# Treating the spinors as Grassmann numbers or as c-number objects

In the literature on supersymmetry, the following spinor summation convention is often used (eg. Wess & Bagger's book Supersymmetry and Supergravity) $$\psi\chi = \psi^{\alpha}\chi_{\alpha} = -\psi_{\alpha}\chi^{\alpha} = \chi^{\alpha}\psi_{\alpha} = \chi\psi$$ where the spinors are treated as Grassmann numbers and anticommute with each other. However, in some articles like arXiv:hep-th/0312171, the angle spinor bracket is defined as $$\langle\lambda_1\lambda_2\rangle = \lambda_1^{\alpha}\lambda_{2\alpha} = \varepsilon_{\alpha\beta} \lambda_1^{\alpha}\lambda_2^{\beta} = -\varepsilon_{\beta\alpha} \lambda_2^{\beta}\lambda_1^{\alpha} = -\lambda_2^{\beta}\lambda_{1\beta} = -\langle\lambda_2\lambda_1\rangle$$ Here the spinors are treated as c-number objects and commute with each other. Why can we use such two different conventions? Are they contradictory?

• The spinors in Wess & Bagger corresponds to fermionic fields and hence have anti-commuting components. The spinors in Witten's papers, on the other hand, are part of the twistor formalism and encode the momentum in a convenient way. They do not represent fields and are not fermionic, so their components still commute. – Olof Dec 8 '14 at 11:52

Lorentz spinors appear as irreducible representations of the group SL(2,C). Elements of the group are 2x2 matrices with complex entries and unity determinant. A Lorentz spinor is a two component vector $\psi^{A},\chi^{A}\in V_{2}$ with $A=1,2$. The Levi-Civita tensor $\epsilon_{AB}$ is an invariant tensor under SL(2,C). This means that if we have an irrep $\psi^{A}$ of SL(2,C) we can transform it with the Levi-Civita tensor and this will give an equivalent irrep. So, the covariant vectors $\psi_{A}\in \tilde{V}_{2}$ made as $\psi_{A}=\psi^{B}\epsilon_{BA}$ are equivalent to the contravariant vectors $\psi^{A}$; it doesn't matter whether we use $\psi^{A}$ or $\psi_{A}$ because both quantities represent the same physical thing.

The situation is the same as in Minkowski spacetime when we use $x^{\mu}$ or $x_{\mu}=\eta_{\mu\lambda}x^{\lambda}$. Covariant and contravariant vectors in Minkowski spacetime are equivalent irreps of the Lorentz group $O(1,3)$ because the metric $\eta_{\mu\lambda}$ is an invariant tensor.

The only problem with using the Levi-Civita tensor to lower a spinor index is that it is antisymmetric so that it matters which index is summed. I've decided to lower spinor indices as $\psi_{A}=\psi^{B}\epsilon_{BA}$ and so, for consistency, I've got to stick to this convention and not be tempted to use $\psi_{A}=\epsilon_{AB}\psi^{B}$.

Having picked a lowering convention, I'm forced to raise a spinor index with $\psi^{A}=\epsilon^{AB}\psi_{B}$ because then both operations are consistent. $$\psi_{A}=\psi^{B}\epsilon_{BA}=\epsilon^{BC}\psi_{C}\epsilon_{BA}=\delta^{C}_{A}\psi_{C}=\psi_{A}$$

Now we can make a SL(2,C) scalar $\psi_{A}\chi^{A}=\chi^{A}\psi_{A}$. The order of the vectors does not matter because the components $\psi^{1},\psi^{2}$ are just complex numbers. The following bit of index gymnastics recovers the property in the second equation in soliton's question. $$\psi_{A}\chi^{A}=\psi^{B}\epsilon_{BA}\chi^{A}=-\psi^{B}\chi^{A}\epsilon_{AB}=-\psi^{A}\chi_{A}$$

Everything so far has been classical. When we go over to quantum theory, the spinors are promoted to operators $\psi^{A}\rightarrow \hat{\psi}^{A}$. These operators represent fermions: as operators, they have to obey anticommutation relations, in this case $[\hat{\psi}^{A},\hat{\chi}^{B}]_{+}=0$. So, repeating the last calculation with operators, $$\hat{\psi}_{A}\hat{\chi}^{A}=\hat{\psi}^{B}\epsilon_{BA}\hat{\chi}^{A}=-\hat{\psi}^{B}\hat{\chi}^{A}\epsilon_{AB}=-\hat{\psi}^{A}\hat{\chi}_{A}=+\hat{\chi}_{A}\hat{\psi}^{A}$$ recovers the first equation in soliton's question.

I have to own up to the fact that I've not yet studied supersymmetry, but I think this is what must be going on based on general principles.

• If I just study the representations of $SL(2,\mathbb{C})$, without involving any quantum theory, do I still need to require $\psi$ to be anticommuting? – Nahc Jan 22 '17 at 21:28