I am trying to do some calculations from Modern Supersymmetry by Terning and I am stuck on how he derived a particular term. Specifically, I am looking at 2.67 on page 27. My current work is below.

$$\mathcal{L}=-\frac{1}{2}W^{jk}\psi^{\alpha}_{j}\psi_{\alpha k} + ...$$

With $\delta \psi_{\alpha k} = -i(\sigma^{\nu}\epsilon^{\dagger})_{\alpha}\partial_{\nu}\phi_{k}$, where $\alpha$ is a spinor index.

Taking the variation and only looking at terms with a derivative

$$\delta \mathcal{L} \propto -\frac{1}{2}W^{jk}\Big[(-i\sigma^{\nu}\epsilon^{\dagger})^{\alpha}\partial_{\nu}\phi_{j}\psi_{\alpha k}-\psi^{a}_{j}(-i\sigma^{\nu}\epsilon^{\dagger})_{\alpha}\partial_{\nu}\phi_{k})\Big] $$

This is supposed to be

$$\delta \mathcal{L} \propto -iW^{jk}\partial_{\nu}\phi_{k}\psi_{j}^{\alpha}(\sigma^{\nu}\epsilon^{\dagger})_{\alpha}$$

But I'm missing a -1 in the first term of my derivation above. I think I might be missing something really obvious but I can't tell what it is.

  • $\begingroup$ I vaguely remember that derivation, the first term requires using some spinor identities to put it into the proper form. I'll have to look for my SUSY notebook for the details... $\endgroup$ Jun 12, 2016 at 2:07
  • $\begingroup$ @AlexNelson The first expression with spin indices is just $\epsilon^{ab}\sigma^{\nu}_{b \dot{b}} \epsilon^{\dagger \dot{b}}\psi_{a}$ with the first ϵ being the raising operator. At least to me, there doesn't seem to be any obvious spinor identity. $\endgroup$
    – ranques
    Jun 12, 2016 at 2:44
  • $\begingroup$ Sorry, I was thinking of the 4-spinor contribution to $\delta\mathcal{L}_{\text{int}}$ on page 26 which required use of the Fierz identity. Although Eq (A11) on page 302 seems particularly relevant to your problem... $\endgroup$ Jun 12, 2016 at 3:58

1 Answer 1


I have talked with people more knowledgeable than I on SUSY, and they agree that something isn't right here. I'm posting my correction to the above in case someone has this problem in the future. This is missing from the errata.

We start with (2.60)

$$\mathcal{L}=-\frac{1}{2}W^{jk}\psi^{\alpha}_{j}\psi_{\alpha k} + W^{j}F_{j} +h.c.$$

We have the following SUSY transformations, ignoring terms without a derivative

$$\delta \psi_{\alpha j}=-i(\sigma^{\mu}\epsilon^{\dagger})_{\alpha}\partial_{\mu}\phi_{k}+... $$ $$\delta F_{j}=-i\epsilon^{\dagger}\bar{\sigma}^{\mu}\partial_{\mu}\psi_{j}= i\partial_{\mu}\psi_{j}\sigma^{\mu}\epsilon^{\dagger}$$

With the last line following from eqn. A.10. We can now take the variation by sending $F_{j} \rightarrow F_{j}+\delta F_{j}$, $\psi_{j}\rightarrow \psi_{j}+ \delta \psi_{j}$ and subtracting the initial $\mathcal{L}$.

$$\delta \mathcal{L}=-\frac{1}{2}W^{jk}\epsilon^{ab}\big[(-i(\sigma^{\mu}\epsilon^{\dagger})_{b}\partial_{\mu}\phi_{j})\psi_{\alpha k}+ \psi_{bj}(-i(\sigma^{\mu}\epsilon^{\dagger})_{a}\partial_{\mu}\phi_{k})\big] + iW^{j}\partial_{\mu}\psi_{j}\sigma^{\mu}\epsilon^{\dagger} +h.c.$$

Notice that $\epsilon^{ab}(\sigma^{\mu}\epsilon^{\dagger})_{b}\psi_{ak}=\psi^{a}_{k}(\sigma^{\mu}\epsilon^{\dagger})_{a}$. This gives us the desired result.

$$\delta \mathcal{L}=iW^{jk}\partial_{\mu}\phi_{j}\psi_{k}\sigma^{\mu}\epsilon^{\dagger}+iW^{j}\partial_{\mu}\psi_{j}\sigma^{\mu}\epsilon^{\dagger}+h.c.$$

Since $W^{jk}$ is symmetric in $j,k$ you can swap the indices to get it into the desired form. Notice that this is the desired eqn. (2.67) without the $-1$'s. This sign error doesn't impact any of the subsequent logic.


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