# A few questions about spinors and gamma matrices

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.

• Are you using any chirality conditions on your gammas? Commented Feb 4, 2019 at 20:17
• Have you reminded yourself how fermion mass terms achieve hermiticity? Commented Feb 4, 2019 at 20:19
• @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 $(\bar{\epsilon}\eta)^{\dagger}=\bar{\epsilon}\eta$. Commented Feb 4, 2019 at 20:19
• @CosmasZachos Oh Thank you for reminding me that, but that is not a mass term. Is the third identity also correct? Commented Feb 4, 2019 at 20:20
• I doubt it. Recall $\gamma^0$ is hermitian and the spacelike ones antihermitian. Commented Feb 4, 2019 at 20:23

• $$\gamma^0$$ 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 $$\bar{\epsilon}\gamma^{\mu}\gamma_{5}\eta=\bar{\eta}\gamma^{\mu}\gamma_{5}\epsilon$$ .