Tuesday, 16 April 2019

conventions - Transpose of fermion bilinears


TL;DR


When we take the transpose of two Grassmann-valued spinors (fermions), should we add a minus sign because we end up anticommutating the two spinors?




More details.


I'm studying the behavior of 1+1-dimensional Majorana fermion. I use the same representation for the gamma matrices as in Lectures on string theory by Lüst and Theisen, chapter 7 appendix B. That is


$$\gamma^0 = i \sigma^2 \qquad \gamma^1 = \sigma^1 \qquad C= \gamma^0 \qquad \eta^{\mu\nu}=diag(-1,+1)$$



Equation B.3 of these lectures says that Majorana fermions obey the following indentity:


$$ \bar{\psi}_1 \gamma^{\alpha_1} \gamma^{\alpha_2} ... \gamma^{\alpha_n} \psi_2 = (-1)^n \bar{\psi}_2 \gamma^{\alpha_n} ... \gamma^{\alpha_2} \gamma^{\alpha_1} \psi_1.$$


However I cannot prove this identity, I always get the wrong sign on the RHS, even for simple cases such as $n=0,1$ or $2$. For instance for $n=1$, knowing that Majorana fermions are real and therefore $\bar{\psi} = \psi^\dagger \gamma^0 = \psi^T \gamma^0$, we can write the LHS as:


$$\psi_1^T \gamma^0 \gamma^\alpha \psi_2.$$


This is just a number, so it should be equal to its transpose. When we take the transpose, $\gamma^0$ takes a minus sign. If $\gamma^\alpha=\gamma^0$ then this one also takes a minus sign which cancels the first one and we get no overall sign. If $\gamma^\alpha=\gamma^1$ then the 2nd minus sign comes when we anticommute the two matrices. In every case I get:


$$ \bar{\psi}_1 \gamma^{\alpha} \psi_2 = \psi_1^T \gamma^0 \gamma^\alpha \psi_2 = \psi_2^T \gamma^0 \gamma^\alpha \psi_1 = \bar{\psi}_2 \gamma^{\alpha} \psi_1,$$


even though the identity above says there should be a minus sign. My guess is that when I took the transpose of the LHS, I should have added a minus sign since I inverted two Grassmann-valued spinors in the process, otherwise I don't know what I did wrong.


On a related note, if the sign changes when we swap to spinors, does that imply that the trace of a fermion bilinear is no longer cyclic?


$$\text{Tr}\ \bar{\psi}\psi \neq \text{Tr}\ \psi \bar{\psi} ?$$





Related but the answer is unclear. This post is actually pretty much the same question, but then why this "transpose antisymmetry" only rarely mentioned? Does it only hold for Majorana fermions?




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