Date: 02/28/15-02:12:15 PM Z

Hi,

first of all, you'll obtain the same result if you do this calculation
with pen and paper.

>From the contraction of the two epsilon tensors you get

2 g^{c,f} g^{d,e} - 2 g^{c,e} g^{d,f}

Now in the first term two polarization vectors get contracted with each
other:

eps(w,la)^d eps*(w,la)^*e g^{d,e} = -1

which is the standard normalization. Hence, in the first term there are
no polarization vectors left.

Now, from the physical point of view, this "discrepancy" is not
surprising, as the polarization sum is by definition a gauge dependent
quantity, i.e. it is not physical. On the other hand, the matrix element
squared, where this sum enters, is a physical quantity and there it is
guaranteed that the gauge dependent terms involving the auxiliary vector
n^mu will cancel.

So I would suggest that you consider the full matrix element from which
you get your bla expression and check there if this difference between
bla and Contract[bla] changes the physical result, which it shouldn't.

P.S. By the way, one can see nicely see how the gauge dependent terms
cancel out in the gg->gg example (QCDGGToGGTree.m) included in the
development version of FeynCalc.

If you replace

ClearAll[pre];
Table[Print[" calculating product of the amplitudes ", i, " and ",
j," (CC), time = ", Timing[pre[i,j]=re[i,j]//polsums[#,k1,k2,
1/2]&//polsums[#,k2,k1,1/2]&//polsums[#,k3,k4,1]&//polsums[#,k4,k3,1]&//Simplify][[1]]];pre[i,j],{i,4},{j,i}];

with

ClearAll[pre];
Table[Print[" calculating product of the amplitudes ", i, " and ",
j," (CC), time = ", Timing[pre[i,j]=re[i,j]//polsums[#,k1,4k2+k3,
1/2]&//polsums[#,k2,6k1+5k4,1/2]&//polsums[#,k3,29k4,1]&//polsums[#,k4,5k3,1]&//Simplify][[1]]];pre[i,j],{i,4},{j,i}];

where the auxiliary vectors are now quite "ugly", then the evaluation
takes slightly longer, but the result (after applying TrickMandelstam)
remains the same, because the gauge dependent terms *must* cancel out in
all physical quantities.

Cheers,

Am 28.02.2015 um 12:07 schrieb Kyrylo Bondarenko:
> Another thing about DoPolarizationSums. Let
> bla = LC[a, b, c, d] FV[w, c] PolarizationVector[w, d] LC[a, b, e, f] FV[w, e] Conjugate[PolarizationVector[w, f]]
> tmp = Contract[bla]
>
> Then DoPolarizationSums[bla, w, n] gives
> 6w^2 (as I expected)
> but DoPolarizationSums[tmp, w, n] gives
> 4w^2 + 2(w^2 - n^2w^2/(nw)^2)
> which is very different answer. For example, for n=w the last answer is 4w^2.
>

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