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.


2 Answers 2


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.

  • $\begingroup$ Thanks for the commentary. How could I eliminate terms with Minkowski Metric using the @ravjotsk approach? $\endgroup$
    – Marcos
    Commented Jun 16, 2018 at 21:48
  • $\begingroup$ I edited the answer to render it more suitable for the problem posed; I think it ought to be stressed that the point is not in the fact that the different terms simplify each others, but that every single term should be zero by himself. $\endgroup$ Commented Jun 17, 2018 at 9:39
  • $\begingroup$ @Marcos To eliminate terms containing the metric just write the term= 1/2 (term + term). Now in one of the term just interchange the dummy indices over the metric tensor for the entire term (like i did for p1 and p3) and then you use the property that to interchange back for just the levicivita, you pick up a negative sign whereas for metric tensor you don’t and you get term=1/2(term-term)=0 $\endgroup$
    – ravjotsk
    Commented Jun 17, 2018 at 11:03
  • $\begingroup$ @FrancescoArnaudo ok, I am removing my downvote and the comment. $\endgroup$
    – TwoBs
    Commented Jun 17, 2018 at 16:23

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


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