I'm trying to understand Eq. 2.6 in this paper. I understand the idea and derivation of the SUSY Ward identity itself and I know how to apply it in the $\mathcal{N}=1$ case. What confuses me here is the particular form of the commutation relations between the supercharges $Q^{(\dagger)}$ and the bosonic/fermionic operators B.

Let's focus on $$[Q_a^\dagger, B^{bcde}] = \langle \epsilon \, p \rangle \,4! \,\delta_a^{[b} B^{cde]}_\phantom{a}.$$ It is said directly below Eq. 2.6 that for the negative helicity gluons, which correspond to $B^{1234}$, this implies $$ [Q_1^\dagger, B^{1234}] = \langle \epsilon \, p \rangle B^{234},$$ but I do not understand how that follows. Here's my working: \begin{aligned} \phantom{}[Q_1^\dagger, B^{1234}] &= \langle \epsilon \, p \rangle \,4! \,\delta_1^{[1} B^{234]}_\phantom{a}\\ &=\langle \epsilon \, p \rangle \,4! \frac{1}{4!}\times(\text{B-terms antisymmetrised in 234})\\ &=\langle \epsilon \, p \rangle (B^{234}+B^{342}+B^{423}-B^{432}-B^{243}-B^{324}) \end{aligned} Out of all terms in the antisymmetrisation of $(1234)$, the $\delta^b_1$ fixes the first index and so we are left with only the 6 terms from antisymmetrising the remaining indices. I do not understand why the equation in the paper contains only $B^{234}$ but not the other 5 terms. Where am I going wrong?


1 Answer 1


Note, that the $B^{abc}$ are already antisymmetric under permutations of indices, as irreducible representations, as claimed in the beginning of the second section of the cited paper.

  • $\begingroup$ Could you please elaborate a bit? How would I expand $\delta_1^{[1} B^{234]}_\phantom{a}$ knowing that $B$ is already antisymmetric? I still don't quite see how to arrive at the correct result. Is it correct to say that $\delta_1^{[1} B^{234]}_\phantom{a} = \frac{1}{4!} (\delta_1^1 B^{234} - \delta_1^2 B^{134} - \delta_1^3 B^{214} - \delta_1^4 B^{231})$ $\endgroup$
    – dzejkob
    Commented Aug 21, 2020 at 17:39
  • $\begingroup$ You have correctly passed to the second line, when writing (B-terms antisymmetrised in 234), and then proceeded to the last line. Now, take into account, that $B^{234} = B^{342} = B^{423} = -B^{432} = -B^{243} = -B^{324}$. However, now there is a factor of 6 missing $\endgroup$ Commented Aug 21, 2020 at 18:22
  • $\begingroup$ Ok, I was thinking along these lines too, but that 6 was precisely what threw me off as well. So anyway, sadly that doesn't solve the problem. $\endgroup$
    – dzejkob
    Commented Aug 21, 2020 at 18:28

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