Graduate Question about commutator involving fermions and Pauli matrices

[Gleeson]

Suppose $\lambda_A$ and $\bar{\lambda}_A$ are fermions (A goes from 1 to N) and $\{ \lambda_{A \alpha}, \bar{\lambda}_B^{\beta}\} = \delta_{AB}\delta_{\alpha}^{\beta}$.

Let $\sigma^i$ denote the Pauli matrices.

Does it follow that $[\bar{\lambda}_A \sigma^i \lambda_A, \bar{\lambda_B} \sigma^j \lambda_B] = 0$? Or should it be $2i \epsilon_{ijk}\bar{\lambda}_C \sigma^k \lambda_C$?

I think it should be the former. $\bar{\lambda_A} \sigma^i \lambda_A$ is a sum over A of a (2 entry row, times a 2x2 matrix, times a 2 entry column), and so is no longer a matrix. And overall it is bosonic. So I think the commutator should be zero. If this is not the case, then why not?

In case it is not clear, $\bar{\lambda}_A \sigma^i \lambda_A = \bar{\lambda}_A^{\alpha}\sigma^{i \beta}_{\alpha}\lambda_{A \beta}$. Repeated indices are summed.

 

[vanhees71]

Just do the calculation! You only have to repeatedly use the identities,
$$[\hat{A} \hat{B},\hat{C}]=\hat{A} \{\hat{B},\hat{C} \} - \{\hat{A},\hat{C} \} \hat{B}$$
and
$$[\hat{A},\hat{B} \hat{C}]=\hat{B} \{\hat{A},\hat{C} \} - \{\hat{A},\hat{B}\} \hat{C}.$$