A Question about commutator involving fermions and Pauli matrices

Gleeson
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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.
 
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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
$$[\hat{A},\hat{B} \hat{C}]=\hat{B} \{\hat{A},\hat{C} \} - \{\hat{A},\hat{B}\} \hat{C}.$$
 
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