# 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?

• I used ^T for the transpose everywhere. The only place I used Tr is when I was actually talking about the trace, to ask if the antisymmetry of spinors had any effect on the trace of fermionic operators. – Jasmeru Feb 2 at 16:23
• Lord... any QFT book would have said, instead, in mainstream notation, $\bar \psi \chi =\bar \chi \psi$ .... – Cosmas Zachos Feb 2 at 16:29

In short, yes there's an extra minus sign when we reorder the terms. I've just encountered this myself and I found going very slowly helps to understand where the minus sign comes in. Let's use index notation - for some anticommuting $$\psi$$ (column) and $$\bar{\psi}$$ (row), and some matrix $$M$$, consider \begin{align} \bar{\psi} M \psi &= \bar{\psi}_i M_{ij} \psi_j\\ &= \bar{\psi}_i \psi_j M_{ij} &&\text{rearranging terms}\\ &= -\psi_j \bar{\psi}_i M_{ij} &&\text{anticommuting } \bar{\psi}, \psi \tag{!}\\ &= -\psi_i \bar{\psi}_j M_{ji} &&\text{relabelling } i \leftrightarrow j\\ &= - \psi_i M_{ji} \bar{\psi}_j &&\text{another rearrangment}\\ &= - \psi_i (M^T)_{ij} \bar{\psi}_j &&\text{definition of transpose}\\ &= -\psi^T M^T \bar{\psi}^T &&`\text{row \times matrix \times column'} \end{align} which reproduces the usual 'transposes in inverse order' but with an extra minus sign because of the anticommuting nature of the $$\bar \psi$$ and $$\psi$$ at $$(!)$$.
When it comes to something being equal to its own transpose, that's still correct - however, we don't have the usual 'transpose of a product' formula and instead we get extra minus signs, e.g. $$\bar{\psi} \psi = (\bar \psi \psi)^T = -\psi^T \bar{\psi}^T$$
• I have some questions about your answer: in the step (!) you use the anticommutation of bar fields with fields, but that is only true if $i \neq j$. Otherwise, you have an extra term with a Dirac delta. The other question is that I don't see very clear that you can write $\bar{\psi}\psi = (\bar{\psi}\psi)^T$ since $\bar{\psi}\psi$ is an operator after all, not a number and the transpose of an operator is not, in general, equal to the operator. I think your reasoning is good if only if $M_{ii} = 0$, as the gamma matrices are, and therefore your last equation does not hold anymore – Vicky May 1 at 12:12
• We're taking $\psi$ and $\bar{\psi}$ as vectors of Grassmann variables. This means that $\bar{\psi} \psi$ is a 1x1 matrix, so it is trivially equal to its transpose. Grassmann variables always satisfy $\psi_i \psi_j = -\psi_j \psi_i$ even when $i = j$ since $\psi_i^2 = 0$. Obviously with fields one has to account for operators, but that isn't what we're doing here. – LtSten May 2 at 13:29