Feynman amplitude and tensor 4-vector multiplication (muon neutrino-electron scattering)

Question

In the calculation of the Feynman Amplitude for the muon neutrino-electron scattering (in the Charged Current way from W boson), or $e + \nu_\mu \rightarrow \nu_e + \mu$ (considering the 4-momentum conservation as $p_1 + p_2 = p_3 + p_4$, in the same sequency of the reaction described) the multiplication appears:

$64 [p_1^\mu p_{3}^\nu - g^{\mu \nu} (p_1 \cdot p_{3}) + p_1^\nu p_{3}^\mu - i\epsilon^{\mu \nu \lambda \sigma} {p_1}_\lambda {p_{3}}_\sigma] [{p_2}_\mu {p_{4}}_\nu - g_{\mu \nu} (p_2 \cdot p_{4}) + {p_2}_\nu {p_{4}}_\mu - i\epsilon_{\mu \nu \kappa \tau} {p_2}^\kappa {p_{4}}^\tau]$

My result is:

$64 [4(p_1 \cdot p_2)(p_3 \cdot p_4) - i\epsilon_{\mu \nu \kappa \tau} p_2^\kappa p_4^\tau p_1^\mu p_3^\nu - i\epsilon_{\mu \nu \kappa \tau} p_2^\kappa p_4^\tau p_1^\nu p_3^\mu - i\epsilon^{\mu \nu \lambda \sigma} {p_1}_\lambda {p_3}_\sigma {p_2}_\mu {p_4}_\nu - i\epsilon^{\mu \nu \lambda \sigma} {p_1}_\lambda {p_3}_\sigma {p_2}_\nu {p_4}_\mu + g^{\mu\nu} (p_1 \cdot p_3) i \epsilon_{\mu\nu\kappa\tau} p_2^\kappa p_4^\tau + g_{\mu\nu} (p_2 \cdot p_4) i \epsilon^{\mu\nu\lambda\sigma} {p_1}_\lambda {p_3}_\sigma]$

But the solution is just:

$64[4(p_1 \cdot p_2)(p_3 \cdot p_4)]$

that is the first term of my result. Can anybody save me? PS: It is similar to Problem 9.3 in Introduction to Elementary Particles (Griffiths) or the Example 9.1.

Answer 1

Your result is correct; the contraction of a completely antisymmetric tensor, such as $\epsilon_{\mu\nu\alpha\beta}$, with a symmetric tensor, such are the products $p_1^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta}$ and $g^{\mu\nu}p_i^{\alpha}p_j^{\beta}$ in a QED or EW tree-level process (where any external momentum can be seen as the linear combination of the others), is always zero. As a matter of fact, for example,

$$\epsilon_{\mu\nu\alpha\beta}p_1^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} = \epsilon_{\mu\nu\alpha\beta}(p_3 + p_4 - p_2)^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} = $$ $$=\epsilon_{\mu\nu\alpha\beta}p_3^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} + \epsilon_{\mu\nu\alpha\beta}p_4^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} - \epsilon_{\mu\nu\alpha\beta}p_2^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta},$$

and every term can be seen to be zero like this:

$$\epsilon_{\mu\nu\alpha\beta}p_2^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} = - \epsilon_{\nu\mu\alpha\beta}p_2^{\nu}p_2^{\mu}p_3^{\alpha}p_4^{\beta} = - \epsilon_{\mu\nu\alpha\beta}p_2^{\mu}p_2^{\nu}p_3^{\alpha}p_4^{\beta} = 0.$$

Besides it always holds true that

$$\epsilon^{\mu\nu\alpha\beta}g_{\mu\nu}p_{\alpha}p_{\beta} = - \epsilon^{\nu\mu\alpha\beta}g_{\nu\mu}p_{\alpha}p_{\beta} = - \epsilon^{\mu\nu\alpha\beta}g_{\mu\nu}p_{\alpha}p_{\beta} = 0$$

The last operation done in both the last two lines is just a change of name for the indexes.

Answer 2

All the other terms in your answer are zero. I will demonstrate for one of the terms as the reasoning is similar The second term and third term in your solution is $$-i \epsilon_{\mu \nu \kappa \tau}p_2^\kappa p_4^\tau p_1^\mu p_3^\nu - i \epsilon_{\mu \nu \kappa \tau}p_2^\kappa p_4^\tau p_1^\nu p_3^\mu$$ Since the indices are summed over, the names I give them don’t matter and so in the last term, I replace the index $\mu$ with $\nu$ and vice versa to give $$-i \epsilon_{\mu \nu \kappa \tau}p_2^\kappa p_4^\tau p_1^\mu p_3^\nu - i \epsilon_{\nu \mu \kappa \tau}p_2^\kappa p_4^\tau p_1^\mu p_3^\nu$$ Now using that the levi-civita tensor is anti-symmetric in its arguments, if I change the order of the index $\mu$ and $\nu$, it will give a negative sign $$-i \epsilon_{\mu \nu \kappa \tau}p_2^\kappa p_4^\tau p_1^\mu p_3^\nu + i \epsilon_{\mu \nu \kappa \tau}p_2^\kappa p_4^\tau p_1^\mu p_3^\nu=0$$

Similarly all your other terms cancel to give the final answer