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I am following BUSSTEPP Lectures on Supersymmetry and trying to show that the Wess-Zumino action is invariant under SUSY transformations. I encountered the following questions about spinors and gamma matrices.

Let $\epsilon$ and $\eta$ be any two Grassmann-valued Majorana spinors. Here, $\bar{\epsilon}$ means the Majorana adjoint, i.e. $\bar{\epsilon}=\epsilon^{T}\mathcal{C}$, where $\mathcal{C}$ is the charge conjugation matrix.

I want to prove the following identities

$$\bar{\epsilon}\eta=\bar{\eta}\epsilon,\quad\bar{\epsilon}\gamma_{5}\eta=\bar{\eta}\gamma_{5}\epsilon,\quad\bar{\epsilon}\gamma^{\mu}\eta=-\eta\gamma^{\mu}\epsilon,\quad\bar{\epsilon}\gamma^{\mu}\gamma_{5}\eta=\bar{\eta}\gamma^{\mu}\gamma_{5}\epsilon$$


I just found that they are equation (3.51) of Supergravity by Daniel Z. Freedman.

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  • $\begingroup$ Are you using any chirality conditions on your gammas? $\endgroup$ – R. Rankin Feb 4 at 20:17
  • $\begingroup$ Have you reminded yourself how fermion mass terms achieve hermiticity? $\endgroup$ – Cosmas Zachos Feb 4 at 20:19
  • $\begingroup$ @R.Rankin What are chirality conditions? I am not assuming anything. If I do a SUSY variation on the Lagrangian, I will end up with "inner products" shown above. But I cannot prove $(\bar{\epsilon}\eta)^{\dagger}=\bar{\epsilon}\eta$. $\endgroup$ – The Last Knight of Silk Road Feb 4 at 20:19
  • $\begingroup$ @CosmasZachos Oh Thank you for reminding me that, but that is not a mass term. Is the third identity also correct? $\endgroup$ – The Last Knight of Silk Road Feb 4 at 20:20
  • $\begingroup$ I doubt it. Recall $\gamma^0$ is hermitian and the spacelike ones antihermitian. $\endgroup$ – Cosmas Zachos Feb 4 at 20:23
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Your spinors are real and your Dirac matrices in the Majorana representation are imaginary, so

  • $\gamma^0$ is Hermitean A(ntisymmetric)
  • $\gamma^i$ are Antihermitean S(ymmetric)
  • $\gamma^5$ is Hermitean A(ntisymmetric) $\leadsto$ check this from the above!

Thus $$ \bar{\epsilon}\eta= i\epsilon^T \gamma^0 \eta =-i\eta^T \gamma^{0~~T} \epsilon =i\eta^T \gamma^{0 }\epsilon = \bar{\eta} \epsilon , $$ So now you got your baseline. Make sure you appreciate every step.

The rest follow trivially from the above properties of the γ matrices and the above properties, so supplanting $\gamma^0$, you now have $$(\gamma^0 \gamma^5)^T= \gamma^5 \gamma^0= -\gamma^0 \gamma^5,$$ just like for the scalar, so $\bar{\epsilon}\gamma_{5}\eta=\bar{\eta}\gamma_{5}\epsilon$. (Anti-)likewise, $$ (\gamma^0 \gamma^\mu)^T= \gamma^0 \gamma^\mu, $$ where you can marvel at how the time-like and space like parts synchronize to yield a uniform answer, hence $\bar{\epsilon}\gamma^{\mu}\eta=-\bar{\eta}\gamma^{\mu}\epsilon$. Finally, for the axial, $\gamma^5$ flips the sign to yield $\bar{\epsilon}\gamma^{\mu}\gamma_{5}\eta=\bar{\eta}\gamma^{\mu}\gamma_{5}\epsilon$ .

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  • $\begingroup$ Thanks for your answer. I had already derived those identities. I thought you were going to prove the generic identity (3.51) from the book Supergravity by Freedman. $\endgroup$ – The Last Knight of Silk Road Feb 4 at 23:06
  • $\begingroup$ I don't have that book! If you want an identity proven, you should at least provide it... If it is the generic string of γ matrices, it should follow easily from the combinatoric nature of their transpose and rearrangement of the resulting strings... $\endgroup$ – Cosmas Zachos Feb 4 at 23:11

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