu], D]]\varepsilon ^{\mu }(q)